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Chapter 15 Circuit Analysis in the s-Domain
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Resistors in the Frequency Domain
Ohm’s law specifies that
v(t) = Ri(t)
Taking the Laplace transform of both sides
V(s) = RI(s)
The impedance Z(s) is defined as
and the admittance Y(s)=I(s)/V(s) is 1/R
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Z(s)V(s)
I(s) R
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Modeling Inductors in the s-Domain
Since v(t) = L di/dt taking the Laplace transform of both sides of this equation yields
V(s) = L[sI(s) − i (0−)]
The impedance is Z(s)=sL and the initial condition is modeled as a voltage source in series.
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Example: Inductors
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Find the voltage v(t) given an initial current i(0−) =1 A.
Answer: v(t) = [3.2e−8t − 1.2e−0.5t ]u(t) volts
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Solution
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Modeling Capacitors in the s-Domain
Using i=C dv/dt, we can find Z(s)=1/sC and
we model the initial condition as in (b) and (c)
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Example: Mesh Currents
Determine the mesh currents i1(t) and i2(t) (assume zero initial energies)
Answer: (using a computer for calculations)
i1(t) =−96.39e−2t − 344.8e−t +
841.2e−0.15t cos 0.8529t + 197.7e−0.15t sin 0.8529t mA
i2(t) =−481.9e−2t − 241.4e−t +
723.3e−0.15t cos 0.8529t + 472.8e−0.15t sin 0.8529t mA
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Example: Nodal Analysis
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Find vx
Answer: vx = [4 + 6.864e−1.707t − 5.864e−0.2929t ]u(t)
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Solution
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Example: Source Transformations
Find v(t) using source transformation (The initial conditions can be determined to be zero).
Answer:
v(t) =[5.590 ×10−5e−0.1023t + 2.098 cos(3t + 3.912◦)
+ 0.1017e−0.04885t cos(0.6573t + 157.9◦)]u(t)
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Example: Thévenin Equivalent
Find the frequency-domain Thévenin equivalent of the highlighted network:
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Method: apply a test current:
The three impedance in parallel on the left and on the right can all be combined.
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Solution
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The Transfer Function H(s)
H(s) is the transfer function of the circuit, defined as the ratio of the output to the input.
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H(s)Vout
Vin
1
1 sRC1/RC
s1/RC
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Impulse Response
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𝛿(𝑡) ℎ(𝑡) Network
Network
Network
Network
𝛿(𝑡 − 𝜆) ℎ(𝑡 − 𝜆)
𝑥 𝜆 𝛿(𝑡 − 𝜆) 𝑥 𝜆 ℎ(𝑡 − 𝜆)
𝑥 𝑡 = 𝑥 𝜆 𝛿(𝑡 − 𝜆)∞
−∞
𝑑𝜆 𝑦 𝑡 = 𝑥 𝜆 ℎ(𝑡 − 𝜆)𝑑𝜆∞
−∞
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Convolution
If we define the impulse response as h(t), then the output y(t) is related to the input x(t) via the convolution integral:
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If x(t)=vi(t)=u(t)-u(t-1) and h(t)=2e-tu(t), then by flip/slide/integrate, we find vo(t):
Graphical Convolution
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t <1 t >1
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Example: Convolution
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Convolution and Laplace
Convolution in the time domain is multiplication in the frequency domain:
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Visualizing Laplace: the s Plane
We can explore the properties of F(s) by graphing |F(s)| on the s plane.
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Interpreting the Complex Frequency (s) Plane
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Example: Y(s) = 1 / (s+3)
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Pole-Zero Constellations
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Z(s) ks2
s2 2s26
s2
(s1)2 52
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Natural Response from the s Plane
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When Vs=0, the output is the
natural response, and it can only be
non-zero at the pole s=−R/L.
Hence in(t)=Ae-Rt/L.
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Implementing a Pole
H (s) R f
R1
1/R fC f
s1/R fC f
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Implementing a Zero
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H (s) R fC1 s1
R1C1
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Designing a Circuit to Achieve a Particular H(s)
Synthesize a circuit that will yield the transfer function H(s) =Vout/Vin =10(s+2)/(s+5).
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