# Quantization Effect MCQ Quiz – Objective Question with Answer for Quantization Effect

1. The effect of round-off errors due to the multiplications performed in the DFT with fixed-point arithmetic is known as Quantization error.

A. True
B. False

Since DFT plays a very important role in many applications of DSP, it is very important for us to know the effect of quantization errors in its computation.

In particular, we shall consider the effect of round-off errors due to the multiplications performed in the DFT with fixed-point arithmetic.

2. What is the model that has been adopted for characterizing a round-off error in multiplication?

A. Multiplicative white noise model
B. Subtractive white noise model
D. None of the mentioned

The additive white noise model is the model that we use in the statistical analysis of round-off errors in IIR and FIR filters.

3. How many quantization errors are present in one complex-valued multiplication?

A. One
B. Two
C. Three
D. Four

We assume that the real and imaginary components of {x(n)} and {WNkn} are represented by ‘b’ bits. Consequently, the computation of product x(n).

WNkn requires four real multiplications. Each real multiplication is rounded from 2b bits to b bits and hence there are four quantization errors for each complex-valued multiplication.

4. What is the total number of quantization errors in the computation of single point DFT of a sequence of length N?

A. 2N
B. 4N
C. 8N
D. 12N

Since the computation of single point DFT of a sequence of length N involves N number of complex multiplications, it contains 4N number of quantization errors.

5. What is the range in which the quantization errors due to rounding off are uniformly distributed as random variables if Δ=2-b?

A. (0,Δ)
B. (-Δ,0)
C. (-Δ/2,Δ/2)
D. None of the mentioned

The Quantization errors due to rounding off are uniformly distributed random variables in the range (-Δ/2, Δ/2) if Δ=2-b. This is one of the assumptions that is made about the statistical properties of the quantization error.

6. The 4N quantization errors are mutually uncorrelated.
A. True
B. False

The 4N quantization errors are mutually uncorrelated. This is one of the assumptions that is made about the statistical properties of the quantization error.

7. The 4N quantization errors are correlated with the sequence {x(n)}.
A. True
B. False

According to one of the assumptions that is made about the statistical properties of the quantization error, the 4N quantization errors are uncorrelated with the sequence {x(n)}.

8. How is the variance of the quantization error related to the size of the DFT?

A. Equal
B. Inversely proportional
C. Square proportional
D. Proportional

We know that each of the quantization has a variance of Δ2/12=2-2b/12.

The variance of the quantization errors from the 4N multiplications is 4N. 2-2b/12=2-2b(N/3).

Thus the variance of the quantization error is directly proportional to the size of the DFT.

9. Every fourfold increase in the size N of the DFT requires an additional bit in computational precision to offset the additional quantization errors.

A. True
B. False

We know that the variance of the quantization errors is directly proportional to the size N of the DFT. So, every fourfold increase in the size N of the DFT requires an additional bit in computational precision to offset the additional quantization errors.

10. What is the variance of the output DFT coefficients |X(k)|?

A. $$\frac{1}{N}$$

B. $$\frac{1}{2N}$$

C. $$\frac{1}{3N}$$

D. $$\frac{1}{4N}$$

We know that the variance of the signal sequence is

(2/N)2/12=$$\frac{1}{3N^2}$$

Now the variance of the output DFT coefficients

|X(k)|=N.$$\frac{1}{3N^2} = \frac{1}{3N}$$.

11. What is the signal-to-noise ratio?

A. σX2.σq2
B. σX2/σq2
C. σX2+σq2
D. σX2-σq2

The signal-to-noise ratio of a signal, SNR is given by the ratio of the variance of the output DFT coefficients to the variance of the quantization errors.

12. How many number of bits are required to compute the DFT of a 1024 point sequence with an SNR of 30db?

A. 15
B. 10
C. 5
D. 20

The size of the sequence is N=210. Hence the SNR is10log10(σX22/σq2)=10 log1022b-20

For an SNR of 30db, we have

3(2b-20)=30=>b=15 bits.

Note that 15 bits are the precision for both addition and multiplication.

13. How many number of butterflies are required per output point in the FFT algorithm?

A. N
B. N+1
C. 2N
D. N-1

We find that, in general, there are N/2 in the first stage of FFT, N/4 in the second stage, N/8 in the third state, and so on, until the last stage where there is only one. Consequently, the number of butterflies per output point is N-1.

