Unit 4 | unit 4 electric circuit analysis using laplace transforms

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Unit - 4

Electric Circuit Analysis using Laplace Transforms

4.1 Review of Laplace transform

Consider an LTI system by a complex exponential signal of the form

x(t) = Gest.

Where s = any complex number = σ+jω

σ = real of s, and

ω = imaginary of s

The response of LTI can be obtained by the convolution of input with its impulse response i.e.

y(t)=x(t)×h(t)=

Where H(S) = Laplace transform of

Similarly, Laplace transform of

4.2 Waveform synthesis

1) Step Function:

U(t)=1 t≥0

=0 t<0

Fig 1 Step Signal

Fig 2 Shifted unit step signal

Fig 3 Unit step signal -u(t)

2) Pulse Function:

Fig 4 Pulse signal

L[u(t)]==

f(t)<—>F(s)

f(t-a)=F(s)

L[u(t-a)]=

Laplace transforms of pulse

Fig 5 Shifted Pulse signal

Fig 6 Shifted Pulse signal

Fig 7 Final required Pulse signal

Examples

Q.1) Write equation for given waveform

Sol:

The final equation is=u(t)+u(t-1)+u(t-2)+u(t-3)-4u(t-4)

As the waveform is stopped at t=4 with amplitude 4,so we have to balance that using -4u(t-4).

Ramp function:

Fig 8 Ramp Function

Fig 9 Shifted Ramp Function

Fig 10 Shifted Ramp Function

L[r(t)] =

R(t)=tu(t)

R(t)=∫u(t)

Impulse Signal:

This signal has zero amplitude everywhere expect at the origin. Fig below shown the representation of Impulse signal.

Fig 11 Impulse Function

The mathematical representations

A(t) = 0 for t ≠0

dt = A e

Where A represent energy or area of the Laplace Transform of Impulse signal is

L [A (t) ]= A

The transfer function of a linear time invariant System is the Laplace transform of the impulse response of the system

Example

Q) Write the equation for the given waveform?

Sol: Calculating slope of above waveform

Slope for (0,0) and (1,2)

a1 slope==2

a2 slope==-4

a3 slope =4 and a4= -2

At t=1 slope changes from 2 to -4 so we need to add -6r(t-1) to 2r(t) to get the slope of -4.

At t=1.5 slope changes from -4 to 4 so adding 8r(t-1.5) to get slope of 4

At t=2 slope changes from +4 to -2 so we add -6r(t-2). But we need to stop the waveform at t=3. Hence, we have to make slope 0. Which can be done by adding +2r(t-3). Final equation is given as 2r(t)-6r(t-1.5)+8r(t-1.5)-6r(t-2)+2r(t-3)

Analysis of electrical circuits using Laplace Transform for standard inputs

LAPALACE TRANSFORM:

If LT[f(t)]= F(s)

Where F(s)=dt

f(t)F(s)

sF(s)-f(0-)

f(t)S2F(s)-sf(0-)-f’(0-)

Now, we find Laplace for

Now if f(t)=ic(t) [current through capacitor]

=[q(0)-q()]+

=[q(0)-0]+

Multiplying both sides by 1/C

Vc(t)=+

Vc(t)=Vc(0)+

By Laplace Transform

Vc(s)=+

Fig 12 Laplace transformed circuit for C

Now by source conversion

Fig 13 Source transformed circuit for C

Now if f(t)=VL(t) [voltage across inductor]

=-]+

=-0]+

Multiplying both sides by 1/L

iL(t)=iL(0)+

By Laplace Transform

IL(s)=+

NOTE: Voltage------->Series circuit

Current------->Parallel circuit

Fig 14 Laplace transformed circuit for L

By source transformation

Fig 15 Source transformed circuit for L

Current through capacitor

iC(t)=

IC(s)=C[sVc(s)-Vc(0)]

Vc(s)=+

Voltage about inductor,

VL(t)=

VL(s)=L[sIL(s)-iL(0)]

IL(s)=+

Examples

Que For given RL circuit find the differential equation for current i?

Fig 16 Series RL circuit

After switch is closed applying KVL

This is first order homogeneous differential equation so

Integrating both sides

Ln i= t+K

Taking antilog of both sides

i=k

At t=0

i(0)==I0

=ke0

The particular solution is given as

i= for t≥0

=for t<0

Ques: Find IL(0) for the given circuit below

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224139.jpg$

Sol: From above circuit

L = 2H

LiL (0) = 25 mv

IL (0) = 12.5 m A

Que: Find value across capacitor?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224353.jpg$

Sol: From above circuit we can write

1/cs = 2/s

C=1/2f

Vc (0)/s = 0.25v/s

Vc(0)= .25v

Ques: Find voltage across capacitor?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191229_224735.jpg$

Sol: 1/cs = 1/2S

C=2F

CVc (0) = 0.02

Vc(0)= 0.01V

Que:

(a) switch is at before moving to position = at t= 0, at t= 0+, i1(t) is

(a) –V/2R (b) –V/R (c) – V/4R

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160151.jpg$

Sol:

I1(t) L.T I1(s)

I2(t) L.T I2(s)

Then express in form

[ I1(s)/I2(s)] [::] = [:]

Drawing the ckl at t = 0-

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160708.jpg$

:. The capacitor is connected for long

Vc1(0-)= V

iL(0-) =0

Vc2(0-)=0

At t= 0+

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161211.jpg$

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161423.jpg$

i1 (0+) = -V/2R

For t>0 the circuit is

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161828.jpg$

Apply KVL

-R i1(t) -1/c dt -L=0

Also, we can deduce i1(t)- i2 (t) = iL(t)

