1.

The quantization inherent in the finite precision arithmetic operations render the
system linear.
a) True
b) False

Answer: b
Explanation: In the realization of a digital filter, either in digital hardware or in software on
a digital computer, the quantization inherent in the finite precision arithmetic operations
render the system linear.
2. In recursive systems, which of the following is caused because of the nonlinearities
due to the finite-precision arithmetic operations?
a) Periodic oscillations in the input
b) Non-Periodic oscillations in the input
c) Non-Periodic oscillations in the output
d) Periodic oscillations in the output

Answer: d
Explanation: In the recursive systems, the nonlinearities due to the finite-precision
arithmetic operations often cause periodic oscillations to occur in the output even when
the input sequence is zero or some non zero constant value.
3. The oscillations in the output of the recursive system are called as ‘limit cycles’.
a) True
b) False

Answer: a
Explanation: In the recursive systems, the nonlinearities due to the finite-precision
arithmetic operations often cause periodic oscillations to occur in the output even when
the input sequence is zero or some non zero constant value. The oscillations thus
produced in the output are known as ‘limit cycles’.
o which of the following?
a) Round-off errors in multiplication
b) Overflow errors in addition
c) Both of the mentioned
d) None of the mentioned

Answer: c
Explanation: The oscillations in the output of the recursive system are called as limit
cycles and are directly attributable to round-off errors in multiplication and overflow
errors in addition.
5. What is the range of values called as to which the amplitudes of the output during a
limit cycle ae confined to?
a) Stop band
b) Pass band
c) Live band
d) Dead band
Answer: d
Explanation: The amplitudes of the output during a limit circle are confined to a range of
values that is called the ‘dead band’ of the filter.
6. Zero input limit cycles occur from non-zero initial conditions with the input x(n)=0.
a) True
b) False

Answer: a
Explanation: When the input sequence x(n) to the filter becomes zero, the output of the
filter then, after a number of iterations, enters into the limit cycle. The output remains in
the limit cycle until another input of sufficient size is applied that drives the system out of
the limit cycle. Similarly, zero input limit cycles occur from non-zero initial conditions with
the input x(n)=0.
7. Which of the following is true when the response of the single pole filter is in the limit
cycle?
a) Actual non-linear system acts as an equivalent non-linear system
b) Actual non-linear system acts as an equivalent linear system
c) Actual linear system acts as an equivalent non-linear system
d) Actual linear system acts as an equivalent linear system

Answer: b
Explanation: We note that when the response of the single pole filter is in the limit cycle,
the actual non-linear system acts as an equivalent linear system with a pole at z=1 when
the pole is positive and z=-1 when the poles is negative.
8. Which of the following expressions define the dead band for a single-pole filter?
a) |v(n-1)| ≥ (1/2).2−b1+|a|
b) |v(n-1)| ≥ (1/2).2−b1−|a|
c) |v(n-1)| ≤ (1/2).2−b1−|a|
d) None of the mentioned

Answer: c
Explanation: Since the quantization product av(n-1) is obtained by rounding, it follows
that the quantization error is bounded as
|Qr[av(n-1)]-av(n-1)| ≤ (12).2−b
Where ‘b’ is the number of bits (exclusive of sign) used in the representation of the pole
‘a’ and v(n). Consequently, we get
|v(n-1)|-|av(n-1)| ≤ (12).2−b
and hence
|v(n-1)| ≤ (12).2−b1−|a|
.
9. What is the dead band of a single pole filter with a pole at 1/2 and represented by 4
bits?
a) (-1/2,1/2)
b) (-1/4,1/4)
c) (-1/8,1/8)
d) (-1/16,1/16)
Answer: d
Explanation: We know that
|v(n-1)| ≤ (12).2−b1−|a|
Given |a|=1/2 and b=4 => |v(n-1)| ≤ 1/16=> The dead band is (-1/16,1/16).
10. The limit cycle mode with zero input, which occurs as a result of rounding the
multiplications, corresponds to an equivalent second order system with poles at z=±1.
a) True
b) False

