TRANSFORM AND ITS APPLICATION

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LAPLACE

TRANSFORM AND

ITS APPLICATION

IN CIRCUIT

ANALYSIS

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12.1 Definition of the Laplace Transform

12.2 Useful Laplace Transform Pairs

12.3 Circuit Analysis in S Domain

12.4 The Transfer Function and the Convolution

Integral

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12.5 The Transfer Function and the Steady state

Sinusoidal Response

12.6 The Impulse Function in Circuit Analysis

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12.1 Definition of the Laplace Transform

Pierre Simon Laplace (1749-1827) :

A French astronomer and mathematician

First presented the Laplace transform and

its applications to differential equations

in 1979.

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12.1 Definition of the Laplace Transform

Definition:

[

( )

]

( ) ₀ ( ) a complex variable st L f t F s f t e dt s σ jω − ∞ ₋ = = = +

∫

The Laplace transform is an integral

transformation of a function f(t) from the time domain into the complex frequency domain, F(s).

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12.1 Definition of the Laplace Transform

[

]

1

1 1

( ) ( ) ( )

2

Look-up table , an easier way for circuit application ( ) ( ) j _st j L F s f t F s e ds j f t F s + − − = = ⇔

∫

σσ ωω π

One-sided (unilateral) Laplace transform Two-sided (bilateral) Laplace transform

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12.1 Definition of the Laplace Transform

Similar to the application of phasor transform to solve the steady state AC circuits , Laplace

transform can be used to transform the time domain circuits into S domain circuits to simplify the

solution of integral differential equations to the manipulation of a set of algebraic equations.

8 8 C.T. Pan C.T. Pan 88 Functions f(t) , t> F(s) impulse 1 step ramp _t exponential sine 0− ( )t δ ( ) u t 2 1 S at e− _S1₊_a 1 S sin tω 2 2 S ω ω +

12.2 Useful Laplace Transform Pairs

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9 9 C.T. Pan C.T. Pan 99 Functions f(t) , t> F(s) cosine damped ramp damped sine damped cosine 0− cos tω 2 2 S S +ω at te− cos at e− ωt ( )2 1 S+a ( )2 ₂ S a ω ω + + sin at e− ωt ( )2 2 S a S a ω + + +

12.2 Useful Laplace Transform Pairs

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12.2 Useful Laplace Transform Pairs

[ ] [ ] [ ] [ ] -0 0 0 ( ) ( ) ( 0 ) ( ) ( ) ( ) ( ) ( ) , 0 ( ) ( ) 1 ( ) ( ) , 0 ( ) ( ) ( ) ( )

l i m

t a s a t s t t s d L f t s F s f d t F s L f d S L f t a u t a e F s a L e f t F s a s L f a t F a a a f t s F s f t s F s + − − − → ∞ → → ∞ →   ∆ _ _ = −     ∆ _ _ =   ∆ − − = >   ∆ _ _ = + ∆ = > ∆ = ∆ =

∫

τ τ

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12.2 Useful Laplace Transform Pairs

Example

Use the Laplace transform to

solve the differential equation.

[

]

2 2 2 6 8 2 ( ) (0) 1 '(0) 2

Take Laplace transfrom

2 ( ) (0) '(0) 6 ( ) (0) 8 ( ) d v dv v u t dt dt v v s V s sv v sV s v V s s + + = = = −  − − + − + =   12 12 C.T. Pan C.T. Pan 1212

12.2 Useful Laplace Transform Pairs

[

]

2 2 2 2 2 2 2 4 2 ( ) (0) '(0) 6 ( ) (0) 8 ( ) 4 2 ( 6 8) ( ) 4 2 4 2 ( ) ( 2)( 4) ( 6 8) 1 ( ) (1 2 ) ( ) 4 t t s V s sv v sV s v V s s s s s s V s s s s s s V s s s s s s s v t e− e− u t  ₋ ₋ ₊ ₋ ₊ ₌   + + + + = + + + + ∴ = = + + + + = + +

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12.3 Circuit Analysis in S Domain

(1) KCL , ( ) 0 , for any node.

