Homework 3 Solutions

Transcription

1 Signals and Systems Profs. Byron Yu and Pulkit Grover Fall 2018 Homework 3 Solutions Part One 1. (25 points) The following systems have x(t) or x[n] as input and y(t) or y[n] as output. For each system state whether the following properties hold and justify: Causal Stable Memoryless Linear Time invariant (a) y(t) = x 2 (t) (b) y(t) = x 2 (τ)dτ (c) y[n] = e jn (x[n + 1] x[n]) (d) y[n] = K Solution: x[n m] where K is a positive integer (a) Causal, stable, memoryless, not linear, and not time invariant Causality: Since y(t) only depends on the current value of x(t), it is causal. Stability: y(t) is just a scaling of x(t). Let x(t) M. Then, y(t) = (x(t)) 2 = y(t) = x(t) x(t) M 2 = y(t) M 2. Hence the system is stable. Memorylessness: y(t) depends only on the current value of x(t). Hence the system is memoryless. Linearity: Let x 1 (t) and x 2 (t) be two inputs, and let x(t) = ax 1 (t) + bx 2 (t) be a third input, where a, b are arbitrary constants. Then,

2 2 Homework 3 Solutions y 1 (t) = x 2 1(t) y 2 (t) = x 2 2(t) Hence the system is not linear. Time invariance Let ˆx(t) = x(t τ). y(t) = (ax 1 (t) + bx 2 (t)) 2 = a 2 x 2 1(t) + 2abx 1 (t)x 2 (t) + b 2 x 2 (t) ay 1 (t) + by 2 (t) Hence the system is time invariant. y(t) = x 2 (t) = y(t τ) = (x(t τ)) 2. ŷ(t) = (ˆx(t)) 2 = (x(t τ)) 2 = y(t τ). (b) Causal, not stable, not memoryless, not linear, time invariant Causality: Since y(t) depends only on present and past values of x(t), the system is causal. Stability: Let x(t) M, M 0, R y(t) = y(t) = y(t) x 2 (τ)dτ x 2 (τ)dτ x 2 (τ) dτ M 2 dτ = M 2 dτ = M 2 t as t.

3 Homework 3 Solutions 3 Hence not stable. Memorylessness: y(t) depends on past values of x(t) and hence is not memoryless. Linearity: Let x 1 (t) and x 2 (t) be two inputs, and let x(t) = ax 1 (t) + bx 2 (t) be a third input, where a, b are arbitrary constants. Then, y 1 (t) = x 2 1(τ)dτ y 2 (t) = y(t) = x 2 2(τ)dτ x 2 (τ)dτ = (ax 1 (τ) + bx 2 (τ)) 2 dτ = (ax 2 1(τ) + bx 2 2(τ) + 2abx 1 (τ)x 2 (τ))dτ y 1 (t) + y 2 (t). Hence the system is not linear. Time invariance y(t) = x 2 (τ)dτ t t 0 = y(t t 0 ) = x 2 (τ)dτ.

4 4 Homework 3 Solutions Let ˆx(t) = x(t t 0 ). Then, ŷ(t) = ˆx 2 (τ)dτ = x 2 (τ t 0 )dτ. Let ˆτ = τ t 0. Then, dˆτ = dτ. When τ =, ˆτ =. When τ = t, ˆτ = t t 0. ŷ(t) = t t 0 ˆ x 2 (ˆτ)dˆτ. Hence the system is time invariant. (c) Not causal, stable, not memoryless, linear, not time invariant Causality: Since y[n] depends on x[n + 1], the system is not causal. Stability: Let x[n] M. y[n] = e jn (x[n + 1] x[n]) y[n] = e jn (x[n + 1] x[n]) e jn ( x[n + 1] + x[n] ) (M + M) 2M. Hence the system is stable. Memorylessness: Since the system is not causal, the system is not memoryless. Linearity: Let x 1 [n] and x 2 [n] be two inputs, and let x[n] = ax 1 [n] + bx 2 [n] be a third input, where a, b are arbitrary constants. Then, y 1 [n] = e jn (x 1 [n + 1] x 1 [n]) y 2 [n] = e jn (x 2 [n + 1] x 2 [n]) y[n] = e jn (x[n + 1] x[n]) = e jn (ax 1 [n + 1] + bx 2 [n + 1] ax 1 [n] bx 2 [n]) = ae jn (x 1 [n + 1] x 1 [n]) + be jn (x 2 [n + 1] x 1 [n] x 2 [n]) = ay 1 [n] + by 2 [n].

