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Consider a discrete-time, linear, shift-invariant system that has unit sample response h[n] and
input x[n].
(a) Sketch the response of this system if x[n] = b[n - no], for some no > 0, and h[n] = (i)"u[n].
(b) Evaluate and sketch the output of the system if h[n] = (I)"u[n] and x[n] = u[n].
(c) Consider reversing the role of the input and system response in part (b). That is

Evaluate the system output y[n] and sketch.
P4.5
(a) Using convolution, determine and sketch the responses of a linear, time-invariant system
with impulse response h(t) = e-t 2 u(t) to each of the two inputs x 1(t), x 2(t) shown in Figures
P4.5-1 and P4.5-2. Use yi(t) to denote the response to x 1(t) and use y2(t) to denote the
response to x 2(t).

(b) x 2(t) can be expressed in terms of x,(t) as
By taking advantage of the linearity and time-invariance properties, determine how y 2(t) can be
expressed in terms of yi(t). Verify your expression by evaluating it with yl(t) obtained in part (a) and
comparing it with y 2(t) obtained in part (a).
Optional Problems
P4.6
Graphically determine the continuous-time convolution of h(t) and x(t) for the cases shown in
Figures P4.6-1 and P4.6-2

P4.7
P4.8
Suppose that h(t) is as shown in Figure P4.8 and x(t) is an impulse train, i.e.,

P4.9
Determine if each of the following statements is true in general. Provide proofs for those
that you think are true and counterexamples for those that you think are false.
(a) x[n] *{h[ng[n]} = {x[n] *h[n]}g[n]
(b) If y(t) = x(t) * h(t), then y(2t) = 2x(2t) * h(2t).
(c) If x(t) and h(t) are odd signals, then y(t) = x(t) * h(t) is an even
signal.
(d) If y(t) = x(t) * h(t), then Ev{y(t)} = x(t) * Ev{h(t)} + Ev{x(t)} * h(t).
P4.10
Let 1 1(t) and 2 2(t) be two periodic signals with a common period To. It is not too difficult to
check that the convolution of 1 1(t) and t 2(t) does not converge. However, it is sometimes
useful to consider a form of convolution for such signals that is referred to as periodic
convolution. Specifically, we define the periodic convolution of t 1(t) and X2(t) as
Note that we are integrating over exactly one period.
(a) Show that q(t) is periodic with period To.
(b) Consider the signal

where a is an arbitrary real number. Show that
(c) Compute the periodic convolution of the signals depicted in Figure P4.10-1, where To = 1.
(d) Consider the signals x1[n] and x 2[n] depicted in Figure P4.10-2. These signals are periodic with
period 6. Compute and sketch their periodic convolution using No = 6.

(e) Since these signals are periodic with period 6, they are also periodic with period 12. Compute
the periodic convolution of xi[n] and x2[n] using No = 12.
P4.11
One important use of the concept of inverse systems is to remove distortions of some type. A
good example is the problem of removing echoes from acoustic signals. For example, if an
auditorium has a perceptible echo, then an initial acoustic impulse is

followed by attenuated versions of the sound at regularly spaced intervals. Consequently, a
common model for this phenomenon is a linear, time-invariant system with an impulse
response consisting of a train of impulses:
Here the echoes occur T s apart, and hk represents the gain factor on the kth echo resulting
from an initial acoustic impulse.
(a) Suppose that x(t) represents the original acoustic signal (the music produced by an
orchestra, for example) and that y(t) = x(t) * h(t) is the actual signal that is heard if no
processing is done to remove the echoes. To remove the distortion introduced by the echoes,
assume that a microphone is used to sense y(t) and that the resulting signal is transduced
into an electrical signal. We will also use y(t) to denote this signal, as it represents the
electrical equivalent of the acoustic signal, and we can go from one to the other via acoustic-
electrical conversion systems.
The important point to note is that the system with impulse response given in eq. (P4.11-1) is
invertible. Therefore, we can find an LTI system with impulse response g(t) such that
and thus, by processing the electrical signal y(t) in this fashion and then converting back to
an acoustic signal, we can remove the troublesome echoes. The required impulse response
g(t) is also an impulse train:
Determine the algebraic equations that the successive gk must satisfy and solve for gi, g2,
and g3 in terms of the hk. [Hint: You may find part (a) of Problem 3.16 of the text (page 136)
useful.]

