Digital Electronic MCQ Number 00928

Answer: a [Reason:] We know that the expression for an DFT is given as

Therefore, we got x(n)=x(n+N)

: c [Reason:] We know that

Therefore, we have X(k)=X(k+N)

Answer: b [Reason:] We know that, the DFT of a signal x(n) is given by the expression

=>X(k)= aX1(k)+bX2(k).

Answer: d [Reason:] Given x(n)=xR(n)+jxI(n)=> xR(n)=1/2(x(n)+x*(n))
Substitute the above equation in the DFT expression
Thus we get,

Answer: d [Reason:] We know that

Therefore,
X(N-k)=X*(k)=X(-k)

Answer: b [Reason:] Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to

Answer: a [Reason:] If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to

Answer: c [Reason:] If x1(n),x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k),X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the multiplication property of DFT we have x3(m) is the circular convolution of x1(n) and x2(n).

Answer: d [Reason:] We know that the circular convolution of two sequences is given by the expression

For m=0,x2((-n))4={1,4,3,2}
For m=1,x2((1-n))4={2,1,4,3}
For m=2,x2((2-n))4={3,2,1,4}
For m=3,x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.

Answer: b [Reason:] Given x1(n)={2,1,2,1}=>X1(k)=[6,0,2,0]
Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2]
when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0]
By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.

Answer: a [Reason:] According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))_N is X(k)e^-j2πkl/N.

Answer: According to the complex conjugate property of DFT, we have if X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).

Answer: c [Reason:] We know that

Answer: a [Reason:] We know that the inverse transform or the synthesis equation of a signal x(n) is given as

By substituting ejω = cosω + jsinω in the above equation and separating the real and imaginary parts we get

Answer: b [Reason:] If the signal x(n) is real, then x_I(n)=0
We know that,

Now substitute xI(n)=0 in the above equation=>xR(n)=x(n)

Answer: d [Reason:] We know that, if x(n) is a real sequence

If we combine the above two equations, we get
X*(ω)=X(-ω)

Answer: a [Reason:] We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that,

Answer: b [Reason:] If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently
XR(ω)=0

Answer: c [Reason:] Given, X(ω)= 1/(1-ae^-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_R(ω)= (1-acosω)/(1-2acosω+a² ).

Answer: d [Reason:] Given, X(ω)= 1/(1-ae^-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_I(ω)= (-asinω)/(1-2acosω+a² ).

Answer: a [Reason:] For the given X(ω)=1/(1-ae^-jω ) ,|a|<1 we obtain
X_I(ω)= (-asinω)/(1-2acosω+a² ) and X_R(ω)= (1-acosω)/(1-2acosω+a² )
We know that |X(ω)|=√(〖X_R (ω)〗²+〖X_I (ω)〗² )
Thus on calculating, we obtain
|X(ω)|= 1/√(1-2acosω+a² )

Answer: c [Reason:] Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that

Answer: b [Reason:] First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= aⁿ, n>0
=0, elsewhere

x2(n)=a^-n, n<0 =0, elsewhere Now applying Fourier transform for the above two signals, we get X1(ω)= 1/(1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )

Now, X(ω)= X1(ω)+ X2(ω)= 1/(1-ae^(-jω) )+(ae^jω)/(1-ae^jω ) = (1-a²)/(1-2acosω+a²).

Answer: d [Reason:] Given

Answer: a [Reason:] Given x1(n)=x2(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ ejω + e -jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}

Answer: b [Reason:] Given x(n)= aⁿu(n), |a|<1
The auto correlation of the above signal is
rxx(l)=1/(1-a² ) a^|l|, -∞< l <∞
According to Wiener-Khintchine Theorem,
S_xx(ω)=F{ r_xx(l)}= [1/(1-a²)].F{a^|l|} = 1/(1-2acosω+a² )

Answer: a [Reason:] 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.

Answer: c [Reason:] Additive white noise model is the model that we use in the statistical analysis of round off errors in IIR and FIR filters.

