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] = x2[n-2] linear?
a) Yes
b) No
Explanation: The system is not linear, as x12 + x22 is not equal to (x1 + x2)2.
5. Is the above system, i.e y[n] = x2[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 cos2(10 πt) is __________
a) 0.2
b) 0.1
c) 0.5
d) No fundamental period exists
Explanation: X (t) = 10 cos2 (10 πt)
cos 2t = 2cos2 t – 1
cos2 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) = ejt is _________________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Explanation: Let Xe (t) represents the even component of X (t)
Now, Xe (t) = \(\frac{1}{2}\)[X (t) + X (-t)]
= \(\frac{1}{2}\)[ejt + e-jt]
= cos t.
9. The odd component of the signal X (t) = ejt is _______________
a) Sin t
b) Cos t
c) Sinh t
d) Cosh t
Explanation: Let Xo (t) represents the odd component of X (t)
Now, Xo (t) = \(\frac{1}{2}\)[X (t) – X (-t)]
= \(\frac{1}{2}\)[ejt + 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.