5 Homework 3 Solutions 5 Hence the system is linear. Time invariance y[n] = e jn (x[n + 1] x[n]) y[n m] = e j(n m) (x[n m + 1] x[n m]). Let ˆx[n] = x[n m]. Then, ŷ[n] = e jn (ˆx[n + 1] ˆx[n]) = e jn (x[n m + 1] x[n m]) y[n m]. Hence the system is not time invariant. (d) Causal, stable, not memoryless, linear, time invariant. Causality: Since y[n] only depends on present and past values of x[n], system is causal. stability: Let x[n] < M. Then, Hence the system is stable. y[n] = x[n m] = y[n] = x[n m] x[n m] M = (K + 1)M. Memorylessness: Since the output depends on past values of x[n], it is not memoryless. Linearity: Let x 1 [n] and x 2 [n] be two inputs, and let x[n] = ax 1 [n] + bx 2 [n] be a third input, where a, b are arbitrary constants. Then,

6 6 Homework 3 Solutions y 1 [n] = y 2 [n] = y[n] = = Hence the system is linear. Time invariance: = a x 1 [n m] x 2 [n m] x[n m] (ax 1 [n m] + bx 2 [n m]) x 1 [n m] + b = ay 1 [n] + by 2 [n]. y[n] = y[n p] = x 2 [n m] x[n m] x[n p m]. Let ˆx[n] = x[n p]. Then, ŷ[n] = ˆx[n m] = x[n p m] = y[n p]. Hence the system is time invariant.

7 Homework 3 Solutions 7 2. (15 points) Determine if each of the following systems is invertible. If it is, construct the inverse system. Otherwise, show why the system is not invertible. (a) y(t) = sin(πx(t)) (b) y(t) = exp ( x(t) ) (c) y[n] = sin ( π 2 n) x[n] (d) y[n] = n k= x[k] Solution: (a) y(t) = sin(πx(t)) Let x(t) = t where t = 1, 2,..., T. We can have y(t) = 0. In this case, multiple x(t) share same y(t), so y(t) is not invertible. (b) y(t) = e x(t) In this case, multiple inputs x(t) can produce the same output y(t), so y(t) is not invertible. For example, let x(t) = u(t) 2u(t 1) + u(t + 2), we have y(t) = exp ( u(t) 2u(t 1) + u(t 2) ) = exp (u(t) u(t 2)). (c) y[n] = sin( πn)x[n] 2 Although x[n] = csc( π n)y[n], x[n] = inf when n =..., 4, 2, 0, 2, 4,... So, y[n] 2 is not invertible. (d) y[n] = n k= x[k] x[n] = y[n] n 1 k= inf x[k] = y[n] y[n 1]. Y [n] is invertible.

8 8 Homework 3 Solutions 3. (15 points) Suppose we have two signals x(t) and h(t) defined as: { t t 0 x(t) = 0 otherwise, h(t) = (a) Plot the signals x(t) and h(t). { 2 0 t 2 0 otherwise. (b) Plot x(t)h( t) and evaluate the integral x(t)h( t)dt. (c) Plot x(t)h( 1 t) and evaluate the integral x(t)h( 1 t)dt. (d) Evaluate the integral x(t)h(τ t)dt. (Hint: the result should include τ.) Solution: (a) plot signal x(t) and y(t). x(t) t 2 y(t) t

