Test: Quantization Effects - Electrical Engineering (EE) MCQ

Explanation: 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.

Explanation: Additive white noise model is the model that we use in the statistical analysis of round off errors in IIR and FIR filters.

Explanation: 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.

Explanation: 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.

Explanation: 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 assumption that is made about the statistical properties of the quantization error.

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

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

Explanation: 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.

Explanation: 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.

Explanation: We know that the variance of the signal sequence is (2/N)²/12=1/(3N²)
Now the variance of the output DFT coefficients |X(k)|=N. 1/(3N^2 ²) = 1/3N.

Explanation: 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.

Explanation: The size of the sequence is N=210. Hence the SNR is
10log₁₀(σX2²/ σq²)=10 log₁₀2^2b-20
For an SNR of 30db, we have
3(2b-20)=30=>b=15 bits.
Note that 15 bits is the precision for both addition and multiplication.

Explanation: 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.

Explanation: 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 FFT algorithm is equal to the variance of the quantization error due to direct computation.

Explanation: The size of the FFT is N=2¹⁰. Hence the SNR is 10 log₁₀2^2b-v-1=30
=>3(2b-11)=30
=>b=21/2=11 bits.