1. CHAPTER 4
Laplace Transform.
EMT Signal Analysis

2. 4.0 Laplace Transform. 4.1 Introduction. 4.2 The Laplace Transform.
4.3 The Unilateral Transform and Properties.
4.4 Inversion of the Unilateral.
4.5 Solving Differential Equation with Initial Conditions.
4.6 Laplace Transform Methods in Circuit Analysis.
4.7 Properties of the Bilateral Laplace Transform.
4.8 Properties of the Region of Convergence.
4.9 Inversion of the Bilateral Laplace Transform.
4.10 The Transfer Function
4.11 Causality and stability
4.12 Determining the Function Response from poles and zeros

3. 4.1 Introduction.
In Chapter 3 we developed representation of signal and LTI by using superposition of complex sinusoids.
In this Chapter 4 we are considering the continuous-time signal and system representation based on complex exponential signals.
The Laplace transform can be used to analyze a large class of continuous-time problems involving signal that are not absolutely integrable, such as impulse response of an unstable system.
Laplace transform come in two varieties;
(i) Unilateral (one sided); is a tool for solving differential equations with initial condition.
(ii) Bilateral (two sided); offer insight into the nature of system characteristic such as stability, causality, and frequency response.

4. 4.2 Laplace Transform.
Let est be a complex exponential with complex frequency
s = s +jw. We may write,
The real part of est is an exponential damped cosine
And the imaginary part is an exponential damped sine as shown in Figure 4.1.
The real part of s is the exponential damping factor s.
And the imaginary part of s is the frequency of the cosine and sine factor, w.

5. Figure 4.1: Real and imaginary parts of the complex exponential est,
Cont’d…
Figure 4.1: Real and imaginary parts of the complex exponential est,
where s =  + j.

6. 4.2.1 Eigen Function Property of est.
Apply an input to the form x(t) =est to an LTI system with impulse response h(t). The system output is given by,
Derivation:
We use x(t) =est to obtain
We define transfer function as

7. Cont’d…
We can write
An eigen function is a signal that passes through the system without being modified except by multiplication by scalar.
The equation below indicates that,
- est is the eigenfunction of the LTI system.
- H(s) is the eigen value.

8. Cont’d… Express complex-value transfer function in Polar Form
Where |H(s)| and f(s) are the magnitude and phase of H(s)

9. 4.2.2 Laplace Transform Representation.
H(s) is the Laplace Transform of h(t) and the h(t) is the inverse Laplace transform of H(s).
The Laplace transform of x(t) is
The Inverse Laplace Transform of X(s) is
We can express the relationship with the notation

10. 4.2.3 Convergence.
The condition for convergence of the Laplace transform is the absolute integrability of x(t)e-σt ,
The range of s for which the Laplace transform converges is termed the region of convergence (ROC)
Figure 4.2: The Laplace transform applies to more general signals than the Fourier transform does. (a) Signal for which the Fourier transform does not exist. (b) Attenuating factor associated with Laplace transform.
(c) The modified signal x(t)e-σt is absolutely integrable for  > 1.
x(t)e-σt
e-σt
σ > 1

11. 4.2.4 The s-Plane.
It is convenience to represent the complex frequency s graphically in termed the s-plane.
(i) the horizontal axis represents the real part of s
(exponential damping factor s).
(ii) The vertical axis represents the imaginary part of s (sinusoidal frequency w)
In s-plane, s =0 correspond to imaginary axis.
Fourier transfrom is given by the Laplace transform evaluated along the imaginary axis.

