Impulse Function-Network And 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: Impulse, Function, Discontinuous, Discontinuity, Abrupt, Derivative, Parameter, Variable, Duration, Amplitude

Typology: Slides

2011/2012

Uploaded on 07/23/2012

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Impulse Function
Derivative of u(t) at t=0 doesn’t
exist
Discontinuous function
Assume function varies linearly
across the discontinuity
As 0 abrupt discontinuity
occurs at origin
Derivative between & -is
0.5/
Derivative for t > is –ae-a(t-)
t
f(t)
1
0.5
5.0t
5.0
)t(a
e
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pf4
pf5
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pf9
pfa
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pff

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Impulse Function

^

Derivative of u(t) at t=0 doesn’texist ^

Discontinuous function

^

Assume function varies linearlyacross the discontinuity ^

As



0 abrupt discontinuity

occurs at origin ^

Derivative between

& -

^

is

0.5/

^

Derivative for t >

is –ae

-a(t-

)

t

f(t) 1

(^5). 0 5 t

. 0

 



) t( a

e^



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Impulse Function ^

As

0 f’(t) between

approaches

^

(t) =du(t)/dt

t

f(t) 1

(^5). 0

5 t

. 0

 



) t( a e



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Impulse Function ^

Impulse function is created from a

variable

parameter

function when that parameter

approaches zero

^

Variable parameter function must exhibit thefollowing three characteristics as the

parameter

approaches zero ^

Amplitude approaches infinity ^

Duration of function approaches zero ^

Area under the variable-parameter function is constant asthe parameter changes

Impulse Function ^

Impulse Functions ^

f(t) = 0.5t/

^

(linear

function) ^

as

approaches 0, function becomes

^

time decays to 0 ^

Area under the function isindependent of

(^5). 0 (^5). t 0

 



) t( a e^

 

Impulse Function ^

Area = ^

Therefore as

0, f(t)

K

(t)

^

Mathematically Impulse Function is defined as

(t) = 0, t

dt 0 e K 2

dt e K 2

/t

0

/t

0

 



^

K
K^2
K^2

e K 2

e / 1 K 2

0 /t

0 /t

  

 

K
dt)
t(
K

  

Impulse Function ^

An impulse that occursat t = a is

K

(t - a)

^

Sifting Property ^

Function f(t) iscontinuous at t = a atlocation of pulse

)t( δK^

)a t(δ K^

)a (f

dt) a t( δ) t( f^

  

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Impulse Function 

Sifting property of the impulse function isused to find its Laplace transform 

{

(t)} =

This is an important Laplace transformused in circuit analysis.

dt)t

dt

e)

t(

0

st

0

^

 

 

Derivatives of the ImpulseFunction ^

The impulse functiongenerates an impulsefunction as



0

^

Defined as derivative of animpulse ^

[’(t)] as



0

^

Derivative of an impulsefunction is referred to as amoment function or unitdoublet

f(t)

t

0



 /^1 (^2) /t

   ^

/ 1 (^2) /t

f(t)

t

0

2 / 1 

2 / 1  

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Functional Transform ^

Laplace transform of a specified function of t

^

Unilateral Laplace Transform ^

All functions zero for t < 0

^

Laplace of unit step function ^

{u(t)} =

^

Laplace transform of a decaying exponentialfunction ^

{e

-at

} =

s/ 1

s e dt e 1

dt e) t( f

0 st

0

st

0

st^

 

)s a/( 1

)s a( e

dt

e

dt e e

0 t)s a(

0

t)s a(

0

st at^

  

 

 

Functional Transform 

Laplace transform of sin

t

^

{sin

t} =

dt e

e j 2

e

dt e) t ω (sin

0

st tωj

tωj

0

st

^  

2

2

0

t) ωj s(

t) ωj s(

ω s

ω

ωj s

ωj s

1 j 2

dt

e j 2

e

^  
^

^ 

 

  