Digital Electronic MCQ Number 00920

Answer: c [Reason:] The design method based on the use of windows to truncate the impulse response h(n) and obtaining the desired spectral shaping, was the first method proposed for designing linear phase FIR filters.

Answer: a [Reason:] The major disadvantage of the window design method is the lack of precise control of the critical frequencies.

Answer: d [Reason:] The values of the cutoff frequencies of a filter in general by windowing technique depend on the type of the filter and the length of the filter.

Answer: b [Reason:] In the frequency sampling technique, the transition band is a multiple of 2π/M.

Answer: a [Reason:] Frequency sampling design method is particularly attractive when the FIR is realized either in the frequency domain by means of the DFT or in any of the frequency sampling realizations.

Answer: c [Reason:] The attractive feature of the frequency sampling design is that the frequency response can take either zero or one at all frequencies, except in the transition band.

Answer: b [Reason:] The chebyshev approximation method provides total control of the filter specifications, and as a consequence, it is usually preferable over the other two methods.

Answer: a [Reason:] By spreading the approximation error over the pass band and stop band of the filter, this method results in an optimal filter design and using this the maximum side lobe level is minimized.

Answer: d [Reason:] The expression for Δf i.e., for the transition band is given as
Δf=(ωs- ωp)/2π.

Answer: a [Reason:] The estimate is used to carry out the design and if the resulting δ exceeds the specified δ2, then the length can be increased until we obtain a side lobe level that meets the specification.

Answer: a [Reason:] The process of converting a continuous-time signal into a discrete-time signal by taking samples of continuous time signal at discrete time instants is known as ‘sampling’.

Answer: b [Reason:] In this process, the value of each signal sample is represented by a value selected from a finite set of possible values. Hence this process is known as ‘quantization’

Answer: d [Reason:] Quantization error is the difference in the signal obtained after sampling i.e., x(n) and the signal obtained after quantization i.e., xq(n) at any instant of time.

Answer: d [Reason:] The process of joining in terms of steps is known as staircase approximation, connecting two samples by a straight line is known as Linear interpolation, connecting three samples by fitting a quadratic curve is called as Quadratic interpolation.

Answer: b [Reason:] Consider an analog signal of frequency ‘F’, which when sampled periodically at a rate Fs=1/T samples per second yields a frequency of f=F/Fs=>f=F*T.

Answer: c [Reason:] From the question F=40Hz, Fs=20Hz
=>f=F/Fs
=>f=40/20
=>f=2Hz
=>x(n)=cos(4*pi*n).

Answer: a [Reason:] According to Nyquist rate, to avoid aliasing the sampling frequency should be equal to twice of the analog frequency.

Answer: d [Reason:] The frequencies present in the given signal are F1=25Hz, F2=150Hz, F3=50Hz
Thus Fmax=150Hz and from the sampling theorem,
nyquist rate=2*Fmax
Therefore, Fs=2*150=300Hz.

Answer: b [Reason:] From the given analog signal, F1=1000Hz F2=2500Hz and Fs=5000Hz
=>f1=F1/Fs and f2=F2/Fs
=>f1=0.2 and f2=0.5
=>x(n)= cos(0.4*pi*n)+sin(pi*n).

Answer: a [Reason:] If it obeys sampling theorem, then the only error in A/D conversion is quantization error. So, the error is same for both analog and discrete-time signal.

Answer: c [Reason:] The quality is measured by taking the ratio of noises of input signal and the quantized signal i.e., SQNR and is measured in terms of dB.

Answer: b [Reason:] To code a signal with L number of levels, we require a coder with (log L/log 2) number of bits. So, log16/log2=4 bit coder is required.

Answer: b [Reason:] We know that, x(n)δ(n-k)=x(k)δ(n-k)
x(-1)=2=2δ(n+1)
x(0)=4=4δ(n)
x(2)=3=3δ(n-2)
Therefore, x(n)= 2δ(n+1)+4δ(n)+3δ(n-2).

Answer: a [Reason:] The input x(n) is convoluted with the impulse response h(n) to yield the output y(n).As we are summing the different values, we call it as Convolution sum.

