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Multiple Choice Questions and Answers :-
1. If x(n) is a
discrete-time signal, then the value of x(n) at non integer value of ‘n’ is:
a) Zero
b) Positive
c) Negative
d) Not defined
Answer: d
Explanation: For a
discrete time signal, the value of x(n) exists only at integral values of n.
So, for a non- integer value of ‘n’ the value of x(n) does not exist.
2. The discrete time
function defined as u(n)=n for n=0;=0 for n<0 is an:
a) Unit sample signal
b) Unit step signal
c) Unit ramp signal
d) None of the mentioned
Answer: c
Explanation: When we
plot the graph for the given function, we get a straight line passing through
origin with a unit positive slope. So, the function is called as unit ramp
signal.
3.The phase function of
a discrete time signal x(n)=an, where a=r.ej? is:
a) tan(n?)
b) n?
c) tan-1(n?)
d) None of the mentioned
Answer: b
Explanation: Given
x(n)=an=(r.ej?)n =rn.ejn?
=>x(n)=rn.(cosn?+jsinn?)
Phase function is
tan-1(cosn?/sinn?)=tan-1(tan n?)=n?
4. The signal given by
the equation4 is known as:
a) Energy signal
b) Power signal
c) Work done signal
d) None of the mentioned
Answer: a
5.Explanation: We have
used the magnitude-squared values of x(n), so that our definition applies to
complex-valued as well as real-valued signals. If the energy of the signal is
finite i.e., 0<e<8 then="" the="" given=""
signal="" is="" known="" as=""
energy="" signal.5.="" x(n)*d(n-k)="?"
a)="" x(n)="" b)="" x(k)=""
c)="" x(k)*d(n-k)="" d)="" x(k)*d(k)=""
answer:="" c="" 6.explanation:=""
defined="" only="" when="" n="k"
by="" definition="" of="" delta=""
function.="" so,="" 6.="" a=""
real="" valued="" called=""
anti-symmetric="" if:="" none="" mentioned=""
b="" explanation:="" according="" to=""
signal,="" should="" be="" symmetric=""
over="" origin.="" for="" symmetric,=""
it="" satisfy="" condition="" 7.=""
odd="" part="" x(t)="" is:=""
x(t)+x(-t)="" x(t)-x(-t)="" (1=""
2)*(x(t)+x(-t))="" 2)*(x(t)-x(-t))="" d=""
let="" =="">x(-t)=xe(-t)-xo(-t)
By subtracting the above
two equations, we get
xo(t)=(1/2)*(x(t)-x(-t))
8. Time scaling
operation is also known as:
a) Down-sampling
b) Up-sampling
c) Sampling
d) None of the mentioned
Answer: a
Explanation: If the signal
x(n) was originally obtained by sampling a signal xa(t), then x(n)=xa(nT). Now,
y(n)=x(2n)(say)=xa(2nT). Hence the time scaling operation is equivalent to
changing the sampling rate from 1/T to 1/2T, that is to decrease the rate by a
factor of 2. So, time scaling is also called as down-sampling.
9. What is the condition
for a signal x(n)=Brn where r=eaT to be called as an decaying exponential
signal?
a) 0<r<8
b)="" 0><r<1="" c)=""
r="">1
d) r<0
Answer: b
Explanation: When the
value of ‘r’ lies between 0 and 1 then the value of x(n) goes on decreasing
exponentially with increase in value of ‘n’. So, the signal is called as
decaying exponential signal.
10. The function given
by the equation x(n)=1, for n=0;=0, for n?0 is a:
a) Step function
b) Ramp function
c) Triangular function
d) Impulse function
Answer: d
Explanation: According
to the definition of the impulse function, it is defined only at n=0 and is not
defined elsewhere which is as per the signal given.
11. Which of the
following should be done in order to convert a continuous-time signal to a
discrete-time signal?
a) Sampling
b) Differentiating
c) Integrating
d) None of the mentioned
Answer: a
Explanation: 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’.
12. The process of
converting discrete-time continuous valued signal into discrete-time discrete
valued(digital) signal is known as:
a) Sampling
b) Quantization
c) Coding
d) None of the mentioned
Answer: b
Explanation: 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’
13. The difference
between the unquantized x(n) and quantized xq(n) is known as:
a) Quantization
coefficient
b) Quantization ratio
c) Quantization factor
d) Quantization error
Answer: d
Explanation:
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.
14. Which of the
following is a digital-to-analog conversion process?
a) Staircase
approximation
b) Linear interpolation
c) Quadratic
interpolation
d) All of the mentioned
Answer: d
Explanation: 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.