9 Homework 3 Solutions 9 (b) Plot x(t)y( t) and compute the integral x(t)u( t)dt. x(t)y( t) t x(t)y( 1 t)dt = 2x(t)[u( t)]dt = 0 2x(t)dt = t+4dt = [t 2 +4t] 0 2 = 4 (c) Plot x(t)y( 1 t) and compute the integral x(t)u( 1 t)dt. x(t)y( 1 t) t

10 10 Homework 3 Solutions x(t)y( 1 t)dt = 1 = = 2x(t)[u( 1 t) u( 3 t)]dt 3 1 (d) Compute the integral x(t)y(τ t)dt. 2 2x(t)dt 2t + 4dt = [t 2 + 4t] 1 2 = 1 x(t)y(τ t)dt = τ 2 = = 2x(t)[u(τ t) u(τ t 2)]dt τ τ 2 0dt + τ 2x(t)dt = τ 2 τ 2x(t)dt + τ 2 τ 2(t + 2)dt = = (t 2 + 4t) τ t=τ 2, for 2 t 0 0dt We need to think about when 2 τ 0, and also when, 2 τ 2 0 to know when to evaluate the upper and lower limit of the integral. τ < 2, i.e. τ 2 < τ < 2. In this case, (t 2 + 4t) t=0 = 0. 2 τ 0, i.e. τ 2 < 2 τ. In this case, (t 2 + 4t) t=τ = τ 2 + 4τ < τ 2, i.e. 2 τ 2 0 < τ. In this case, (t 2 + 4t) t=τ 2 = 4 τ 2. τ > 2, i.e. 0 < τ 2 < τ. In this case, (t 2 + 4t) t=0 = 0.

11 Homework 3 Solutions 11 Part Two 4. (15 points) Consider the system with input x[n] and output y[n] defined below: y[n] = x[n] (g[n] + g[n 1] + g[n 2] + g[n 3]). For each of the following definitions of g[n], show whether or not the system is time invariant. (a) (5 points) If g[n] = 1 for all n. (b) (5 points) If g[n] = n. (c) (5 points) If g[n] = 1 n. Solution: g[n] = 1: g[n] = n: g[n] = 1 n: y[n] = x[n]( ) = 4x[n] y[n n 0 ] = 4x[n n 0 ] y[n] = x[n](n + n 1 + n 2 + n 3) = (4n 6)x[n] y S [n] = (4n 6)x[n n 0 ] y[n n 0 ] = (4(n n 0 ) 6)x[n n 0 ] y S [n] y[n] = x[n](1 n + 2 n + 3 n + 4 n) = (10 4n)x[n] y[n n 0 ] = (10 4(n n 0 ))x[n n 0 ] (10 4n)x[n n 0 ]

12 12 Homework 3 Solutions 5. (14 points) (MATLAB) For all the questions below, assume the values before and after the given vectors are zeros. (a) (4 points) Create a system that takes in a signal represented by a vector x and reverses it, where x can be any length. In other words, y[n] = x[ n], for all n Create the system as the following function: y = myreverse(x). Call your function on x = [ ], which corresponds to n = -5:5. This vector x represents the signal x[n] such that x[ 5] = 0, x[0] = 1, x[5] = 0, etc. Plot the output y for n = -5:5 using stem, and submit it along with your source code. (b) (5 points) Create a system that shifts a vector x to the right by k. In other words, y[n] = x[n k], for all n Create the system as the following function: y = rshift(x,k). Vectors have finite length when represented in MATLAB, so elements could be shifted out of the bounds of a vector. For this function, you can simply drop the values shifted outside the boundary and add zeros on the other side. For example, if x = [1 2 3] and k= 1, the function should shift every element to the right by one, drop the rightmost element, and make the leftmost element zero, which gives y = [0 1 2]. Call your function on x = [ ] and k = 1, where n = -5:5. Plot the output y for n = -5:5 using stem, and submit it along with your source code. (c) (5 points) Follow the steps below to implement a system that takes vectors A and B, and outputs C: (1) Reverse A and get a new vector denoted as A r. (2) Perform an element-wise multiplication of A r and B. (3) Sum all the elements in the vector obtained in the previous step to get one element of C. (4) Shift A r to the right by 1. (5) Repeat starting from step (2) until all elements in A r are shifted out of its bounds.

