Signal and System Frequency Analysis MCQ Quiz – Objective Question with Answer for Signal and System Frequency Analysis

Given H(z) = \(\frac{1}{1-0.8z^{-1}}\) = z/(z-0.8)
Clearly, H(z) has a zero at z = 0 and a pole at p = 0.8. Hence the frequency response of the system is given as

H(ω) = \(\frac{e^{jω}}{e^{jω}-0.8}\).

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.

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.

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.

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.

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.

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

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

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

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).

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).