15. The relation between
analog frequency ‘F’ and digital frequency ‘f’ is:
a) F=f*T(where T is
sampling period)
b) f=F*T
c) No relation
d) None of the mentioned
Answer: b
Explanation: 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.
16. What is output
signal when a signal x(t)=cos(2*pi*40*t) is sampled with a sampling frequency
of 20Hz?
a) cos(pi*n)
b) cos(2*pi*n)
c) cos(4*pi*n)
d) cos(8*pi*n)
Answer: c
Explanation: From the
question F=40Hz, Fs=20Hz
=>f=F/Fs
=>f=40/20
=>f=2Hz
=>x(n)=cos(4*pi*n)
17. If ‘F’ is the
frequency of the analog signal, then what is the minimum sampling rate required
to avoid aliasing?
a) F
b) 2F
c) 3F
d) 4F
Answer: a
Explanation: According
to Nyquist rate, to avoid aliasing the sampling frequency should be equal to
twice of the analog frequency.
18. What is the nyquist
rate of the signal x(t)=3cos(50*pi*t)+10sin(300*pi*t)-cos(100*pi*t)?
a) 50Hz
b) 100Hz
c) 200Hz
d) 300Hz
Answer: d
Explanation: 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.
19. What is the
discrete-time signal obtained after sampling the analog signal
x(t)=cos(2000*pi*t)+sin(5000*pi*t) at a sampling rate of 5000samples/sec?
a)
cos(2.5*pi*n)+sin(pi*n)
b) cos(0.4*pi*n)+sin(pi*n)
c)
cos(2000*pi*n)+sin(5000*pi*n)
d) None of the mentioned
Answer: b
Explanation: 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)
20. If the sampling rate
Fs satisfies the sampling theorem, then the relation between quantization
errors of analog signal(eq(t)) and discrete-time signal(eq(n)) is:
a) eq(t)=eq(n)
b) eq(t)eq(n)
d) Not related
Answer: a
Explanation: 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.
21. The quality of
output signal from a A/D converter is measured in terms of:
a) Quantization error
b) Quantization to
signal noise ratio
c) Signal to
quantization noise ratio
d) Conversion constant
Answer: c
Explanation: 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.
22. Which bit coder is
required to code a signal with 16 levels?
a) 8 bit
b) 4 bit
c) 2 bit
d) 1 bit
Answer: b
Explanation: 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.
23. Which of the
following is done to convert a continuous time signal into discrete time
signal?
a) Modulating
b) Sampling
c) Differentiating
d) Integrating
Answer: b
Explanation: A discrete
time signal can be obtained from a continuous time signal by replacing t by nT,
where T is the reciprocal of the sampling rate or time interval between the
adjacent values. This procedure is known as sampling.
24. The deflection voltage
of an oscilloscope is a ‘deterministic’
signal. True or False?
a) True
b) False
Answer: a
Explanation: The
behavior of the signal is known and can be represented by a saw tooth wave
form. So, the signal is deterministic.
25. The even part of a
signal x(t) is:
a) x(t)+x(-t)
b) x(t)-x(-t)
c) (1/2)*(x(t)+x(-t))
d) (1/2)*(x(t)-x(-t))
Answer: c
Explanation: Let
x(t)=xe(t)+xo(t)
=>x(-t)=xe(-t)-xo(-t)
By adding the above two
equations, we get
xe(t)=(1/2)*(x(t)+x(-t))
26. Which of the
following is the odd component of the signal x(t)=e(jt)?
a) cost
b) j*sint
c) j*cost
d) sint
Answer: b
Explanation: Let
x(t)=e(jt)
Now,
xo(t)=(1/2)*(x(t)-x(-t))
=(1/2)*(e(jt) – e(-jt))
=(1/2)*(cost+jsint-cost+jsint)
=(1/2)*(2jsint)
=j*sint
27. For a continuous
time signal x(t) to be periodic with a period T, then x(t+mT) should be equal
to:
a) x(-t)
b) x(mT)
c) x(mt)
d) x(t)
Answer: d
Explanation: If a signal
x(t) is said to be periodic with period T, then x(t+mT)=x(t) for all t and any
integer m.
