# Electronics and Communication Engineering - Signals and Systems

### Exercise :: Signals and Systems - Section 1

16.

Let x(t) and y(t) with F.T. x(f) and y(f) respectively be related as shown in figure Then y(f) is

 A. B. C. -X(f/2)ej2pf D. -X(f/2)e- j2pf

Explanation:

By applying time shifting and scaling property.

17.

Which one is a linear system?

 A. y[n] = x[n] x x[n - 1] B. y[n] = x[n] + x[n - 10] C. y[n] = x2[n] D. (a) and (c)

Explanation:

For linearity

y1[n] = x1[n] + x2[n - 10] ...(1)

y2 = x2[n] + x2[n - 10] ...(2)

y1[n] = x1[n] + x2[n] + x2[n - 10] + x2[n - 10] ...(3)

Now find y1[n] + y2[n]

Corresponding to x1[n] + x2[n]

It is same as equation (3) hence linear.

But in part (c) y[n] = x2[n]

y1[n] = x21[n], y2[n] = x22[n]⇒ y1[n] + y2[n] = x22[n]...3

But y1[n] + y2[n]

Corresponing x1[n] + x2[n] is y1[n] + y2[n] = {x1[n] + x2[n]}2

= x12[n] + x22[n] + 2x1[n] x2[n]....4

Equations (3) and (4) are not same hence not linear.

18.

Laplace transform of a pulse function of magnitude E and duration from t = 0 to t = a is

 A. E / s B. C. D. Explanation: .

19.

The continuous time system with impulse response h(t) = is stable, for n is even, when

 A. a = 0 B. a > 0 C. a < 0 D. a ≤ 0

Explanation:    For stable system exponential must be -ve so that a < 0.

20.

If I (s) , initial value of i(t) is

 A. 5A B. 12.5 A C. 0.05 A D. 1250 A .