2 order transients - cvut.czamber.feld.cvut.cz/vyu/eo2/english/files/8_transients2.pdf• from last...

2nd order transientsin time domain

Electrical circuits 2 – Lecture 8

Pavel MášaXE31EO2 -

Pav

el Máš

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XE31EO2 - Pavel Máša - Lecture 8

INTRODUCTION

• On last lectures we familiarize with general approach of transient solution, namely 1st

order transient solution.• Now we will investigate procedure of 2nd order circuits analysis.• From last lectures we know, higher order transients differs solely in number of exponential functions / exponentially damped harmonic functions, so that we may finish our study of higher order transients just by 2nd order.

RLC defibrillatorIt is the simplest example of 2nd order circuitWhat waveform has the current, passing the human body?How distinct circuit parameters (L, C, R) affects current waveform?

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2ND ORDER TRANSIENT SOLUTION IN TIME DOMAIN

1. Circuit equations• In contrast to 1st order circuits, in 2nd order circuits is not preferred any method of

network analysisWe may choose any method (mesh analysis, nodal analysis)

In this example, mesh analysis is more suitable

Ldi(t)

dt+ Ri(t) +

1

C

Z t

0

i(¿ ) d¿ ¡ uC(0) = 0Ldi(t)

dt+ Ri(t) +

1

C

Z t

0

i(¿ ) d¿ ¡ uC(0) = 0

2. separation of variablesin this example is only one circuit variable, we omit this step

3. Differentiate circuit equation (if only we did not proceed separation of variables in 2nd step)

d2i(t)

dt2+

R

L

di(t)

dt+

1

LCi(t) = 0

d2i(t)

dt2+

R

L

di(t)

dt+

1

LCi(t) = 0

4. Equation will be solved by method of variation of parameterswith zero right‐hand side of an equation, thus without sources

¸2 +R

L¸ +

1

LC= 0¸2 +

R

L¸ +

1

LC= 0

The form of general solution of a differential equation consist in the roots of an quadratic equation

Circuit, whereon the decision procedure is

illustrated– RLC defibrillator

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Distinct real roots

i(t) = K1 e¸1t + K2 e¸2t + ip(t)i(t) = K1 e¸1t + K2 e¸2t + ip(t)

¸2 + 4000¸ + 106 = 0 ) ¸1;2 = ¡2000§p

20002 ¡ 106 = ¡2000§ 1732:1 = ¡3732:1= ¡267:9

¸2 + 4000¸ + 106 = 0 ) ¸1;2 = ¡2000§p

20002 ¡ 106 = ¡2000§ 1732:1 = ¡3732:1= ¡267:9

Example:

i(t) = K1 e¡3732:1t + K2 e¡267:9t + ip(t)i(t) = K1 e¡3732:1t + K2 e¡267:9t + ip(t)

Double real roots

i(t) = (K1 + K2 t) e¸t + ip(t)i(t) = (K1 + K2 t) e¸t + ip(t)

¸2+2000¸+106 = 0 ) ¸1;2 = ¡1000§p

10002 ¡ 106 = ¡1000§0 = ¡1000¸2+2000¸+106 = 0 ) ¸1;2 = ¡1000§p

10002 ¡ 106 = ¡1000§0 = ¡1000

Example:

i(t) = (K1 + K2t) e¡1000t + ip(t)i(t) = (K1 + K2t) e¡1000t + ip(t)

Complex conjugate roots

i(t) = (K1 cos !t + K2 sin !t) e¡®t + ip(t)i(t) = (K1 cos !t + K2 sin !t) e¡®t + ip(t)

¸1;2 = ¡® § jp

!2r ¡ ®2 = ¡® § j!¸1;2 = ¡® § j

p!2

r ¡ ®2 = ¡® § j!

¸2+1000¸+106 = 0 ) ¸1;2 = ¡500§p

5002 ¡ 106 = ¡500§ jp

10002 ¡ 5002 = ¡500§866j¸2+1000¸+106 = 0 ) ¸1;2 = ¡500§p

5002 ¡ 106 = ¡500§ jp

10002 ¡ 5002 = ¡500§866j

Example:

i(t) = (K1 cos(866t) + K2 sin(866t)) e¡500t + ip(t)i(t) = (K1 cos(866t) + K2 sin(866t)) e¡500t + ip(t)XE31

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5. Solve particular solutionSteady state in the circuit, after transient dies away

According to the kind of exciting sources DC / AC / periodical steady state analysis