28. Let x1(t) and x2(t)
be periodic signals with fundamental periods T1 and T2 respectively. Which of
the following must be a rational number for x(t)=x1(t)+x2(t) to be periodic?
a) T1+T2
b) T1-T2
c) T1/T2
d) T1*T2
Answer: c
Explanation: Let T be
the period of the signal x(t)
=>x(t+T)=x1(t+mT1)+x2(t+nT2)
Thus, we must have
mT1=nT2=T
=>(T1/T2)=(k/m)= a
rational number
29. Let x1(t) and x2(t)
be periodic signals with fundamental periods T1 and T2 respectively. Then the
fundamental period of x(t)=x1(t)+x2(t) is:
a) LCM of T1 and T2
b) HCF of T1and T2
c) Product of T1 and T2
d) Ratio of T1 to T2
Answer: a
Explanation: For the sum
of x1(t) and x2(t) to be periodic the ratio of their periods should be a
rational number, then the fundamental period is the LCM of T1 and T2.
30. All energy signals
will have an average power of:
a) Infinite
b) Zero
c) Positive
d) Cannot be calculated
Answer: b
Explanation: For any
energy signal, the average power should be equal to 0 i.e., P=0.
31. x(t) or x(n) is
defined to be an energy signal, if and only if the total energy content of the
signal is a:
a) Finite quantity
b) Infinite
c) Zero
d) None of the mentioned
Answer: a
Explanation: The energy
signal should have total energy value that lies between 0 and infinity.
32. What is the period
of cos2t+sin3t?
a) pi
b) 2*pi
c) 3*pi
d) 4*pi
Answer: b
Explanation: Period of
cos2t=(2*pi)/2=pi
Period of sin3t=(2*pi)/3
LCM of pi and (2*pi)/3
is 2*pi.
33. Which of the
following justifies the linearity property of z-transform?[x(n)?X(z)] a)
x(n)+y(n) ?X(z)Y(z)
b) x(n)+y(n) ?X(z)+Y(z)
c) x(n)y(n) ?X(z)+Y(z)
d) x(n)y(n) ?X(z)Y(z)
Answer: b
Explanation: According
to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms
of x(n) and y(n) respectively then, the z-transform of x(n)+y(n) is X(z)+Y(z).
34. What is the
z-transform of the signal x(n)=[3(2n)-4(3n)]u(n)?
a) 3/(1-2z-1)-4/(1-3z-1)
b) 3/(1+2z-1)-4/(1+3z-1)
c) 3/(1-2z)-4/(1-3z)
d) None of the mentioned
Answer: a
Explanation: Let us
divide the given x(n) into x1(n)= 3(2n)u(n) and x2(n)= 4(3n)u(n) and
x(n)=x1(n)-x2(n)From the definition of z-transform X1(z)= 3/(1-2z-1) and X2(z)=
4/(1-3z-1) So, from the linearity property of z-transform X(z)=X1(z)-X2(z)=>
X(z)= 3/(1-2z-1)-4/(1-3z-1)
35. According to Time
shifting property of z-transform, if X(z) is the z-transform of x(n) then what
is the z-transform of x(n-k)?
a) zkX(z)
b) z-kX(z)
c) X(z-k)
d) X(z+k)
Answer: b
Explanation: According
to the definition of Z-transform
36. If X(z) is the
z-transform of the signal x(n) then what is the z-transform of anx(n)?
a) X(az)
b) X(az-1)
c) X(a-1z)
d) None of the mentioned
Answer: c
Explanation: We know
that from the definition of z-transform
37. If the ROC of X(z)
is r1<|z|<|z|<|a|r2="" b)=""
|a|r1="">|z|>|a|r2
c) |a|r1<|z|>|a|r2
d) |a|r1>|z|<|a|r2
Answer: a
Explanation: Given ROC
of X(z) is r1<|z|<|a-1z="" |> 5a
42. What is the value of
XR(?) given X(?)=1/(1-ae-j? ) ,|a|<1? a) asin?/(1-2acos?+a2 )
b)
(1+acos?)/(1-2acos?+a2 )
c)
(1-acos?)/(1-2acos?+a2 )
d) (-asin?)/(1-2acos?+a2
)
Answer: c
Explanation: Given,
X(?)= 1/(1-ae-j? ) ,|a|<1 By multiplying both the numerator and denominator
of the above equation by the complex conjugate of the denominator, we obtain
X(?)= (1-ae^j?)/((1-ae^(-j?) )(1-ae^j?)) = (1-acos?-jasin?)/(1-2acos?+a^2 ) This
expression can be subdivided into real and imaginary parts, thus we obtain
XR(?)= (1-acos?)/(1-2acos?+a2 ).
