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)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

=> 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

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)=a^{n}u(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) \(\frac{1}{1-2acosω+a^2}\)

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

X(ω)=\(\frac{1}{1-ae^{-jω}}\)

The energy density spectrum is given by

S

_{xx}(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) Ae^{j(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

Explanation: The Fourier transform of this signal is

X(ω)=\(\sum_{n=0}^{L-1} Ae^{-jωn}\)

=A.\(\frac{1-e^{-jωL}}{1-e^{-jω}}\)

=\(Ae^{-j(ω/2)(L-1)}\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})}\)

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)}=\(\sum_{n=-∞}^∞ x(n)z^{-n} and X(ω)=\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

Now, represent the ‘z’ in the polar form

=> z=r.e

^{jω}

=>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

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 c

_{x}(n), which is the inverse z-transform of C

_{x}(z), where C

_{x}(z)=ln X(z)

=> c

_{x}(z)= X

^{-1}(ln X(z))

8. If c_{x}(n) is the complex cepstrum sequence obtained from the inverse Fourier transform of ln X(ω), then what is the expression for c_{θ}(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) \(\frac{1}{2π} \int_{-π}^π \theta(ω) e^{jωn} dω\)

Explanation: We know that,

c

_{x}(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

X(ω)=|X(ω)|e

^{jθ(ω)}

Then ln X(ω)=ln |X(ω)|+jθ(ω)

=> c

_{x}(n)=\(\frac{1}{2π} \int_{-π}^π[ln|X(ω)|+jθ(ω)]e^{jωn} dω\) => c

_{x}(n)=c

_{m}(n)+jc

_{θ}(n)(say)

=> c

_{θ}(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) \(\frac{1}{2sin(ω/2)} e^{j(ω-π)/2}\)

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

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