Answer: a
Explanation: There is an possible limit cycle mode with zero input, which occurs as a
result of rounding the multiplications, corresponds to an equivalent second order system
with poles at z=±1. In this case the two pole filter exhibits oscillations with an amplitude
that falls in the dead band bounded by 2-b/(1-|a1|-a2).
11. What is the necessary and sufficient condition for a second order filter that no
zero-input overflow limit cycles occur?
a) |a1|+|a2|=1
b) |a1|+|a2|>1
c) |a1|+|a2|<1
d) None of the mentioned

Answer: c
Explanation: It can be easily shown that a necessary and sufficient condition for ensuring
that no zero-input overflow limit cycles occur is |a1|+|a2|<1
which is extremely restrictive and hence an unreasonable constraint to impose on any
second order section.
12. An effective remedy for curing the problem of overflow oscillations is to modify the
adder characteristic.
a) True
b) False

Answer: a
Explanation: An effective remedy for curing the problem of overflow oscillations is to
modify the adder characteristic, so that it performs saturation arithmetic. Thus when an
overflow is sensed, the output of the adder will be the full scale value of ±1.
13. What is the dead band of a single pole filter with a pole at 3/4 and represented by 4
bits?
a) (-1/2,1/2)
b) (-1/8,1/8)
c) (-1/4,1/4)
d) (-1/16,1/16)

Answer: b
Explanation: We know that
|v(n-1)| ≤ (12).2−b1−|a|
Given |a|=3/4 and b=4 => |v(n-1)| ≤ 1/8=> The dead band is (-1/8,1/8).
. The decimal numbers represented in the computer are called as floating point
numbers, as the decimal point floats through the number.
a) True
b) False

Answer: a
Explanation: By doing this the computer is capable of accommodating the large float
numbers also.
2. The numbers written to the power of 10 in the representation of decimal numbers are
called as _____
a) Height factors
b) Size factors
c) Scale factors
d) None of the mentioned

Answer: c
Explanation: These are called as scale factors cause they’re responsible in determining
the degree of specification of a number.
3. If the decimal point is placed to the right of the first significant digit, then the number is
called ________
a) Orthogonal
b) Normalized
c) Determinate
d) None of the mentioned

Answer: b
Explanation: None.
4. ________ constitute the representation of the floating number.
a) Sign
b) Significant digits
c) Scale factor
d) All of the mentioned

Answer: d
Explanation: The following factors are responsible for the representation of the number.
5. The sign followed by the string of digits is called as ______
a) Significant
b) Determinant
c) Mantissa
d) Exponent

Answer: c
Explanation: The mantissa also consists of the decimal point.
6. In IEEE 32-bit representations, the mantissa of the fraction is said to occupy ______
bits.
a) 24
b) 23
c) 20
d) 16

Answer: b
Explanation: The mantissa is made to occupy 23 bits, with 8 bit exponent.
7. The normalized representation of 0.0010110 * 2 9 is _______
a) 0 10001000 0010110
b) 0 10000101 0110
c) 0 10101010 1110
d) 0 11110100 11100

Answer: b
Explanation: Normalized representation is done by shifting the decimal point.
8. The 32 bit representation of the decimal number is called as ___________
a) Double-precision
b) Single-precision
c) Extended format
d) None of the mentioned

Answer: b
Explanation: None.
9. In 32 bit representation the scale factor as a range of ________
a) -128 to 127
b) -256 to 255
c) 0 to 255
d) None of the mentioned

Answer: a
Explanation: Since the exponent field has only 8 bits to store the value.
10. In double precision format, the size of the mantissa is ______
a) 32 bit
b) 52 bit
c) 64 bit
d) 72 bit