14. What is the value of the variance of quantization error in FFT algorithm, compared to that of direct computation?

A. Greater
B. Less
C. Equal
D. Cannot be compared

If we assume that the quantization errors in each butterfly are uncorrelated with the errors in the other butterflies, then there are 4(N-1) errors that affect the output of each point of the FFT.

Consequently, the variance of the quantization error due to FFT algorithm is given by

4(N-1)(Δ2/12)=N(Δ2/3)(approximately)

Thus, the variance of quantization error due to the FFT algorithm is equal to the variance of the quantization error due to direct computation.

15. How many number of bits are required to compute the FFT of a 1024 point sequence with an SNR of 30db?

A. 11
B. 10
C. 5
D. 20

The size of the FFT is N=210. Hence the SNR is 10

log1022b-v-1=30

=>3(2b-11)=30

=>b=21/2=11 bits.

16. Which of the following is true regarding the number of computations required to compute an N-point DFT?

A. N2 complex multiplications and N(N-1) complex additions
B. N2 complex additions and N(N-1) complex multiplications
C. N2 complex multiplications and N(N+1) complex additions
D. N2 complex additions and N(N+1) complex multiplications

N point DFT is given as

X(k)=$$\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}$$

From the formula given at every step of computing, we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N2 complex multiplications and N(N-1) complex additions.

17. Which of the following is true regarding the number of computations required to compute DFT at any one value of ‘k’?

A. 4N-2 real multiplications and 4N real additions
B. 4N real multiplications and 4N-4 real additions
C. 4N-2 real multiplications and 4N+2 real additions
D. 4N real multiplications and 4N-2 real additions

The formula for calculating N point DFT is given as

X(k)=$$\sum_{n=0}^{N-1} x(n)e^{-j2πkn/N}$$

From the formula given at every step of computing, we are performing N complex multiplications and N-1 complex additions. So, it requires 4N real multiplications and 4N-2 real additions for any value of ‘k’ to compute the DFT of the sequence.

18. WNk+N/2=?

A. WNk
B. -WNk
C. WN-k
D. None of the mentioned

According to the symmetry property, we get WNk+N/2=-WNk.

19. What is the real part of the N point DFT XR(k) of a complex valued sequence x(n)?

A. $$\sum_{n=0}^{N-1} [x_R (n) cos⁡\frac{2πkn}{N} – x_I (n) sin⁡\frac{2πkn}{N}]$$

B. $$\sum_{n=0}^{N-1} [x_R (n) sin⁡\frac{2πkn}{N} + x_I (n) cos⁡\frac{2πkn}{N}]$$

C. $$\sum_{n=0}^{N-1} [x_R (n) cos⁡\frac{2πkn}{N} + x_I (n) sin⁡\frac{2πkn}{N}]$$

D. None of the mentioned

For a complex-valued sequence x(n) of N points, the DFT may be expressed as

XR(k)=$$\sum_{n=0}^{N-1} [x_R (n) cos⁡\frac{2πkn}{N} + x_I (n) sin⁡\frac{2πkn}{N}]$$

20. The computation of XR(k) for a complex-valued x(n) of N points requires ________

A. 2N2 evaluations of trigonometric functions
B. 4N2 real multiplications
D. All of the mentioned

The expression for XR(k) is given as

XR(k)=$$\sum_{n=0}^{N-1} [x_R (n) cos⁡\frac{2πkn}{N} + x_I (n) sin⁡\frac{2πkn}{N}]$$

So, from the equation, we can tell that the computation of XR(k) requires 2N2 evaluations of trigonometric functions, 4N2 real multiplications, and 4N(N-1) real additions.

21. Divide-and-conquer approach is based on the decomposition of an N-point DFT into successively smaller DFTs. This basic approach leads to FFT algorithms.

A. True
B. False

The development of computationally efficient algorithms for the DFT is made possible if we adopt a divide-and-conquer approach.

This approach is based on the decomposition of an N-point DFT into successively smaller DFTs. This basic approach leads to a family of computationally efficient algorithms known collectively as FFT algorithms.

22. If the arrangement is of the form in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on, then which of the following mapping represents the above arrangement?

A. n=l+mL
B. n=Ml+m
C. n=ML+l
D. none of the mentioned

If we consider the mapping n=Ml+m, then it leads to an arrangement in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on.

23. If N=LM, then what is the value of WNmqL?
A. WMmq
B. WLmq
C. WNmq
D. None of the mentioned

We know that if N=LM, then WNmqL = WN/Lmq = WMmq.