Take Laplace Transform

( ) 0 , for any node.

(2) KVL , ( ) 0 , for any loop.

Take Laplace Transform ( ) 0 , k n k n k m k m i t I s v t V s = = = =

∑

for any loop.

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12.3 Circuit Analysis in S Domain

(3) Circuit Component Models resistor ( ) ( ) ( ) ( ) ( ) ( ) R R R R R R v t Ri t V s RI s I s GV s = = =

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12.3 Circuit Analysis in S Domain

0 inductor ( ) 1 ( ) (0 ) ( ) ( ) ( ) (0 ) ( ) (0 ) ( ) L L t L L L L L L L L di v t L dt i t i v d L V s sLI s Li V s i I s sL s − − − − = = + = − = +

∫

τ τ LiL(0 ) − (0 ) L i s − 16 16 C.T. Pan C.T. Pan 1616

12.3 Circuit Analysis in S Domain

0 capacitor 1 ( ) (0 ) ( ) ( ) ( ) (0 ) (0 ) 1 ( ) ( ) C C t C C C C C c C C C dv i C dt v t v i d C I s sCV s Cv v V s I s s sC − − − − = = + = − = +

∫

τ τ 1 sC (0 ) C v s − 1 sC CvC(0 ) −

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12.3 Circuit Analysis in S Domain

M i2 i1 L1 _L2 v1 v2 + - -+ Coupling inductors V₁(s)=L₁SI₁(s)-L₁i₁(0)+MSI₂(s)-Mi₂(0) V₂(s)= MSI₁(s)- Mi₁(0)+L₂SI₂(s)-L₂i₂(0) dt d L dt d M dt d M dt d L 2 2 1 2 2 1 1 1 i i v i i v + = + = 18 18 C.T. Pan C.T. Pan 1818

12.3 Circuit Analysis in S Domain

For zero initial conditions

( ) impedance ( ) ( ) ( ) 1 admittance ( ) ( ) ( ) ( ) ( ) ( ) ' V s Z s I S I S Y s V s Z s

V s Z s I s ohm s law in s domain

=

= =

= −

@ @

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12.3 Circuit Analysis in S Domain

The elegance of using the Laplace transform in circuit analysis lies in the automatic inclusion of the initial conditions in the transformation process, thus providing a complete (transient and steady state) solution.

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12.3 Circuit Analysis in S Domain

Circuit analysis in s domain

nStep 1 : Transform the time domain circuit into s-domain circuit.

nStep 2 : Solve the s-domain circuit.

e.g. Nodal analysis or mesh analysis. nStep 3 : Transform the solution back into time

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12.3 Circuit Analysis in S Domain

Example Find v_o(t) given v_o(0)=5V

S-domain equivalent circuit

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12.3 Circuit Analysis in S Domain

2 Nodal analysis 10 ( ) ( ) ( ) 1 _{2 0.5} 10 10 10 / 25 35 10 15 ( ) ( 1)( 2) 1 2 ( ) (10 15 ) ( ) V o o o o t t o V s V s V s s s s V s s s s s v t e− e− u t − + _{+ +} ₌ ₊ + ∴ = = + + + + + ∴ = + 10 1 s+ 10 s

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12.4 The Transfer Function and the

Convolution Integral

Given a linear circuit N in s domain as shown below

Transfer function H(s) is defined as

) s ( X ) s ( Y ) s ( H = 24 24 C.T. Pan C.T. Pan 2424

12.4 The Transfer Function and the

Convolution Integral

( ) ( ) , ( ) ( ) ; ( ) ( ) ( ) , ( ) ( ) ; ( ) ( ) ( ) , ( ) ( ) ; ( ) ( ) ( ) , ( ) ( ) ; ( ) o i o i

If Y s V s X s V s then H s voltage gain

If Y s I s X s I s then H s current gain

If Y s V s X s I s then H s impedance If Y s I s X s V s then H s admittance = = = = = = = = = = = =

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12.4 The Transfer Function and the

Convolution Integral

Given the transfer funtion H(s) and input

X(s) , then Y(s)=H(s)X(s)

If the input is δ(t) , then X(s)=1 and

Y(s)=H(s)

Hence , the physical meaning of H(s) is in

fact the Laplace transform of the impulse

response of the corresponding circuit.