Answer: b
Explanation: The double precision format is also called as 64 bit representation.
1. For a given number of bits, the power of quantization noise is proportional to the
variance of the signal to be quantized.
a) True
b) False

Answer: a
Explanation: The dynamic range of the signal, which is proportional to its standard
deviation σx, should match the range R of the quantizer, it follows that ∆ is proportional to
σx. Hence for a given number of bits, the power of the quantization noise is proportional
to the variance of the signal to be quantized.
2. What is the variance of the difference between two successive signal samples, d(n) =
x(n) – x(n-1)?
a) σ2d=2σ2x[1+γxx(1)]
b) σ2d=2σ2x[1−γxx(1)]
c) σ2d=4σ2x[1−γxx(1)]
d) σ2d=3σ2x[1−γxx(1)]

Answer: b
Explanation: σ2d=E[d2(n)]=E[x(n)−x(n−1)]2
= E[x2(n)]−2Ex(n)x(n−1)+E[x2(n−1)]
= 2σ2x[1+γxx(1)].
3. What is the variance of the difference between two successive signal samples, d(n) =
x(n)–ax(n-1)?
a) σ2d=2σ2x[1−a2]
b) σ2d=σ2x[1+a2]
c) σ2d=σ2x[1−a2]
d) σ2d=2σ2x[1+a2]

Answer: c
Explanation: An even better approach is to quantize the difference, d(n) = x(n)–ax(n-1),
w here a is a parameter selected to minimize the variance in d(n).
Therefore σ2d=σ2x[1−a2] .
4. If the difference d(n) = x(n)–ax(n-1), then what is the optimum choice for a = ?
a) γxx(1)σ2x
b) γxx(0)σ2x
c) γxx(0)σ2d
d) γxx(1)σ2d

Answer: a
Explanation: An even better approach is to quantize the difference, d(n) = x(n)–ax(n-1),
w here a is a parameter selected to minimize the variance in d(n). This leads to the
result that the optimum choice of a is γxx(1)γxx(0)=γxx(1)σ2x.
5. What is the quantity ax(n-1) is called?
a) Second-order predictor of x(n)
b) Zero-order predictor of x(n)
c) First-order predictor of x(n)
d) Third-order predictor of x(n)
View Answer
Answer: c
Explanation: In the equation d(n) = x(n)–ax(n-1), the quantity ax(n-1) is called a
First-order predictor of x(n).
6. The differential predictive signal quantizer system is known as?
a) DCPM
b) DMPC
c) DPCM
d) None of the mentioned
Answer: c
Explanation: A differential predictive signal quantizer system. This system is used in
speech encoding and transmission over telephone channels and is known as differential
pulse code modulation (DPCM).
7. What is the expansion of DPCM?
a) Differential Pulse Code Modulation
b) Differential Plus Code Modulation
c) Different Pulse Code Modulation
d) None of the mentioned

Answer: a
Explanation: A differential predictive signal quantizer system. This system is used in
speech encoding and transmission over telephone channels and is known as differential
pulse code modulation (DPCM ).
8. What are the main uses of DPCM?
a) Speech Decoding and Transmission over mobiles
b) Speech Encoding and Transmission over mobiles
c) Speech Decoding and Transmission over telephone channels
d) Speech Encoding and Transmission over telephone channels

Answer: d
Explanation: A differential predictive signal quantizer system. This system is used in
speech encoding and transmission over telephone channels and is known as differential
pulse code modulation (DPCM ).

