Homework 5 Solutions

Transcription

1 Signals and Systems Profs. Byron Yu and Pulkit Grover Fall 2018 Homework 5 Solutions. Part One 1. (12 points) Calculate the following convolutions: (a) x[n] δ[n n 0 ] (b) 2 n u[n] u[n] (c) 2 n u[n] u[n 4]. (Hint: Use the results from parts (a) and (b).) Solution: (a) x[n] δ[n n 0 ] (b) 2 n u[n] u[n] x[n] δ[n n 0 ] = k= x[k]δ[n n 0 k] = x[n n 0 ] 2 n u[n] u[n] = 2 k u[k]u[n k] = k= =(2 n+1 1)u[n] (c) 2 n u[n] u[n 4]. (Hint: Use the results from parts (a) and (b).) 2 n u[n] u[n 4] = 2 n u[n] u[n] δ[n 4] = (2 n 3 1)u[n 4] n k=0 2 k

2 2 Homework 5 Solutions 2. (14 points) Determine the step response of a LTI system represented by each of the following impulse responses: (a) h(t) = ( 1 2 )t 1 u(t 1) (b) h[n] = u[n] u[n 5] Solution: (a) s(t) = t h(τ)dτ Case 1: t < 1, s(t) = 0 Case 2: t 1. s(t) = t (b) s[n] = k = n h[n] Case 1: n < 0, s[n] = 0 Case 2: 0 n 3, s[n] = n k=0 1 = n + 1 Case 3: n 4, s[n] = 4 k=0 1 = 5 1 ( 1 2 )τ 1 dτ = 1 2 t+1 ln2

3 Homework 5 Solutions 3 3. (7 points) Determine the frequency response of a LTI system represented by the impulse response h(t) = e αt+β u(t). Solution: (a) h(t) = e αt+β u(t) H(jω) = = = τ= τ= τ=0 h(τ)e jωτ dτ e ατ+β u(τ)e jωτ dτ e ατ+β e jωτ dτ = e β e α e jωτ dτ τ=0 = e β e (α jω)τ dτ τ=0 This integral converges only of e αt < 1, t > 0, a < 0. H(jω) = e β e(α jω)τ α jω τ=0 = eβ α jω

4 4 Homework 5 Solutions 4. (13 points) Consider the cascade of two LTI systems depicted below, where: h 1 [n] = sin(n) h 2 [n] = u[n] = { 1 n 0 0 n < 0 Suppose the input is x[n] = δ[n] δ[n 1]. Compute the output y[n]. (Hint: use the commutative property of convolution.) Solution: y[n] = h 2 [n] (h 1 [n] x[n]) = h 2 [n] (x[n] h 1 [n]) = (h 2 [n] x[n]) h 1 [n] h 2 [n] x[n] = h 2 [n] (δ[n] δ[n 1]) = h 2 [n] h 2 [n 1] = u[n] u[n 1] = δ[n] y[n] = δ[n] h 1 [n] = h 1 [n] = sin(n)

5 Homework 5 Solutions 5 5. (13 points) Consider the cascade of causal LTI systems depicted below: where h 2 [n] = δ[n] + δ[n 1] is an impulse response and the overall impulse response h[n] is depicted below: h[n] n Find the impulse response h 1 [n]. Solution: h[n] = h 2 [n] h 1 [n] h 2 [n] = h 2 [n] h 2 [n] h 1 [n] h 2 [n] h 2 [n] = (δ[n] + δ[n 1])h 2 [n] = h 2 [n] + h 2 [n 1] = δ[n] + δ[n 1] + δ[n 1] + δ[n 2] = δ[n] + 2δ[n 1] + δ[n 2] h[n] = (δ[n] + 2δ[n 1] + δ[n 2]) h 1 [n] = h 1 [n] + 2h 1 [n 1] + h 1 [n 2] h 1 [n] = h[n] 2h 1 [n 1] h 1 [n 2]

6 6 Homework 5 Solutions Since h 1 [n] is causal, we have that h 1 [n] = 0 for n < 0, so that: Hence, h 1 [n] is as specified above for n = 0,, 7, and h 1 [n] = 0 for all other values of n.

7 Homework 5 Solutions 7 6. (12 points) Given each impulse response below, determine whether the corresponding LTI system is: Stable Memoryless Causal and justify. (a) h(t) = e (1 2j)t u(t) (b) h[n] = n cos( π 4 n)u[n] Solution: (a) h(t) = e (1 2j)t u(t) i. Stable. h(t) dt = = 0 0 e t e 2jt dt e t dt = e t 0 = ( 1) = 1 Note that e jα has a magnitude of 1 regardless of α. ii. It is not memoryless, as it is not in the format of h(t) = cδ(t). iii. It is causal, as t < 0, h(t) = 0 (b) h[n] = n cos( π 4 n)u[n] i. Not stable. n= h[n] = n= n cos( π 4 n)u[n] = n cos( π 4 n) ii. It is not memoryless, as it is not in the format of h[n] = cδ[n]. iii. It is causal, as n < 0, h[n] = 0 n=0