12. Cont’d…
The jw-axis divides the s-plane in half. (i) The region to the left of the jw-axis is termed the left half of the s-plane. (ii) The region to the right of the jw-axis is termed the right half of the s-plane.
The real part of s is negative in the left half of the s-plane and positive in the right half of the s -plane..
Figure 4.3: The s-plane. The horizontal axis is Re{s}and the vertical axis is Im{s}. Zeros are depicted at s = –1 and s = –4  2j, and poles are depicted at s = –3, s = 2  3j, and s = 4. jω σ

13. 4.2.5 Poles and Zeros.
Zeros. The ck are the root of the numerator polynomial and are termed the zeros of X(s). Location of zeros are denoted as “o”.
Poles. The dk are the root of the denominator polynomial and are termed the poles of X(s). Location of poles are denoted as “x”.
The Laplace transform does not uniquely correspond to a signal x(t) if the ROC is not specified.
Two different signal may have identical Laplace Transform, but different ROC. Below is the example.
Figure 4.4a Figure 4.4b
Figure 4.4a. The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is located at s = a.
Figure 4.4b. The ROC for y(t) = –eatu(–t) is depicted by the shaded region. A pole is located at s = a. jω σ jω σ

14. Example 4.1: Laplace Transform of a Causal Exponential Signal.
Determine the Laplace transform of x(t)=eatu(t).
Solution:
Step 1: Find the Laplace transform.
To evaluate e-(s-a)t, Substitute s=s + jw

15. Cont’d…
If s > 0, then e-(σ-a)t goes to zero as t approach infinity,
*The Lapalce transform does not exist for s=<a because the integral does not converge.
*The ROC is at s>a, the shade region of the s-plane in Figure below. The pole is at s=a. jω σ
Figure 4.5: The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is located at s = a.

16. 4.3 The Unilateral Laplace Transform and Properties.
The Unilateral Laplace Transform of a signal x(t) is defined by
The lower limit of 0- implies that we do include discontinuities and impulses that occur at t = 0 in the integral. X(s) depends on x(t)for t >= 0.
The relationship between X(s) and x(t) as
The unilateral and bilateral Laplace transforms are equivalent for signals that are zero for time t<0.

17. Cont’d… Properties of Unilateral Laplace Transform. Scaling Linearity,
For a>0
Time Shift
for all t such that x(t - t)u(t) = x(t - t)u(t - t)
A shift in t in time correspond to multiplication of the Laplace transform by the complex exponential e-st.

18. Cont’d… s-Domain Shift
Multiplication by a complex exponential in time introduces a shift in complex frequency s into the Laplace transform.
Figure 4.6: Time shifts for which the unilateral Laplace transform time-shift property does not apply. (a) A nonzero portion of x(t) that occurs at times t  0 is shifted to times t < 0. (b) A nonzero portion x(t) that occurs at times t < 0 is shifted to times t  0.

19. Cont’d…
Convolution.
Convolution in time corresponds to multiplication of Laplace transform. This property apply when x(t)=0 and y(t) = 0 for t < 0.
Differentiation in the s-Domain.
Differentiation in the s-domain corresponds to multiplication by -t in the time domain.

20. Cont’d… Differentiation in the Time Domain.
Initial and Final Value Theorem.
The initial value theorem allow us to determine the initial value, x(0+), and the final value, x(infinity), of x(t) directly from X(s).
The initial value theorem does not apply to rational functions X(s) in which the order of the numerator polynomial is greater than or equal to hat of the denominator polynomial.

21. Example 4.2: Applying Properties. Solution:
Find the unilateral Laplace Transform of x(t)=(-e3tu(t))*(tu(t)).
Solution:
Find the Unilateral Laplace Transform.
And
Apply s-domain differentiation property,
Use the convolution property, . 1

22. 4.4 Inversion of the Unilateral Laplace Transform.
We can determine the inverse Laplace transforms using one-to-one relationship between the signal and its unilateral Laplace transform.
Appendix 4.1 consists of the table of Laplace Transform.
X(s) is the sum of simple terms,
Using the residue method, solve for a system linear equation.
Then sum the Inverse Laplace transform of each term.

23. Example 4.3: Inversion by Partial-Fraction Expansion.
Find the Inverse Laplace Transform of
Solution:
Step 1: Use the partial fraction expansion of X(s) to write
Solving the A, B and C by the method of residues

24. Cont’d…

25. Cont’d…
A=1, B=-1 and C=2
Step 2: Construct the Inverse Laplace transform from the above partial-fraction term above.
- The pole of the 1st term is at s = -1, so
- The pole of the 2nd term is at s = -2, so
-The double pole of the 3rd term is at s = -1, so
Step 3: Combining the terms. .