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12.4 The Transfer Function and the

Convolution Integral

in time domain

Y(s)=H(s)X(s) , in s-domain

( ) ( ) ( ) ( ) ( ) y t h t τ τ τx d h t x t ∞ −∞ =

∫

− @ ∗

Geometrical interpretation of finding the

convolution integral value at t=t

_k

is based on :

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12.4 The Transfer Function and the

Convolution Integral

(1)Approximating the input function by using a series of impulse functions.

(2)Shifting property of linear systems input x(t)→output y(t)

x(t-τ)→output y(t- τ)

(3)Superposition theorem for linear systems (4)Definition of integral : finding the area

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12.4 The Transfer Function and the

Convolution Integral

(1)Input x(τ) is approximated using impulse functions , x(τ)=0 , for τ<0 0 1 1 1 2 2 2 3 3 3 , ( ) , 0 , (0) , (0) , ( ) , ( ) 2 , ( ) , ( ) 3 , ( ) , ( ) k k

instant x value area

x x x x x x x x τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ = = = = V V V V V V V 0 τ1 τ2 τ3 τ4 τ5 (0) x ( ) xτ 0 τ1 ( ) xτ 2 τ 0 1 1 2 2 0 ( ) ( ) ( ) ( ) ( ) ( ) k x f f f x k k ∞ = ≅ + − + − + −

∑

L @ V V V τ δ τ δ τ τ δ τ τ τ τδ τ τ

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12.4 The Transfer Function and the

Convolution Integral

(2) Use the linearity property

1 1 1 1 2 2 2 2 ( ) (0) ( ) (0) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

input output response

x x h x x h x x h up to τδ τ τ τ τ τδ τ τ τ τ τ τ τ τδ τ τ τ τ τ τ τ → → − → − − → − V V V V V V M M k t = 30 30 C.T. Pan C.T. Pan 3030

12.4 The Transfer Function and the

Convolution Integral

(3) Use superposition theorem to find the total approximate response $ 0 ( ) ( ) ( ) [ ] n k k k k y t x k h t k t n integer = = − =

∑

V V V V τ τ τ τ

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12.4 The Transfer Function and the

Convolution Integral

(4) Take the limit , Δτ→dτ,

( )_k ( ) ( _k )

y t x h t d

∞

−∞

=

∫

τ −τ τ

Due to causality principle , h(t-τ)=0 for t≥t_k and x(τ)=0 for τ<0 0 ( ) ( ) k t k h t x d =

∫

−τ τ τ $( )y t_k →y t( )_k 32 32 C.T. Pan C.T. Pan 3232

12.4 The Transfer Function and the

Convolution Integral

( ) ( ) , ( ) ( ) , (4) , ( ) ( ) ( ) (4) (4 ) ( ) t Given x t u t h t e u t find y y t h t x d y h x d − ∞ −∞ ∞ −∞ = = = − ∴ = −

∫

τ τ τ τ τ τ Example:

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12.4 The Transfer Function and the

Convolution Integral

Step1.

