EE 303 SIGNAL & SYSTEMS
LAB 4: CONVOLUTION
Objective: The objective of this experiment is to learn how to convolve two signals using
Matlab.
Preliminary Work: Please review material about convolution from textbook or your
notes.
Part I: Signals Perspective
For two discrete-time signals x[n] and h[n] that are both o f finite length, you can
implement the convolution operation y[n] = x[n] * h[n] in Matlab using the command
conv. The resulting Matlab array that holds the values of y[n] will have a length equal to
the length o f x[n] plus the length of h[n] minu s o ne.
Here is an example pro ble m: suppose that
x[ n] = δ[ n] - 2δ [ n - 1] + 3δ[n - 2] - 4δ[ n - 3] + 2δ[n - 5] + δ[n - 6]
an d
h[ n] = 3δ [n] + 2 δ [n - 1] + δ[ n - 2 ] - 2 δ[n - 3] + δ[ n - 4 ] - 4 δ[n - 6] + 3δ [n - 8] .
The problem is to use Matlab to co mpute the convolutions y1[n] = x[n] * h[n] and
y2[n] = x[n — 3] * h[n].
Here is t he solut ion for the example problem:
The signal x[n] "goes" from n = 0 t o n = 6, so the shifted signal x[n — 3] "goes" from
n = 0 t o n = 9. This means that we need a Matlab array with 10 elements to hold the
values of x[n — 3] (for n = 0,1,..., 9). To mak e the plott ing easy, we will store both
x[n] and x[n — 3] in 10-element Matlab arrays.
Since h[n] "goes" fro m n = 0 to n = 8, we w ill need a nine-element Matlab array to
hold the values of h[n]. Each convolution result will have a length of 10 + 9 - 1 = 18,
corresponding to the "times" n = 0 , 1, . . ., 17 .
Her e is sa mple Mat lab code t o co mput e the two convolutio ns and plot them both in
the same figur e using the subp lot command:
x = [1 -2 3 -4 0 2 1 0 0 0]; % Input signal zero-padded to length 10
xm3 = [0 0 0 1 -2 3 -4 0 2 1]; % Shifted input signal x[n-3]
h = [3 2 1 -2 1 0 -4 0 3]; % Impulse response
y1 = conv(x,h); % Make y1[n] = x[n] * h[n]
y2 = conv(xm3,h); % Make y2[n] = x[n-3] * h[n]
n = 0:17; % "time" values for x-axis of plot