26. Example 4.4: Inversion An Improper Rational Laplace Transform.
Find the Inverse Laplace Transform of
Solution:
Step 1: Use the long division to espress X(s) as sum of rational polynomial function.
We can write,

27. Cont’d… Use partial fraction to expand the rational function,
Step 2: Construct the Inverse Laplace transform from the above partial-fraction term above. Refer to the Laplace transform Table. .

28. 4.5 Solving Differential Equation with Initial Condition.
Primary application of the unilateral Laplace transform in system analysis, solving differential equations with nonzero initial conditions.
Refer to the example.

29. Figure 4.7: RC circuit for Examples 6.4 and 6.10. Note that RC = 1/5.
Example 4.5: RC Circuit Analysis (Initial condition)
Use the Laplace transform to find the voltage across the capacitor , y(t), for the RC circuit shown in Figure 4.7 in response to the applied voltage x(t)=(3/5)e-2tu(t) and the initial condition y(0-) = -2.
Solution:
Step 1: Derive differential equation from the circuit.
KVL around the loop.
Figure 4.7: RC circuit for Examples 6.4 and Note that RC = 1/5.

30. Cont’d… Step 2: Get the unilateral Laplace Transform.
Apply the differential property,
Step 3: Substitute Unilateral Laplace Transform of x(t) into Y(s).
Given initial condition y(0-)=-2.
(2) Into (3),

31. Cont’d… Step 4: Expand Y(s) into partial fraction.
Step 5: Take Inverse Unilateral Laplace Transform of Y(s). .

32. 4.6 LT Methods in Circuit Analysis.
We can use the differentiation and integration properties to transform circuits involving capacitive and inductive elements so that we can solve using Laplace transforms.
Replace resistive, capacitive, and inductive elements by the Laplace transform equivalent.
Figure 4.8: Laplace transform circuit models for use with Kirchhoff’s voltage law.
Resistor. (b) Inductor with initial current iL(0–).
(c) Capacitor with initial voltage vc(0–).

33. Cont’d… (a) Resistor  Eq (4.29) Transforming the above equation
Representation of the transform resistor element.
(b) Inductor.
Eq (4.30)
Representation of the transform inductor element.

34. Cont’d… (c) Capacitor. Transforming the above equation Eq (4.31)
Representation of the transform capacitor element.

35. Cont’d…
When apply Kirchhoff’s voltage law to solve a circuit, use equation (4.29),(4.30) and (4.31).
When apply Kirchhoff’s current law to solve a circuit, rewrite equation (4.29),(4.30) and (4.31) to express current as a function of voltage.
Figure 4.9: Laplace transform circuit models for use wit Kirchhoff’s current law.
Resistor. (b) Inductor with initial current iL(0–).
(c) Capacitor with initial voltage vc(0–).

36. Figure 4.10 Electrical circuit (a) Original circuit.
Example 4.6: Solving the Second Order Circuit.
Use the Laplace transform circuit model to determine the voltage y(t) in the circuit of Figure 4.10 for the applied voltage x(t)=3e-10tu(t) V. the voltage across the capacitor at time t=0- is 5V.
Figure 4.10 Electrical circuit (a) Original circuit.
Solution:
Step 1: Transform the circuit (a) to (b).
I1(s) and I2(s) are the currents through the branch.
Figure 4.11: Electrical circuit (a) Original circuit. (b) Transformed circuit.

37. Cont’d… Step 2: Derive equation from the Transformed circuit. KVL .
(2) and (3) into(1)

38. 4.7 Properties of Bilateral Laplace Transform.
The Bilateral Lapalace Transform is suitable to problems involving noncausal signals and systems.
The properties of linearity, scaling, s-domain shift, convolution and differentiation in the s-domain is identical fort the bilateral and unilateral LT, the operations associated y these properties may change the ROC.
Example; a linearity property.
ROC of the sum of the signals is an intersection of the individual ROCs.

