15: transients (a) - imperial college london · 2017-09-13 · 15: transients (a) 15: transients...
TRANSCRIPT
![Page 1: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/1.jpg)
15: Transients (A)
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 1 / 11
![Page 2: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/2.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):
![Page 3: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/3.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysis
![Page 4: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/4.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysis
![Page 5: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/5.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
![Page 6: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/6.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt
![Page 7: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/7.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR
![Page 8: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/8.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
![Page 9: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/9.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
![Page 10: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/10.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
![Page 11: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/11.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).
![Page 12: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/12.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).
Complementary Function: Solution to RC dydt + y = 0
![Page 13: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/13.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).
Complementary Function: Solution to RC dydt + y = 0
Does not depend on x(t), only on the circuit.
![Page 14: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/14.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).
Complementary Function: Solution to RC dydt + y = 0
Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−
t/τ
where τ = RC is the time constant of the circuit.
![Page 15: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/15.jpg)
Differential Equation
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 2 / 11
To find y(t):x(t) constant: Nodal analysisx(t) sinusoidal: Phasors + nodal analysisx(t) anything else: Differential equation
i(t) = C dydt =
x−yR ⇒ RC dy
dt + y = x
General Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to RC dydt + y = x
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution for y(t).
Complementary Function: Solution to RC dydt + y = 0
Does not depend on x(t), only on the circuit.Solution is y(t) = Ae−
t/τ
where τ = RC is the time constant of the circuit.
The amplitude, A, is determined by the initial conditions at t = 0.
![Page 16: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/16.jpg)
Piecewise steady state inputs
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 3 / 11
We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.
![Page 17: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/17.jpg)
Piecewise steady state inputs
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 3 / 11
We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.
Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)
![Page 18: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/18.jpg)
Piecewise steady state inputs
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 3 / 11
We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.
Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)
Steady state, ySS(t), is the same frequency as theinput; use phasors + nodal analysis.
![Page 19: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/19.jpg)
Piecewise steady state inputs
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 3 / 11
We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.
Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)
Steady state, ySS(t), is the same frequency as theinput; use phasors + nodal analysis.
Transient is always yTr(t) = Ae−t
τ at each change.
![Page 20: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/20.jpg)
Piecewise steady state inputs
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 3 / 11
We will consider input signals that are sinusoidal or constant for a particulartime interval and then suddenly change in amplitude, phase or frequency.
Output is the sum of the steady state and a transient:y(t) = ySS(t) + yTr(t)
Steady state, ySS(t), is the same frequency as theinput; use phasors + nodal analysis.
Transient is always yTr(t) = Ae−t
τ at each change.
![Page 21: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/21.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
-1 0 1 2 3
-2
0
2
4
t (ms)
![Page 22: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/22.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
![Page 23: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/23.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
![Page 24: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/24.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
![Page 25: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/25.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
![Page 26: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/26.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
y(0+) = 4 +A and y(0−) = 1
![Page 27: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/27.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1
![Page 28: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/28.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3
![Page 29: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/29.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
yTr
y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3
So transient: yTr(t) = −3e−t/τ
![Page 30: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/30.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
yTr
y
y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3
So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ
![Page 31: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/31.jpg)
Step Input
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 4 / 11