Answer: d [Reason:] First the signal h(k) is folded to get h(-k). Then it is shifted by n to get h(n-k). Then it is multiplied by x(k) and then summed over -∞ to ∞.

Answer: c [Reason:] Let y(n)=x(n)*h(n)(‘*’ symbol indicates convolution symbol)
From the formula of convolution we get,
y(0)=x(0)h(0)=1.1=1
y(1)=x(0)h(1)+x(1)h(0)=1.1+2.1=3
y(2)=x(0)h(2)+x(1)h(1)+x(2)h(0)=1.1+2.1+3.1=6
y(3)=x(1)h(2)+x(2)h(1)=2.1+3.1=5
y(4)=x(2)h(2)=3.1=3
Therefore, y(n)=x(n)*h(n)={1,3,6,5,3}.

Answer: a [Reason:] Now fold the signal x(n) and shift it by one unit at a time and sum as follows
y(0)=x(0)h(0)=1
y(1)=h(0)x(1)+h(1)x(0)=1.1+a.1=1+a
y(2)=h(0)x(2)+h(1)x(1)+h(2)x(0)=1.1+a.1+a².1=1+a+a²
Similarly, y(n)=1+a+a²+….aⁿ= (1-a⁽ⁿ⁺¹⁾)/(1-a).

Answer: a [Reason:] According to the properties of convolution, Convolution of three signals obeys Associative property.

Answer: b [Reason:] Let h2(n) be shifted and folded.
so, h(k)=h1(n)*h2(n)=
For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.

Answer: a [Reason:] According to the properties of the convolution, convolution exhibits distributive property.

Answer: d [Reason:] Let us consider a LTI system having an output at time n=n0given by the convolution formula

=(h(0)x(n0)+h(1)x(n0-1)+h(2)x(n0-2)+….)+(h(-1)x(n0+1)+h(-2)x(n0+2)+…)
As per the definition of the causality, the output should depend only on the present and past values of the input. So, the coefficients of the terms x(n0+1), x(n0+2)…. should be equal to zero.
that is, h(n)=0 for n<0 .

Answer: b [Reason:]

Answer: b [Reason:]

So, the system is not stable.

Answer: b [Reason:] The system function is H(z)=1/(1-0.9z^-1+0.81z^-2 )
The system has two complex-conjugate poles at p1=0.9e^jπ/3 and p2=0.9e ^-jπ/3
The z-transform of the unit step sequence is
X(z)=1/(1-z^-1 )
Therefore,
Y_zs(z) = 1/((1-0.9e^(jπ/3) z^-1)(1-0.9e^-jπ/3 z^-1 )(1-z^-1))
= (0.542-j0.049)/((1-0.9e^jπ/3 z^-1) ) + (0.542-j0.049)/((1-0.9e^(jπ/3) z^-1 ) ) + 1.099/(1-z^-1 )
and hence the zero state response is y_zs(n)= [1.099+1.088(0.9)ⁿ.cos(πn/3-5.2^o)]u(n)
Since the initial conditions are zero in this case, we can conclude that y(n)= y_zs(n).

Answer: d [Reason:] If all the poles of H(z) are outside an unit circle, it means that the system is neither causal nor BIBO stable.

Answer: a [Reason:] If |p_k| < 1 for all k, then y_nr(n) decays to 0 as n approaches infinity. In such a case we refer to the natural response of the system as the transient response.

Answer: c [Reason:] If the magnitudes of the poles of the response of any system is very small i.e., almost equal to zero, then the system decays very rapidly.

Answer: a [Reason:] If the magnitudes of the poles of the response of any system is almost equal to one, then the system decays very slowly or the transient will persist for a relatively long time.

Answer: d [Reason:] The system function for the system is
and therefore the system has a pole at z=0.5. The z-transform of the input signal is

The natural or transient response is
y_nr(n)= 6.3(0.5)ⁿu(n)

Answer: c [Reason:] The system function for the system is

Answer: b [Reason:] A linear time invariant system is said to be causal if and only if the ROC of the system function is the exterior of a circle of radius r < ∞, including the point z = ∞.

Answer: a [Reason:] For an LTI system, if the ROC of the system function includes the unit circle, then the systm is said to be BIBO stable.

