rlc circuit transient response solutions

8/9/2019 RLC Circuit Transient Response Solutions

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EE305 Lecture 6

Aaron Mueller

February 11, 2013

Aaron Mueller EE305 Lecture 6 February 11, 2013 1 / 15


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Roots for various damping ratios

s 1,2 = (−ζ ± ζ 2−1)ω 0

(−ζ ± j

1−ζ 2)ω 0 = σ ± j ω d if ζ


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Location of roots in complex plane



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Location of roots in complex plane



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Total responses for various damping ratios

x (t ) =

( s 2s 1−s 2

)e s 1t + ( s 1s 2−s 1

)e s 2t + 1 if ζ > 1,

−(ω 0t + 1)e −ω 0t + 1 if ζ = 1,−e −σ t (cos(ω d t ) +

σ ω d

sin(ω d t )) + 1 if ζ


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Total responses for various damping ratios

x (t ) =

( s 2s 1−s 2

)e s 1t + ( s 1s 2−s 1

)e s 2t + 1 if ζ > 1,

−(ω 0t + 1)e −ω 0t + 1 if ζ = 1,−e −σ t (cos(ω d t ) +

σ ω d

sin(ω d t )) + 1 if ζ


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General procedure for solving for second order responses

1 Find the homogeneous solution (natural response) to the differential

equation. This will generally contain two unknown coefficients.

x n (t ) = A1x n ,1(t ) +A2x n ,2(t )

2 Find the particular solution (forced response) to the differential equation.

This can often be done using the method of undetermined coefficients.

For a constant forcing function (DC voltage and/or current source), this

solution will be a constant. x F (t ) = x f 3 Construct the total solution, which is the sum of the homogeneous and

particular solutions.

x (t ) = x n (t ) + x F (t ) = A1x n ,1(t ) +A2x n ,2(t ) + x f

4 Solve for the unknown coefficients (A1 and A2) by, for example, using

known the known values for x (t ) and/or its derivative x (t ) at a given timeor times. If these values are not directly given. They can usually be

obtained by using the original (integro-)differential equation and/or other

circuit relations.Aaron Mueller EE305 Lecture 6 February 11, 2013 7 / 15


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Overdamped CaseThe natural response is of the form x n (t ) = A1e

s 1t +A2e s 2t . Assume

x F (t ) = x f , x (0) = x 0, and x (0) = x 1 (all constant) are given. Then we can

solve for A1

and A2

.

x (t ) = A1e s 1t +A2e

s 2t +x f

x (0) = A1 +A2 +x f = x 0

x

(t ) = A1s 1e s 1t

+A2s 2e s 2t

x (0) = A1s 1 +A2s 2 = x 1

We have two equations with two unknowns. Solving, we get

A1 = x 1 + (x f −x 0)s 2

s 1− s 2 , A2 =

x 1 + (x f − x 0)s 1s 1−s 1

General overdamped solution

x (t ) = x 1+(x f −x 0)s 2s 1−s 2

e s 1t + x 1+(x f −x 0)s 1s 1−s 1

e s 2t



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Critically Damped CaseThe natural response is of the form x n (t ) = (A3 +A4t )e

st . Assume x F (t ) = x f ,x (0) = x 0, and x

(0) = x 1 (all constant) are given. Then we can solve for A3and A

4.

x (t ) = (A3 +A4t )e st +x f

x (0) = A3 +x f = x 0

x

(t ) = e st

[s (A3 +A4t ) +A4]

x (0) = A3s +A4 = x 1


A3 = x 0−x f , A4 = x 1− (x 0−x f )s

General critically damped solution

x (t ) = {(x 0− x f ) + [x 1− (x 0−x f )s ]t }e st +x f



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Underdamped CaseThe natural response is of the form x n (t ) = e

−σ t [A5cos(ω d t ) +A6sin(ω d t )].Assume x F (t ) = x f , x (0) = x 0, and x

(0) = x 1 (all constant) are given. Thenwe can solve for A

5 and A

6.

x (t ) = e −σ t [A5cos(ω d t ) +A6sin(ω d t )] +x f

x (0) = [A5 + 0] + x f = x 0

x (t ) = e −σ t [(A6ω d −A5σ )cos(ω d t )− (A5ω d +A6σ )sin(ω d t )]

x (0) = A6ω d −A5σ = x 1


A5 = x 0−x f , A6 = x 1 + σ (x 0− x f )

ω d

General underdamped solution

x (t ) = e −σ t [(x 0−x f )cos(ω d t ) + x 1+σ (x 0−x f )

ω d sin(ω d t )] + x f



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Problem 7.101 (1)

Given the circuit below, solve for v 0(t ) for t > 0.



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Problem 7.101 (2)

t = 0−

By inspection,

i L(0−) = 0

v C (0−) = −12V

t > 0

We can construct a Thévenin equivalent for the

portion of the circuit to the right of the inductor.

Here

V OC = 4V , R Th = 4k Ω



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Problem 7.101 (3)The integro-differential equation that governs the circuit behavior for t > 0 is(using KCL)

(R 2 +R Th )i (t ) +Ldi dt

+ 1C

t

0i (t )dt +v C (0) = V OC .

Differentiating to get the purely differential form,

d 2i

dt 2 + (

R 2 +R Th L )

di

dt +

1

LC i = 0.

The characteristic equation is

s 2 + (R 2 +R Th

L)s +

1

LC = 0, or

s 2 + 4×106s + 3×1012 = 0

⇒ s 1 = −1×106,s 2 = −3×10

6



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Problem 7.101 (4)

i (t ) = A1e s 1t +A2e s 2t + i (∞)

Since the capacitor acts as an open circuit at steady state, i (∞) = v 0(∞) = 0.We know that i (0+) = i L(0

−) = 0 = A1 +A2, but how do we get a value fordi dt t =0 (so that we can use

di dt

= A1s 1e s 1t +A2s 2e

s 2t )?

Use the original (integro-differential) equation at t = 0:

(R 2 +R Th )i (0) +Ldi dt

t =0

+v C (0) = V OC

i (0) = V OC −v C (0)L

= (4− (−12))/(2.5×10−3) = 6400 A/sA1 +A2 = 0 & (−1×10

6A1−3×106A2 = 6400)

⇒ A1 = 0.0032,A2 = −0.0032⇒ i (t ) = 3.2(e −1×10

6t −e 3×106t ) mA

⇒ v 0(t ) = R 2i (t ) = 19.2(e −1×106t −e 3×10

6t ) V


rlc circuit transient response solutions

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