For t < 0, y(t) = x(t) = 1For t ≥ 0, RC dy
dt + y = x= 4Time Const: τ = RC = 1ms
Steady State (Particular Integral)ySS(t) = x(t) = 4 for t ≥ 0
Transient (Complementary Function)yTr(t) = Ae−
t/τ
Steady State + Transienty(t) = ySS + yTr = 4 +Ae−
t/τ
To find A, use capacitor property:Capacitor voltage never changes abruptly
-1 0 1 2 3
-2
0
2
4
t (ms)
-1 0 1 2 3
-2
0
2
4
t (ms)
ySS
yTr
y
y(0+) = 4 +A and y(0−) = 1⇒ 4 +A = 1⇒ A = −3
So transient: yTr(t) = −3e−t/τ and total y(t) = 4− 3e−t/τ
Transient amplitude ⇐ capacitor voltage continuity: vC(0+) = vC(0−)
![Page 32: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/32.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et
0 1 2 3 4 5
-10
0
10 et
t
![Page 33: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/33.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4
0 1 2 3 4 5
-10
0
10 et3e¼t
t
![Page 34: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/34.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
![Page 35: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/35.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t
0 2 4 6 8-2
0
22e-t
t
![Page 36: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/36.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4
0 2 4 6 8-2
0
22e-t
e-¼t
t
![Page 37: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/37.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
![Page 38: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/38.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
![Page 39: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/39.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
a 2a 3a 4a 5a0
0.5
1
t
![Page 40: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/40.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
37% after 1 time constant
a 2a 3a 4a 5a0
0.5
1
0.37
t
![Page 41: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/41.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
37% after 1 time constant5% after 3
a 2a 3a 4a 5a0
0.5
1
0.37
0.05
t
![Page 42: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/42.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
37% after 1 time constant5% after 3, <1% after 5
a 2a 3a 4a 5a0
0.5
1
0.37
0.05 0.01
t
![Page 43: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/43.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
37% after 1 time constant5% after 3, <1% after 5
a 2a 3a 4a 5a0
0.5
1
0.37
0.05 0.01
t
Gradient of e−t/a
Gradient at t hits zero at t+ a.
a 2a 3a0
0.5
1
t
![Page 44: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/44.jpg)
Negative exponentials
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 5 / 11
Positive exponentials grow to ±∞:et, 3et/4, −2et/2
0 1 2 3 4 5
-10
0
10 et3e¼t
-2e½t
t
Negative exponentials decay to 0:2e−t, e
−t/4, −2e−t/2
Transients are negative exponentials.
0 2 4 6 8-2
0
22e-t
e-¼t
-2e-½t
t
Decay rate of e−t/a
37% after 1 time constant5% after 3, <1% after 5
a 2a 3a 4a 5a0
0.5
1
0.37
0.05 0.01
t
Gradient of e−t/a
Gradient at t hits zero at t+ a.True for any t.
a 2a 3a0
0.5
1
t
![Page 45: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/45.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
![Page 46: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/46.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
How long does it take to go from A to B ?
![Page 47: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/47.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
How long does it take to go from A to B ?
At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ
![Page 48: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/48.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
How long does it take to go from A to B ?
At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ
B−FA−F = e
−(TB−TA)/τ
![Page 49: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/49.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
How long does it take to go from A to B ?
At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ
B−FA−F = e
−(TB−TA)/τ
Hence TB − TA = τ ln(
A−FB−F
)
= τ ln(
initial distance toFfinal distance toF
)
![Page 50: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/50.jpg)
Exponential Time Delays
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 6 / 11
Negative exponential with a finalvalue of F .
y(t) = F + (A− F ) e−(t−TA)/τ
How long does it take to go from A to B ?
At t = TB :y(TB) = B = F + (A− F ) e−(TB−TA)/τ
B−FA−F = e
−(TB−TA)/τ
Hence TB − TA = τ ln(
A−FB−F
)
= τ ln(
initial distance toFfinal distance toF
)
Useful formula - worth remembering.
![Page 51: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/51.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
![Page 52: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/52.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt
![Page 53: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/53.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt
![Page 54: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/54.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
![Page 55: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/55.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
![Page 56: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/56.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
![Page 57: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/57.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
![Page 58: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/58.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
![Page 59: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/59.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
Complementary Function: Solution to LR
dydt + y = 0
![Page 60: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/60.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
Complementary Function: Solution to LR
dydt + y = 0
Does not depend on x(t), only on the circuit.
![Page 61: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/61.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
Complementary Function: Solution to LR
dydt + y = 0
Does not depend on x(t), only on the circuit.Solution is yTr(t) = Ae−
t/τ
where τ = LR is the time constant of the circuit.
