4. Deals with Aperiodic Signals
Input signal changing often at t=0
Stability analysis
Region of Convergence
Time Domain to complex
frequency domain(S-Domain)
4
Laplace Transform
5. Formula for Laplace Transform
5
▰ It is used to transform a time domain to complex frequency domain signal (s-domain)
▰ Two Sided Laplace transform (or) Bilateral Laplace transform
Let 𝑥(𝑡) be a continuous time signal defined for all values of 𝑡.
Let 𝑋(𝑆) be Laplace transform of 𝑥(𝑡).
▰ One sided Laplace transform (or) Unilateral Laplace transform
Let 𝑥(𝑡) be a continuous time signal defined for 𝑡≥0 (ie If 𝑥(𝑡) is causal)
then,
6. ▰ Inverse Laplace transform (S-domain signal 𝑋(𝑆) Time domain signal x(t) )
▰ Transform: x(t) X(s), where t is integrated and s is variable
▰ Conversely X(s) x(t), t is variable and s is integrated
▰ The Laplace transform helps to scan exponential signal and sinusoidal signal
6
Complex variable, S= α + jω
Formula for Laplace Transform
7. Laplace transform for elementary signals
1)
Solution
2)
Solution
7
Impulse signal
L[δ(t)]
Step signal
L[u(t)]
8. Laplace transform for elementary signals
3)
Solution
4)
Solution
8
Constant
Exponential signal
10. Laplace transform for elementary signals
7)
8)
9)
10
Hint
x(t) = cos ω0 t u(t)
x(t) = sin ω0 t u(t)
11. Laplace transform for elementary signals
8)
Solution
Using Euler’s Formula
11
----> (1)
(𝒔 + 𝒊𝒂)
(𝒔 + 𝒊𝒂)
----> (2)
Compare (1) and (2)
Real part Imaginary part
13. Advantages of Laplace Transform
▰ Signal which are not convergent on Fourier
transform, will converge in Laplace transform
13
14. Complex S Plane
▰ The most general form of Laplace
transform is
▰ L[x(t)]= X(s) =
𝑵(𝑺)
𝑫(𝑺)
14
LHS RHS
- ∞ 0 ∞
jω
σ
Complex variable, S= α + jω
The zeros are found by setting the numerator polynomial to Zero
The Poles are found by setting the Denominator polynomial to Zero
15. Region of Convergence
The range variation of complex variable ‘s’ (σ) for which the Laplace transform
converges(Finite) is called region of convergence.
Properties of ROC of Laplace Transform
ROC contains strip lines parallel to jω axis in s-plane.
ROC doesn’t contain any poles
If x(t) is absolutely integral and it is of finite duration, then ROC is entire s-
plane.
If x(t) is a right sided signal(causal) then ROC : Re{s} > σ of X(s) extends
to right of the rightmost pole
If x(t) is a left sided signal then ROC : Re{s} < σ of X(s) extends to left of
the leftmost pole
If x(t) is a two sided sequence then ROC is the combination of two regions.
15
16. Properties of ROC of Laplace Transform
▰ ROC doesn’t contain any poles
▰ If x(t) is absolutely integral and it is of finite duration, then ROC is entire
s-plane.
16
L[e-2t u(t)] 1/(s+2)
1/(-2+2)
Poles, S=-2 = 1/0 = ∞
-a a
x(t)
ROC includes
imaginary axis jω
- ∞ 0 ∞
jω
σ
Impulse signal have ROC is entire S plane
17. ▰ If x(t) is a right sided signal(causal) then
ROC : Re{s} > σ of X(s) extends to right
of the rightmost pole
▰ If x(t) is a left sided signal then ROC :
Re{s} < σ of X(s) extends to left of the
leftmost pole
17
- ∞ 0 ∞
jω
σ
- ∞ 0 ∞
jω
σ
18. ▰ If x(t) is a two sided sequence then ROC is the combination of two
regions.
