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EEEB123EEEB123Circuit Analysis 2Circuit Analysis 2

Chapter 16Chapter 16

Applications of the Applications of the Laplace TransformLaplace Transform

Materials from Fundamentals of Electric Circuits (4th Edition), Alexander & Sadiku, McGraw-Hill Companies, Inc.

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Application of the Laplace TransformApplication of the Laplace TransformChapter 16Chapter 16

16.2 Circuit Element Models 16.3 Circuit Analysis 16.4 Transfer Functions

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16.2 Circuit Element Models (1)16.2 Circuit Element Models (1)

Steps in Applying the Laplace Transform:

1. Transform the circuit from the time domain to the s-domain

2. Solve the circuit using nodal analysis, mesh analysis, source transformation, superposition, or any circuit analysis technique with which we are familiar

3. Take the inverse transform of the solution and thus obtain the solution in the time domain.

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16.2 Circuit Element Models (2)16.2 Circuit Element Models (2)

Representation of an inductorRepresentation of an inductor , at initial conditions : :

(a)time-domain and (b,c) s-domain equivalents.

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16.2 Circuit Element Models (2a)16.2 Circuit Element Models (2a)

Representation of a capacitorRepresentation of a capacitor , at initial conditions : :

(a)time-domain and (b,c) s-domain equivalents.

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16.2 Circuit Element Models (2b)16.2 Circuit Element Models (2b) Assume zero initial condition for

the inductor and capacitor,

Resistor : V(s)=RI(s)

Inductor: V(s)=sLI(s)

Capacitor: V(s) = I(s)/sC

The impedance in the s-domain is defined as Z(s) = V(s)/I(s)

Resistor : Z(s)=R

Inductor: Z(s)=sL

Capacitor: Z(s) = 1/sC

The admittance in the s-domain is defined as Y(s) = I(s)/V(s)

Time-domain and s-domain representations of passive elements under zero initial conditions.

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16.2 Circuit Element Models (3)16.2 Circuit Element Models (3)

Example 16.1:Example 16.1:

Find v0(t) in the circuit shown below, assuming zero initial conditions.

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16.2 Circuit Element Models (4)16.2 Circuit Element Models (4)

Solution:Solution:

Transform the circuit from the time domain to the s-domain, we have

s

L

tu

3

sC

1 F

3

1

ss H 1s

1 )(

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Solution:Solution:

Apply mesh analysis, on solving for V0(s)

Taking the inverse transform give

220)2()4(

2

2

3)(V

ss

16.2 Circuit Element Models (5)16.2 Circuit Element Models (5)

0 V, )2sin(2

3)( 4

0 ttetv t

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16.2 Circuit Element Models (6)16.2 Circuit Element Models (6)

Practice Problem 16.1:Practice Problem 16.1:

Determine v0(t) in the circuit shown below, assuming zero initial conditions.

V )()21(8 :Ans 22 tutee tt

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16.2 Circuit Element Models (7)16.2 Circuit Element Models (7)

Example 16.2:Example 16.2:

Find v0(t) in the circuit shown below. Assume v0(0)=5V .

V )()1510()( v:Ans 20 tueet tt

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16.2 Circuit Element Models (8)16.2 Circuit Element Models (8)

Example 16.3:Example 16.3:

The switch shown below has been in position b for a long time. It is moved to position a at t=0. Determine v(t) for t > 0.

where0, t ,I)IV()( v:Ans 0/

00 RCReRt t

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16.3 Circuit Analysis (1)16.3 Circuit Analysis (1)

• Circuit analysis is relatively easy to do in the s-domain.

• By transforming a complicated set of mathematical relationships in the time domain into the s-domain where we convert operators (derivatives and integrals) into simple multipliers of s and 1/s.

• This allow us to use algebra to set up and solve the circuit equations.

• In this case, all the circuit theorems and relationships developed for dc circuits are perfectly valid in the s-domain.

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16.3 Circuit Analysis (2)16.3 Circuit Analysis (2)

Example 16.4:Example 16.4:

Consider the circuit (a).

Find the value of the voltage across the capacitor assuming that the value of vs(t)=10u(t) V.

Assume that at t=0, -1A flows through the inductor and +5 is across the capacitor.

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Solution:Solution:

Transform the circuit from time-domain (a) into s-domain (b) using Laplace Transform. On rearranging the terms, we have

By taking the inverse transform, we get

2

30

1

35V1

ss

16.3 Circuit Analysis (3)16.3 Circuit Analysis (3)

V )()3035()(v 21 tueet tt

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16.3 Circuit Analysis (4)16.3 Circuit Analysis (4)

Practice Problem 16.6:Practice Problem 16.6:

The initial energy in the circuit below is zero at t=0. Assume that vs=5u(t) V. (a) Find V0(s) using the Thevenin theorem. (b) Apply the initial- and final-value theorem to find v0(0) and v0(∞). (c) Obtain v0(t).

Ans: (a) V0(s) = 4(s+0.25)/(s(s+0.3)) (b) 4,3.333V, (c) (3.333+0.6667e-0.3t)u(t) V.

*Refer to in-class illustration, textbook

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• The transfer function H(s) is the ratio of the output response Y(s) to the input response X(s), assuming all the initial conditions are zero.

, h(t) is the impulse response function.

• Four types of gain:

1. H(s) = voltage gain = V0(s)/Vi(s)

2. H(s) = Current gain = I0(s)/Ii(s)

3. H(s) = Impedance = V(s)/I(s)

4. H(s) = Admittance = I(s)/V(s)

16.4 Transfer Functions (1)16.4 Transfer Functions (1)

)(

)()(

sX

sYsH

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Example 16.7:Example 16.7:

The output of a linear system is y(t)=10e-tcos4t when the input is x(t)=e-tu(t). Find the transfer function of the system and its impulse response.

Solution:Solution:

Transform y(t) and x(t) into s-domain and apply H(s)=Y(s)/X(s), we get

Apply inverse transform for H(s), we get

16)1(

44010

16)1(

)1(10

)(

)()(

22

2

ss

s

sX

sYsH

16.4 Transfer Function (2)16.4 Transfer Function (2)

)()4sin(40)(10)( tutetth t

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Practice Problem 16.7:Practice Problem 16.7:

The transfer function of a linear system is

Find the output y(t) due to the input e-3tu(t) and its impulse response.

6

2)(

s

ssH

16.4 Transfer Function (3)16.4 Transfer Function (3)

)(12e-(t)2 0; t,42 :Ans -6t63 tuee tt