Fourier Transforms Properties MCQ Quiz – Objective Question with Answer for Fourier Transforms Properties

First we observe x(n) can be expressed as

x(n)=x1(n)+x2(n)

where
x1(n)= an, n>0

=0, elsewhere

x2(n)=a-n, n<0

=0, elsewhere

Now applying Fourier transform for the above two signals, we get

X1(ω)=\(\frac{1}{1-ae^{-jω}}\) and X2(ω)=\(\frac{ae^{jω}}{1-ae^{jω}}\)

X(ω)=X1(ω)+ X2(ω)=\(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}}=\frac{1-a^2}{1-2acosω+a^2}\).

Given

F{x(n)}= X(ω)=\(\sum_{n=-∞}^∞ x(n)e^{-jωn}\)

=>F{x(n-k)}=\(\sum_{n=-∞}^∞ x(n-k)e^{-jωn}=e^{-jωk}.\sum_{n=-∞}^∞ x(n-k)e^{-jω(n-k)}\)

=>F{x(n-k)}= e-jωk. X(ω)

Given x1(n)=x2(n)={1,1,1}
By calculating the Fourier transform of the above two signals, we get
X1(ω)= X2(ω)=1+ ejω + e-jω = 1+2cosω
From the convolution property of Fourier transform we have,
X(ω)= X1(ω). X2(ω)=(1+2cosω)2=3+4cosω+2cos2ω
By applying the inverse Fourier transform of the above signal, we get
x1(n)*x2(n)={1,2,3,2,1}

Given x(n)= anu(n), |a|<1

The auto correlation of the above signal is

rxx(l)=\(\frac{1}{1-a^2}\) a|l|, -∞< l <∞

According to Wiener-Khintchine Theorem,

Sxx(ω)=F{rxx(l)}=\([\frac{1}{1-a^2}]\).F{a|l|} = \(\frac{1}{1-2acosω+a^2}\)

If x(n)= Aejωn is the input and h(n) is the response o the system, then we know that

y(n)=\(\sum_{k=-∞}^∞ h(k)x(n-k)\)

=>y(n)=\(\sum_{k=-∞}^∞ h(k)Ae^{jω(n-k)}\)

= A \([\sum_{k=-∞}^∞ h(k) e^{-jωk}] e^{jωn}\)

= A. H(ω). ejωn

= H(ω)x(n)

An Eigenfunction of a system is an input signal that produces an output that differs from the input by a constant multiplicative factor known as the Eigenvalue of the system.

First we evaluate the Fourier transform of the impulse response of the system h(n)

H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-1/2 e^{-jω}}\)

At ω=π/2, the above equation yields,

H(π/2)=\(\frac{1}{1+j 1/2}=\frac{2}{\sqrt{5}} e^{-j26.6°}\)

We know that if the input signal is a complex exponential signal, then y(n)=x(n) . H(ω)

=>y(n)=\(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)

H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-\frac{1}{2} e^{-jω}}\)

At ω=π, the above equation yields,

H(π)=\(\frac{1}{1+\frac{1}{2}}\)=2/3

If the input signal is a complex exponential signal, then the input is known as the Eigen function and H(ω) is called the Eigenvalue of the system. So, the Eigenvalue of the system mentioned above is 2/3.

From the definition of H(ω), we have

H(ω)=\(\sum_{k=-∞}^∞h(k) e^{-jωk}\)

=\(\sum_{k=-∞}^∞h(k) cos⁡ωk-j\sum_{k=-∞}^∞h(k) sin⁡ωk\)

= HR(ω)+j HI(ω)

=> HI(ω)=-\(\sum_{k=-∞}^∞h(k) sin⁡ωk\)

If h(n) is the real valued impulse response sequence of an LTI system, then H(ω) can be represented as HR(ω)+j HI(ω).

=> tanθ=\(\frac{H_I (ω)}{H_R (ω)}\) => Phase of H(ω)=\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

For a three point moving average system, we can define the output of the system as

y(n)=\(\frac{1}{3}[x(n+1)+x(n)+x(n-1)]=>h(n)=\{\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\}\)

it follows that H(ω)=\(\frac{1}{3}(e^{jω}+1+e^{-jω})=\frac{1}{3}(1+2cosω)\)

=>| H(ω)|=\(\frac{1}{3}\)|1+2cosω|

The frequency response of the system is

H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-\frac{1}{2} e^{-jω}}\)

For first term, ω=0=>H(0)=2

For second term, ω=π/2=>H(π/2)=\(\frac{1}{1+j\frac{1}{2}} = \frac{2}{\sqrt{5}} e^{-j26.6°}\)

For third term, ω=π=> H(π)=\(\frac{1}{1+\frac{1}{2}}\) = 2/3

Hence the response of the system to x(n) is

y(n)=20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.6^0)+ \frac{40}{3}cosπn\)

Given y(n)=ay(n-1)+bx(n)

=>H(ω)=\(\frac{|b|}{1-ae^{-jω}}\)

By calculating the magnitude of the above equation we get

|H(ω)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

We know that,

|H(ω)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)
Since the parameter ‘a’ is positive, the denominator of |H(ω)| becomes minimum at ω=0. So, |H(ω)| attains its maximum value at ω=0. At this frequency we have,

\(\frac{|b|}{1-a}\) = 1 => b=±(1-A..