9. To reduce the dynamic range of the difference signal d(n) = x(n) – x^(n), thus a
predictor of order p has the form?
a) x^(n)=∑pk=1akx(n+k)
b) x^(n)=∑pk=1akx(n−k)
c) x^(n)=∑pk=0akx(n+k)
d) x^(n)=∑pk=0akx(n−k)

Answer: b
Explanation: The goal of the predictor is to provide an estimate x^(n) of x(n) from a
linear combination of past values of x(n), so as to reduce the dynamic range of the
difference signal d(n) = x(n)-x^(n). Thus a predictor of order p has the
form x^(n)=∑pk=1akx(n−k).
10. The simplest form of differential predictive quantization is called?
a) AM
b) BM
c) DM
d) None of the mentioned

Answer: c
Explanation: The simplest form of differential predictive quantization is called delta
modulation (DM).
11. What is the abbreviation of DM?
a) Diameter Modulation
b) Distance Modulation
c) Delta Modulation
d) None of the mentioned

Answer: c
Explanation: The simplest form of differential predictive quantization is called delta
modulation (DM).
12. In DM, the quantizer is a simple ________ bit and ______ level quantizer.
a) 2-bit, one-level
b) 1-bit, two-level
c) 2-bit, two level
d) 1-bit, one level

Answer: b
Explanation: The simplest form of differential predictive quantization is called delta
modulation (DM). In DM, the quantizer is a simple 1-bit (two-level) quantizer.
13. In DM, What is the order of predictor is used?
a) Zero-order predictor
b) Second-order predictor
c) First-order predictor
d) Third-order predictor

Answer: c
Explanation: In DM, the quantizer is a simple 1-bit (two-level) quantizer and the predictor
is a first-order predictor.
14. In the equation xq(n)=axq(n-1)+dq(n), if a = 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator

Answer: d
Explanation: In the equation xq(n)=axq(n-1)+dq(n), if a = 1, we have an ideal accumulator
(integrator).
15. In the equation xq(n)=axq(n-1)+dq(n), if a < 1 then integrator is called?
a) Leaky integrator
b) Ideal integrator
c) Ideal accumulator
d) Both Ideal integrator & accumulator

Answer: a
Explanation: In the equation xq(n)=axq(n-1)+ dq(n), a < 1 results in a ”leaky integrator”.
1. If the input analog signal is within the range of the quantizer, the quantization error
eq (n) is bounded in magnitude i.e., |eq (n)| < Δ/2 and the resulting error is called?
a) Granular noise
b) Overload noise
c) Particulate noise
d) Heavy noise
Answer: a
Explanation: In the statistical approach, we assume that the quantization error is random
in nature. We model this error as noise that is added to the original (unquantized) signal.
If the input analog signal is within the range of the quantizer, the quantization error eq (n)
is bounded in magnitude
i.e., |eq (n)| < Δ/2 and the resulting error is called Granular noise.
2. If the input analog signal falls outside the range of the quantizer (clipping), eq (n)
becomes unbounded and results in _____________
a) Granular noise
b) Overload noise
c) Particulate noise
d) Heavy noise

Answer: b
Explanation: In the statistical approach, we assume that the quantization error is random
in nature. We model this error as noise that is added to the original (unquantized) signal.
If the input analog signal falls outside the range of the quantizer (clipping), eq (n)
becomes unbounded and results in overload noise.
3. In the mathematical model for the quantization error eq (n), to carry out the analysis,
what are the assumptions made about the statistical properties of eq (n)?
i. The error eq (n) is uniformly distributed over the range — Δ/2 < eq (n) < Δ/2.
ii. The error sequence is a stationary white noise sequence. In other words, the error
eq (m) and the error eq (n) for m≠n are uncorrelated.
iii. The error sequence {eq (n)} is uncorrelated with the signal sequence x(n).
iv. The signal sequence x(n) is zero mean and stationary.
a) i, ii & iii
b) i, ii, iii, iv
c) i, iii
d) ii, iii, iv

Answer: b
Explanation: In the mathematical model for the quantization error eq (n). To carry out the
analysis, the following are the assumptions made about the statistical properties of
eq (n).
i. The error eq (n) is uniformly distributed over the range — Δ/2 < eq (n) < Δ/2.
ii. The error sequence is a stationary white noise sequence. In other words, the error
eq (m)and the error eq (n) for m≠n are uncorrelated.
iii. The error sequence {eq (n)} is uncorrelated with the signal sequence x(n).
iv. The signal sequence x(n) is zero mean and stationary.
4. What is the abbreviation of SQNR?
a) Signal-to-Quantization Net Ratio
b) Signal-to-Quantization Noise Ratio
c) Signal-to-Quantization Noise Region
d) Signal-to-Quantization Net Region