(b) Suppose that ho = 1, hi = i, and hi = 0 for all i > 2. What is g(t) in this case?
(c) A good model for the generation of echoes is illustrated in Figure P4.11. Each successive echo
represents a fedback version of y(t), delayed by T s and scaled by a. Typically 0 < a < 1 because
successive echoes are attenuated.
(i) What is the impulse response of this system? (Assume initial rest, i.e., y(t) = 0 for t < 0 if x(t) =
0 for t < 0.)
(ii) Show that the system is stable if 0 < a < 1 and unstable if a > 1.
(iii) What is g(t) in this case? Construct a realization of this inverse system using adders, coefficient
multipliers, and T-s delay elements.
Although we have phrased this discussion in terms of continuous-time systems because of
the application we are considering, the same general ideas hold in discrete time. That is, the
LTI system with impulse response
is invertible and has as its inverse an LTI system with impulse response

It is not difficult to check that the gi satisfy the same algebraic equations as in part (a).
(d) Consider the discrete-time LTI system with impulse response
This system is not invertible. Find two inputs that produce the same output.
P4.12
Our development of the convolution sum representation for discrete-time LTI systems was
based on using the unit sample function as a building block for the representation of arbitrary
input signals. This representation, together with knowledge of the response to 5[n] and the
property of superposition, allowed us to represent the system response to an arbitrary input in
terms of a convolution. In this problem we consider the use of other signals as building blocks
for the construction of arbitrary input signals.
Consider the following set of signals:
(a) Show that an arbitrary signal can be represented in the form

by determining an explicit expression for the coefficient ak in terms of the values of the signal
x[n]. [Hint:What is the representation for 6[n]?]
(b) Let r[n] be the response of an LTI system to the input x[n] = #[n]. Find an expression for the
response y[n] to an arbitrary input x[n] in terms of r[n] and x[n].
(c) Show that y[n] can be written as

Solutions to Recommended Problems
S4.1
The given input in Figure S4.1-1 can be expressed as linear combinations of xi[n], x 2[n], X3[n].
Matlab Assignment He

(c) The system is not time-invariant because an input xi[n] + xi[n - 1] does not produce an output yi[n] +
yi[n - 1]. The input x,[n] + xi[n - 11 is xi[n] + xi[n - 1] = x2[n] (shown in Figure S4.1-3), which we are told
produces y 2[n]. Since y 2[n] # yi[n] + yi[n - 1], this system is not time-invariant.
S4.2
The required convolutions are most easily done graphically by reflecting x[n] about the origin and
shifting the reflected signal.
(a) By reflecting x[n] about the origin, shifting, multiplying, and adding, we see that y[n] = x[n] *
h[n] is as shown in Figure S4.2-1.

(b) By reflecting x[n] about the origin, shifting, multiplying, and adding, we see that y[n] = x[n] *
h[n] is as shown in Figure S4.2-2.
Notice that y[n] is a shifted and scaled version of h[n].
S4.
3
(a) It is easiest to perform this convolution graphically. The result is shown in Figure S4.3-1.

(b) The convolution can be evaluated by using the convolution formula. The limits can be
verified by graphically visualizing the convolution.
(c) The convolution can be evaluated graphically or by using the convolution formula.

S4.4
We note that this is merely a shifted version of h[n].

(c) Reversing the role of the system and the input has no effect because
The output and sketch are identical to those in part (b).
(a) (i) Using the formula for convolution, we have

(ii) Using the formula for convolution, we have

(b) Since x 2(t) = 2[xl(t) - xl(t - 3)] and the system is linear and time-invariant, y2(t) =
2[yi(t) - y1(t - 3)].
We see that this result is identical to the result obtained in part (a)(ii).
Solutions to Optional Problems

For t < 0 or t > 4, there is no contribution, so y(t) is as shown in Figure S4.6-12.

S4.9
(a) False. Counterexample: Let g[n] = b[n]. Then
and x[n] may in general differ from b[n].

S4.10
S4-12

(b) If r[n] is the response to #[n], we can use superposition to note that if

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