1. 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

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

2. 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

We know that, after passing the signal through an 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.

3. The finite observation interval for the signal places a limit on the frequency resolution.

A. True
B. False

Answer: A

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.

4. 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

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.

5. What is the Fourier transform of rectangular window of length L?

A. \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{jω(L+1)/2}\)

B. \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{jω(L-1)/2}\)

C. \(\frac{sin(\frac{ωL}{2})}{sin(\frac{ω}{2})} e^{-jω(L-1)/2}\)

D. None of the mentioned

Answer: C

We know that the equation for the rectangular window w(n) is given as

w(n)=1, 0≤ n≤ L-1 =0, otherwise

We know that the Fourier transform of a signal x(n) is given as

6. If x(n)=cosω0n and W(ω) is the Fourier transform of the rectangular signal w(n), then what is the Fourier transform of the signal x(n).w(n)?

A. 1/2[W(ω-ω0)- W(ω+ω0)]
B. 1/2[W(ω-ω0)+ W(ω+ω0)]
C. [W(ω-ω0)+ W(ω+ω0)]
D. [W(ω-ω0)- W(ω+ω0)]

Answer: B

According to the exponential properties of Fourier transform, we get

Fourier transform of x(n).w(n)= 1/2[W(ω-ω0)+ W(ω+ω0)]

7. 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

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 “Leakage”.

8. Which of the following is the advantage of the Hanning window over the rectangular window?

A. More side lobes
B. Less side lobes
C. More width of the main lobe
D. None of the mentioned

Answer: B

The Hanning window has less side lobes and the leakage is less in this windowing technique.

9. Which of the following is the disadvantage of the Hanning window over the rectangular window?

A. More side lobes
B. Less side lobes
C. More width of main lobe
D. None of the mentioned

Answer: C

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

10. The condition with less number of samples L should be avoided.
A. True
B. False

Answer: A

When the number of samples L is small, the window spectrum masks the signal spectrum, and, consequently, the DFT of the data reflects the spectral characteristics of the window function. So, this situation should be avoided.