8 8 Homework 5 Solutions Part Two 7. (15 points) MATLAB. In this problem, we will use convolution to decompose and approximately recompose a signal in MATLAB. Please download hw5 matlab.zip from the course website ( matlab.zip). Unzip it in your working directory and you will see two files, hw5 q1.mat and view frequencies.m. hw5 q1.mat contains several variables that you will work with: i. a signal sig ii. impulse response of 3 filters IR1,IR2 and IR3 iii. fs = which is the sampling frequency (44.1 khz) of sig The function view frequencies takes in an impulse response and the sampling frequency, and outputs the frequency response. To obtain the frequency response of IR1 you can call the function by view frequencies(ir1, fs). The built-in MATLAB function sound(sig, fs) will play back the signal sig at frequency fs. (a) (3 points) Observe the frequency response of the three filters by calling view - frequencies. The script below gives you an example of how to plot the frequency response of IR1. clear all load ( hw5_q1. mat ) % load data into the workspace [f,amp ] = view_frequencies (IR1,fs ); plot (f,amp ) title ( Frequency response of IR1 ) xlabel ( Frequencies (Hz ) ) ylabel ( Magnitude ) Submit your answer to the question below (you don t need to submit your code or plots for this part): Of the three filters which one is a low-pass filter, high-pass filter and bandpass filter? (b) (4 points) Convolve each of the three impulse responses separately with the signal sig. Listen to sig and each of the three filtered outputs. Submit the following: Based only on what you hear, briefly describe whether each of the 3 filtered outputs sounds different from sig and if so, how they are different. (c) (4 points) Calling view frequencies(sig,fs) can be used to examine the frequency components contained in the signal sig. Generally speaking, music played by a cello will contain more low frequency components than that played by a flute.

9 Homework 5 Solutions 9 So from what you hear in part (a), you should have an expectation of what kinds of frequency components are contained in each filtered output. Verify your prediction by plotting the frequency components of sig, as well as those of the 3 filtered outputs. Submit the following: Plots of the frequency components of sig and each of the 3 filtered outputs and your code for generating them (including the convolution step for filtering the signals). Title the plots with the corresponding type of filter used to obtain the resulting signal (or simply Original Signal for sig). For example, title the plot for the signal obtained by convolving sig with a low-pass filter as Convolution with Low-Pass. Briefly explain how each filter modifies the frequency components of sig after convolving with it. (d) (4 points) We have separated the signal sig into three components, i.e., we decomposed it into 3 bands based on frequency: low, middle, and high frequencies. Now, let s try to put the 3 components back together. Add the three filtered signals together and listen to the resulting signal. Submit the following: Solution: Plots of the original signal sig and the reconstructed signal (remember to truncate the excess section of the reconstructed signal due to convolution). Do the plots look exactly the same? Do the signals sound similar or different? (a) IR1 is high-pass, IR2 is low-pass and IR3 is band-pass. (b) All three filtered outputs sound different from the original signal. The one filtered by IR1 has the high-pitched instruments left. The one filtered by IR2 has the lowpitched instruments left. The one filtered by IR3 is left with instruments that are at the middle of the pitch range. (c) IR1 attenuated the low frequency components, IR2 attenuated the high frequency components, and IR3 attenuated both high and low frequency components. % convolve the filters with the original signal sig1 = conv (sig, IR1, same ); sig2 = conv (sig, IR2, same ); sig3 = conv (sig, IR3, same ); % view the frequency components of the original signal and filtere [f,amp ] = view_frequencies (sig, fs ); figure ; plot (f, amp )

10 10 Homework 5 Solutions title ( The Original Signal ) xlabel ( Frequencies (Hz ) ) ylabel ( Magnitude ) % repeat the following and replace with the % other two signals sig2 and sig3 [f1, amp ] = view_frequencies (sig1, fs ); figure ; plot (f1, amp ) title ( Convolution with High - Pass ) xlabel ( Frequencies (Hz ) ) ylabel ( Magnitude )

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12 12 Homework 5 Solutions (d) The plots seem similar while the amplitude of the reconstructed signal seems to be greater. Both signals sound quite similar to each other, but the reconstructed signal sounds like it s played by a speaker covered by a thin sheet. % convolve the filters with the original signal recomp_sig = sig1 + sig2 + sig3 ; figure plot ( sig ) xlabel ( Time ) ylabel ( Amplitude ) title ( The Original Signal ) figure plot ( recomp_sig ) % or plot ( recomp_sig (501: end -500)) % if used default convolution result length xalbel ( Time ) ylabel ( Amplitude ) title ( The Reconstructed Signal )