24. How many complex multiplications are performed in computing the N-point DFT of a sequence using the divide-and-conquer method if N=LM

A. N(L+M+2)
B. N(L+M-2)
C. N(L+M-1)
D. N(L+M+1)

The expression for N point DFT is given as

X(p,q)=$$\sum_{l=0}^{L-1}\{W_N^{lq}[\sum_{m=0}^{M-1}x(l,m) W_M^{mq}]\} W_L^{lp}$$

The first step involves L DFTs, each of M points. Hence this step requires LM2 complex multiplications, the second requires LM and finally third requires ML2. So, Total complex multiplications = N(L+M+1).

25. How many complex additions are performed in computing the N-point DFT of a sequence using the divide-and-conquer method if N=LM?

A. N(L+M+2)
B. N(L+M-2)
C. N(L+M-1)
D. N(L+M+1)

The expression for N point DFT is given as

X(p,q)=$$\sum_{l=0}^{L-1}\{W_N^{lq}[\sum_{m=0}^{M-1}x(l,m) W_M^{mq}]\} W_L^{lp}$$

The first step involves L DFTs, each of M points. Hence this step requires LM(M-1) complex additions, the second step does not require any additions and finally, the third step requires ML(L-1) complex additions.

So, a total number of complex additions=N(L+M-2).

26. Which is the correct order of the following steps to be done in one of the algorithms of the divide and conquer method?

A. Store the signal column-wise
B. Compute the M-point DFT of each row
C. Multiply the resulting array by the phase factors WNlq.
D. All of the above

The steps to be done in one of the algorithms of the divide and conquer method is

i) Store the signal column-wise
ii) Compute the M-point DFT of each row
iii) Multiply the resulting array by the phase factors WNlq.
iv) Compute the L-point DFT of each column.
v) Read the result array row-wise.

According to one of the algorithms describing the divide and conquer method, if we store the signal column-wise, then compute the M-point DFT of each row and multiply the resulting array by the phase factors WNlq and then compute the L-point DFT of each column and read the result row-wise.

27. If we store the signal row-wise then the result must be read column-wise.

A. True
B. False

According to the second algorithm of the divide and conquer approach, if the input signal is stored row-wise, then the result must be read column-wise.

28. If we store the signal row-wise and compute the L point DFT at each column, the resulting array must be multiplied by which of the following factors?

A. WNlq
B. WNpq
C. WNlq
D. WNpm

According to the second algorithm of the divide and conquer approach, if the input signal is stored row-wise, then we calculate the L point DFT of each column and we multiply the resulting array by the factor WNpm.

29. If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even-numbered and odd-numbered samples of x(n), then such an FFT algorithm is known as a decimation-in-time algorithm.

A. True
B. False

Let us consider the computation of the N=2v point DFT by the divide and conquer approach. We select M=N/2 and L=2. This selection results in a split of N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even-numbered and odd-numbered samples of x(n), respectively, that is

f1(n)=x(2n)

f2(n)=x(2n+1), n=0,1,2…N/2-1

Thus f1(n) and f2(n) are obtained by decimating x(n) by a factor of 2, and hence the resulting FFT algorithm is called a decimation-in-time algorithm.

30. If we split the N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even-numbered and odd-numbered samples of x(n) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)

A. F1(k)+F2(k)
B. F1(k)-WNk F2(k)
C. F1(k)+WNk F2(k)
D. None of the mentioned

From the question, it is given that

f1(n)=x(2n)

f2(n)=x(2n+1), n=0,1,2…N/2-1

X(k)=$$\sum_{n=0}^{N-1} x(n) W_N^{kn}$$, k=0,1,2..N-1

=$$\sum_{n \,even} x(n) W_N^{kn}+\sum_{n \,odd} x(n) W_N^{kn}$$

=$$\sum_{m=0}^{(\frac{N}{2})-1} x(2m)W_N^{2km}+\sum_{m=0}^{(\frac{N}{2})-1} x(2m+1) W_N ^{k(2m+1)}$$

=$$\sum_{m=0}^{(\frac{N}{2})-1} f_1(m) W_{N/2}^{km} + W_N^k \sum_{m=0}^{(N/2)-1} f_2(m) W_{(\frac{N}{2})}^{km}$$

X(k)=F1(k)+ WNk F2(k).

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