![Page 62: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/62.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
Complementary Function: Solution to LR
dydt + y = 0
Does not depend on x(t), only on the circuit.Solution is yTr(t) = Ae−
t/τ
where τ = LR is the time constant of the circuit.
1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L
R
![Page 63: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/63.jpg)
Inductor Transients
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 7 / 11
We know i = x−yR
y(t) = L didt =
LR ×
d(x−y)dt = L
Rdxdt − L
Rdydt
⇒ LR
dydt + y = L
Rdxdt
Solution: Particular Integral + Complementary Function
Particular Integral: Any solution to LR
dydt + y = L
Rdxdt
If x(t) is piecewise constant or sinusoidal, we will usenodal/phasor analysis to find the steady state solution, ySS(t).
Complementary Function: Solution to LR
dydt + y = 0
Does not depend on x(t), only on the circuit.Solution is yTr(t) = Ae−
t/τ
where τ = LR is the time constant of the circuit.
1st order transient is always yTr(t) = Ae−t/τ where τ = RC or L
RAmplitude A ⇐ no abrupt change in capacitor voltage or inductor current.
![Page 64: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/64.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.
![Page 65: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/65.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C .
![Page 66: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/66.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
![Page 67: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/67.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
![Page 68: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/68.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
![Page 69: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/69.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
![Page 70: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/70.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.
![Page 71: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/71.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.
Linearity: y = ax+ bv = ax+ bvSS + bvTr
![Page 72: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/72.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.
Linearity: y = ax+ bv = ax+ bvSS + bvTr = ySS + bvTr
![Page 73: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/73.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.
Linearity: y = ax+ bv = ax+ bvSS + bvTr = ySS + bvTr
All voltages and currents in a circuit have the same transient (but scaled).
![Page 74: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/74.jpg)
Linearity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 8 / 11
1st order circuit has only one C or L.Make a Thévenin equivalent of the networkconnected to the terminals of C . Rememberx is a voltage source but y is not.
Now v(t) = vSS(t) + vTr(t)= vSS(t) +Ae
−t/τ
Time constant is τ = RThC
where RTh is the Thévenin resistance.
Replace the capacitor with a voltage sourcev(t); all voltages and currents in the circuitwill remain unchanged.
Linearity: y = ax+ bv = ax+ bvSS + bvTr = ySS + bvTr
All voltages and currents in a circuit have the same transient (but scaled).
The circuit’s time constant is τ = RThC or LRTh
where RTh is theThévenin resistance of the network connected to C or L.
![Page 75: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/75.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)
![Page 76: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/76.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
![Page 77: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/77.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
![Page 78: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/78.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
![Page 79: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/79.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+
![Page 80: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/80.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1
![Page 81: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/81.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
![Page 82: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/82.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time Constant
![Page 83: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/83.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2k
![Page 84: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/84.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
![Page 85: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/85.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
Result
![Page 86: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/86.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
![Page 87: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/87.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 3 + (5− 3) e−t/τ
![Page 88: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/88.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 3 + (5− 3) e−t/τ
= 3 + 2e−t/τ
![Page 89: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/89.jpg)
Transient Amplitude
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 9 / 11
Find Steady State (DC ⇒ ZL = 0)Potential divider: ySS = 1
2x
ySS(0−) = 1, ySS(0+) = 3
Inductor Current ContinuityiSS(0−) = 1mA ⇒ iL(0+) = 1mA
At t = 0+x− y = 1mA× 1 k = 1y(0+) = x(0+)− 1 = 5
Time ConstantSet x ≡ 0 → RTh = 2kτ = L
RTh= 2µs
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 3 + (5− 3) e−t/τ
= 3 + 2e−t/τ
![Page 90: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/90.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
