This set of Signals & Systems Interview Questions and Answers for freshers focuses on “Basic Operations on Signals – 2”.

1. Considering Figure 1, sketch y= 2* x (t).

a)

b)

c)

d)

View Answer

Explanation: Y (t) = 2*x (t) is an example for amplitude scaling. Here amplitude is scaled by a factor 2.

2. Considering Figure 1, sketch y= -3* x (t).

a)

b)

c)

d)

View Answer

Explanation: Y (t) = -3*x (t) is an example for amplitude scaling. Here amplitude is scaled by a factor -3.

3. In the following diagram, X [n] and y [n] are related by ______

a) Y [n] = 2*x [n]

b) Y [n] = -2*x [n]

c) Y [n] = x [2n]

d) Y [n] = x [-2n]

View Answer

Explanation: Y [n] = 2*x [n] is an example for amplitude scaling of discrete time signal. The given figure is an example for 2*x [n] hence Y [n] = 2*x [n] is correct.

4. X [n] and y [n] is as shown below, the relationship between x [n] and y [n] is given by ______

a) X [n] = y [n]/3

b) X [n] = 3* y [n]

c) Y [n] = x [n]/3

d) Y [n] = 3*x [n]

View Answer

Explanation: The given y [n] is amplitude scaling of a discrete time signal by a factor 1/3.

Hence the amplitude is reduced by 1/3.

5. Considering figure 3 below, is the following figure true for y [n] = x [2n]?

a) True

b) False

View Answer

Explanation: X [2n] is an example of time scaling. For discrete time signal x [k*n], k>1 the samples will be lost.

6. Considering figure 3 below, is the following figure true for y [n] = x [n/2]?

a) True

b) False

View Answer

Explanation: X [n/2] is an example for time scaling by factor ½ and it will be a stretched signal. The discrete time signal should extend from -10 to 10.

7. Consider figure 4, is the given y (t) an integration of x (t)?

a) Y (t) = ∫x (t).dt

b) Y (t) = ∫x^{2} (t).dt

c) Y (t) = 3* ∫x (t).dt

d) Y (t) = 3* ∫x^{2} (t).dt

View Answer

Explanation: The given y (t) is integral of x (t) and amplitude 3 remains constant for t>1.

It is because of the properties of integration.

8. Consider figure 4, is the given y (t) a differentiation of x (t)?

a) Y (t) = \(\frac{dx(t)}{dt}\)

b) Y (t) = \(\frac{-2dx(t)}{dt}\)

c) Y (t) = \(\frac{dx(-t)}{dt}\)

d) Y (t) = ∫x (t).dt

View Answer

Explanation: The given y (t) is differentiation of x (t) and hence we have impulses at -1, 0 and 1.

9. The given pair x (t) and y (t) is _______

a) Y (t) = d/dt (x (t))

b) Y (t) = ∫x (t).dt

c) Y (t) = x (t) -1

d) Y (t) = x (t) /2

View Answer

Explanation: The given pair x (t) and y (t) is related by y (t) = d/dt (x (t)). From -2 to 2 we have Y (t) is zero because differentiation of constant is zero.

10. The given pair x (t) and y (t) is related by _______

a) Y (t) = d/dt (x (t))

b) Y (t) = x (t) + 1

c) Y (t) = ∫x (t) .dt

d) Not related

View Answer

Explanation: The given pair x (t) and y (t) is related by Y (t) = ∫x (t) .dt. The integral of x (t) gives the Y (t). Y (t) = 0 for t > 1.

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