43. What is the value of
XI(?) given X(?)=1/(1-ae-j? ) ,|a|<1? a) asin?/(1-2acos?+a2 )
b)
(1+acos?)/(1-2acos?+a2 )
c)
(1-acos?)/(1-2acos?+a2 )
d) (-asin?)/(1-2acos?+a2
)
Answer: d
Explanation: Given,
X(?)= 1/(1-ae-j? ) ,|a|<1 By multiplying both the numerator and denominator
of the above equation by the complex conjugate of the denominator, we obtain
X(?)= (1-ae^j?)/((1-ae^(-j?) )(1-ae^j?)) = (1-acos?-jasin?)/(1-2acos?+a^2 )
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(?)= (-asin?)/(1-2acos?+a2 ).
44. What is the value of
|X(?)| given X(?)=1/(1-ae-j? ) ,|a|<1? a) 1/v(1-2acos?+a2 )
b) 1/v(1+2acos?+a2)
c) 1/(1-2acos?+a2 )
d) 1/(1+2acos?+a2 )
Answer: a
Explanation: For the
given X(?)=1/(1-ae-j? ) ,|a|<1 we obtain XI(?)= (-asin?)/(1-2acos?+a2 ) and
XR(?)= (1-acos?)/(1-2acos?+a2 )
We know that
|X(?)|=v(?X_R (?)?^2+?X_I (?)?^2 )
Thus on calculating, we
obtain
|X(?)|= 1/v(1-2acos?+a2
)
45. What is the Fourier
transform of the signal x(n)=a|n|, |a|<1? a) (1+a2)/(1-2acos?+a2)
b) (1-a2)/(1-2acos?+a2)
c) 2a/(1-2acos?+a2 )
d) None of the mentioned
Answer: b
Explanation: 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(?)= 1/(1-ae^(-j?) ) and X2(?)= (ae^j?)/(1-ae^j? ) Now, X(?)= X1(?)+ X2(?)=
1/(1-ae^(-j?) )+(ae^j?)/(1-ae^j? ) = (1-a2)/(1-2acos?+a2)
46. If X(?) is the
Fourier transform of the signal x(n), then what is the Fourier transform of the
signal x(n-k)?
a) ej?k. X(-?)
b) ej?k. X(?)
c) e-j?k. X(-?)
d) e-j?k. X(?)
Answer: d
Explanation: Given
47. What is the
convolution of the sequences of x1(n)=x2(n)={1,1,1}?
a) {1,2,3,2,1}
b) {1,2,3,2,1}
c) {1,1,1,1,1}
d) {1,1,1,1,1}
Answer: a
Explanation: Given
x1(n)=x2(n)={1,1,1}
By calculating the
Fourier transform of the above two signals, we get
X1(?)= X2(?)=1+ ej? + e
-j? = 1+2cos?
From the convolution
property of Fourier transform we have,
X(?)= X1(?).
X2(?)=(1+2cos?)2=3+4cos?+2cos2?
By applying the inverse
Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}
48. What is the energy
density spectrum of the signal x(n)=anu(n), |a|<1? a) 1/(1+2acos?+a2 )
b) 1/(1-2acos?+a2 )
c) 1/(1-2acos?-a2 )
d) 1/(1+2acos?-a2 )
Answer: b
Explanation: Given x(n)=
anu(n), |a|<1 The auto correlation of the above signal is rxx(l)=1/(1-a2 )
a|l|, -8< l <8 According to Wiener-Khintchine Theorem, Sxx(?)=F{ rxx(l)}=
[1/(1-a2)].F{a|l|} = 1/(1-2acos?+a2 )
49. When the frequency
band is selected we can specify the sampling rate and the characteristics of
the pre filter, which is also called as __ filter?
a) Analog filter
b) Anti aliasing filter
c) Both a& b
d) None of the mentioned
Answer: b
Explanation: Once the
desired frequency band is selected w e can specify the sampling rate and the
characteristics of the pre filter, which is also called an anti aliasing
filter. The anti aliasing filter is an analog filter which has a twofold
purpose.
50. What are the main
characteristics of Anti aliasing filter?
a) Ensures that
bandwidth of signal to be sampled is limited to frequency range.
b) To limit the additive
noise spectrum and other interference, which corrupts the signal.
c) Both a& b
d) None of the mentioned
Answer: c
Explanation: T he anti
aliasing filter is an analog filter which has a twofold purpose. First, it
ensures that the bandwidth of the signal to be sampled is limited to the
desired frequency range. Using an antialiasing filter is to limit the additive
noise spectrum and other interference, which often corrupts the desired signal.
Usually, additive noise is wideband and exceeds the bandwidth of the desired
signal.
