Unit – 4
Laplace transforms and properties
It is a technique for solving differential equations. The time domain differential equations are transformed into algebraic equation of frequency domain. After solving the algebraic equation in frequency domain, the result then is finally transformed to time domain form to achieve the ultimate solution of the differential equation.
The Laplace Transform is given as
F(s)=
dt
The advantages of Laplace Transform are:
1) It is systematic.
2) It gives total solution (transient and sustained solution) in one operation.
3) The initial conditions are automatically specified in the transformed equations.
Properties of Laplace Transform:
1) Linearity Property:
If f1(t) and f2(t) are two functions of time. Then, in domain of convergence
L[a f1(t)+b f2(t)]=a
+ b
=aF1(s)+bF2(s)
2) Differentiation Property:
If x(t) is function of time then Laplace transform of nth derivative is given as
3) Integration Property:
The Laplace of nth order integral is given as
L[
]=
+
L[f-n(t)]=
+
+
+……….
As 
……..=0. Hence
L[f-n(t)]=
Inverse Laplace transform
The Laplace inverse of any function H(s) is given by
L-1 H(s) = h(t)
Properties of Inverse Laplace transform are
Linearity property
L-1 {aH1(s)+b H2(s)} = ah1(t)+bh2(t)
Shifting property
L-1 H(s) = h(t)
L-1 H(s-a) = eat h(t)
If L-1 H(s) = h(t)
Then, L-1 [H(s)/s] = 
If L-1 H(s) = h(t)
L-1 e-at H(s) = u(t-a). h(t-a)
Inverse Laplace transform using partial fraction expansion:
- The roots of D(s) (the values of s that make D(s) = 0) are called poles.
- The roots of N(s) (the values of s that make N(s) = 0) are called zeros.
Examples
Que) Find the inverse Laplace transform of
Sol:
We know that
Hence,
Que) Find inverse Laplace transform of
Sol:
Taking inverse Laplace of above we get
Que) Find Inverse Laplace transform of
Sol:
Que) Find Inverse Laplace Transform of
Sol:
Key takeaway
Laplace Transform of some common functions
Function, f(t) | Laplace Transformation, F(s) | Function, f(t) | Laplace Transformation, F(s) |
         
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Singularity functions are discontinuous functions or their derivatives are discontinuous. A singularity is a point at which a function does not possess a derivative. In other words, a singularity function is discontinuous at its singular points. Hence a function that is described by polynomial in t is thus a singularity function. The commonly used singularity functions are
Step Function,
Ramp Function, and
Impulse Function
The convolution theorem states that f1(t) and f2(t) are two time functions which are zero for t<0 and if Laplace transform are F1(s) and F2(s)then the transform of convolution of f1(t) and f2(t) is the product of individual transforms F1(s) and F2(s).
f(t) = f1(t)*f2(t) = 
= f2(t)*f1(t) = 
He convolution theorem of
Laplace transform is
L[f1(t)*f2(t)] = L[f2(t)*f1(t)] = F1(s)F2(s)
The Laplace Transformation transforms convolution into multiplication.
The graphical presentation of the convolution integral helps in the understanding of every step in the convolution procedure. According to the definition integral, the convolution procedure involves the following steps:
1) Apply the convolution duration property to identify intervals in which the convolution is equal to zero.
2) Flip about the vertical axis one of the signals (the one that has a simpler form (shape) since the commutativity holds), that is, represent one of the signals in the time scale.
3) Vary the parameter from -
to +
, that is, slide the flipped signal from the left to the right, look for the intervals where it overlaps with the other signal, and evaluate the integral of the product of two signals in the corresponding intervals.
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.3) 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)=
+
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)=
+
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
=0
This is first order homogeneous differential equation so
dt
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
Sol: From above circuit
L = 2H
LiL (0) = 25 mv
IL (0) = 12.5 m A
Que: Find value across capacitor?
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?
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
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-
:. The capacitor is connected for long
Vc1(0-)= V
iL(0-) =0
Vc2(0-)=0
At t= 0+
i1 (0+) = -V/2R
For t>0 the circuit is
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)?
Sol: Drawing Laplace circuit
(2+S) 11 2/S
Zeq = [(2+S)2/S]/[2+S+2/S] = [2(S+2)]/[S2+2s+2]
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
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-
Fig 17 circuit at t=0-
:. The capacitor is connected for long
Vc1(0-) = V
iL(0-) =0
Vc2(0-) =0
At t= 0+
Fig 18 Circuit at t=0+
i1 (0+) = -V/2R
For t>0 the circuit is
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)?
Sol: Drawing Laplace circuit
(2+S) 11 2/S
Zeq = [(2+S)2/S]/[2+S+2/S]=[2(S+2)]/[S2+2s+2]
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
Que) Find v0 (t) for t > 0.
Sol: Taking Laplace transform of above circuit we have,
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 
When 
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?
Sol: By KVL we have
Taking LT of above equations, we get
Solving above equation we get
By inverse Laplace transform we get
The voltage across 5 Ohm resistor is
The initial value of i1(t) is
The final value of i2(t) is
The initial value of i2(t) is
The Final value of i2(t) is
References:
M. E. Van Valkenburg, “Network Analysis”, Prentice Hall, 2006.
• D. RoyChoudhury, “Networks and Systems”, New Age International Publications, 1998.
• W. H. Hayt and J. E. Kemmerly, “Engineering Circuit Analysis”, McGraw Hill
Education, 2013.
• C. K. Alexander and M. N. O. Sadiku, “Electric Circuits”, McGraw Hill Education, 2004.
• K. V. V. Murthy and M. S. Kamath, “Basic Circuit Analysis”, Jaico Publishers, 1999.
