Discrete Fourier Transform Properties MCQ Quiz – Objective Question with Answer for Discrete Fourier Transform Properties

From the figure given, we can notice that each data block consists of the last M-1 data points of the previous data block followed by L new data points to form a data sequence of length N+L+M-1 which is the same as in the case of Overlap save method.

From the figure given, it is clear that the last M-1 points of the first sequence and the first M-1 points of the next sequence are added and nothing is discarded because there is no aliasing in the input sequence. This is the same as in the case of the Overlap add method.

Since M=3, we chose the transform length for DFT and IDFT computations as L=2M=23=8.

Since L=M+N-1, we get N=6.

According to Overlap add method, we get

x1‘(n)={1,2,0,-3,4,2,0,0} and h'(n)={1,1,1,0,0,0,0,0}
y1(n)=x1‘(n)*N h'(n) (circular convolution)={1,3,3,-1,1,3,6,2}
x2‘(n)={-1,1,-2,3,2,1,0,0} and h'(n)={1,1,1,0,0,0,0,0}
y2(n) = x2‘(n)*N h'(n)={-1,0,-2,2,3,6,3,1}

Thus we get, y(n) = {1,3,3,-1,1,3,5,2,-2,2,3,6}.

Since M=3, we chose the transform length for DFT and IDFT computations as L=2M=23=8.

Since L=M+N-1, we get N=6.

According to Overlap save technique, we get

x1‘(n)={0,0,1,2,0,-3,4,2} and h'(n)={1,1,1,0,0,0,0,0}
=>y1(n)={1,3,3,-1,1,3}
x2‘(n)={4,2,-1,1,-2,3,2,1} and h'(n)={1,1,1,0,0,0,0,0}
=>y2(n)={5,2,-2,2,3,6}
=>y(n)= {1,3,3,-1,1,3,5,2,-2,2,3,6}.

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

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.

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.

The equation of the rectangular window w(n) is given as
w(n)=1, 0≤ n≤ L-1 =0

Thus, we can limit the duration of the signal x(n) to L samples by multiplying it with a rectangular window of length L.

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

X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

=>W(ω)=\(\sum_{n=0}^{L-1} e^{-jωn}=\frac{sin⁡(\frac{ωL}{2})}{sin⁡(\frac{ω}{2})} e^{-jω(L-1)/2}\)

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)]

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”.

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

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.

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.