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

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

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## 10 Questions MCQ Test Signals and Systems - Test: DFT Computation Filtering Approach

Test: DFT Computation Filtering Approach for Electronics and Communication Engineering (ECE) 2024 is part of Signals and Systems preparation. The Test: DFT Computation Filtering Approach questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Test: DFT Computation Filtering Approach MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: DFT Computation Filtering Approach below.
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Test: DFT Computation Filtering Approach - Question 1

### By means of the DFT and IDFT, determine the response of the FIR filter with impulse response h(n)={1,2,3} to the input sequence x(n)={1,2,2,1}?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 1

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}.

Test: DFT Computation Filtering Approach - Question 2

### What is the sequence y(n) that results from the use of four point DFTs if the impulse response is h(n)={1,2,3} and the input sequence x(n)={1,2,2,1}?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 2

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.

Test: DFT Computation Filtering Approach - Question 3

### Overlap add and Overlap save are the two methods for linear FIR filtering a long sequence on a block-by-block basis using DFT. a) True

Detailed Solution for Test: DFT Computation Filtering Approach - Question 3

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.

Test: DFT Computation Filtering Approach - Question 4

In Overlap save method of long sequence filtering, what is the length of the input sequence block?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 4

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.

Test: DFT Computation Filtering Approach - Question 5

In Overlap save method of long sequence filtering, how many zeros are appended to the impulse response of the FIR filter?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 5

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.

Test: DFT Computation Filtering Approach - Question 6

The first M-1 values of the output sequence in every step of Overlap save method of filtering of long sequence are discarded.

Detailed Solution for Test: DFT Computation Filtering Approach - Question 6

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.

Test: DFT Computation Filtering Approach - Question 7

In Overlap add method, what is the length of the input data block?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 7

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.

Test: DFT Computation Filtering Approach - Question 8

Which of the following is true in case of Overlap add method?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 8

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.

Test: DFT Computation Filtering Approach - Question 9

What is the value of x(n)*h(n), 0≤n≤11 for the sequences x(n)={1,2,0,-3,4,2,-1,1,-2,3,2,1,-3} and h(n)={1,1,1} if we perform using overlap add fast convolution technique?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 9

Explanation: 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}.

Test: DFT Computation Filtering Approach - Question 10

What is the value of x(n)*h(n), 0≤n≤11 for the sequences x(n)={1,2,0,-3,4,2,-1,1,-2,3,2,1,-3} and h(n)={1,1,1} if we perform using overlap save fast convolution technique?

Detailed Solution for Test: DFT Computation Filtering Approach - Question 10

Explanation: 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}.

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