This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Round Off Effects in Digital Filters”.
1. The quantization inherent in the finite precision arithmetic operations render the system linear.
a) True
b) False
View Answer
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
View Answer
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
View Answer
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’.
4. Limit cycles in the recursive are directly attributable to 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
View Answer
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
View Answer
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
View Answer
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
View Answer
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)| ≥ \(\frac{(1/2).2^{-b}}{1+|a|}\)
b) |v(n-1)| ≥ \(\frac{(1/2).2^{-b}}{1-|a|}\)
c) |v(n-1)| ≤ \(\frac{(1/2).2^{-b}}{1-|a|}\)
d) None of the mentioned
View Answer
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)| ≤ \((\frac{1}{2}).2^{-b}\)
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)| ≤ \((\frac{1}{2}).2^{-b}\)
and hence
|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)
.
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)
View Answer
Explanation: We know that
|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)
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
View Answer
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
View Answer
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
View Answer
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)
View Answer
Explanation: We know that
|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\)
Given |a|=3/4 and b=4 => |v(n-1)| ≤ 1/8=> The dead band is (-1/8,1/8).
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