Convolution MATLAB has a function called conv(x,h) that you can use to convolve two discrete-time functions x(n) and h(n). It assumes that the time steps are the same in both cases. The input signals must be finite length, and the result of the convolution has a length that is the sum of the lengths of the two signals you are convolving (actually L1+L2-1). Recall that a linear time-invariant system is completely described by its impulse function. In MATLAB, the impulse response must be discrete. For example, consider the system with impulse response

h = [1 zeros(1,20) .5 zeros(1,10)];

Plot the impulse response using the plot command. Consider an input to the system,

x = [0 1:10 ones(1,5)*5 zeros(1,40)];

Plot the input with the plot command. Use the command conv to convolve x and h like this,

y = conv(x, h);

Use subplot to show the impulse response, input, and output of the convolution. Note that you need to add zeros to the end of x and h (to make them the same length as y) or define a time vector for each signal in order to make the timing comparable in the different subplots. Every non-zero coefficient of the impulse response h, acts as an echo. When you convolve the input x and impulse response h, you add up all the time-shifted and scaled echoes. Try making the second coefficient negative. How does this change the final result? Convolution and Echo Create a new script for this problem. Download the trumpet jazz lick "fall" here, and then load it into MATLAB using load('fall') and plot it. Use whos to see that the variables fall and Fs are created for you. (The sampling rate (Fs) for this signal should be 8000 Hz.) Use the following commands to convolve the following impulse response h, with the trumpet sound.

Fs = 8000     % for this example
h = [1 zeros(1,10000) .25 zeros(1,1000)];
y = conv(fall, h);
plot(y)
soundsc(y, Fs)

What if the second echo (in h) is a negative coefficient? When you play it, it should not sound different since your ear is not sensitive to that sort of modification (simple phase change). Now let's build a system that delays the echo for a quarter second by inserting Fs/4 zeros before the second impulse:


h = [1 zeros(1, round(Fs/4)) 0.25 zeros(1,1000)];

Pass the fall input signal through the system to get the output y:


y = conv(h, fall);

How do the input and output signals compare in the above step? (Look and listen). Experiment with different numbers of zeros, and try repeating this with some of the built-in MATLAB sounds. Some built-in sounds in MATLAB are chirp, gong, handel, laughter, splat, and train. Load them with the load command and the sound data will be loaded into the variable y and the sampling rate in Fs. Show the TA your script file. You should be able to run it and have it generate any plots and sounds. You can use the pause command to pause MATLAB until a key is pressed to prevent it from playing all your sounds at once. Convolution and Smoothing Build a box impulse response:


h2=[ones(1,50)/50 zeros(1,20)];

Create a new signal y2 by convolving "fall" with h2 How does the output sound different from the input signal? Visually, a difference is that the input signal fall looks like it's centered around value 0, and the system output y2 looks like it's more positive. Let's look more closely. Find the average value of the signal fall (use sum(fall)/length(fall)), and you should see that in fact the fall signal isn't really centered around 0. Next, to see what this system does to the input signal, zoom in on part of the signal:

subplot(2,1,1), plot(6400:6500, fall(6400:6500))
subplot(2,1,2), plot(6400:6500, y2(6400:6500))

The convolved signal should look a little smoother to you. This is because this impulse response applies a low-pass filter to the signal. We'll learn more about filters a bit later, but basically the idea is that the original signal is made up of sounds at many different frequencies, and the lower frequencies pass through the system, but the higher frequencies are attenuated. This affects how it sounds as well as how it looks. Box Function Create a new function called unitstep.m in MATLAB. The function should take two parameters, a time vector that specifies the finite range of the signal and a time shift value. Calling unitstep([time],ts) should be equivalent to u(t + ts) Use the unitstep function to create a box-shaped time signal. Write a new function called boxt.m that creates a box with specified start and end times t1 and t2. In other words, your function should take three inputs: scalars t1 and t2, and a time vector t, and should output a vector of the same size as t, which contains the values of u(t-t1)-u(t-t2) evaluated at each point in t. Create a script file called boxtscript.m that uses the function to create a box that starts at time t = -1 and ends at time t = 1, where the signal lasts from time t = -3 to t = 3. Generate three different versions of this box using three different time granularities, where the finest granularity has very sharp edges similar to the ideal box and the coarsest granularity has a step size of 0.5. The different versions should all have the same time span; the difference in the plots should only be at the edges of the box because of artifacts in continuous plotting of a discrete-time signal. Plot all three versions in one figure using subplot and save it as boxtscript.tif. Time: If u is a vector of length n with time span tu = t1:del:t2, and v is a vector of length m with time span tv = t3:del:t4, and both have the same time step del, then the result of conv(u,v) will be a vector of length n + m - 1 with a time span tc = (t1+t3):del:(t2+t4). Using the box function that you wrote in step 2 with a sufficiently fine grained step size (for example, del = 0.01), find and plot the result of the convolution of box(0,4) and box(-1,1) and save it as convplot.tif. Verify that the timing of signal rising and falling matches what you expect in theory. Amplitude: In the resulting plot from the previous step, you should notice that the amplitude is much higher than the max of 2 that you would expect from analytically computing the convolution. This is because it is thinking that the length of the box is n rather n del, which impacts the area computation in convolution. To get the correct height, you need to scale by del. Scale and plot the resulting function, and verify that the height is now 2. Save the figure as scaled.tif Triangle: Design the impulse response for a system h and a system input x such that you get a perfectly symmetric triangle of length 100 as the system output y. Use subplot to plot x, h, and y, and save the plot as tri.tif. Be able to demonstrate your code, show your plots, and play sounds.