ECE2111 review questions for topic 1–3
The following 10 questions cover concepts introduced in topics 1–3. They are indicative of the style of problems on the first mid-semester test for ECE2111. Any of the problems on tutorial sheets 1–4 are also useful for review, as are many problems at the end of chapter 1 and 2 of the recommended text A. V. Oppenheim and A. S. Willsky, ‘Signals and Systems’.
1. Let x be a discrete-time signal defined by
x[n] = 2 cos(4πn/5) + 1 for all n.
Which of the following is true?
(a) x is odd and periodic with fundamental period 5 samples. (b) x is even and periodic with fundamental period 5/2 samples.
(c) x is even and periodic with fundamental period 5 samples. (d) x is odd and periodic with fundamental period 5/2 samples.
Solution: x even and periodic with fundamental period 5 samples. x is even because
x[−n] = 2 cos(−4πn/5) + 1 = 2 cos(4πn/5) + 1 = x[n] for all n.
The fundamental period of x is the smallest positive integer N such that N = k(2π/(4π/5)) = 5/2 for some integer k. We can see that the fundamental period is N = 5. We could have deduced this simply by noting that the period of a discrete-time signal must be an integer (ruling out the answer 5/2).
2. Thegraphofxfor−1≤t≤4isshownbelow: x(t)
Which of the following is the graph of |x(3 − t)| restricted to −1 ≤ t ≤ 4? (a)
|x(3 − t)| 1
−1 1 2 4 1
|x(3 − t)| 1
−1 34 |x(3 − t)|
−1 4 |x(3 − t)|
Solution: The correct graph is obtained by first reflecting the graph of x in the vertical axis and then shifting to the right by 3 units. This gives
|x(3 − t)| 1
3. Suppose x is a continuous-time bounded signal with period T. If S is a BIBO stable, time-invariant, continuous-time system then which of the following statements is true of the output y = S(x)?
(a) The output satisfies y(t + T ) = y(t) for all t but we cannot determine whether the output is bounded.
(b) The output is bounded and we cannot determine whether it is periodic.
(c) The output is bounded and y(t + T ) = y(t) for all t.
(d) We cannot determine whether the output is bounded or whether it is periodic.
Solution: The output is bounded and y(t + T ) = y(t) for all t. Because the system
is time-invariant we know that if y = S(x) then Delay−T (y) = S(Delay−T (x)). Because
x is periodic with periodic T we know that x(t + T ) = x(t) for all t. In other words, δ−T (x) = x. Combining these two facts we see that
y = S(x) = S(Delay−T (x)) = Delay−T (y). This is equivalent to y(t) = y(t + T ) for all t.
Because the system input is bounded, and the system is BIBO stable, we know (by definition) that the system output must be bounded.
4. Let S be a linear system. Let h0 be the output of the system when the input is the unit impulse. Let h1 be the output when the input is Delay1(δ), the unit impulse delayed by one sample. Which of the following is true?
(a) h1[n]=h0[n−1]foralln
(b) If the system input is x[n] = δ[n] + 2δ[n − 1] then the system output is h0[n] + 2h1[n]
(c) The output of the system is the convolution of the input and h0. (d) The output of the system is never the zero signal.
Solution: If the system input is x[n] = δ[n] + 2δ[n − 1] then the system output is h0[n] + 2h1[n]. This follows from the definition of linearity of a system. Note that the system is not time-invariant, so there is not necessarily any relationship between h1 and h0. Similarly, since the system in not time-invariant, the output is not necessarily the convolution of the input and the inpulse response. Finally, since the system is linear, if the input is x[n] = 0 for all n, then the output will be y[n] = 0 for all n, so it is possible for the output to be the zero signal.
5. A discrete-time linear time invariant system with input x and output y is described by the following difference equation
y[n]=1.1y[n−1]+x[n] foralln
subject to the condition of initial rest. Which of the following statements are true?
(a) The impulse response of the system is h[n] = (1.1)n for all n and the system is not BIBO stable.
(b) The impulse response of the system is h[n] = (1.1)nu[n] for all n and the system is BIBO stable.
(c) The impulse response of the system is h[n] = (1.1)nu[n] for all n and the system is not BIBO stable.
(d) The impulse response of the system is h[n] = (0.1)nu[n] for all n and the system is BIBO stable.
Solution: The impulse response of the system is h[n] = (1.1)nu[n] for all n and the system is not BIBO stable. The impulse response is the output of the system if the input is the unit impulse. The unit impulse satisfies δ[n] = 0 for all n < 0. Since the system satisfies the condition of initial rest, h[n] = 0 for all n < 0. This immediately rules out h[n] = (1.1)n for all n as a correct form for the impulse response. We can then see that y[0] = δ[0] = 1 and that for n ≥ 1, we have that y[n] = 1.1y[n−1]. From this we can see that y[n] = (1.1)ny[0] = (1.1)n for all n ≥ 0. Hence h[n] = (1.1)nu[n] for all n. Finally, to decide whether the system is or is not BIBO stable, we can see that the unit impulse δ is a bounded input, and yet the corresponding output h is unbounded. This tells us that the system is not BIBO stable.
