hw1_sol

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Oct 30, 2023

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ECE 102, Fall 2023 Homework #1 Department of Electrical and Computer Engineering Prof. Jonathan C. Kao University of California, Los Angeles TAs: Yang, Bruce, Shreyas Due Monday, 16 Oct 2023, by 11:59pm on Gradescope Covers material up to lecture 3. 100 points total 1. (15 points) Scaling and shifting (a) (10 points) Sketch the following signals for x ( t ) as shown in the figure below. Figure 1: x ( t ) i. 1 2 x (2 t − 6) Solution: • Right shift by 6: x ( t − 6) • Horizontal compress by 2: x (2 t − 6) • Downscale by 2: 1 2 x (2 t − 6) Figure 2: x (2 t − 6) ii. x ( 1 10 − 1 5 t ) Solution: • Left shift by 1 10 : x ( t + 1 10 ) • Horizontal stretch by 5: x ( 1 5 t + 1 10 ) • Time reverse: x ( 1 10 − 1 5 t ) 1

Figure 3: x ( 1 10 − 1 5 t ) (b) (5 points) Express y ( t ), as shown in the figure below, in terms of x ( t ). Figure 4: y ( t ) Solution: y ( t ) = x ( − 1 2 t − 3 2 ) • Right shift by 3 2 : x ( t − 3 2 ) • Horizontal stretch by 2: x ( 1 2 t − 3 2 ) • Time reverse: x ( − 1 2 t − 3 2 ) 2. (10 points) Even and odd signals (a) (3 points) Show that the product of two odd signals is even. Solution: Let x 1 ( t ) and x 2 ( t ) be two odd signals. Then x 1 ( t ) = − x 1 ( − t ) and x 2 ( t ) = − x 2 ( − t ). Let y ( t ) be their product. Then y ( t ) = x 1 ( t ) x 2 ( t ) = [ − x 1 ( − t )][ − x 2 ( − t )] = x 1 ( − t ) x 2 ( − t ) = y ( − t ) is even. 2

(b) (3 points) Show that the product of an even signal and an odd signal is odd. Solution: Let x 1 ( t ) and x 2 ( t ) be an even and an odd signal respectively. Then x 1 ( t ) = x 1 ( − t ) and x 2 ( t ) = − x 2 ( − t ). Let y ( t ) be there product. Then y ( t ) = x 1 ( t ) x 2 ( t ) = x 1 ( − t )[ − x 2 ( − t )] = − x 1 ( − t ) x 2 ( − t ) = − y ( − t ) is odd. (c) (4 points) Find the even and the odd components of x ( t ) = t sin 2 ( t ) + t 3 ( e t + e − t 2 ) + t 5 sin( t ) + 2023 Hint: Use the properties you proved in previous parts. You can also use the fact that the sum of odd signals is odd and the sum of even signals is even without proving. Solution: • t sin 2 ( t ) is odd because t is odd and sin 2 ( t ) = 1 − cos(2 t ) 2 is even. • t 3 ( e t + e − t 2 ) is odd because t 3 is odd and e t + e − t 2 is even. • t 5 sin( t ) is even because both t 5 and sin( t ) are odd. Therefore, x o ( t ) = t sin 2 ( t ) + t 3 ( e t + e − t 2 ) x e ( t ) = t 5 sin( t ) + 2023 3. (22 points) Periodic signals (a) (12 points) For each of the following signals, determine whether it is periodic or not. If the signal is periodic, determine its fundamental frequency. i. x ( t ) = 24 + 50 sin(60 πt ) ii. x ( t ) = 10 cos 2 ( π 2 t ) iii. x ( t ) = sin(5 πt ) + cos(16 π 2 t ) iv. x ( t ) = ( 24 + 50 sin(60 πt ) t < 0 10 cos 2 (10 πt ) t ≥ 0 (1) Solution: i. Yes, 2 πf = 60 π = ⇒ f = 30. ii. x ( t ) = 5 + 5 cos( πt ). Yes, 2 πf = π = ⇒ f = 0 . 5. 3

iii. No. sin(5 πt ): 2 πf 1 = 5 π = ⇒ f 1 = 5 2 cos(16 π 2 t ): 2 πf 2 = 16 π 2 = ⇒ f 2 = 8 π Since f 1 is rational while f 2 is irrational, T 1 is rational while T 2 is irrational. Therefore, there doesn’t exist a T 0 where the cycles of the two component signals overlap. iv. No, due to the discontinuity at t = 0. (b) (5 points) Suppose x ( t ) is odd and periodic with period T 0 . What is the value of x ( T 0 )? Solution: x ( T 0 ) = x ( T 0 − T 0 ) since x ( t ) is periodic = x (0) = 0 since x ( t ) is odd (c) (1 point) What is the effect of time shifting on the fundamental frequency of a signal? Solution: There is no effect. (d) (4 points) If x ( t ) is periodic, is x (5 t + 2) periodic? Solution: Yes. If x ( t ) has a period of T 0 , then x (5 t + 2) has a period of T 0 / 5 using the fact in part (c). 4. (21 points) Energy and power signals (a) (15 points) Determine whether the following signals are energy or power signals. If the signal is an energy signal, determine its energy. If the signal is a power signal, determine its power. i. x ( t ) = e −| t | ii. x ( t ) = ( 1 √ t , if t ≥ 1 0 , otherwise iii. x ( t ) = ( 1 + e − t , if t ≥ 0 0 , otherwise Solution: 4

i. E = Z ∞ −∞ | x ( t ) | 2 dt = Z ∞ −∞ | e −| t | | 2 dt = Z ∞ −∞ e − 2 | t | dt = Z ∞ 0 e − 2 t dt + Z 0 −∞ e 2 t dt = 2 Z ∞ 0 e − 2 t dt = − e − 2 t ∞ t =0 = 1 Therefore, x ( t ) is an energy signal and its power is 0. ii. E = Z ∞ −∞ | x ( t ) | 2 dt = Z 1 −∞ 0 dt + Z ∞ 1 1 t dt = ln ( t ) ∞ t =1 which goes to infinity Since the energy of x ( t ) is infinite, let’s calculate its power. P = lim T →∞ 1 2 T Z T − T | x ( t ) | 2 dt = lim T →∞ 1 2 T Z 1 − T 0 dt + 1 2 T Z T 1 1 t dt = lim T →∞ ln ( T ) 2 T = 0 The power of x ( t ) is zero. Therefore, x ( t ) is neither an energy signal nor a power signal. iii. E = Z ∞ −∞ | x ( t ) | 2 dt = Z ∞ 0 | 1 + e − t | 2 dt = Z ∞ 0 1 + 2 e − t + e − 2 t dt which goes to infinity 5

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