In the example of RLC defibrillator:

uc(0) = 4216 Vuc(0) = 4216 V

¸2+50

0:3¸+

1

0:3 ¢ 45 ¢ 10¡6= 0 ) ¸ = ¡83:3§

q83:3

2 ¡ 74074 = ¡83:3§259:1j¸2+50

0:3¸+

1

0:3 ¢ 45 ¢ 10¡6= 0 ) ¸ = ¡83:3§

q83:3

2 ¡ 74074 = ¡83:3§259:1j

i(t) = (K1 cos(259:1 t) + K2 sin(259:1 t)) e¡83:3t + i(p)i(t) = (K1 cos(259:1 t) + K2 sin(259:1 t)) e¡83:3t + i(p)

ip(t) = 0ip(t) = 0

Now we “just” have to find constants of integration K1, K2

• We have just one circuit equation, but two constants of integration K1, K2

• It is not enough to set t = 0, herewith we got just one equation

iL(0) = 0 AiL(0) = 0 A

6. Differentiate solution of a differential equationform of the solution results from the kind of roots of an quadratic equationXE31

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Distinct real roots

i0(t) =

d

dt

hK1 e¸1t + K2 e¸2t + ip(t)

i= K1¸1e

¸1t + K2¸2e¸2t + i

0

p(t)i0(t) =

d

dt

hK1 e¸1t + K2 e¸2t + ip(t)

i= K1¸1e

¸1t + K2¸2e¸2t + i

0

p(t)

Example:

i0(t) = ¡3732:1 K1 e¡3732:1t ¡ 267:9 K2 e¡267:9t + i

0

p(t)i0(t) = ¡3732:1 K1 e¡3732:1t ¡ 267:9 K2 e¡267:9t + i

0

p(t)

Double real roots

i0(t) =

d

dt

³(K1 + K2 t) e¸t + ip(t)

´= K2 e¸t + (K1 + K2 t) ¸e¸t + i

0

p(t)i0(t) =

d

dt

³(K1 + K2 t) e¸t + ip(t)

´= K2 e¸t + (K1 + K2 t) ¸e¸t + i

0

p(t)

Example:

i0(t) = K2 e¡1000t + (K1 + K2t) (¡1000)e¡1000t + i

0

p(t)i0(t) = K2 e¡1000t + (K1 + K2t) (¡1000)e¡1000t + i

0

p(t)

Complex conjugate roots

i0(t) =

d

dt

h(K1 cos !t + K2 sin !t) e¡®t + ip(t)

i=

=h! (¡K1 sin !t + K2 cos !t)¡ ® (K1 cos !t + K2 sin !t)

ie¡®t + i

0p(t)

i0(t) =

d

dt

h(K1 cos !t + K2 sin !t) e¡®t + ip(t)

i=

=h! (¡K1 sin !t + K2 cos !t)¡ ® (K1 cos !t + K2 sin !t)

ie¡®t + i

0p(t)

¸1;2 = ¡® § jp

!2r ¡ ®2 = ¡® § j!¸1;2 = ¡® § j

p!2

r ¡ ®2 = ¡® § j!

Example:

i0(t) =

h866 (¡K1 sin(866t) + K2 cos(866t))¡500 (K1 cos(866t) + K2 sin(866t))

¤e¡500t+i

0

p(t)i0(t) =

h866 (¡K1 sin(866t) + K2 cos(866t))¡500 (K1 cos(866t) + K2 sin(866t))

¤e¡500t+i

0

p(t)XE31

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But if we substitute for t = 0 in differentiated equation, we need to know its value at this time

7. Mathematical initial conditionat this time we return back to primary equation, to which we substitute• t = 0• energetic initial conditions

Li0(0)+RiL(0)+

1

C

Z 0

0

i(¿ ) d¿| {z }=0

¡uC(0) = 0 ) i0(0) =

uC(0)¡RiL(0)

LLi

0(0)+RiL(0)+

1

C

Z 0

0

i(¿ ) d¿| {z }=0

¡uC(0) = 0 ) i0(0) =

uC(0)¡RiL(0)

L

Specific form of solution of mathematical initial condition depends on equation (equations), describing the circuit

8. Into solution of a differential equation and its derivative substitute t = 0, initial conditions and solve system of linear equations

Distinct real roots

i(0) = K1 + K2 + ip(0)

i0(0) = K1¸1 + K2¸2 + i

0

p(0)

)!

K1 =i

0(0)¡ i

0p(0)¡ ¸2

¡i(0)¡ ip(0)

¢¸1 ¡ ¸2

K2 =i

0(0)¡ i

0p(0)¡ ¸1

¡i(0)¡ ip(0)

¢¸2 ¡ ¸1

i(0) = K1 + K2 + ip(0)

i0(0) = K1¸1 + K2¸2 + i

0

p(0)

)!

K1 =i

0(0)¡ i

0p(0)¡ ¸2

¡i(0)¡ ip(0)

¢¸1 ¡ ¸2

K2 =i

0(0)¡ i

0p(0)¡ ¸1

¡i(0)¡ ip(0)

¢¸2 ¡ ¸1

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Double real roots

i(0) = K1 + ip(0)

i0(0) = K2 + K1¸ + i

0

p(0)

)!

