Test: DFT Computation Filtering Approach - Electronics and Communication Engineering (ECE) MCQ

Explanation: The input sequence has a length N=4 and impulse response has a length M=3. So, the response must have a length of 6(4+3-1).
We know that, Y(k)=X(k).H(k)
Thus we obtain Y(k)={36,-14.07-j17.48,j4,0.07+j0.515,0,0.07-j0.515,-j4,-14.07+j17.48}
By applying IDFT to the above sequence, we get y(n)={1,4,9,11,8,3,0,0}
Thus the output of the system is {1,4,9,11,8,3}.

Explanation: The four point DFT of h(n) is H(k)=1+2e^-jkπ/2+3 e^-jkπ (k=0,1,2,3)
Hence H(0)=6, H(1)=-2-j2, H(3)=2, H(4)=-2+j2
The four point DFT of x(n) is X(k)= 1+2e^-jkπ/2+2 e^-jkπ+3e^-3jkπ/2(k=0,1,2,3)
Hence X(0)=6, X(1)=-1-j, X(2)=0, X(3)=-1+j
The product of these two four point DFTs is
Y ̂(0)=36, Y ̂(1)=j4, Y ̂(2)=0, Y ̂(3)=-j4
The four point IDFT yields y ̂(n)={9,7,9,11}
We can verify as follows
We know that from the previous question y(n)={1,4,9,11,8,3}
y ̂(0)=y(0)+y(4)=9
y ̂(1)=y(1)+y(5)=7
y ̂(2)=y(2)=9
y ̂(3)=y(3)=11.

Explanation: In these two methods, the input sequence is segmented into blocks and each block is processed via DFT and IDFT to produce a block of output data. The output blocks are fitted together to form an overall output sequence which is identical to the sequence obtained if the long block had been processed via time domain convolution. So, Overlap add and Overlap save are the two methods for linear FIR filtering a long sequence on a block-by-block basis using DFT.

Explanation: In this method, 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.

Explanation: The impulse of the FIR filter is increased in length by appending L-1 zeros and an N-point DFT of the sequence is computed once and stored.

Explanation: Since the data record of length N, the first M-1 points of ym(n) are corrupted by aliasing and must be discarded. The last L points of ym(n) are exactly as same as the result from linear convolution.

Explanation: In this method the size of the input data block is L points and the size of the DFTs and IDFT is N=L+M-1.

Explanation: In Overlap add method, to each data block we append M-1 zeros at last and compute N point DFT, so that the length of the input sequence is L+M-1=N.

Explanation: Since M=3, we chose the transform length for DFT and IDFT computations as L=2^M=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}.

Explanation: Since M=3, we chose the transform length for DFT and IDFT computations as L=2^M=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}.