![Page 91: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/91.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
![Page 92: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/92.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
![Page 93: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/93.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 94: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/94.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 95: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/95.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 96: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/96.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0 -RC 0 RC 2RC 3RC
-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 97: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/97.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 98: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/98.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
Time Constantτ = RThC = 2RC (from earlier slide)
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 99: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/99.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
Time Constantτ = RThC = 2RC (from earlier slide)
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 100: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/100.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
Time Constantτ = RThC = 2RC (from earlier slide)
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 134 +
(
− 54 − 13
4
)
e−t/τ
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
ySS
![Page 101: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/101.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
Time Constantτ = RThC = 2RC (from earlier slide)
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 134 +
(
− 54 − 13
4
)
e−t/τ
= 134 − 18
4 e−t/τ = 3 1
4 − 4 12e
−t/2RC
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
yTr
ySS
![Page 102: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/102.jpg)
Capacitor Voltage Continuity
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 10 / 11
Find Steady State (DC ⇒ ZC = ∞)KCL @ V: v−x
4R + v8R + v−y
2R = 0
KCL @ Y: y−v2R + y−x
6R = 0
vSS = 34x, ySS = 13
16x
Capacitor Voltage ContinuityvSS(0−) = −3 ⇒ v(0+) = −3
At t = 0+: x = 4 and v = −3KCL @ Y: y−(−3)
2R + y−46R = 0
y(0+) = −9+44 = − 5
4
Time Constantτ = RThC = 2RC (from earlier slide)
Resulty = ySS + (y (0+)− ySS (0+)) e−t/τ
= 134 +
(
− 54 − 13
4
)
e−t/τ
= 134 − 18
4 e−t/τ = 3 1
4 − 4 12e
−t/2RC
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
-RC 0 RC 2RC 3RC-4
-2
0
2
4
t
yTr
ySS
y
![Page 103: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/103.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
![Page 104: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/104.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
• Circuit time constant: τ = RThC or LRTh
◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.
![Page 105: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/105.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
• Circuit time constant: τ = RThC or LRTh
◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.
• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise
constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.
![Page 106: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/106.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
• Circuit time constant: τ = RThC or LRTh
◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.
• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise
constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.
◦ Transient: Find vC(0−) or iL(0−): unchanged at t = 0+Find y(0+) assuming source of vC(0+) or iL(0+)Amplitude never complex, never depends on t.
![Page 107: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/107.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
• Circuit time constant: τ = RThC or LRTh
◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.
• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise
constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.
◦ Transient: Find vC(0−) or iL(0−): unchanged at t = 0+Find y(0+) assuming source of vC(0+) or iL(0+)Amplitude never complex, never depends on t.
◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ
![Page 108: 15: Transients (A) - Imperial College London · 2017-09-13 · 15: Transients (A) 15: Transients (A) •Differential Equation •Piecewise steady state inputs •Step Input •Negative](https://reader030.vdocuments.us/reader030/viewer/2022040210/5e5815ab2b5f59108a789944/html5/thumbnails/108.jpg)
Summary
15: Transients (A)
• Differential Equation
• Piecewise steady stateinputs
• Step Input
• Negative exponentials
• Exponential Time Delays
• Inductor Transients
• Linearity
• Transient Amplitude
• Capacitor VoltageContinuity
• Summary
E1.1 Analysis of Circuits (2017-10110) Transients (A): 15 – 11 / 11
• 1st order circuits: include one C or one L.◦ vC or iL never change abruptly. The output, y, is not necessarily
continuous unless it equals vC .
• Circuit time constant: τ = RThC or LRTh
◦ RTh is the Thévenin resistance seen by C or L.◦ Same τ for all voltages and currents.
• Output = Steady State + Transient◦ Steady State: use nodal/Phasor analysis when input is piecewise
constant or piecewise sinusoidal. The steady state has the samefrequency as the input signal.
◦ Transient: Find vC(0−) or iL(0−): unchanged at t = 0+Find y(0+) assuming source of vC(0+) or iL(0+)Amplitude never complex, never depends on t.
◦ y(t) = ySS(t) + (y(0+)− ySS(0+)) e−t/τ
See Hayt Ch 8 or Irwin Ch 7.