4 2 0 2 4  0.2 0.4 0.6 0.8 1.0 x 4 2 0 2 4  0.2 0.4 0.6 0.8 1.0 h ( ) xτ τ τ ( ) h−τ 2 − 2 − 4 − 4 − 34 34 C.T. Pan C.T. Pan 3434

12.4 The Transfer Function and the

Convolution Integral

Step2. Shift to t_k=4 4 2 0 2 4  0.2 0.4 0.6 0.8 1.0 h(44  ) h − τ 4 − −2 τ

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12.4 The Transfer Function and the

Convolution Integral

Step3. Find the product h(4-τ)x(τ)

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12.4 The Transfer Function and the

Convolution Integral

Step4. Find the integral (area)

4 ₍₄ ₎ ₄ 4 ₄ 4 0 0 0 4 4 4 (4) = = ( 1) (1 ) y e d e e d e e e e e τ _τ τ _τ τ − − − − − − = = − = −

∫

Step5. Check 4 1 1 1 1 ( ) ( ) ( ) 1 ( ) (1 ) , 1 (4) (1 ) t a a Y s H s X s s a s s a s y t e a a y e − − − = = × = + + + ∴ = − = ∴ = −

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

From definition of transfer function

( ) ( ) ( ) ( ) ( ) ( ) Y s H s X s Y s H s X s = ⇒ =

Assume input X(t)=Acos(wt+Ф) and H(s) is given, then one can get the steady state solution without needing a separate phasor analysis.

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

2 2 * 1 1 * 1 1 1

: ( ) cos cos sin sin

(cos sin ) ( ) ( ) ( ) ( ) : ( ) 1 ( ) 2 ( ) ( cos sin ) ss j proof X t A t A t A s X s s Y s H s X s K K s j s j

other terms due to poles under steady state

K K Y s s j s j K H j Ae s j H s A s s j φ φ φ φ φ φ φ = ω − ω − ω ∴ = + ω ∴ = = + − ω + ω +∑ ∴ = + − ω + ω = = ω = ω −ω + ω

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

( ) ( ) | ( ) | ( ) | ( ) | cos[ ( )] . . ( ( )) ( ( )) | ( ) | ( ) j ss ss Let H j H j e

and take inverse Laplace transform

Then y t A H j t i e P X t A then P y t A H j θ φ θ φ φ θ ω ω = ω = ω ω + + ω = ∠ = ω ∠ + ω 40 40 C.T. Pan C.T. Pan 4040

Example : The transfer function H(s) of the circuit given below is known.

Find the steady state solution of V_o(t) for the given V_g(t).

12.5 The Transfer Function and the

Steady State Sinusoidal Response

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

2 6 1000( 5000) ( ) 6000 25 10 ( ) 120 cos(5000 30 ) g s H s s s V t t ° V + = + + ⋅ = + 42 42 C.T. Pan C.T. Pan 4242

12.5 The Transfer Function and the

Steady State Sinusoidal Response

6 6 : 5000 1000( 5000 5000) ( 5000) 25 10 5000(6000) 25 10 2 45 6 2 ( ) 120 cos(5000 30 45 ) 6 20 2 cos(5000 15 ) ss o Solution Let s j j j Evaluate H j j Then V t t t V ° ° ° ° = ω= + = − × + + × = ∠− = × + − = −

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nIn theory , the relationship between H(s) and H(jw) provides a link between the time domain and the frequency domain.

nIn some cases , we can determine H(jw)

experimentally and then construct H(s) from the data.

12.5 The Transfer Function and the

Steady State Sinusoidal Response

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

Example: Find the impulse response of the

following circuit. following circuit. in

v

R

+

_

o

v

+

_

C

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

(a) Time domain solution

0 ( ) 0 , (0 ) 0 1 ( ) 0 , (0 ) 1 o o o t o dv RC v t dt At t v t At t v dt C R V RC δ δ − − + + + = = = = = =

∫

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

0 , ( ) 0 For t> + δ t = R 1 ( ) , (0 ) o v t v RC + ₌ + _ C 0 1 ( ) ( ) o o t RC o dv RC v dt v t e u t RC − + = ∴ =

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12.5 The Transfer Function and the

Steady State Sinusoidal Response

(b) s

(b) s--domain solutiondomain solution

. . ( ) ( ) ( ) o in _{zero I C} V s H s V s =

Find the transfer function

Transform into s

Transform into s--domain circuitdomain circuit

( ) in V s R + _ ( ) o V s 1 Cs 48 48 C.T. Pan C.T. Pan 4848

12.5 The Transfer Function and the

Steady State Sinusoidal Response

-1 1 ( ) 1 ( ) 1 ( ) 1 ( ) [ ( )] 1 ( ) o in t RC V s _Cs H s V s _R RCs Cs h t H s e u t RC − ∴ = = = + + ∴ = = L ( ) in V s R + _ ( ) o V s 1 Cs Same answer. Same answer.