Answer: d [Reason:] 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.

Answer: b [Reason:] 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.

Answer: c [Reason:] 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.

Answer: a [Reason:] The 4N quantization errors are mutually uncorrelated. This is one of the assumption that is made about the statistical properties of the quantization error.

Answer: b [Reason:] 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)}.

Answer: d [Reason:] 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.

Answer: a [Reason:] 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.

Answer: c [Reason:] 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.

Answer: b [Reason:] 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.

Answer: a [Reason:] 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.

Answer: d [Reason:] 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.

Answer: c [Reason:] 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.

Answer: a [Reason:] 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.

Answer: a [Reason:] If ak and bk are the filter coefficients, then the transfer function of a general IIR filter is given by the expression

Answer: c [Reason:] The quantized coefficient a ̅k can be related to the un-quantized coefficient ak by the relation
a ̅k = ak + Δak
where Δak represents the quantization error.

Answer: d [Reason:] We know that the system function of a general IIR filter is given by the equation

Answer: b [Reason:] We know that p ̅k = pk + Δpk, k=1,2…N and Δpk that is the error resulting from the quantization of filter coefficients, which is called as perturbation error.

Answer: a [Reason:]

Answer: c [Reason:]

Answer: b [Reason:] If the poles are tightly clustered as they are in a narrow band filter, the lengths of |pi-pl| are small for the poles in the vicinity of pi. These small lengths will contribute to large errors and hence a large perturbation error results.

Answer: a [Reason:] The perturbation error can be minimized by maximizing the lengths of |pi-pl|. This can be accomplished by realizing the high order filter with either single pole or double pole filter sections.

Answer: b [Reason:] The sensitivity analysis made on the poles of a system results on the zeros of the IIR filters.

Answer: d [Reason:] We know that the system function of a general IIR filter is given by the equation

Answer: d [Reason:] (101.01)₂=1*2²+0*2¹+1*2⁰+0*2^-1+1*2^-2=(5.25)₁₀
=>x=5.25.

Answer: a [Reason:] A real number X can be represented as where b_i represents the digit, ‘r’ is the radix or base, A is the number of integer digits, and B is the number of fractional digits.

Answer: b [Reason:] The binary point between the digits b0 and b1 does not exist physically in the computer. Simply, the logic circuits of the computer are designed such that the computations result in numbers that correspond to the assumed location of this point.

Answer: c [Reason:] A fixed point representation of numbers allows us to cover a range of numbers, say,xmax-xmin with a resolution
Δ=(xmax-xmin)/(m-1)
where m=2^b is the number of levels and ‘b’ is the number of bits.

Answer: d [Reason:] We can represent 5 as
5=0.625*8=0.625*2³
The above number can be represented in binary float point representation as 0.101000*2⁰¹¹
Thus Mantissa=0.101000, Exponent=011.

Answer: a [Reason:] Let us consider two numbers X=M.2^E and Y=N.2^F
If we multiply both X and Y, we get X.Y=(M.N).2^E+F
Thus if we multiply two numbers, the mantissas are multiplied and the exponents are added.

Answer: d [Reason:] Let the mantissa be represented by 23 bits plus a sign bit and let the exponent be represented by 7 bits plus a sign bit.
Thus, the smallest floating point number that can be represented using the 32 bit number is
(1/2)*2^-127=0.3*10^-38
Thus, the smallest floating point number that can be represented using the 32 bit number is
(1-2^-23)*2 ^{127/sup>=1.7*1038.}

Answer: c [Reason:] According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)c(M).
From the above equation we can interpret that,
If 0<E<255, then X=(-1)^s.2^E-127(1.M)=>X is a mixed number.

Answer: b [Reason:] For a two’s complement representation, the truncation error is always negative and falls in the range
-(2^-b-2^-bm) ≤ E_{t ≤ 0.}

Answer: a [Reason:] In floating point representation, the mantissa is either rounded or truncated. Due to non-uniform resolution, the corresponding error in a floating point representation is proportional to the number being quantized.