51. In general, a
digital system designer has better control of tolerances in a digital signal
processing system than an analog system designer who is designing an equivalent
analog system.
a) True
b) False
Answer: a
Explanation: Analog
signal processing operations cannot be done very precisely either, since electronic
components in analog systems have tolerances and they introduce noise during
their operation. In general, a digital system designer has better control of
tolerances in a digital signal processing system than an analog system designer
who is designing an equivalent analog system.
52. The selection o f
the sampling rate Fs=1/T, where T is the sampling interval, not only determines
the highest frequency (Fs/2) that is preserved in the analog signal, but also
serves as a scale factor that influences the design specifications for digital
filters
a) True
b) False
Answer: a
Explanation: Once we
have specified the pre filter requirements and have selected the desired
sampling rate, w e can
proceed with the design of the digital signal processing operations to be
performed on the discrete-time signal. The selection of the sampling rate
Fs=1/T , where T is the sampling interval, not only determines the highest
frequency (Fs/2) that is preserved in the analog signal, but also serves as a
scale factor that influences the design specifications for digital filters and
any other
discrete-time systems
through which the signal is processed.
53. What is the
configuration of system for digital processing of an analog signal?
a) Analog signal||
Pre-filter ->D/A Converter -> Digital Processor -> A/D Converter ->
Post-filter
b) Analog signal||
Pre-filter ->A/D Converter -> Digital Processor -> D/A Converter ->
Post-filter
c) Analog signal||
Post-filter ->D/A Converter -> Digital Processor -> A/D Converter
-> Pre-filter
d) None of the mentioned
Answer: b
Explanation: The
anti-aliasing filter is an analog filter which has a twofold purpose.
Analog signal||
Pre-filter ->A/D Converter -> Digital Processor -> D/A Converter ->
Post-filter
54. In DM, further the
two integrators at encode are replaced by one integrator placed before
comparator, and then such system is called?
a) System-delta
modulation
b) Sigma-delta
modulation
c) Source-delta
modulation
d) None of the mentioned
Answer: b
Explanation: In DM,
Furthermore, the two integrators at the encoder can be replaced by a single
integrator placed before the comparator. This system is known as sigma-delta
modulation (SDM ).
55. In IIR Filter design
by the Bilinear Transformation, the Bilinear Transformation is a mapping from
a) Z-plane to S-plane
b) S-plane to Z-plane
c) S-plane to J-plane
d) J-plane to Z-plane
Answer: b
Explanation: From the
equation,
56. it is clear that
transformation occurs from s-plane to z-plane
2. In Bilinear
Transformation, aliasing of frequency components is been avoided.
a) True
b) False
Answer: a
Explanation: The
bilinear transformation is a conformal mapping that transforms the j?-axis into
the unit circle in the z-plane only once, thus avoiding the aliasing.
57. Is IIR Filter design
by Bilinear Transformation is the advanced technique when compared to other
design techniques?
a) True
b) False
Answer: True
Explanation: Because in
other techniques, only lowpass filters and limited class of bandpass filters
are been supported. But this technique overcomes the limitations of other
techniques and supports more.
58. In the Bilinear
Transformation mapping, which of the following are correct?
a) All points in the LHP
of s are mapped inside the unit circle in the z-plane
b) All points in the RHP
of s are mapped outside the unit circle in the z-plane
c) Both a & b
d) None of the mentioned
Answer: c
Explanation: The
bilinear transformation is a conformal mapping that transforms the j?-axis into
the unit circle in the z-plane and all the points are linked as mentioned
above.
59. In equation 10if r
< 1 then o < 0 and then mapping from s-plane to z-plane occurs in which
of the following order? a) LHP in s-plane maps into the inside of the unit
circle in the z-plane b) RHP in s-plane maps into the outside of the unit
circle in the z-plane c) None of the mentioned d) Both a & b [expand
title="View Answer"]Answer: a Explanation: In the above equation, if
we substitute the values of r, o then we get mapping in the required
way[/expand] 11. In equation 10 if r < 1 then o > 0 and then mapping from
s-plane to z-plane occurs in which of the following order?
a) LHP in s-plane maps
into the inside of the unit circle in the z-plane
b) RHP in s-plane maps
into the outside of the unit circle in the z-plane
c) None of the mentioned
d) Both a & b
Answer: b
Explanation: In the
above equation, if we substitute the values of r, o then we get mapping in the
required way
60. The lower and upper
limits on the convolution sum reflect the causality and finite duration
characteristics of the filter.
a) True
b) False
Answer: a
Explanation: We can
express the output sequence as the convolution of the unit sample response h(n)
of the system with the input signal. The lower and upper limits on the
convolution sum reflect the causality and finite duration characteristics of
the filter.