Answer: c [Reason:] If all the poles of H(z) are inside an unit circle, then it follows the condition that |z|>r < 1, it means that the system is both causal and BIBO stable.

Answer: d [Reason:] The system has poles at z=0.5 and at z=3.
Since the system is stable, its ROC must include unit circle and hence it is 0.5<|z|<3 . Consequently, h(n) is non causal and is given as h(n)= (0.5)ⁿu(n)-2(3)ⁿu(-n-1).

Answer: b [Reason:] The system has poles at z=0.5 and at z=3.
Since the system is causal, its ROC is |z|>0.5 and |z|>3. The common region is |z|>3. So, ROC of given H(z) is |z|>3.

Answer: c [Reason:] The system has poles at z=0.5 and at z=3.
If the system is anti causal, then the ROC is |z|<0.5.Hence
h(n)= -[(0.5)ⁿ+2(3)ⁿ]u(-n-1).

Answer: a [Reason:] The FFT algorithm is designed to perform complex multiplications and additions, even though the input data may be real valued. The basic reason for this is that the phase factors are complex and hence, after the first stage of the algorithm, all variables are basically complex valued.

Answer: b [Reason:] Given x(n)=x1(n)+jx2(n)
=>x*(n)= x1(n)-jx2(n)
Upon adding the above two equations, we get x1(n)= (x(n)+x*(n))/2.

Answer: d [Reason:] Given x(n)=x1(n)+jx2(n)
=>x*(n)= x1(n)-jx2(n)
Upon subtracting the above two equations, we get x2(n)= (x(n)-x*(n))/2j.

Answer: a [Reason:] We know that if x(n)=x1(n)+jx2(n) then x1(n)= (x(n)+x*(n))/2
On applying DFT on both sides of the above equation, we get
X1(k)= 1/2 {DFT[x(n)]+DFT[x*(n)]}
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)
=>X1(k)= 1/2 [X*(k)+X*(N-k)].

Answer: c [Reason:] We know that if x(n)=x1(n)+jx2(n) then x2(n)= (x(n)-x^* (n))/2j.
On applying DFT on both sides of the above equation, we get
X2(k)= (1/2j) {DFT[x(n)]-DFT[x*(n)]}
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)
=>X2(k)= (1/2j) [X*(k)-X*(N-k)].

Answer: b [Reason:] Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided into two N point sequences x1(n) and x2(n)
Let x(n)= x1(n)+jx2(n)
=> X1(k)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)]
We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)+W₂^kNX2(k), k=0,1,2…N-1.

Answer: d [Reason:] Given g(n) is a real valued 2N point sequence. The 2N point sequence is divided into two N point sequences x1(n) and x2(n)
Let x(n)= x1(n)+jx2(n)
=> X1(k)= 1/2 [X*(k)+X*(N-k)] and X2(k)= 1/2j [X*(k)-X*(N-k)]
We know that g(n)= x1(n)+x2(n)
=>G(k)= X1(k)-W₂^kNX2(k), k= N,N-1,…2N-1.

Answer: a [Reason:] The N-point DFT of h(n), which is padded by L-1 zeros, is denoted as H(k). This computation is performed once via the FFT and resulting N complex numbers are stored. To be specific we assume that the decimation-in frequency FFT algorithm is used to compute H(k). This yields H(k) in the bit-reversed order, which is the way it is stored in the memory.

Answer: c [Reason:] The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one point sequences. For N=2^v, this decimation can be performed v=log₂N times. Thus the total number of complex multiplications is reduced to (N/2)log₂N.

Answer: b [Reason:] The number of additions to be performed in FFT are Nlog₂N. But in linear filtering of a sequence, we calculate DFT which requires Nlog₂N complex additions and IDFT requires Nlog₂N complex additions. So, the total number of complex additions to be performed in linear filtering of a sequence using FFT algorithm is 2Nlog₂N.

Answer: b [Reason:] In the overlap add method, the N-point data block consists of L new data points and additional M-1 zeros and the number of complex multiplications required in FFT algorithm are (N/2)log₂N. So, the number of complex multiplications per output data point is [Nlog₂2N]/L.