K1 = i(0)¡ ip(0)

K2 = i0(0)¡ i

0

p(0)¡ ¸¡i(0)¡ ip(0)

¢i(0) = K1 + ip(0)

i0(0) = K2 + K1¸ + i

0

p(0)

)!

K1 = i(0)¡ ip(0)

K2 = i0(0)¡ i

0

p(0)¡ ¸¡i(0)¡ ip(0)

¢Complex conjugate roots

i(0) = K1 + ip(0)

i0(0) = !K2 + ®K1 + i

0

p(0)

)!

K1 = i(0)¡ ip(0)

K2 =i

0(0)¡ i

0p(0)¡ ®

¡i(0)¡ ip(0)

¢!

i(0) = K1 + ip(0)

i0(0) = !K2 + ®K1 + i

0

p(0)

)!

K1 = i(0)¡ ip(0)

K2 =i

0(0)¡ i

0p(0)¡ ®

¡i(0)¡ ip(0)

¢!

Now we will finish solution of our example of RLC defibrillator:

i(t) = (K1 cos(259:1 t) + K2 sin(259:1 t)) e¡83:3t + 0i(t) = (K1 cos(259:1 t) + K2 sin(259:1 t)) e¡83:3t + 0

i0(t) = 259:1 (¡K1 sin 259:1t + K2 cos 259:1t)¡83:3 (K1 cos 259:1t + K2 sin 259:1t)

ie¡83:3ti

0(t) = 259:1 (¡K1 sin 259:1t + K2 cos 259:1t)¡83:3 (K1 cos 259:1t + K2 sin 259:1t)

ie¡83:3t

i0(0) =

4216¡ 50 ¢ 00:3

= 14053i0(0) =

4216¡ 50 ¢ 00:3

= 14053

0 = K1 + 0

14053 = 259:1K2 + 83:3K1

)!

K1 = 0

K2 =14053¡ 83:3 ¢ 0

259:1= 54:24

0 = K1 + 0

14053 = 259:1K2 + 83:3K1

)!

K1 = 0

K2 =14053¡ 83:3 ¢ 0

259:1= 54:24

i(t) = 54:24 sin(259:1 t) ¢ e¡83:3t Ai(t) = 54:24 sin(259:1 t) ¢ e¡83:3t A

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Voltage waveform across resistor

Current waveform

Instantaneous power delivered into the resistorp [W]p [W]

I [A]I [A]

U [V]U [V]

t[s]t[s]

Waveform of 2nd order transient – RLC defibrillator

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WHAT AFFECTS THE WAVEFORM?

¸2 +R

L¸ +

1

LC= 0¸2 +

R

L¸ +

1

LC= 0

¸1;2 =¡R

2L§

sμR

2L

¶2

¡ 1

LC¸1;2 =

¡R

2L§

sμR

2L

¶2

¡ 1

LC

We have characteristic equation:

... and its roots

If:μR

2L

¶2

>1

LC

μR

2L

¶2

>1

LC Aperiodic transient• The bigger the R is, the larger a time constant is

The bigger the R is, the transient last longer

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μR

2L

¶2

=1

LC) R = 2

rL

C

μR

2L

¶2

=1

LC) R = 2

rL

CLimit of aperiodicity

Transient last shortest possible time

Limit of aperiodicityAperiodic transientQuasi‐periodic transient

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μR

2L

¶2

<1

LC

μR

2L

¶2

<1

LC Quasi‐periodic transient• With decreasing R the dumping factor α also decrease

Settling time is increasing

¸1;2 =¡R

2L§

sμR

2L

¶2

¡ 1

LC= ¡®§

p®2 ¡ !2

r| {z }j!

¸1;2 =¡R

2L§

sμR

2L

¶2

¡ 1

LC= ¡®§

p®2 ¡ !2

r| {z }j!

Dumping factor

Frequency of natural oscillations

Resonant frequency

®®

!!

!r!r

What affects:

R Dumping factor α, but also magnitude of the current I

L Dumping factor α, but also magnitude of the current I (ω in K2 expression) and frequency of natural oscillations of the circuit (together with resonant frequency)

C Magnitude of the current I (ω in K2 expression) and frequency of natural oscillations of the circuit (together with resonant frequency)

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What happens, if the patient will not have the correct resistivity...Typical resistivity of human body while defibrillation is 50 ΩResistivity of human body exposed to line voltage is indicated in interval of 500 Ω– 10 kΩ, ordinary value is 2 kΩ

The waveform of the current in the case, the human body has 2 kΩ

The waveform of the current in the case, the human body has 10 Ω

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2 order transients - cvut.czamber.feld.cvut.cz/vyu/eo2/english/files/8_transients2.pdf• from last...

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