18
- ∞ 0 ∞
jω
σ
19. Problem using ROC
19
1. Find the Laplace transform and ROC of x(t)=e-at u(t)
Solution
2. Find the Laplace transform and ROC of x(t)=eat u(−t)
Solution
Right sided
signal
Left sided
signal
20. 20
Find the Laplace transform and ROC of x(t)=e−at u(t)+eat u(−t)
Solution
Problem using ROC
Both sided signal
Referring to the diagram, combination region lies from
–a to a. Hence,
21. Shortcut for ROC
▰ Step 1: Compare real part of S complex variable (σ) with real part of
coefficient of power of e
▰ Step 2: Check if the signal is left sided or right sided, then decide < or >
21
Consider L[e-2t u(t)]
Step 1 : σ = -2
Step 2 : σ > -2
ROC is
22. Roc helps to check the impulse response is absolutely
integrable or not
22
Shortcut for ROC
Find Roc of following signals
1. x(t) = e-2t u(-t)
2. x(t) = e3t u(t)
3. x(t) = e(4+3j)t u(-t)
4. x(t) = e-4t u(t)
5. x(t) = e3t u(t) + e-2t u(t)
6. x(t) = e3t u(-t) + e-2t u(-t)
7. x(t) = e-3t u(t) + e2t u(-t)
- ∞ 0 ∞
jω
σ
23. Causality and Stability
▰ For a system to be causal, all poles of its transfer
function must be left half of s-plane.
▰ For causal system: A system is said to be stable
all poles of its transfer function must be left half of
s-plane, (ROC include Imaginary axis jω)
▰ For Anticausal system: A system is said to be
stable all poles of its transfer function must be
RHS of s-plane, (ROC include Imaginary axis jω)
▰ A system is said to be unstable when at least one
pole of its transfer function is shifted to the right
half of s-plane.(ROC doesn’t include Imaginary
axis jω)
23
σ
σ
σ
jω
Poles
24. 24
Problems
Check causality and stability
1. x(t) = e-2t u(-t)
2. x(t) = e3t u(t)
3. x(t) = e(4+3j)t u(-t)
4. x(t) = e-4t u(t)
5. x(t) = e-3t u(t) + e-2t u(t)
For a system to be causal, all poles of its transfer function must be right half of s-plane.
If signal is causal, then ROC Re{s} >a
If signal is Non causal, then ROC Re{s} <a
25. Causality and Stability
25
5. Find the LT and ROC of x(t)=e−3t u(t)+e-2t u(t), Check causality and stability
Solution:
L[x(t)= L[e−3t u(t)+e-2t u(t)]
X(s) =
1
(𝑠+3)
+
1
(𝑠+2)
ROC: Re{s} =σ >-3, Re{s} =σ >-2
ROC: Re{s} =σ >-2
- ∞ -3 -2 0 ∞
jω
σ
Both will converged if Re{s} =σ >-2
Causal and stable
29. Properties of Laplace Transform
▰ Linearity
▰ Time Scaling
▰ Time shifting
▰ Frequency or s-plane shift
▰ Multiplication by tn
▰ Integration
29
▰ Differentiation
▰ Convolution
▰ Initial Value Theorem and
Final value Theorem
36. Frequency Shifting(s- Shifting) or
Modulation in frequency
36
Proof:
Hence Proved
x(t) X(s)
Problem:
Solution:
1. Find the Laplace transform of and
Solution:
37. 2. solve
37
Solution:
= e-6 L[e-3(t-2) u(t-2)]
Given : t0 =2
= e-6 L[e-3t] e-2s
Wkt, L[e-at ] = (1/s+a)
= e-2s e-6(1/s+3)
=
e−(2s+6)
(s+3)
=
e−2(s+3)
(s+3)
x(t) X(s)
= L[e-3(t-2+2) u(t-2)]
L[u(t-2)] =
𝑒−2𝑠
𝑠
Sub : s by s+3
=
𝑒−2(𝑠+3)
(𝑠+3)
Problem : Frequency Shifting+ Time Shifting