61. Which of the
following condition should the unit sample response of a FIR filter satisfy to
have a linear phase?
a) h(M-1-n) n=0,1,2…M-1
b) ±h(M-1-n) n=0,1,2…M-1
c) -h(M-1-n) n=0,1,2…M-1
d) None of the mentioned
Answer: b
Explanation: An FIR
filter has an linear phase if its unit sample response satisfies the condition
h(n)= ±h(M-1-n)
n=0,1,2…M-1
62. The roots of the
polynomial H(z) are identical to the roots of the polynomial H(z -1).
a) True
b) False
Answer: a
Explanation: We know
that 5. This result implies that the roots of the polynomial H(z) are identical
to the roots of the polynomial H(z -1).
63. The roots of the
equation H(z) must occur in:
a) Identical
b) Zero
c) Reciprocal pairs
d) Conjugate pairs
Answer: c
Explanation: We know
that the roots of the polynomial H(z) are identical to the roots of the
polynomial H(z -1). Consequently, the roots of H(z) must occur in reciprocal
pairs.
64. If the unit sample
response h(n) of the filter is real, complex valued roots need not occur in
complex conjugate pairs.
a) True
b) False
Answer: b
Explanation: We know
that the roots of the polynomial H(z) are identical to the roots of the
polynomial H(z -1). This implies that if the unit sample response h(n) of the
filter is real, complex valued roots must occur in complex conjugate pairs.
65. What is the value of
h(M-1/2) if the unit sample response is anti-symmetric?
a) 0
b) 1
c) -1
d) None of the mentioned
Answer: a
Explanation: When
h(n)=-h(M-1-n), the unit sample response is anti-symmetric. For M odd, the
center point of the anti-symmetric is n=M-1/2. Consequently, h(M-1/2)=0.
66. What is the number
of filter coefficients that specify the frequency response for h(n) symmetric?
a) (M-1)/2 when M is odd
and M/2 when M is even
b) (M-1)/2 when M is
even and M/2 when M is odd
c) (M+1)/2 when M is
even and M/2 when M is odd
d) (M+1)/2 when M is odd
and M/2 when M is even
Answer: d
Explanation: We know
that, for a symmetric h(n), the number of filter coefficients that specify the
frequency response is (M+1)/2 when M is odd and M/2 when M is even.
67. What is the number
of filter coefficients that specify the frequency response for h(n)
anti-symmetric?
a) (M-1)/2 when M is
even and M/2 when M is odd
b) (M-1)/2 when M is odd
and M/2 when M is even
c) (M+1)/2 when M is
even and M/2 when M is odd
d) (M+1)/2 when M is odd
and M/2 when M is even
Answer: b
Explanation: We know
that, for a anti-symmetric h(n) h(M-1/2)=0 and thus the number of filter
coefficients that specify the frequency response is (M-1)/2 when M is odd and
M/2 when M is even.
68. Which of the
following is not suitable either as low pass or a high pass filter?
a) h(n) symmetric and M
odd
b) h(n) symmetric and M
even
c) h(n) anti-symmetric
and M odd
d) h(n) anti-symmetric
and M even
Answer: c
Explanation: If
h(n)=-h(M-1-n) and M is odd, we get H(0)=0 and H(p)=0. Consequently, this is
not suitable as either a low pass filter or a high pass filter.
69. The anti-symmetric
condition with M even is not used in the design of which of the following
linear-phase FIR filter?
a) Low pass
b) High pass
c) Band pass
d) Bans stop
Answer: a
Explanation: When
h(n)=-h(M-1-n) and M is even, we know that H(0)=0. Thus it is not used in the
design of a low pass linear phase FIR filter.
70. The anti-symmetric
condition is not used in the design of low pass linear phase FIR filter.
a) True
b) False
Answer: a
Explanation: We know
that if h(n)=-h(M-1-n) and M is odd, we get H(0)=0 and H(p)=0. Consequently,
this is not suitable as either a low pass filter or a high pass filter and when
h(n)=-h(M-1-n) and M is even, we know that H(0)=0. Thus it is not used in the
design of a low pass linear phase FIR filter. Thus the anti-symmetric condition
is not used in the design of low pass linear phase FIR filter.
71. Sampling rate
conversion by the rational factor I/D is accomplished by what connection of interpolator
and decimator?
a) Parallel
b) Cascade
c) Convolution
d) None of the mentioned
Answer: b
Explanation: A sampling
rate conversion by the rational factor I/D is accomplished by cascading an
interpolator with a decimator.