13 Homework 3 Solutions 13 Here s an example of computing C, given A = [ ] and B = [ ] corresponding to n = -2:2. First, reverse A and get A r = [ ]. Perform an element-wise multiplication with B and sum up all the elements. This gives C[0] = 2 n= 2 A r [n]b[n] = ( 1) = 2 The sum is from n = -2 to n = 2 because both A r and B are zeros outside this range. Now you have the first element in vector C. Next, shift A r to the right by 1, repeat the element-wise multiplication and add all the resulting elements to get C[1]. Keep shifting A r one by one and performing the calculations until all the elements in A r are shifted out of its bounds.the result should be C = [ ]. Solution: Now consider A = [ ] and B = [ ] for n = -5:5 i. Calculate C[4] and C[6] by hand. Show all your work. ii. Create function C = CSum(A,B), using the functions you wrote in parts (a) and (b). iii. Call CSum on vectors A and B and plot the output C for n = 0:10 using stem. You may use.* for element-wise multiplication and sum to sum all the elements of a vector. Submit your plot and code. iv. Describe the relationship between signals C and A. (Hint: For what values of n is B equal to 1?) (a) myreverse function y = myreverse ( x) n = length (x); y = x(n : -1:1); end Plotting x = [ ]; y = myreverse (x); n = -5:5; stem (n,y) xlabel ( n ) ylabel ( y[n] ) title ( myreverse )

14 14 Homework 3 Solutions Plotting (b) rshift function y = rshift (x, k) y = [ zeros (1,k) x (1: end -k )]; end Plotting x = [ ]; y = rshift (x,1); n = -5:5; stem (n, y) xlabel ( n ) ylabel ( y[n] ) title ( rshift ) Plotting

15 Homework 3 Solutions 15 (c) i. 5 C[4] = A r[n]b[n] = = 0 n=0 ii. 5 C[6] = A r[n]b[n] = = 2 n=0 function C = CSum (A, B) A_r = myreverse (A); C = zeros ( size (A )); for i = 1: length (A) C(i) = sum ( A_r.*B); A_r = rshift ( A_r,1); end end iii. A = [ ]; B = [ ]; C = CSum (A, B);

16 16 Homework 3 Solutions n1 = 0: length (A) -1; stem (n1, C); xlabel ( n ) ylabel ( C[n] ); title ( CSum ) iv. C consists of two copies of A, one shifted by 0 and the other by 5.

17 Homework 3 Solutions (15 points) (MATLAB) Let s now implement the following system: y[n] = x[n] + x[n 1] + x[n 2]. Write a function y=simplesys(x) that takes a sequence x as input and returns an output y. function y = simplesys ( x) %% SIMPLESYS returns output y for a given % input x, such that y[n] = x[n] + x[n -1] + x[n -2] < your code goes here > end Let us now find out whether the system is linear and time invariant. (a) (5 points) Linearity Let s define two signals x 1 [n] = [1 : 10] for n = 1 : 10, and x 2 [n] = [ 8 : 2 : 10] for n = 1 : 10. Use the function simplesys to generate two outputs y 1 [n] and y 2 [n]. Now, generate another output y 3 [n] using the input signal 2x 1 [n] + 3x 2 [n]. Plot y 3 [n]. Create a new figure using Matlab command figure without arguments. In this new figure plot 2y 1 [n] + 3y 2 [n]. Compare these two plots. Does this result suggest that the system is linear and why? Please submit the following: Code of simplesys MATLAB function. Plots of y 3 [n] and 2y 1 [n] + 3y 2 [n]. Is the system linear? Why? (b) (5 points) Time-invariance For a time-invariant system, if the input is delayed by δ time steps, then the output is the same as before, but delayed by δ time steps. Let s check if our system satisfies this property. Let s say the system is H. Plot the output y[n] generated by the signal x[n] = [1, 3, 5, 7, 15, 7, 5, 3, 1, 0, 0]. Now, generate an output using a delayed version of x with δ = 3. Hence x[n 3] = [0, 0, 0, 1, 3, 5, 7, 15, 7, 5, 3]. Plot the output by using x[n 3] as input to function simplesys, note the output as simplesys(x[n 3]). Compare simplesys(x[n 3]) to y[n 3] delayed by 3. Does this result suggest the system is time-invariant? Please submit the following: Plots of y[n 3] and simplesys(x[n 3]). Is the system time-invariant? Why?