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14 14 Homework 5 Solutions 8. (13 points) MATLAB. Convolutions are also defined for 2D signals, such as grayscale images. In this problem we will explore image filtering based on 2D convolutions. Consider a grayscale image I(x, y), which is a discrete 2D signal, and a discrete 2D impulse response h(x, y). A discrete 2D convolution is defined as: J(x, y) = W U w= W u= U I(x u, y w)h(u, w) The impulse response h is defined on the 2D interval { U u U, W w W }, which is represented in a matrix of size (2U + 1) (2W + 1). (a) (4 points) Gaussian blur impulse response We will experiment with the Gaussian blur impulse response. In the 1D case, the Gaussian impulse response smooths out the signal and has the following form h(t) = e t2 2σ 2, where σ controls the degree of smoothing. In 2D, we can smooth in both directions. If we smooth the same amount in all directions, the impulse response will have the following expression: h(u, w) = e ( u2 +w 2 2σ 2 ) Our impulse response has a finite size of (2U + 1) (2W + 1) since it is easier to implement in Matlab. Below is the function to build a symmetric 2D Guassian impulse response for a given size and given σ. function h = gaussian2d ( sigma, U, W) %% GAUSSIAN2D generates a symmetric normalized 2 D Gauss. % of size is [2* U +1 x 2* W +1] h = zeros (2* W+1, 2*U +1); for y = 1 : 2* W +1 for x = 1 : 2* U + 1 w = y - W - 1; u = x - U - 1; h(y, x) = exp (-(w^2 + u ^2) /(2* sigma ^2)); end end end Using this function, generate three impulse responses for three combinations of parameters (σ, U, W ) = (1, 5, 5), (5, 5, 5), and (5, 10, 10). Submit the following: Plots of the three impulse responses using surf(h).

15 Homework 5 Solutions 15 (b) (2 points) Normalization Our impulse response has to be normalized so that its values sum up to one. Change the function gaussian2d appropriately. The result should satisfy: h = gaussian2d ( sigma, U, W); % sum of all the values should be 1 assert ( abs ( sum (h (:)) -1) < 10^( -12)) Submit the following: Your code for the modified function. Explain why this normalization is necessary for image blur (Hint: think about what the blur should do to a one-colored gray image, i.e. I(x, y) = I 0 for all x and y). (c) (4 points) Convolution We will run experiments on saturn.png image, which is built in to Matlab. Load it and transform the image to grayscale: I = imread ( saturn.png ); I = rgb2gray (I); To display the image in a figure, use Matlab command imshow: imshow (I); Instead of implementing our own 2D convolution function, we will use Matlab function conv2 for this exercise. Just like with 1D convolution, we always need to pay attention to the boundary conditions, i.e. how to treat pixels near image edges. Matlab offers several different options. In this exercise we will use the one named same, which does not change the size of the input image: I = double (I); I2 = conv2 (I, h, same ); imshow ( uint8 (I2 )); Create two impulse responses with different values of σ, i.e (σ, U, W ) = (3, 9, 9) and (10, 30, 30). Use conv2 to blur the saturn image and submit two output blurred images. You can use function imwrite(uint8(i2), filepath ) to save an image. Function uint8 transforms floating point values into integers in range [0, 255]. Submit the following: The two blurred images. Does the level of blurring increase or decrease with higher values of σ?

16 16 Homework 5 Solutions (d) (3 points) Directional blur Instead of blurring uniformly, we can perform a directional blur. To blur an image in the X-direction, we need to create an impulse response, such that its values change only in the X-direction. Fortunately, we can accomplish this without making any changes to the gaussian2d function! The following code generates an impulse response with σ= 10. Note that W is set to 0, which means the impulse response is of size (2U + 1) 1. That is, the impulse response only exists along the X-direction. h_x = gaussian2d (10, 30, 0); Perform convolution of the saturn image with h x and submit the following The blurred image. How can you tell the result is blurred only along the X-direction? Solution: (a) Three figures shown below:

17 Homework 5 Solutions 17 (b) The blur must have no effect if applied to a one-colored gray image because there is nothing to blur. Let image intensity be I 0. Then convolution operation applied at every pixel can be written as: W U W U I(x u, y w)h(u, w) = I 0 h(u, w) = I 0 w= W u= U w= W u= U

18 18 Homework 5 Solutions W U w= W u= U h(u, w) = 1 function h = gaussian2d ( sigma, U, W) %% GAUSSIAN2D generates a symmetric normalized 2 D Gauss. % of size is X x Y = [2* U +1 x 2* W +1] h = zeros (2* W+1, 2*U +1); for y = 1 : 2* W +1 for x = 1 : 2* U + 1 w = y - W - 1; u = x - U - 1; h(y, x) = exp (-(w^2 + u ^2) /(2* sigma ^2)); end end h = h / sum (h (:)); (c) As σ increases, blurring increases. (3, 9, 9) (10, 30, 30)

19 Homework 5 Solutions 19 (d) The image is blurred only along the X-direction because the more vertical the edges in the image the more blurred they are.

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