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

a) Time invariant

b) Time variant

c) Both Time variant and Time invariant

d) None of the mentioned

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

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

a) Yes

b) No

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

3. How is a linear function described as?

a) Zero in Finite out

b) Zero in infinite out

c) Zero in zero out

d) Zero in Negative out

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

4. Is the system y[n] = x^{2}[n-2] linear?

a) Yes

b) No

Explanation: The system is not linear, as x1

^{2}+ x2

^{2}is not equal to (x1 + x2)

^{2}.

5. Is the above system, i.e y[n] = x^{2}[n-2] time invariant?

a) Yes

b) No

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

6. Given the signal

X (t) = cos t, if t<0

Sin t, if t≥0

The correct statement among the following is?

a) Periodic with fundamental period 2π

b) Periodic but with no fundamental period

c) Non-periodic and discontinuous

d) Non-periodic but continuous

Explanation: From the graphs of cos and sin, we can infer that at t=0, the function becomes discontinuous.

Since, cos 0 = 1, but sin 0 = 0

As 1 ≠ 0, so, the function X (t) is discontinuous and therefore Non-periodic.

7. The fundamental period of the signal X (t) = 10 cos^{2}(10 πt) is __________

a) 0.2

b) 0.1

c) 0.5

d) No fundamental period exists

Explanation: X (t) = 10 cos

^{2}(10 πt)

cos 2t = 2cos

^{2}t – 1

cos

^{2}t = \(\frac{1+cos2t}{2}\)

X (t) = 5 + 5 cos 20πt

Now, Y (t) = cos 20πt

Fundamental period of the signal is = \(\frac{2π}{20π} = \frac{1}{10}\) = 0.1.

8. The even component of the signal X (t) = e^{jt} is _________________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

Explanation: Let X

_{e}(t) represents the even component of X (t)

Now, X

_{e}(t) = \(\frac{1}{2}\)[X (t) + X (-t)]

= \(\frac{1}{2}\)[e

^{jt}+ e

^{-jt}]

= cos t.

9. The odd component of the signal X (t) = e^{jt} is _______________

a) Sin t

b) Cos t

c) Sinh t

d) Cosh t

Explanation: Let X

_{o}(t) represents the odd component of X (t)

Now, X

_{o}(t) = \(\frac{1}{2}\)[X (t) – X (-t)]

= \(\frac{1}{2}\)[e

^{jt}+ e

^{-jt}]

= sin t.

10. The period of the signal X (t) = 24 + 50 cos 60πt is _______________

a) \(\frac{1}{30}\) s

b) 60 π s

c) \(\frac{1}{60π}\) s

d) Non-periodic

Explanation: Period of cos t = 2π

Period of cos at = \(\frac{2π}{a}\)

Here, a = 60π

So, period of cos 60πt = \(\frac{2π}{60π}\)

= \(\frac{1}{30}\) s.