18 18 Homework 3 Solutions (c) (5 points) Now, let s look more closely at our system: y[n] = x[n] + x[n 1] + x[n 2] Dividing the right side of this equation by some constant number does not change the behavior of the system (though it does scale the magnitude of the output). You can verify this using the above exercise to convince yourself it is true. Let s divide by 3: y[n] = x[n] + x[n 1] + x[n 2] 3 This results is a moving average using three consecutive elements in x. Since averaging essentially smooths out a signal, what we have here is a simple low-pass filter! (We will see eventually that sharp changes in a signal correspond to high frequencies and removing those sharp changes and smoothing the signal over time is performed using low-pass filters.) We can increase the level of smoothing by increasing the number of elements being averaged over: y[n] = x[n] + x[n 1] + x[n 2] + x[n 3] 4 Recall that this is a function you wrote in the previous homework called moveaverage and you saw the effects of increasing k. You will now implement another kind of low pass filter in MATLAB. Implement a first order low-pass filter with an output as follows. The function file should look like : y[n] = x[n] + ry[n 1] where 0 < r 1 function [y] = order1lp (x,r) %% low pass filter of first order with parameter r < your code goes here > end Try values of r = 0.1, 0.3, 0.6 and comment on how r affects the output in terms of smoothing of the following signal s[n]= sin(n/4) + 1/2*cos(n/5) + sin(n) with n=1:100. Please submit the following: Code of order1lp MATLAB function. Plots of the output of order1lp with r = 0.1, 0.3, 0.6. Comment on how r affects the in terms of the amount of smoothing.

19 Homework 3 Solutions 19 Solution: The code for the function: function y = simplesys ( x) %% SIMPLESYS returns output y for a given % input x, such that y[n] = x[n] + x[n -1] x= cat (2, zeros (1,2), x); for i = 3: length (x) y(i -2) = x(i) + x(i -1) + x(i -2); end (a) (x points) Linearity x1 = 1:10; x2 = -8:2:10; y1 = simplesys ( x1 ); y2 = simplesys ( x2 ); y3 = simplesys (2* x1 + 3* x2 ); y4 = 2* y1 + 3* y2; The system satisfies the property of linearity hence we can say that it is linear. (b) (x points) Time Invariance

20 20 Homework 3 Solutions x = [ ]; x_delay = [ ]; y = simplesys (x); y_delay = simplesys ( x_delay ); figure plot (y) hold on plot ( y_delay ) The output y delay is delayed by the same amount and does not change in shape otherwise. Hence system is time-invariant by definition of time invariance. (c) (x points) Simple Low Pass Filter For the first order LP filter, code is: function [ out ] = order1lp (x,r) out (1)=0; ctr = 2; for i =1: length (x) out ( ctr ) = x(i) + r* out ( ctr -1); ctr = ctr + 1; end

21 Homework 3 Solutions 21 or function [ out ] = order1lp (x,r) out (1) = x (1); for i =2: length (x) out (i) = x(i) + r* out (i -1); end end As value of r increases, the level of smoothing is increased.