72. Which of the following
has to be performed in sampling rate conversion by rational factor?
a) Interpolation
b) Decimation
c) Either interpolation
or decimation
d) None of the mentioned
Answer: a
Explanation: We
emphasize that the importance of performing the interpolation first and
decimation second, is to preserve the desired spectral characteristics of x(n).
73. Which of the
following operation is performed by the blocks given the figure below?
3
a) Sampling rate
conversion by a factor I
b) Sampling rate
conversion by a factor D
c) Sampling rate
conversion by a factor D/I
d) Sampling rate
conversion by a factor I/D
Answer: d
Explanation: In the
diagram given, a interpolator is in cascade with a decimator which together
performs the action of sampling rate conversion by a factor I/D.
74. The Nth root of
unity WN is given as:
a) ej2pN
b) e-j2pN
c) e-j2p/N
d) ej2p/N
Answer: c
Explanation: We know
that the Discrete Fourier transform of a signal x(n) is given as
75. Which of the
following is true regarding the number of computations requires 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
Answer: a
Explanation: The formula
for calculating N point DFT is given as5 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.
76. What is the DFT of
the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2+2j,-2,-2-2j}
d) {6,-2-2j,-2,-2+2j}
Answer: c
Explanation: The first
step is to determine the matrix W4. By exploiting the periodicity property of
W4 and the symmetry property
77. If X(k) is the N
point DFT of a sequence whose Fourier series coefficients is given by ck, then
which of the following is true?
a) X(k)=Nck
b) X(k)=ck/N
c) X(k)=N/ck
d) None of the mentioned
Answer: a
Explanation: The Fourier
series coefficients are given by the expression
78. What is the DFT of
the four point sequence x(n)={0,1,2,3}?
a) {6,-2+2j-2,-2-2j}
b) {6,-2-2j,2,-2+2j}
c) {6,-2-2j,-2,-2+2j}
d) {6,-2+2j,-2,-2-2j}
Answer: d
Explanation: Given
x(n)={0,1,2,3}
We know that the 4-point
DFT of the above given sequence is given by the expression
79. If W4100=Wx200, then
what is the value of x?
a) 2
b) 4
c) 8
d) 16
Answer: c
Explanation: We know
that according to the periodicity and symmetry property,
100/4=200/x=>x=8.
80. There is no
requirement to process the various signals at different rates commensurate with
the corresponding bandwidths of the signals.
a) True
b) False
Answer: b
Explanation: In
telecommunication systems that transmit and receive different types of signals,
there is a requirement to process the various signals at different rates
commensurate with the corresponding bandwidths of the signals.
81. What is the process
of converting a signal from a given rate to a different rate?
a) Sampling
b) Normalizing
c) Sampling rate
conversion
d) None of the mentioned
Answer: c
Explanation: The process
of converting a signal from a given rate to a different rate is known as
sampling rate conversion.
82. The systems that
employ multiple sampling rates are called multi-rate DSP systems.
a) True
b) False
Answer: a
Explanation: Systems
that employ multiple sampling rates in the processing of digital signals are
called multi rate digital signal processing systems.
83. Which of the
following methods are used in sampling rate conversion of a digital signal?
a) D/A convertor and A/D
convertor
b) Performing entirely
in digital domain
c) None of the mentioned
d) Both of the mentioned
Answer: d
Explanation: Sampling
rate conversion of a digital signal can be accomplished in one of the two
general methods. One method is to pass the signal through D/A converter, filter
it if necessary, and then to resample the resulting analog signal at the
desired rate. The second method is to perform the sampling rate conversion
entirely in the digital domain.
84. Which of the
following is the advantage of sampling rate conversion by converting the signal
into analog signal?
a) Less signal
distortion
b) Quantization effects
c) New sampling rate can
be arbitrarily selected
d) None of the mentioned
Answer: c
Explanation: One
apparent advantage of the given method is that the new sampling rate can be
arbitrarily selected and need not have any special relationship with the old
sampling rate.
85. Which of the
following is the disadvantage of sampling rate conversion by converting the
signal into analog signal?
a) Signal distortion
b) Quantization effects
c) New sampling rate can
be arbitrarily selected
d) Both a & b
Answer: d
Explanation: The major
disadvantage by the given type of conversion is the signal distortion
introduced by the D/A converter in the signal reconstruction and by the
quantization effects in the A/D conversion.
