EC8352-Signals and Systems - Laplace transform

Laplace transform - Properties and Problems

1. Hello! Nimitha N Assistant Professor/ECE [email protected] 1 Pierre-Simon Laplace One of the first scientists to suggest the existence of black holes

2. Laplace Transform

3. Process 3 Time domain x(t) Freq domain X(s) Freq domain X(s) Time domain x(t) Differential Equation Algebraic Equation LT ILT

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 + j ω 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

9. Laplace transform for elementary signals 5) 6) Solution W.k.t 9 Exponential signal

10. Laplace transform for elementary signals 7) 8) 9) 10 Hint x(t) = cos ω t u(t) x(t) = sin ω t u(t)

11. Laplace transform for elementary signals 8) Solution Using Euler’s Formula 11 ----> (1)   ----> (2) Compare (1) and (2) Real part Imaginary part

12. Summary Impulse Step 12 L[ δ (t)] 1 L[u(t)] 1/s a= ω L[1]

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) =   14 LHS RHS - ∞ ∞ j ω σ Complex variable, S= α + j ω 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 ) f or 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 j ω - ∞ ∞ j ω 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 - ∞ ∞ j ω σ - ∞ ∞ j ω σ

18. If x(t) is a two sided sequence then ROC is the combination of two regions. 18 - ∞ ∞ j ω 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)=e at u(−t) Solution Right sided signal Left sided signal

20. 20 Find the Laplace transform and ROC of x(t)=e− at u(t)+e at 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 x(t) = e -2t u(-t) x(t) = e 3t u(t) x(t) = e (4+3j)t u(-t) x(t) = e -4t u(t) x(t) = e 3t u(t) + e -2t u(t) x(t) = e 3t u(-t) + e -2t u(-t) x(t) = e -3t u(t) + e 2t u(-t) - ∞ ∞ j ω 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 j ω ) 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 j ω ) 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 j ω ) 23 σ σ σ j ω Poles

24. 24 Problems Check causality and stability x(t) = e -2t u(-t) x(t) = e 3t u(t) x(t) = e (4+3j)t u(-t) x(t) = e -4t u(t) 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− 3 t u(t)+e -2 t u(t), Check causality and stability Solution: L[x(t)= L[ e −3t u(t)+e -2t u(t)] X(s) = +   ROC: Re{s} = σ >-3, Re{s} = σ >-2 ROC: Re{s} = σ >-2 - ∞ -3 -2 ∞ j ω σ Both will converged if Re{s} = σ >-2 Causal and stable

26. Summary 26

27. 27 Summary

28. LT for Elementary signals 28

29. Properties of Laplace Transform Linearity Time Scaling Time shifting Frequency or s-plane shift Multiplication by t n Integration 29 Differentiation Convolution Initial Value Theorem and Final value Theorem

30. Linearity 30 Proof: x(t) X(s) Problem Hence Proved 1. Find the Laplace transform of x(t) = 2 δ (t)+ 3 u(t) L[ δ (t)] 1 L[u(t)] 1/s L[x(t)] = L[2 δ (t)+ 3 u(t)] = X(s) = 2 L[ δ (t)] + 3 L[u(t)] X(s) = 2+ 3(1/s) y(t) Y(s)

31. Time Scaling 31 Proof: x(t) X(s) Consider Dummy variable τ = at t = τ/ a dt = dτ / a Hence Proved

32. Time Shifting 32 Proof: x(t) X(s) Consider Dummy variable τ = t-t t = τ + t dt = dτ Hence Proved

33. Problem : Time shifting 1. Using Time shifting property, solve 33 Solution: x(t) X(s) L[u(t-3)] =   L[u(t-3)] Using Time shifting property Given : t =3 L[u(t-3)] = X(s) =   =   =   L[u(t-3)] = X(s) =   =   = 0   L[u(t-3)] =   L[u(t)] =   Wkt

34. Time Reversal 34 Proof: X(s) x(t)

35. Time Differentiation 35 Proof: X(s) x(t) General Form Repeat

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 : t =2 = e -6 L[e -3t ] e -2s Wkt , L[e -at ] = (1/ s+a ) = e -2s e -6 (1/s+3) =   =   x(t) X(s) = L[e -3(t-2+2) u(t-2)] L[u(t-2)] =   Sub : s by s+3 =   Problem : Frequency Shifting+ Time Shifting