Answer: b
Explanation: The effect of the additive noise eq (n) on the desired signal can be
quantified by evaluating the signal-to-quantization noise (power) ratio (SQNR).
5. What is the scale used for the measurement of SQNR?
a) DB
b) db
c) dB
d) All of the mentioned

Answer: c
Explanation: The effect of the additive noise eq (n) on the desired signal can be
quantified by evaluating the signal-to-quantization noise (power) ratio (SQNR), which
can be expressed on a logarithmic scale (in decibels or dB).
6. What is the expression for SQNR which can be expressed in a logarithmic scale?
a) 10 log10PxPn
b) 10 log10PnPx
c) 10 log2PxPn
d) 2 log2PxPn

Answer: a
Explanation: The signal-to-quantization noise (power) ratio (SQNR), which can be
expressed on a logarithmic scale (in decibels or dB) :
SQNR = 10 log10PxPn.

7. In the equation SQNR = 10 log10PxPn. what are the terms Px and Pn are called ___
respectively.
a) Power of the Quantization noise and Signal power
b) Signal power and power of the quantization noise
c) None of the mentioned
d) All of the mentioned

Answer: b
Explanation: In the equation SQNR = 10log10PxPn then the terms Px is the signal power
and Pn is the power of the quantization noise

8. In the equation SQNR = 10 log10PxPn, what are the expressions of Px and Pn?
⁡
a) Px=σ2=E[x2(n)]andPn=σ2e=E[e2q(n)]
b) Px=σ2=E[x2(n)]andPn=σ2e=E[e3q(n)]
c) Px=σ2=E[x3(n)]andPn=σ2e=E[e2q(n)]
d) None of the mentioned

Answer: a
Explanation: In the equation SQNR = 10log10PxPn, then the
terms Px=σ2=E[x2(n)] and Pn=σ2e=E[e2q(n)].
9. If the quantization error is uniformly distributed in the range (-Δ/2, Δ/2), the mean
value of the error is zero then the variance Pn is?
a) Pn=σ2e=Δ2/12
b) Pn=σ2e=Δ2/6
c) Pn=σ2e=Δ2/4
d) Pn=σ2e=Δ2/2
Answer: a
Explanation: Pn=σ2e=∫Δ/2−Δ/2e2p(e)de=1/Δ∫Δ2−Δ2e2de=Δ212.

10. By combining Δ=R2b+1 with Pn=σ2e=Δ2/12 and substituting the result into SQNR =
10 log10PxPn, what is the final expression for SQNR = ?
a) 6.02b + 16.81 + 20log10Rσx
b) 6.02b + 16.81 – 20log10Rσx
c) 6.02b – 16.81 – 20log10Rσx
d) 6.02b – 16.81 – 20log10Rσx

Answer: b
Explanation: SQNR = 10log10PxPn=20log10σxσe
= 6.02b + 16.81 – 20log10RσxdB.
⁡
11. In the equation SQNR = 6.02b + 16.81 – 20log10Rσx, for R = 6σx the equation
becomes?
a) SQNR = 6.02b-1.25 dB
b) SQNR = 6.87b-1.55 dB
c) SQNR = 6.02b+1.25 dB
d) SQNR = 6.87b+1.25 dB

Answer: c
Explanation: For example, if we assume that x(n) is Gaussian distributed and the range
o f the quantizer extends from -3σx to 3σx (i.e., R = 6σx), then less than 3 out o f every
1000 input signal amplitudes would result in an overload on the average. For R = 6σx,
then the equation becomes
SQNR = 6.02b+1.25 dB.