Digital Signal Processing Questions and Answers Part-13

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

Answer: b
Explanation: We know that,
X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)
=> X*(ω)=\([\sum_{n=-∞}^∞ x(n)e^{-jωn}]^*\)
Given the signal x(n) is real. Therefore,
X*(ω)=\(\sum_{n=-∞}^∞ x(n)e^{jωn}\)
=> 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

Answer: a
Explanation: We know that, if a signal x(n) is real, then
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) \(\frac{1}{1+2acosω+a^2}\)
b) \(\frac{1}{1+2asinω+a^2}\)
c) \(\frac{1}{1-2asinω+a^2}\)
d) \(\frac{1}{1-2acosω+a^2}\)

Answer: d
Explanation: Since |a|<1, the sequence x(n) is absolutely summable, as can be verified by applying the geometric summation formula.
\(\sum_{n=-∞}^∞|x(n)| = \sum_{n=-∞}^∞ |a|^n = \frac{1}{1-|a|} \lt ∞\)
Hence the Fourier transform of x(n) exists and is obtained as
X(ω) = \(\sum_{n=-∞}^∞ a^n e^{-jωn}=\sum_{n=-∞}^∞ (ae^{-jω})^n\)
Since |ae-jω|=|a|<1, use of the geometric summation formula again yields
The energy density spectrum is given by
Sxx(ω)=|X(ω)|2= X(ω).X*(ω)=\(\frac{1}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1}{1-2acosω+a^2}\).

4. What is the Fourier transform of the signal x(n) which is defined as below?
a) Ae-j(ω/2)(L)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
b) Aej(ω/2)(L-1)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
c) Ae-j(ω/2)(L-1)\(\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})}\)
d) None of the mentioned

Answer: c
Explanation: The Fourier transform of this signal is
X(ω)=\(\sum_{n=0}^{L-1} Ae^{-jωn}\)

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

Answer: a
Explanation: Let us consider the signal to be x(n)
Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)z^{-n} and X(ω)=\sum_{n=-∞}^∞ x(n)e^{-jωn}\)
Now, represent the ‘z’ in the polar form
=> z=r.e
=>Z{x(n)}=\(\sum_{n=-∞}^∞ x(n)r^{-n} e^{-jωn}\)
Now Z{x(n)}= X(ω) only when r=1=>|z|=1.

6. The sequence x(n)=\(\frac{sin⁡ ω_c n}{πn}\) does not have both z-transform and Fourier transform.
a) True
b) False

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

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) \(\frac{1}{2π} \int_0^π \theta(ω) e^{jωn} dω\)
b) \(\frac{1}{2π} \int_{-π}^π \theta(ω) e^{-jωn} dω\)
c) \(\frac{1}{2π} \int_0^π \theta(ω) e^{jωn} dω\)
d) \(\frac{1}{2π} \int_{-π}^π \theta(ω) e^{jωn} dω\)

Answer: d
Explanation: We know that,
cx(n)=\(\frac{1}{2π} \int_{-π}^π ln⁡(X(ω))e^{jωn} dω\)
If we express X(ω) in terms of its magnitude and phase, say
Then ln X(ω)=ln |X(ω)|+jθ(ω)
=> cx(n)=\(\frac{1}{2π} \int_{-π}^π[ln|X(ω)|+jθ(ω)]e^{jωn} dω\) => cx(n)=cm(n)+jcθ(n)(say)
=> cθ(n)=\(\frac{1}{2π} \int_{-π}^πθ(ω) e^{jωn} dω\)

9. What is the Fourier transform of the signal x(n)=u(n)?
a) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)}\)
b) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)}\)
c) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)/2}\)
d) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)

Answer: d
Explanation: Given x(n)=u(n)
We know that the z-transform of the given signal is X(z)=\(\frac{1}{1-z^{-1}}\) 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) = \(\frac{e^{jω/2}}{2jsin(ω/2)} = \frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)

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 above

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