1. Which of the following relation is true if the signal x(n) is real?

a) X*(ω)=X(ω)

b) X*(ω)=X(-ω)

c) X*(ω)=-X(ω)

d) None of the mentioned

View Answer

Answer: b

Explanation: We know that,

X(ω)=∑∞n=−∞x(n)e−jωn

=> X*(ω)=[∑∞n=−∞x(n)e−jωn]∗

Given the signal x(n) is real. Therefore,

X*(ω)=∑∞n=−∞x(n)ejωn

=> X*(ω)=X(-ω).

2. For a signal x(n) to exhibit even symmetry, it should satisfy the condition |X(-ω)|=| X(ω)|.

a) True

b) False

View Answer

Answer: a

Explanation: We know that, if a signal x(n) is real, then

X*(ω)=X(-ω)

If the signal is even symmetric, then the magnitude on both the sides should be equal.

So, |X*(ω)|=|X(-ω)| => |X(-ω)|=|X(ω)|.

3. What is the energy density spectrum Sxx(ω) of the signal x(n)=anu(n), |a|<1?
a) 11+2acosω+a2
b) 11+2asinω+a2
c) 11−2asinω+a2
d) 11−2acosω+a2
View Answer
Answer: d
Explanation: Since |a|<1, the sequence x(n) is absolutely summable, as can be verified by applying the geometric summation formula.
∑∞n=−∞|x(n)|=∑∞n=−∞|a|n=11−|a|<∞
Hence the Fourier transform of x(n) exists and is obtained as
X(ω) = ∑∞n=−∞ane−jωn=∑∞n=−∞(ae−jω)n
Since |ae-jω|=|a|<1, use of the geometric summation formula again yields
X(ω)=11−ae−jω
The energy density spectrum is given by
Sxx(ω)=|X(ω)|2= X(ω).X*(ω)=1(1−ae−jω)(1−aejω)=11−2acosω+a2.
4. What is the Fourier transform of the signal x(n) which is defined as shown in the graph below?
a) Ae-j(ω/2)(L)sin(ωL2)sin(ω2)
b) Aej(ω/2)(L-1)sin(ωL2)sin(ω2)
c) Ae-j(ω/2)(L-1)sin(ωL2)sin(ω2)
d) None of the mentioned
View Answer
Answer: c
Explanation: The Fourier transform of this signal is
X(ω)=∑L−1n=0Ae−jωn
=A.1−e−jωL1−e−jω
=Ae−j(ω/2)(L−1)sin(ωL2)sin(ω2)
5. Which of the following condition is to be satisfied for the Fourier transform of a sequence to be equal as the Z-transform of the same sequence?
a) |z|=1
b) |z|<1
c) |z|>1

d) Can never be equal

View Answer

Answer: a

Explanation: Let us consider the signal to be x(n)

Z{x(n)}=∑∞n=−∞x(n)z−nandX(ω)=∑∞n=−∞x(n)e−jωn

Now, represent the ‘z’ in the polar form

=> z=r.ejω

=>Z{x(n)}=∑∞n=−∞x(n)r−ne−jωn

Now Z{x(n)}= X(ω) only when r=1=>|z|=1.

6. The sequence x(n)=sinωcnπn does not have both z-transform and Fourier transform.

a) True

b) False

View Answer

Answer: b

Explanation: The given x(n) do not have Z-transform. But the sequence have finite energy. So, the given sequence x(n) has a Fourier transform.

7. If x(n) is a stable sequence so that X(z) converges on to a unit circle, then the complex cepstrum signal is defined as ____________

a) X(ln X(z))

b) ln X(z)

c) X-1(ln X(z))

d) None of the mentioned

View Answer

Answer: c

Explanation: Let us consider a sequence x(n) having a z-transform X(z). We assume that x(n) is a stable sequence so that X(z) converges on to the unit circle. The complex cepstrum of the signal x(n) is defined as the sequence cx(n), which is the inverse z-transform of Cx(z), where Cx(z)=ln X(z)

=> cx(z)= X-1(ln X(z))

8. If cx(n) is the complex cepstrum sequence obtained from the inverse Fourier transform of ln X(ω), then what is the expression for cθ(n)?

a) 12π∫π0θ(ω)ejωndω

b) 12π∫π−πθ(ω)e−jωndω

c) 12π∫π0θ(ω)ejωndω

d) 12π∫π−πθ(ω)ejωndω

View Answer

Answer: d

Explanation: We know that,

cx(n)=12π∫π−πln(X(ω))ejωndω

If we express X(ω) in terms of its magnitude and phase, say

X(ω)=|X(ω)|ejθ(ω)

Then ln X(ω)=ln |X(ω)|+jθ(ω)

=> cx(n)=12π∫π−π[ln|X(ω)|+jθ(ω)]ejωndω => cx(n)=cm(n)+jcθ(n)(say)

=> cθ(n)=12π∫π−πθ(ω)ejωndω

9. What is the Fourier transform of the signal x(n)=u(n)?

a) 12sin(ω/2)ej(ω+π)

b) 12sin(ω/2)ej(ω−π)

c) 12sin(ω/2)ej(ω+π)/2

d) 12sin(ω/2)ej(ω−π)/2

View Answer

Answer: d

Explanation: Given x(n)=u(n)

We know that the z-transform of the given signal is X(z)=11−z−1 ROC:|z|>1

X(z) has a pole p=1 on the unit circle, but converges for |z|>1.

If we evaluate X(z) on the unit circle except at z=1, we obtain

X(ω) = ejω/22jsin(ω/2)=12sin(ω/2)ej(ω−π)/2

10. If a power signal has its power density spectrum concentrated about zero frequency, the signal is known as ______________

a) Low frequency signal

b) Middle frequency signal

c) High frequency signal

d) None of the mentioned

View Answer

Answer: a

Explanation: We know that, for a low frequency signal, the power signal has its power density spectrum concentrated about zero frequency.