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12.6 The Impulse Function in Circuit

Analysis

Example 1: Impulse voltage source excitation

( ) O V δ t ( ) i t

(0 )

0 i

−

=

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12.6 The Impulse Function in Circuit

Analysis

0 0, (0 ) 0 1 (0 ) ( ) ( ) t O O At t i i V x dx L V A L δ − − + = = ∴ = =

∫

V_Oδ( )t (0 ) i +

The impulse voltage source has stored energy,

, in the inductor as an initial current

in an infinitesimal moment. in an infinitesimal moment. 2 1 2L(i(0 )) +

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12.6 The Impulse Function in Circuit

Analysis

0 , ( ) 0 For t> + δ t = ( ) i t / 0 , (0 ) ( ) ( ) , O t O di L Ri natural response dt V i A L V L i t e u t L R τ _τ + − + = = ∴ = =

Note that the impulse source just builds up an

initial inductor current but does not contribute

to any forced response.

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12.6 The Impulse Function in Circuit

Analysis

(b) s

(b) s--domain solutiondomain solution

/ / ( ) ( ) ( ) O O t O V V L I s R R sL _s L V i t e u t L τ − ∴ = = + ₊ ∴ = O V R sL ( ) I s

Same answer but much easier.

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12.6 The Impulse Function in Circuit

Analysis

Example 2: Impulse current source excitation

R C + _ ( ) , (0 ) 0 v t v − = ( ) o I δ t 54 54 C.T. Pan C.T. Pan 5454

12.6 The Impulse Function in Circuit

Analysis

0, (0 ) 0 (0) 0 , At t v v short circuit − = = ∴ = 0 1 (0 ) ( ) t o o v I x dx C I C δ + ₌ =

∫

The impulse current source has stored energy,

, in the capacitor as an initial voltage

in an infinitesimal moment. in an infinitesimal moment. 2 1 2C(v(0 )) + ( ) o I δ t _R

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12.6 The Impulse Function in Circuit

Analysis

0 , (0 ) ( ) ( ) , o t o dv v C natural response dt R I v C I v t e u t RC C τ _τ + − + = = ∴ = = 0 , ( ) 0,

For t> + δ t = open circuit

( )

v t

Note that the impulse current just builds up an

initial capacitor voltage but does not contribute

to any forced response.

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12.6 The Impulse Function in Circuit

Analysis

(b) s

(b) s--domain solutiondomain solution Transform into s

Transform into s--domain circuitdomain circuit

R C + _( ) V s o I 1 / V( ) 1 1 ( ) ( ) o o t o R I C Cs s I R s Cs RC I v t e u t C τ − × ∴ = = + + ∴ =

Same answer but much easier.

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12.6 The Impulse Function in Circuit

Analysis

Example 3: Impulse caused by switching operation

1(0 ) 2(0 )

Note that v − ≠v −

The switch is closed at t=0 in the following circuit.

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12.6 The Impulse Function in Circuit

Analysis

Transform into s

Transform into s--domaindomain

1 2 1 2 1 2 1 2 2 1 2 ( ) 1 1 , ( ) o o e o e o e V s I s V C sC sC C C V C V C C V s C C sC s C C = ⋅ + ⋅ ⋅ = = = ⋅ + + @ 1 2 1 2 ( ) ( ) ( ) o e o i t V C t C v t V C C δ ∴ = = +

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12.6 The Impulse Function in Circuit

Analysis

At t=0 , a finite charge of C

At t=0 , a finite charge of C₁₁is transferred to Cis transferred to C₂₂ instantaneously.

instantaneously.

