Test: Discrete Time Systems In The Time Domain


10 Questions MCQ Test Signal and System | Test: Discrete Time Systems In The Time Domain


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QUESTION: 1

 Is the function y[n] = cos(x[n]) periodic or not?

Solution:

 ‘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].

QUESTION: 2

 If n tends to infinity, is the accumulator function an unstable one?

Solution:

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

QUESTION: 3

Comment on the causality of the following discrete time system: y[n] = x[-n].

Solution:

For positive time, the output depends on the input at an earlier time, giving causality for this portion. However, at a negative time, the output depends on the input at a positive time, i.e. at a time in the future, rendering it non causal.

QUESTION: 4

Comment on the causality of the discrete time system: y[n] = x[n+3].

Solution:

The output always depends on the input at a time in the future, rendering it anti-causal.

QUESTION: 5

Consider the system y[n] = 2x[n] + 5. Is the function linear?

Solution:

As we give two inputs, x1 and x2, and give an added input x1 and x2, we do not get the corresponding y1 and y2. Thus, additive rule is disturbed and hence the system is not linear.

QUESTION: 6

Comment on the time invariance of the following discrete system: y[n] = x[2n+4].

Solution:

A time shift in the input scale gives double the time shift in the output scale, and hence is time variant.

QUESTION: 7

Is the function y[2n] = x[2n] linear in nature?

Solution:

The function obeys both additivity and homogeneity properties. Hence, the function is linear.

QUESTION: 8

How is a linear function described as?

Solution:

The system needs to give a zero output for a zero input so as to conserve the law of additivity, to ensure linearity.

QUESTION: 9

 Is the system y[n] = x2[n-2] linear?

Solution:

The system is not linear, as x12 + x22 is not equal to (x1 + x2)2.

QUESTION: 10

 Is the above system, i.e y[n] = x2[n-2] time invariant?

Solution:

A time shift of t0 will still result in an equivalent time shift of t0 in the output, and hence will be time invariant.