6. The following block diagram describes the relationship between the input x and output y of a discrete-time system.
Which of the following is a plot of x[n] vs n for −3 ≤ n ≤ 4? (a)
Which of the following difference equations describes the same input-output relationship?
(a) y[n]=x[n]+2x[n+1]+0.5x[n+2]foralln. (b) y[n]=x[n]+2x[n−1]+0.5x[n−2]foralln. (c) y[n]=x[n]+2y[n−1]+0.5x[n−2]foralln. (d) y[n]=x[n]+2y[n+1]+0.5x[n+2]foralln.
Solution: y[n] = x[n] + 2y[n − 1] + 0.5x[n − 2] for all n. We can see that the output signal y must satisfy
y = x + Delay1(2y + Delay1(0.5x)) = x + 2Delay1(y) + 0.5Delay2(x)
where the second equality holds because delay and gain blocks commute and because
Delay1 is linear. This is equivalent to y[n]=x[n]+2y[n−1]+0.5x[n−2] foralln.
7. A discrete-time signal x is defined by
x[n] = 2δ[n + 2] + δ[n] + 3δ[n − 2] + δ[n − 3].
−3 −2 −1 0 1 2 3 4 4
−3 −2 −1 0 1 2 3 4 x[n]
7 6 5 4 3 2 1
−3 −2 −1 0 1 2 3 4 x[n]
−3 −2 −1 0 1 2 3 4
Solution: We can see that
The correct plot is then
2 ifn=−2
x[n]=3 ifn=2
0 otherwise. x[n]
subject to y[n] = 0 for all n ≤ 0.
(b) A discrete-time LTI system with impulse response h[n] = cos(πn)u[n + 1] for all n
(c) The continuous-time system with input x and output y related by y(t) = (x(t − 1))2 for all t.
(d) The continuous-time system with input x and output y related by y(t) = tx(t) for all t.
Solution: The continuous-time system with input x and output y related by y(t) = (x(t − 1))2 for all t. This system is causal because the output at any time t depends only on the input at the past time t − 1, and certainly not on future inputs. The system is time-invariant because if the input is z(t) = x(t−τ) for some τ, then the output is
−3 −2 −1 0 1 2 3 4
8. Which of the following systems are causal and time-invariant? (a) The discrete-time system defined by the difference equation
y[n]=0.5y[n−2]+x[n] foralln
The system defined by
(z(t−1))2 =(x(t−1−τ))2 =y(t−τ). y[n]=0.5y[n−2]+x[n] foralln
subject to y[n] = 0 for all n ≤ 0, is not time-invariant because of the initial condition. The discrete-time LTI system with impulse response h[n] = cos(πn)u[n + 1] for all n is not causal because h[−1] = cos(−π)u[0] = −1 ̸= 0. The continuous-time system with input x and output y related by y(t) = tx(t) for all t is not time-invariant. To see this, consider the unit step input. The corresponding output is y1(t) = tu(t). If we delay the input by one unit of time, the output is y2(t) = tu(t−1) for all t. Note that if t = 2 then y1(t − 1) = y1(1) = 1 and yet y2(t) = y2(2) = 2. Hence Delay1(y1) and y2 are not the same signal, so the system is not time-invariant.
9. Which of the following is true for a discrete-time infinite impulse response LTI system?
(a) The system is never BIBO stable. (b) The system is always causal.
(c) The system is always BIBO stable.
(d) A possible output of the system is the zero signal, i.e., y[n] = 0 for all n.
Solution: A possible output of the system is the zero signal, i.e., y[n] = 0 for all n. Since the system is linear, if the input signal is x[n] = 0 for all n then the output signal is y[n] = 0 for all n. (This follows from the scaling/homogeneity property.) To see why the other answers are incorrect, we note that the system with (infinite) impulse response h[n] = (0.5)nu[n + 1] is not causal and is BIBO stable (because ∞k=−∞ |h[k]| = 4 and so the impulse response is absolutely summable). This means that IIR systems can be non-causal, and that IIR systems can be BIBO stable. Similarly the system with (infinite) impulse response h[n] = 2nu[n] is not BIBO stable, because the impulse response is not absolutely summable.
10. Let x be the discere-time signal defined by x[n] = (−1)n for all n. If x is the input to the discrete-time LTI system with impulse response defined by
h[n] = n(u[n] − u[n − 4]) for all n, which of the following describes the output y of the system.
(a) y[n] = 2(−1)n−1 for all n. (b) y[n] = 2(−1)n for all n.
(c) y[n] = 3(−1)n−1 for all n. (d) y[n] = 3(−1)n for all n.
Solution: y[n] = 2(−1)n−1 for all n. We first observe that h[n] = 0 unless n = 1, 2, 3. We then use the fact that the output of an LTI system is the convolution of the input
and the impulse response. x ∗ h = h ∗ x, to see that
We use the convolution formula, together with the fact that
(h ∗ x)[n] ∞
h[k]x[n−k] k=−∞
h[1]x[n − 1] + h[2]x[n − 2] + h[3]x[n − 3] x[n − 1] + 2x[n − 2] + 3x[n − 3]
(−1)n−1 + 2(−1)n−2 + 3(−1)n−3 (−1)n−1(1 + 2(−1)−1 + 3(−1)−2) 2(−1)n−1 for all n.