Digital Signal Processing Multiple Choice Questions on “LTI System as Frequency Selective Filters”.
1. An ideal filter should have unity gain in their stop band.
A. True
B. False
Answer: B
Clarification: For an ideal filter, in the magnitude response plot at the stop band it should have a sudden fall which means an ideal filter should have a zero gain at stop band.
2. Which filter has a magnitude frequency response as shown in the plot given below?
A. Low pass Filter
B. High pass Filter
C. Band pass Filter
D. Band stop Filter
Answer: D
Clarification: In the magnitude response shown in the question, the system is stopping a particular band of signals. Hence the filter is called as Band stop filter.
3. An ideal filter should have zero gain in their stop band.
A. True
B. False
Answer: A
Clarification: For an ideal filter, in the magnitude response plot at the stop band it should have a sudden fall which means an ideal filter should have a zero gain at stop band.
4. The ‘Envelope delay’ or ‘Group delay’ is the time delay that the signal component of frequency ω undergoes as it passes from the input to the output of the system.
A. True
B. False
Answer: A
Clarification: The time delay taken to reach the output of the system from the input by a signal component is called as envelope delay or group delay.
5. If the phase ϴ(ω) of the system is linear, then the group delay of the system?
A. Increases with frequency of signal
B. Constant
C. Decreases with frequency of signal
D. Independent of frequency of signal
Answer: B
Clarification: We know that the group delay of the system with phase ϴ(ω) is defined as
Tg(ω)=(frac{dϴ(ω)}{dω})
Given the phase is linear=> the group delay of the system is constant.
6. A two pole low pass filter has a system function H(z)=(frac{b_0}{(1-pz^{-1})^2}), What is the value of ‘p’ such that the frequency response H(ω) satisfies the condition |H(π/4)|2=1/2 and H(0)=1?
A. 0.46
B. 0.38
C. 0.32
D. 0.36
Answer: C
Clarification: Given
H(z)=(frac{b_0}{(1-pz^{-1})^2}) and we know that z=rejω. Here in this case r=1. So z=ejω.
Given at ω=0, H(0)=1=>b0=(1-p)2
Given at ω=π/4, |H(π/4)|2=1/2
=>(frac{(1-p)^2}{(1-pe^{-jπ/4})^2}) = 1/2
=>(frac{(1-p)^4}{((1-p/sqrt 2)^2+p^2/2)^2}) = 1/2
=> √2(1-p)2=1+p2-√2p
Upon solving the above quadratic equation, we get the value of p as 0.32.
7. A two pole low pass filter has a system function H(z)=(frac{b_0}{(1-pz^{-1})^2}), What is the value of ‘b0‘ such that the frequency response H(ω) satisfies the condition |H(π/4)|2=1/2 and H(0)=1?
A. 0.36
B. 0.38
C. 0.32
D. 0.46
Answer: D
Clarification: Given
H(z)=(frac{b_0}{(1-pz^{-1})^2}) and we know that z=rejω. Here in this case r=1. So z=ejω.
Given at ω=0, H(0)=1=>b0=(1-p)2
Given at ω=π/4, |H(π/4)|2=1/2
=>(frac{(1-p)^2}{(1-pe^{-jπ/4})^2}) = 1/2
=>(frac{(1-p)^4}{((1-p/sqrt 2)^2+p^2/2)^2}) = 1/2
=> √2(1-p)2=1+p2-√2p
Upon solving the above quadratic equation, we get the value of p as 0.32.
Already we have
b0=(1-p)2=(1-0.32)2
=>b0 = 0.46
8. What is the system function for a two pole band pass filter that has the centre of its pass band at ω=π/2, zero its frequency response characteristic at ω=0 and at ω=π, and its magnitude response is 1/√2 at ω=4π/9?
A. (0.15frac{1-z^{-2}}{1+0.7z^{-2}})
B. (0.15frac{1+z^{-2}}{1-0.7z^{-2}})
C. (0.15frac{1-z^{-2}}{1-0.7z^{-2}})
D. (0.15frac{1+z^{-2}}{1+0.7z^{-2}})
Answer: A
Clarification: Clearly, the filter must have poles at P1,2=re±jπ/2 and zeros at z=1 and z=-1. Consequently the system function is
H(z)=(Gfrac{(z-1)(z+1)}{(z-jr)(z+jr)} = G frac{(z^2-1)}{(z^2+r^2)})
The gain factor is determined by evaluating the frequency response H(ω) of the filer at ω=π/2. Thus we have,
H(π/2) = (G frac{2}{1-r^2} = 1=>G = frac{1-r^2}{2})
The value of r is determined by evaluating the H(ω) at ω=4π/9. Thus we have
|H(4π/9)|2=(frac{(1-r^2)^2}{4}frac{2-2cos(8π/9)}{1+r^4+2r^2 cos(8π/9)})=1/2
On solving the above equation, we get r2=0.7.Therefore the system function for the desired filter is
H(z)=(0.15frac{1-z^{-2}}{1+0.7z^{-2}})
9. If hlp(n) denotes the impulse response of a low pass filter with frequency response Hlp(ω), then what is the frequency response of the high pass filter in terms of Hlp(ω)?
A. Hlp(ω-π/2)
B. Hlp(ω+π/2)
C. Hlp(ω-π)
D. Hlp(ω+π)
Answer: C
Clarification: The impulse response of a high pass filter is simply obtained from the impulse response of the low pass filter by changing the signs of the odd numbered samples in hlp(n). Thus
hhp(n)=(-1)n hlp(n)=(ejπ)n hlp(n)
Thus the frequency response of the high pass filter is obtained as Hlp(ω-π).
10. If the low pass filter described by the difference equation y(n)=0.9y(n-1)+0.1x(n) is converted into a high pass filter, then what is the frequency response of the high pass filter?
A. 0.1/(1+0.9ejω)
B. 0.1/(1+0.9e-jω)
C. 0.1/(1-0.9ejω)
D. None of the mentioned
Answer: B
Clarification: The difference equation for the high pass filter is
y(n)=-0.9y(n-1)+0.1x(n)
and its frequency response is given as
H(ω)=0.1/(1+0.9e-jω).
11. A digital resonator is a special two pole band pass filter with the pair of complex conjugate poles located near the unit circle.
A. True
B. False
Answer: A
Clarification: The magnitude response of a band pass filter with two complex poles located near the unit circle is as shown below.
The filter gas a large magnitude response at the poles and hence it is called as digital resonator.