-Ri1(t) -1/c - L = 0

-RI1(s)-[-]-L[sI1(s)-iL(0-)]=0

- RI1(s)- - -s L I2(s)=0

- s L I2(s)= - (1)

Again

-Ri2(t) ) -1/c dt -L=0

-Ri1(t) -1/c - L = 0

Taking Laplace Transform

-RI2(s)- [ -]- L[s IL(s)- iL(0-)]=0

-RI2(s)- -Ls[ I1(s)- I2(s)]=0

- s L I1(s)=0 (2)

Hence in matrix representation

Ques: All initial conditions are zero i(t) L.T, I(s), then find I(s)?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162652.jpg$

Sol: Drawing Laplace circuit

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162827.jpg$

(2+S) 11 2/S

Zeq = [(2+S)2/S]/[2+S+2/S]=[2(S+2)]/[S2+2s+2]

$C:\Users\TARUNA\Desktop\internship\29 dec 2019\IMG_20191228_132621.jpg$

V1(s) = [5/S{2(S+2)/S2+2S+2}]/[2+{2(S+2)/ S2+2S+2}]

V1(s) = [5(S+2)/S]/[S2+ 3S +4]

I(s) = V1(s)/(S+2) =(5/S)/(S2+35+4)

Initial and Final Value Theorem

The initial value theorem

This theorem is valid if and only if f(t) has no impulse functions.

The final value theorem

This theorem is valid if and only if all but one of the poles of F(s) are in the left-half of the complex plane, and the one that is not can only be at the origin.

Steps for Laplace transform circuit analysis:

1. Redraw the circuit (nothing about the Laplace transform changes the types of elements or their interconnections).

2. Any voltages or currents with values given are Laplace transformed using the functional and operational tables.

3. Any voltages or currents represented symbolically, using i(t) and v(t), are replaced with the symbols I(s) and V(s).

4. All component values are replaced with the corresponding complex impedance, Z(s).

5. Use DC circuit analysis techniques to write the s-domain equations and solve them. Check your solutions with IVT and FVT.

6. Inverse-Laplace transform s-domain solutions to get time domain solutions. Check your solutions at t = 0 and t = .

Transformed network with initial conditions

Solved Examples

Que: The switch is at before moving to position = at t= 0, at t= 0+, i1(t) is

(b) –V/2R (b) –V/R (c) – V/4R (d) 0

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160151.jpg$

Sol:

I1(t) L.T I1(s)

I2(t) L.T I2(s)

Then express in form

[ I1(s)/I2(s)] [::] .[:]

Drawing the ckl at t = 0-

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_160708.jpg$

Fig 17 circuit at t=0-

:. The capacitor is connected for long

Vc1(0-)= V

iL(0-) =0

Vc2(0-)=0

At t= 0+

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161211.jpg$

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161423.jpg$

Fig 18 Circuit at t=0+

i1 (0+) = -V/2R

For t>0 the circuit is

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_161828.jpg$

Fig 19 KVL in Circuit

Apply KVL

-R i1(t) -1/c dt -L=0

Also, we can deduce i1(t)- i2 (t) = iL(t)

-Ri1(t) -1/c - L = 0

-RI1(s)-[-]-L[sI1(s)-iL(0-)]=0

- RI1(s)- - -s L I2(s)=0

- s L I2(s)= - (1)

Again

-Ri2(t) ) -1/c dt -L=0

-Ri1(t) -1/c - L = 0

Taking Laplace Transform

-RI2(s)- [ -]- L[s IL(s)- iL(0-)]=0

-RI2(s)- -Ls[ I1(s)- I2(s)]=0

- s L I1(s)=0 (2)

Hence in the matrix representation

Ques: All initial conditions are zero i(t) I(s), then find I(s)?

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162652.jpg$

Sol: Drawing Laplace circuit

$C:\Users\TARUNA\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\IMG_20191231_162827.jpg$

(2+S) 11 2/S

Zeq = [(2+S)2/S]/[2+S+2/S]=[2(S+2)]/[S2+2s+2]

$C:\Users\TARUNA\Desktop\internship\29 dec 2019\IMG_20191228_132621.jpg$

V1(s) = [5/S{2(S+2)/S2+2S+2}]/[2+{2(S+2)/ S2+2S+2}]

V1(s) = [5(S+2)/S]/[S2+ 3S +4]

I(s) = V1(s)/(S+2) =(5/S)/(S2+35+4)

Que) There is no initial energy stored in this circuit. Find i1(t) and i2(t) for t>0?

Sol:

When circuit is closed the network is shown

(10s+90)I2 - 42 I1 = 0

I1 =

Que) Find v0 (t) for t > 0.

Sol: Taking Laplace transform of above circuit we have,

V(s) = (350 + 0.2s)I(s) - 0.04

Que) There is no initial energy stored in this circuit. Find vo if ig = 5u(t) mA.

Sol: Laplace Transform of above circuit we get

By voltage division we get

Que) Find value of current i(t) for the given circuit?

Sol:

Applying KVL we get

Taking Laplace transform of above equation we get

There is no initial current in inductor before closing the switch so i(0+)=0.

When α ≠

When α =

Taking ILT we get

Que) For the circuit shown below find the value of i1(t) and i2(t) and output voltage across 5 Ohm resistor when the switch is closed also find initial and final values of current?