Fundamentals of Electric CircuitsChapter 16

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Overview

• This chapter will apply the Laplace transform to circuit analysis.

• The equivalent models for the resistor, capacitor, and inductor will be introduced.

• Setting up proper initial conditions will be covered.

• Transfer functions and state variables are also discussed.

• Finally, circuit stability and network synthesis will be covered.

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Circuit Element Models• We will now look at how to apply Laplace

transforms to circuit.

1. Transform the circuit from 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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Circuit Element Models II

• Only the first step in the process is new.• As was done in phasor analysis, we

transform a circuit from time domain to the frequency or s-domain.

• For a resistor, the voltage current relationship in the time domain is:

• Taking the Laplace transform:

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v t Ri t

V s RI s

Inductors

• For an inductor

• Taking the Laplace Transform:

• Note here that the initial conditions can be represented either as a voltage or as a current source

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di tv t L

dt

01 iI s V s

sL s

Inductors

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• Here are the equivalent circuit elements including the sources for the initial conditions.

Capacitors

• For a Capacitor

• Taking the Laplace Transform:

• Note here that the initial conditions can also be represented either as a voltage or as a current source

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dv ti t C

dt

01 vV s I s

sC s

Capacitors

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• Here are the equivalent circuit elements including the sources for the initial conditions.

Initial Conditions

• One significant advantage of the Laplace transform is that it includes both stead-state and initial conditions.

• This allows for obtaining both the steady-state response as well as the transient response.

• The s-domain equivalent elements can be readily used in first and second order circuits.

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Zero Initial Conditions

• In many applications, we will assume the initial conditions are zero.

• The resistor of course stays the same.• For the inductor and capacitor, the equations

are simplified:

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Resistor

Inductor

Capacitor 1

V s RI s

V s sLI s

V s I ssC

Impedances

• We can define the impedance in the s-domain as:

• Thus, the three elements will have the following impedances:

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V s

Z sI s

Resistor

Inductor

Capacitor 1

Z s R

Z s sL

Z ssC

Admittance• In the s-domain, the admittance is also

simply the reciprocal of the impedance:

• Use of the Laplace transform in circuit analysis facilitates the use of various signals sources, such as:– Impulse– Step– Ramp– Exponential– Sinusoidal

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1 I sY s

Z s V s

Dependent Sources

• The models for dependent sources and op-amps are easy to develop.

• Dependent sources have a linear output based on the input.

• In the Laplace transform, multiplying the time-domain equation by a constant results in the same constant multiplied against the Laplace transform.

• Op amps get treated like resistors; they multiply a voltage times a constant.

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Circuit Analysis

• Using the s-domain to analyze circuits is relatively easy.

• Converting from time domain to s-domain results in complicated equations based on derivatives and integrals into simple multiples of s and 1/s.

• One need only use algebra to set-up and solve the circuit equations.

• All theorems and relationships developed for DC work in the s-domain.

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Transfer Functions

• The transfer function is a key concept in signal processing because it indicates how a signal is processed as it passes through a network.

• The transfer function relates the output of a circuit to a given input, assuming zero initial conditions.

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Y sH s

X s

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Example 1:

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

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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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Apply mesh analysis, on solving for V0(s)

Taking the inverse transform give

220)2()4(

2

2

3)(V

ss

0 V, )2sin(2

3)( 4

0 ttetv t

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Example 2:

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

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Example 3:

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

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

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Example 4:

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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Example 5:

Consider the circuit below. Find the value of the voltage across the capacitor assuming that the value of vs(t)=10u(t) V and assume that at t=0, -1A flows through the inductor and +5 is across the capacitor.

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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

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

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Example 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.

*

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Transfer Functions II

• There are four possible transfer functions:

• It is possible for a circuit to have many transfer functions.

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Voltage gain

Current gain

Impedance

Admittance

o

i

o

i

o

i

o

i

V sH s

V s

I sH s

I s

V sH s

I s

I sH s

V s

Transfer Functions III

• Each transfer function can be found in two ways:

• The first is to assume any convenient input X(s).

• Then use any circuit analysis technique (current or voltage division, nodal analysis, etc…)

• Find the output Y(s)• Take the ratio.

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Example 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:

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

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

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Example 8:

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

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

*Refer to in-class illustration, textbook

State Variables

• So far we have only looked at techniques for analyzing system with only one input and only one output.

• Many systems have more than one input or output.

• The state variable method is a very important tool for analyzing such highly complex systems.

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State Variables

• In the state variable model, we specify a collection of variables that describe the internal behavior of the system.

• These variables are known as the state variables of the system.

• They are variables that determine the future behavior of a system.

• They are a reflection of the current state of the system, without requiring the history of the system.

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State Variables

• The standard way to represent a state variable is to arrange them as a set of first order differential equations.

• And the output equation is:

• The transfer equation is found by taking the Laplace transform of the first equation.

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x Ax Bz

y Cx Dz

1H s C sI A B

State Variables• There are three steps to applying the state variable

method:

1. Select the inductor current i and capacitor voltage v as the state variables, making sure they are consistent with the passive sign convention.

2. Apply KCL and KVL to the circuit and obtain circuit variables (voltages and currents) in terms of the state variables. This should lead to a set of first-order differential equations necessary and sufficient to determine all state variables.

3. Obtain the output equation and put the final result in state-space representation.

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16.4 State Variables (2)

Example 9:

Obtain the state variable model for the circuit shown below. Let R1=1, R2=2 ,C=0.5F and L=0.2H and obtain the transfer function.

3012

20)H ; 0 ,

0

1

1

11

Ans2201

2

1

ss

(si

vRvvCR

i

v

L

R

L

CCRi

vs

*Refer to in-class illustration, textbook

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Applications

• The Laplace transform certainly plays an important role in circuit analysis, however that is not the only area where it has applications.

• It also has applications in signal processing and control systems.

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