# Signal processing interview questions and answers pdf

By John

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

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

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

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

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

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

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

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