# LTI System Frequency Selective Filter MCQ Quiz – Objective Question with Answer for Frequency Selective Filter

1. An ideal filter should have unity gain in its stopband.

A. True
B. False

For an ideal filter, in the magnitude response plot at the stopband, it should have a sudden fall which means an ideal filter should have a zero gain at the stopband.

2. Which filter has a magnitude frequency response as shown in the plot given below?

A. Low pass Filter
B. High pass Filter
C. Bandpass Filter
D. Bandstop Filter

In the magnitude response shown in the question, the system is stopping a particular band of signals. Hence the filter is called a Bandstop filter.

3. An ideal filter should have zero gain in its stopband.

A. True
B. False

For an ideal filter, in the magnitude response plot at the stopband, it should have a sudden fall which means an ideal filter should have a zero gain at the stopband.

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

The time delay taken to reach the output of the system from the input by a signal component is called 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 the frequency of the signal
B. Constant
C. Decreases with the frequency of the signal
D. Independent of frequency of the signal

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

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

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

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.
b0=(1-p)2=(1-0.32)2

=>b0 = 0.46

8. What is the system function for a two-pole bandpass filter that has the center of its passband 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. $$0.15\frac{1-z^{-2}}{1+0.7z^{-2}}$$

B. $$0.15\frac{1+z^{-2}}{1-0.7z^{-2}}$$

C. $$0.15\frac{1-z^{-2}}{1-0.7z^{-2}}$$

D. $$0.15\frac{1+z^{-2}}{1+0.7z^{-2}}$$

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)=$$G\frac{(z-1)(z+1)}{(z-jr)(z+jr)} = G \frac{(z^2-1)}{(z^2+r^2)}$$

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)=$$0.15\frac{1-z^{-2}}{1+0.7z^{-2}}$$

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)

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