Note that , as the switch is closed , the voltage

across C

across C₂₂does not jump to Vdoes not jump to V_o_oof Cof C₁₁but to its final but to its final value of the two paralleled capacitors.

value of the two paralleled capacitors.

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12.6 The Impulse Function in Circuit

Analysis

2 1 2 2 2 2 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 , 0 , 0 + , 0 , 0 , , 0 . o o o o o C C N o te Q C V V t C C C Q C V V t C C Q Q C V t A ls o a t t Q C V Q Q Q C V C o n s e r v a tio n o f c h a r g e + + + − ⋅ = ⋅ = > + = ⋅ = > + = > = = = ∴ + =

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12.6 The Impulse Function in Circuit

Analysis

If we consider charged capacitors as voltage sources ,

then we should not connect two capacitors with

unequal voltages in parallel.

Due to violation of KVL , an impulse will occur which

may damage the components.

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12.6 The Impulse Function in Circuit

Analysis

Example 4: Impulse caused by switching operation

The switch is opened at

The switch is opened at t t = 0 in the following circuit.= 0 in the following circuit.

100V 10Ω 3H 1 i L1 2 i 2 L 2H 15Ω 0 t= vo + _ 1 2 0 , 100 (0 ) 10 10 (0 ) 0

At t steady state solution V i A i A − − − = = = Ω =

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12.6 The Impulse Function in Circuit

Analysis

For t > 0, the S

For t > 0, the S--domain circuit isdomain circuit is

10Ω 3s ( ) I s L1 2 L 2s 15Ω ( ) o V s 100 s 30 5 5 (100 / ) 30 4 2 ( ) 5 25 5 60 10 ( ) ( ) (2 15) 12 5 ( ) (4 2 ) ( ) ...( ) ( ) 12 ( ) (60 10 ) ( ) ...( ) o t t s I s s s s V s I s s s s i t e u t A v t δ t e u t B − − + = = + + + = + = + + + ∴ = + = + + 64 64 C.T. Pan C.T. Pan 6464

12.6 The Impulse Function in Circuit

Analysis

1 2 1 2 0 , (0 ) 10 , (0 ) 0 0 , ( ), (0 ) 6 , (0 ) 6 , ( ), 12 ( ) ( ). L L L L o Note t i A i A t from A i A i A

Also from B there exists δ t at v t

− − + + + = = = = = = 10Ω 3s ( ) I s L1 2 L 2s 15Ω ( ) o V s 100 s 30

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12.6 The Impulse Function in Circuit

Analysis

Thus, if we consider an inductor current as a

current source, then two inductors with unequal

currents should not be connected in series.

Due to violation of KCL, it will result in impulse

voltage which may damage the components.

10Ω 3s ( ) I s L1 2 L 2s 15Ω ( ) o V s 100 s 30

SUMMARY

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Objective 1 : Know the component models in s

Objective 1 : Know the component models in s--domain.domain.

Objective 2 : Be able to transform a time domain circuit

into the s

into the s--domain circuit.domain circuit.

Objective 3 : Know how to analyze the s

Objective 3 : Know how to analyze the s--domain circuitdomain circuit and transform the solution back to time

and transform the solution back to time

domain.

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SUMMARY

Objective 4 : Understand the significance of transfer

function and be able to calculate the

transfer function from the s

transfer function from the s--domain circuit.domain circuit.

Objective 5 : Know the geometrical interpretation of

convolution integral and be able to

calculate the integral.

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SUMMARY

Objective 6 : Know the relation between the

Objective 6 : Know the relation between the phasorphasor solution technique for finding sinusoidal

solution technique for finding sinusoidal

steady state solution and the s

steady state solution and the s--domain domain solution technique .

solution technique .

Objective 7 : Know how to use s

Objective 7 : Know how to use s--domain solution domain solution technique to solve a circuit containing