39. Cont’d…
Time Shift
The bilateral Laplace Transform is evaluated over both positive and negative values of time. ROC is unchanged by a time shift.
Differentiation in the Time Domain.
Differentiation in time corresponds to multiplication by s.

40. Cont’d… Integration with Respect to Time.
Integration corresponds to division by s
Pole is at s=0, we ae integrating to the right the ROC must lie to the right of s=0.

41. 4.8 Properties of Region of Converges.
Note that ROC cannot contain any pole. If Laplace transform converges, then X(s) is finite over the entire ROC. Suppose d is a pole of X(s). This implies that X(d) = ±∞. So the Lpalace transform does not converges at d. Thus, s=d cannot lie in ROC.
Converges of bilateral Lplace transform for a signal x(t) implies that
for some value of σ.
Refer example.

42. 4.9 Inversion of Bilateral Laplace Transform.
The inversion of Bilateral Laplace transforms are expressed as a ratio of polynomial in s.
Compare to the unilateral, in the bilateral Laplace transform we must use the ROC to determine the unique inverse transform in bilateral case.

43. Example 4.7: Inverting a Proper Rational Laplace Transform.
Find the Inverse bilateral Laplace Transform of
With ROC -1<Re(s)<1.
Solution:
Step 1: Use the partial fraction expansion of X(s) to write
Solving the A, B and C by the method of residues

44. Cont’d…
Step 2: Construct the Inverse Laplace transform from the above partial-fraction term above.
- The pole of the 1st term is at s = -1, the ROC lies to the right of this pole, choose the right-sided inverse Laplace Transform.
- The pole of the 2nd term is at s = 1, the ROC is to the left of the pole, choose the right-sided inverse Laplace Transform.
-The pole of the 3rd term is at s = -2, the ROC is to the right of the pole, choose the right-sided inverse Laplace Transform.

45. Figure 4.12 : Poles and ROC for Example 6.17.
Cont’d…
Step 3: Combining the terms.
Combining this three terms we obtain,
Figure 4.12 : Poles and ROC for Example 6.17. jω σ

46. 4.10 Transfer Function.
The transfer function of an LTI system is defined as the Laplace transform of the impulse response.
Take the bilateral Laplace transform of both sides of the equation and use the convolution properties result in,
Rearrange the above equation result in the ratio of Laplace transform of the output signal to the Laplace transform of the input signal. (X(s) is nonzero)

47. 4.10.1 Transfer Function and Differential-Equation System Description.
Given a differential equation.
Step 1: Substitute y(t) = estH(s) into the equation.
y(t) = estH(s), substitute to the above equation result in,
Step 2: Solve for H(s).
H(s) is a ratio of polynomial and
s is termed a rational transfer function.

48. Example 4.8: Find the Transfer Function.
Find the transfer function of the LTI system described by the differential equation below
Solution:
Step 1: Substitute y(t) = estH(s) into the equation.
Step 2: Solve for H(s).

49. Cont’d…
The transfer function is, .

50. Appendix 4.1: Table of Laplace Transform

51. 4.11 Causality and stability
The impulse response of causal system is zero for t<0.
The system is a causal : The impulse response is determined from the transfer function by using the right-sided inverse Laplace transforms.
A system pole at s = dk in the left half of the s-plane [Re(dk)< 0) contribute an exponentially decaying term to impulse response, while a pole in the right half of the s-plane [Re(dk)> 0] contributes an increasing exponential term to the impulse response.

52. 4.12 Inverse Systems
Given an LTI system with impulse response h(t), the impulse response of the inverse system hinv(t):
hinv(t)*h(t) = δ(t)
The inverse system transfer function Hinv(s):
Hinv(s)H(s) = 1
Hinv(s) = 1/H(s)
If H(s) is written in pole-zero form:
The zeros of the inverse system are the poles of H(s) and poles of the inverse sytem are the zeros of H(s)

53. 4.13 Determining the Function Response from poles and zeros