# Signals & Systems Questions and Answers – Discrete Time Signals

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This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Discrete Time Signals”.

1. Is the function y[n] = sin(x[n]) periodic or not?
a) True
b) False

Explanation: ‘y’ will be periodic only if x attains the same value after some time, T. However, if x is a one-one discrete function, it may not be possible for some x[n].

2. What is the time period of the function x[n] = exp(jwn)?
a) pi/2w
b) pi/w
c) 2pi/w
d) 4pi/w

Explanation: Using Euler’s rule, exp(2pi*n) = 1 for all integer n. Thus, the answer can be derived.

3. What is the nature of the following function: y[n] = y[n-1] + x[n]?
a) Integrator
b) Differentiator
c) Subtractor
d) Accumulator

Explanation: If the above recursive definition is repeated for all n, starting from 1,2.. then y[n] will be the sum of all x[n] ranging from 1 to n, making it an accumulator system.
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4. Is the above function defined, causal in nature?
a) True
b) False

Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.

5. Is the function y[n] = x[n-1] – x[n-4] memoryless?
a) True
b) False

Explanation: Since the function needs to store what it was at a time 4 units and 1 unit before the present time, it needs memory.

6. Is the function y[n] = x[n-1] – x[n-56] causal?
a) The system is non causal
b) The system is causal
c) Both causal and non causal
d) None of the mentioned

Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.

7. Is the function y[n] = y[n-1] + x[n] stable in nature?
a) It is stable
b) It is unstable
c) Both stable and unstable
d) None of the mentioned

Explanation: It is BIBO stable in nature, i.e. bounded input-bounded output stable.

8. If n tends to infinity, is the accumulator function a stable one?
a) The function is marginally stable
b) The function is stable
c) The function is unstable
d) None of the mentioned

Explanation: The system would be unstable, as the output will grow out of bound at the maximally worst possible case.

9. We define y[n] = nx[n] – (n-1)x[n]. Now, z[n] = z[n-1] + y[n], is z[n] stable?
a) Yes
b) No

Explanation: As we take the sum of y[n], terms cancel out and deem z[n] to be BIBO stable.

10. We define y[n] = nx[n] – (n-1)x[n]. Now, z[n] = z[n-1] + y[n]. Is z[n] a causal system?
a) No
b) Yes

Explanation: As the value of the function depends solely on the value of the input at a time presently and/or in the past, it is a causal system.

11. Discrete-time signals are _________________
a) Continuous in amplitude and continuous in time
b) Continuous in amplitude and discrete in time
c) Discrete in amplitude and discrete in time
d) Discrete in amplitude and continuous in time

Explanation: A discrete-time signal is continuous in amplitude and discrete in time. It can either be present in nature or is sampled from an analog signal. A digital signal is discrete in amplitude and time.

12. Determine the discrete-time signal: x(n)=1 for n≥0 and x(n)=0 for n<0
a) Unit ramp sequence
b) Unit impulse sequence
c) Exponential sequence
d) Unit step sequence

Explanation: Unit step is defined by: x(n)=1 for n≥0 and x(n)=0 for n<0.

13. Determine the value of the summation: ∑n= -∞δ(n-1)sin2n.
a) 1
b) 0
c) sin2
d) sin4

Explanation: ∑n= -∞δ(n-1)sin2n
⇒ We know, δ(n) is impulse function which means δ(n)=1 when n=0
⇒ δ(n-1)=1 when n=1 otherwise it is 0.
Therefore, the summation’s limit reduces to n=1
⇒ ∑n= -∞δ(n-1)sin2n = sin2n|n=1 = sin2.

14. Determine the value of the summation: ∑n= -∞ δ(n+3)(n2+n).
a) 3
b) 6
c) 9
d) 12

Explanation: ∑n= -∞ δ(n+3)(n2+n)
⇒ δ(n+3)=1 when n= -3 otherwise 0.
Therefore, the limit reduces to n = -3
⇒ ∑n = -∞ δ(n+3)(n2+n) = (n2+n)|n = -3 = (-3)2-3 = 9 – 3 = 6.

15. Determine the product of two signals: x1 (n) = {2,1,1.5,3}; x2 (n) = { 1,1.5,0,2}.
a) {2,1.5,0,6}
b) {2,1.5,6,0}
c) {2,0,1.5,6}
d) {2,1.5,0,3}

Explanation: Product of discrete-time signals is computed element by element.
⇒ x(n) = x1 (n) * x2 (n) = {2×1, 1×1.5, 1.5×0, 3×2} = {2,1.5,0,6}.

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