The Transfer Function-Advanced Circuit Analysis-Lecture Slides, Slides of Electrical Circuit Analysis

This lecture is part of lecture series on Electrical Circuit Analysis course. It was delivered by Prof. Mursleen Sayed at Bengal Engineering and Science University. It includes: Transfer, Function, Initial, Conditions, Rlc, Circuit, Series, Input, Output, Capacitor, Multiple

Typology: Slides

2011/2012

Uploaded on 07/23/2012

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The Transfer Function
Transfer function H(s) = Y(s)/X(s)
Y(s) is the laplace of output signal
X(s) is the laplace of input signal
In computing the transfer function, circuits
where all initial conditions are zero are
considered
Transfer function depends on what is defined
as the output signal
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The Transfer Function ^

Transfer function H(s) = Y(s)/X(s) ^

Y(s) is the laplace of output signal ^

X(s) is the laplace of input signal ^

In computing the transfer function,

circuits

where all initial conditions are zero

are

considered

^

Transfer function depends on what is definedas the output signal

The Transfer Function ^

Transfer function of a series RLC circuit ^

Input signal is voltage V

g

^

Output signal is current I ^

H(s) = I/V

= 1/(R + sL + 1/sC)g = sC/(s

2 LC + sRC + 1)

^

If voltage across the capacitor is the output signal ^

H(s) = V/V

=(1/sC)/(R + sL + 1/sC)g = 1/(s

2 LC + sRC + 1)

Example ^

H(s) =? ^

Poles & Zeros =?   ^

H(s)=

1000 ohm V g

(^250) ohm 50 mH

μ^ 1 F

Vo

sV 10 s 05

. 0 250

V

V

V

o 6

o

g o^

6

2

g

o^

(^10) x 25 s 6000 s

V)

s( 1000 V^

6

2

(^10) x 25 s 6000 s

s( 1000

docsity.com

Example ^

H(s) = ^

Poles ^

-p1 = -3000 –j ^

-p2 = -3000 +j ^

Zeros ^

-z1 = -

1000 ohm Vg

(^250) ohm 50 mH

μ^ 1 F

Vo

6

2

(^10) x 25 s 6000 s

s( 1000

Transfer function in Partialfraction expansion ^

Y(s) = H(s)X(s) ^

Writing the equation in the form of sum of partialfractions ^

Produces a term for each pole of H(s) ^

Produces a term for each pole of X(s) ^

Terms generated by poles of H(s) gives rise totransient component of total response ^

Terms generated by poles of X(s) gives rise tosteady-state component of total response

Example ^

H(s) = ^

Source voltage is v

= 50tu(t) Vg^

^

V(s) = 50/sg

2

^

V(s) = H(s)Vo^

(s) =g

= ^

K^1

= 5

5x

-4^ 

0

*K 1 = 5

5x

-4^ 

-79.

(^0)

^

K^2

= 10

K^3

= -4x

6

2

(^10) x 25 s 6000 s

) 5000 s( 1000

 

2 6

2

(^50) xs (^10) x 25 s 6000 s

) 5000 s( 1000

 

K^ s K s (^4000) j 3000 s

K

(^4000) j 3000 s

K^

3 (^22)

  • 1

1

      

Response of a circuit ^

Response of a circuit is related to H(s) through apartial fraction expansion ^

Practically driving a circuit with an increasing rampvoltage leads to failure of components ^

Ramp function should only increase to a defined maximumvalue within a finite time interval ^

If time constant of the circuit is small compared to the timetaken by the signal to reach the maximum value. A solutionassuming an unbounded ramp is valid for this finite timeinterval

Response of a circuit with delayedinput ^

ℒ{x(t-a)u(t-a)} = e

-as

X(s)

^

Y(s) = H(s)X(s)e

-as

^

y(t-a)u(t-a) =

-1^ {H(s)X(s)e

-as

^

Delaying the input signal by ‘a’ secdelays the response by ‘a’ sec ^

Circuit is time-invariant