Discrete-Time Systems 489

8.3 Discrete-Time Systems 489

or the convolution sum of the input and the impulse response. This is a very important result, indicating that the output of FIR systems is obtained by means of the convolution sum rather than difference equations, which gives great significance to the efficient computation of the convolution sum.

Considering the convolution sum as an operator—that is,

y[n] = [h ∗ x][n] =

x[k]h[n − k]

k =−∞

it is easily shown to be linear. Indeed, whenever the input is ax 1 [n] + bx 2 [n], and {y i [n] } are the outputs corresponding to {x i [n] } for i = 1, 2, then we have that

X X [h X ∗ (ax

1 + bx 2 ) ][n] =

( ax 1 [k] + bx 2 [k])h[n − k] = a

x 1 [k]h[n − k] + b

x 2 [k]h[n − k]

= a[h ∗ x 1 ][n] + b[h ∗ x 2 ][n] = ay 1 [n] + by 2 [n]

as expected, since the system was assumed to be linear when the expression for the convolution sum was obtained. We will then have that if the output corresponding to x[n] is y[n], given by the convolution sum, then the output corresponding to a shifted version of the input, x[n − N], should be y[n − N]. In

fact, if we let x 1 [n] = x[n − N], the corresponding output is

X [h X ∗x

1 ][n] =

x 1 [n − k]h[k] =

x[n − N − k]h[k]

= [h ∗ x][n − N] = y[n − N]

Again, this result is expected given that the system was considered time invariant when the convolution sum was obtained.

From the equivalent representations for the convolution sum we have that

X [h X ∗ x][n] = x[k]h[n − k] = x[n − k]h[k]

= [x ∗ h][n]

indicating that the convolution commutes with respect to the input x[n] and the impulse response h[n].

Just as with continuous-time systems, when conecting two LTI discrete-time systems (with impulse responses h 1 [n] and h 2 [n]) in cascade or in parallel, their respective impulse responses are given by

[h 1 ∗h 2 ][n] and h 1 [n] +h 2 [n]. See Figure 8.9 for block diagrams.

There are situations when instead of giving the input and the impulse response to compute the output, the information that it is available is, for instance, the input and the output and we wish to find the impulse response of the system, or we have the output and the impulse response and wish to find the input. This type of problem is called deconvolution . We consider this problem later in this chapter after considering causality, and in Chapter 9 where we show that it can be easily solved using the Z-transform.

The computation of the convolution sum is typically difficult. It is made easier when the Z-transform is used, as we will see. MATLAB provides the function conv which greatly simplifies the computation.

C H A P T E R 8: Discrete-Time Signals and Systems

y [n] (h 1 ∗h 2 ) [n] FIGURE 8.9

x [n]

y [n]

(a) Cascade and

(b) parallel connections of LTI

(a)

systems with impulse responses h 1 [n] and

h 1 [n]

h 2 [n]. Equivalent

y [n] systems on the right.

y [n]

x [n]

h 1 [n] + h 2 [n] Notice the interchange of

x [n]

h 2 [n]

systems in the cascade connection.

(b)

■ Example 8.23

Consider a moving-averaging filter where the input is x[n] and the output is y[n]:

1 y[n] = ( x[n] + x[n − 1] + x[n − 2])

Find the impulse response h[n] of this filter. Then, (a) Let x[n] = u[n]. Find the output of the filter y[n] using the input–output relation and the

convolution sum. (b) If the input of the filter is x[n] = A cos(2πn/N)u[n], determine the values of A and N, so that

the steady-state response of the filter is zero. Explain. Use MATLAB to verify your results.