86. In which of the
following, sampling rate conversion are used?
a) Narrow band filters
b) Digital filter banks
c) Quadrature mirror
filters
d) All of the mentioned
Answer: d
Explanation: There are
several applications of sampling rate conversion in multi rate digital signal
processing systems, which include the implementation of narrow band filters,
quadrature mirror filters and digital filter banks.
87. Which of the
following use quadrature mirror filters?
a) Sub band coding
b) Trans-multiplexer
c) Both of the mentioned
d) None of the mentioned
Answer: c
Explanation: There are
many applications where quadrature mirror filters can be used. Some of these
applications are sub-band coding, trans-multiplexers and many other
applications.
88. The sampling rate
conversion can be as shown in the figure below.
a) True
b) False
Answer: a
Explanation: The process
of sampling rate conversion in the digital domain can be viewed as a linear
filtering operation as illustrated in the given figure.
89. If Fx and Fy are the
sampling rates of the input and output signals respectively, then what is the
value of Fy/Fx?
a) D/I
b) I/D
c) I.D
d) None of the mentioned
Answer: b
Explanation: The input
signal x(n) is characterized by the sampling rate Fx and he output signal y(m)
is characterized by the sampling rate Fy, then
Fy/Fx= I/D
where I and D are
relatively prime integers.
90. What is the process
of reducing the sampling rate by a factor D?
a) Sampling rate
conversion
b) Interpolation
c) Decimation
d) None of the mentioned
Answer: c
Explanation: The process
of reducing the sampling rate by a factor D, i.e., down-sampling by D is called
as decimation.
91. What is the process
of increasing the sampling rate by a factor I?
a) Sampling rate
conversion
b) Interpolation
c) Decimation
d) None of the mentioned
Answer: b
Explanation: The process
of increasing the sampling rate by a integer factor I, i.e., up-sampling by I
is called as interpolation.
92. The reconstruction o
f the signal from its samples as a linear filtering process in which a
discrete-time sequence of short pulses (ideally impulses) with amplitudes equal
to the signal samples, excites an analog filter.
a) True
b) False
Answer: a
Explanation: The
reconstruction o f the signal from its samples as a linear filtering process in
which a discrete-time sequence of short pulses (ideally impulses) with
amplitudes equal to the signal samples, excites an analog filter.
93. The ideal
reconstruction filter is an ideal low pass filter and its impulse response
extends for all time.
a) True
b) False
Answer: a
Explanation: The ideal
reconstruction filter is an ideal low pass filter and its impulse response
extends for all time. Hence the filter is noncausal and physically
nonrealizable. Although the interpolation filter with impulse response given
can be approximated closely with some delay, the resulting function is still
impractical for most applications where D /A conversion are required.
94. D /A conversion is
usually performed by combining a D /A converter with a sample-and-hold (S/H )
and followed by a low pass (smoothing) filter.
a) True
b) False
Answer: a
Explanation: D /A
conversion is usually performed by combining a D /A converter with a sample-and
hold (S/H) and followed by a low pass (smoothing) filter. T he D /A converter
accepts at its input, electrical signals that correspond to a binary word, and
produces an output voltage or current that is proportional to the value o f the
binary word.
95. The time required
for the output o f the D /A converter to reach and remain within a given
fraction of the final value, after application of the input code word is
called?
a) Converting time
b) Setting time
c) Both a& b
d) None of the mentioned
Answer: b
Explanation: An
important parameter o f a D /A converter is its settling time, which is defined
as the time required for the output o f the D /A converter to reach and remain
within a given fraction (usually,±1/2 LSB) of the final value, after
application of the input code word.
96. In D/A converter,
the application of the input code word results in a high-amplitude transient,
called?
a) Glitch
b) Deglitch
c) Glitter
d) None of the mentioned
Answer: a
Explanation: The
application o f the input code word results in a high-amplitude transient,
called a “glitch.” This is especially the case when two consecutive code words
to the A /D differ by several bits.
97. In a D/A converter,
the usual way to solve the glitch is to use deglitcher. How is the Deglitcher
designed?
a) By using a low pass
filter
b) By using a S/H
circuit
c) Both a& b
d) None of the mentione
Answer: b
Explanation: The usual
way to remedy this problem is to use an S/H circuit designed to serve as a
“deglitcher”. Hence the basic task of the S/H is to hold the output of the D /A
converter constant at the previous output value until the new sample at the
output o f the D /A reaches steady state, and then it samples and holds the new
value in the next sampling interval. Thus the S/H approximates the analog
signal by a series of rectangular pulses whose height is equal to the
corresponding value of the signal pulse.
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