38. Frequency Differentiation 38 X(s) x(t) 1.Determine LT of Solution: n=1 L[t u(t)] = (-1) L[u(t)]   Wkt L[u(t)] = 1/s = (-1)   = (-1)   = (-1)   = 1/   L[t u(t)] L[t 2 u(t)] = 2/   Similarly Ramp signal Parabolic signal

39. Convolution Theorem 39 L[x 1 (t) * x 2 (t)] X 1 (s) . X 2 (s) 1.Consider x 1 (t) = u(t) and x 2 (t) = u(t)   Y(s) = X 1 (s) . X 2 (s) L[x 1 (t)] =   L[x 2 (t)] =   Y(s) = +   Y(s) = .   =   Y(s) =   Sub s=0 A= 1/5 Sub s=-5 B= -1/5 Y(s) = +   Sub A and B in (1) (1) ILT [Y(s)] = y(t) y (t) = u(t) - u(t)  

40. Initial and Final Value Theorem 40 Initial value Theorem Final value Theorem x(∞+) =   x(0+) =   1. Determine the Initial value and final value of x(t) whose Laplace transform is X(s) =   x(0+) =   Initial value Theorem Solution: x(0+) =   x(0+) =   x(0+) =   Final value Theorem x(∞+) =   x(∞+) =   x(∞+) =  

41. Initial and Final Value Theorem 41 Initial value Theorem Final value Theorem x(∞+) =   x(0+) =   2. Determine the Initial value and final value of x(t) whose Laplace transform is X(s) =   x(0+) =   Initial value Theorem Solution: x(0+) =   Final value Theorem x(∞+) =   x(∞+) = 0 x(0+) =   x(0+) =   x(∞+) =  

42. Inverse Laplace transform using Partial fraction 42 Partial fraction types 01 Poles are distinct L[e at ]u(t) = Roc, Re{s} σ > a L[-e -at ]u(-t) = Roc, Re{s} σ < a 02 Multiple Poles 03 Complex Poles

43. Problems ( Simple Poles) 43 1.Determine Inverse LT of Solution:     Sub S= 0   1   Sub S= -1   -2   (1)

44. 44 Sub S= -2   1   Sub A, B and C in (1)     ILT [X(s)] = x(t)    

45. 45 Problems ( Multiple Poles) 2.Determine Inverse LT of Solution: =     1   Sub S= 0 1/2   Sub S= -1 -2   Sub S= -2 (1)

46. 46       Equating   0= A+B+D 0= 0.5-2+D D= 1.5 or 3/2 +   Sub A,B,C and D in (1) +     ILT [X(s)] = x(t)

47. 47 Problems ( Complex Poles) 3.Determine Inverse LT of X(s) =   Solution: X(s) =             3   -1   Equating   1   (1)

48. 48   (1) Sub A,B and C in (2)   (2)     +   ILT [X(s)] = x(t)   L[     L[     Problems ( Complex Poles)

49. Problem Using RoC 49 4.Determine Inverse LT of X(s) = if ROC ( i ) -2 >Re(s) > -4 , (ii) Re(s) < -4   =   X(s) =   = +     2     X(s) = -   Sub in (1) (1)  

50. 50 ( i ) ROC -2 >Re(s) > -4 (ii) ROC Re(s) < -4     - ∞ ∞ j ω σ Re(s)<-2 and Re(s)> -4 L[e at ]u(t) = Roc, Re{s} σ > a L[-e -at ]u(-t) = Roc, Re{s} σ < a   - ∞ ∞ j ω σ   Problem Using RoC

51. 51 (iii) ROC Re(s) > 2     - ∞ ∞ j ω σ Problem Using RoC

52. THANK YOU

53. 53 https://www.tutorialspoint.com/signals_and_systems/region_of_convergence.htm http://jntuhsd.in/uploads/programmes/Module16_LT_14.0_.2017_.PDF (Properties)