Test: Fourier Transforms Properties - Electronics and Communication Engineering (ECE) MCQ

Explanation: Given x(n)= aⁿu(n), |a|<1
The auto correlation of the above signal is
rxx(l)=1/(1-a² ) a^|l|, -∞< l <∞
According to Wiener-Khintchine Theorem,
S_xx(ω)=F{ r_xx(l)}= [1/(1-a²)].F{a^|l|} = 1/(1-2acosω+a² )

Explanation: 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}

Explanation: 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)= 1/(1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )

Now, X(ω)= X1(ω)+ X2(ω)= 1/(1-ae^(-jω) )+(ae^jω)/(1-ae^jω ) = (1-a²)/(1-2acosω+a²).

Explanation: For the given X(ω)=1/(1-ae^-jω ) ,|a|<1 we obtain
X_I(ω)= (-asinω)/(1-2acosω+a² ) and X_R(ω)= (1-acosω)/(1-2acosω+a² )
We know that |X(ω)|=√(〖X_R (ω)〗²+〖X_I (ω)〗² )
Thus on calculating, we obtain
|X(ω)|= 1/√(1-2acosω+a² )

Explanation: Given, X(ω)= 1/(1-ae^-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_I(ω)= (-asinω)/(1-2acosω+a² ).

Explanation: Given, X(ω)= 1/(1-ae^-jω ) ,|a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a² )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_R(ω)= (1-acosω)/(1-2acosω+a² ).

Explanation: Given

We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that,

Explanation: We know that, if x(n) is a real sequence

If we combine the above two equations, we get
X*(ω)=X(-ω)