Digital Signal Processing Questions and Answers Part-20

1. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}?
a) {14,14,16,16}
b) {16,16,14,14}
c) {2,3,6,4}
d) {14,16,14,16}

Answer: d
Explanation: We know that the circular convolution of two sequences is given by the expression
x(m)= \(\sum_{n=0}^{N-1}x_1 (n) x_2 (m-n)_N\)
For m=0, x2((-n))4={1,4,3,2}
For m=1, x2((1-n))4={2,1,4,3}
For m=2, x2((2-n))4={3,2,1,4}
For m=3, x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.

2. What is the circular convolution of the sequences X1(n)={2,1,2,1} and x2(n)={1,2,3,4}, find using the DFT and IDFT concepts?
a) {16,16,14,14}
b) {14,16,14,16}
c) {14,14,16,16}
d) None of the mentioned

Answer: b
Explanation: Given X1(n)={2,1,2,1}=>X1(k)=[6,0,2,0]
Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2]
when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0]
By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.

3. If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that N-point DFT of x((n-l))N is X(k)e-j2πkl/N.
a) True
b) False

Answer: a
Explanation: According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))N is X(k)e-j2πkl/N.

4. If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?
a) X(N-k)
b) X*(k)
c) X*(N-k)
d) None of the mentioned

Answer: c
Explanation: According to the complex conjugate property of DFT, we have if X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).

5. If the signal to be analyzed is an analog signal, we would pass it through an anti-aliasing filter with B as the bandwidth of the filtered signal and then the signal is sampled at a rate __________
a) Fs ≤ 2B
b) Fs ≤ B
c) Fs ≥ 2B
d) Fs = 2B

Answer: c
Explanation: The filtered signal is sampled at a rate of Fs≥ 2B, where B is the bandwidth of the filtered signal to prevent aliasing.

6. What is the highest frequency that is contained in the sampled signal?
a) 2Fs
b) Fs/2
c) Fs
d) None of the mentioned

Answer: b
Explanation: We know that, after passing the signal through anti-aliasing filter, the filtered signal is sampled at a rate of Fs≥ 2B=>B≤ Fs/2.Thus the maximum frequency of the sampled signal is Fs/2.

7. The finite observation interval for the signal places a limit on the frequency resolution.
a) True
b) False

Answer: a
Explanation: After sampling the signal, we limit the duration of the signal to the time interval T0=LT, where L is the number of samples and T is the sample interval. So, it limits our ability to distinguish two frequency components that are separated by less than 1/T0=1/LT in frequency. So, the finite observation interval for the signal places a limit on the frequency resolution.

8. If {x(n)} is the signal to be analyzed, limiting the duration of the sequence to L samples, in the interval 0≤ n≤ L-1, is equivalent to multiplying {x(n)} by?
a) Kaiser window
b) Hamming window
c) Hanning window
d) Rectangular window

Answer: d
Explanation: The equation of the rectangular window w(n) is given as
w(n)=1, 0≤ n≤ L-1
=0, otherwise
Thus, we can limit the duration of the signal x(n) to L samples by multiplying it with a rectangular window of length L.

9. The characteristic of windowing the signal called “Leakage” is the power that is leaked out into the entire frequency range.
a) True
b) False

Answer: a
Explanation: We note that the windowed spectrum \(\hat{X}\)(w) is not localized to a single frequency, but instead it is spread out over the whole frequency range. Thus the power of the original signal sequence x(n) that was concentrated at a single frequency has been spread by the window into the entire frequency range. We say that the power has been leaked out into the entire frequency range and this phenomenon is called as “Leakage”.

10. Which of the following is the disadvantage of Hanning window over rectangular window?
a) More side lobes
b) Less side lobes
c) More width of main lobe
d) None of the mentioned

Answer: c
Explanation: In the magnitude response of the signal windowed using Hanning window, the width of the main lobe is more which is the disadvantage of this technique over rectangular windowing technique.