1 Crore+ students have signed up on EduRev. Have you? 
What is the energy density spectrum of the signal x(n)=a^{n}u(n), a<1?
Explanation: Given x(n)= a^{n}u(n), a<1
The auto correlation of the above signal is
rxx(l)=1/(1a^{2} ) a^{l}, ∞< l <∞
According to WienerKhintchine Theorem,
S_{xx}(ω)=F{ r_{xx}(l)}= [1/(1a^{2})].F{a^{l}} = 1/(12acosω+a^{2} )
What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?
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}
What is the Fourier transform of the signal x(n)=a^{n}, a<1?
Explanation: First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= a^{n}, 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/(1ae^{jω})/((1ae^{(jω)} )(1ae^{jω})) = (1acosωjasinω)/(12acosω+a^{2} )
Now, X(ω)= X1(ω)+ X2(ω)= 1/(1ae^(jω) )+(ae^jω)/(1ae^jω ) = (1a^{2})/(12acosω+a^{2}).
If x(n)=A, M<n<M, then what is the Fourier transform of the signal?
=0, elsewhere
Explanation: Clearly, x(n)=x(n). Thus the signal x(n) is real and even signal. So, we know that
What is the value of X(ω) given X(ω)=1/(1ae^{jω} ) ,a<1?
Explanation: For the given X(ω)=1/(1ae^{jω} ) ,a<1 we obtain
X_{I}(ω)= (asinω)/(12acosω+a^{2} ) and X_{R}(ω)= (1acosω)/(12acosω+a^{2} )
We know that X(ω)=√(〖X_R (ω)〗^{2}+〖X_I (ω)〗^{2} )
Thus on calculating, we obtain
X(ω)= 1/√(12acosω+a^{2} )
What is the value of X_{I}(ω) given X(ω)=1/(1ae^{jω} ) ,a<1?
Explanation: Given, X(ω)= 1/(1ae^{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)= (1ae^{jω})/((1ae^{(jω)} )(1ae^{jω})) = (1acosωjasinω)/(12acosω+a^{2} )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_{I}(ω)= (asinω)/(12acosω+a^{2} ).
What is the value of X_{R}(ω) given X(ω)=1/(1ae^{jω} ) ,a<1?
Explanation: Given, X(ω)= 1/(1ae^{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)= (1ae^{jω})/((1ae^{(jω)} )(1ae^{jω})) = (1acosωjasinω)/(12acosω+a^{2} )
This expression can be subdivided into real and imaginary parts, thus we obtain
X_{R}(ω)= (1acosω)/(12acosω+a^{2} ).
If X(ω) is the Fourier transform of the signal x(n), then what is the Fourier transform of the signal x(nk)?
Explanation: Given
We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that,
Which of the following relations are true if x(n) is real?
Explanation: We know that, if x(n) is a real sequence
If we combine the above two equations, we get
X*(ω)=X(ω)
40 videos42 docs33 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
40 videos42 docs33 tests









