1. If x(n) is a real signal, then x(n)=\(\frac{1}{π}\int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω.
a) True
b) False
Explanation: We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)
We know that, xR(n)=x(n)=\(\frac{1}{2π}\int_0^{2π}\)[XR(ω) cosωn- XI(ω) sinωn] dω
Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even
=> x(n)=\(\frac{1}{π} \int_0^π\)[XR(ω) cosωn- XI(ω) sinωn] dω
2. If x(n) is a real and odd sequence, then what is the expression for x(n)?
a) \(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω
b) –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω
c) \(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω
d) –\(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω
Explanation: If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently
XR(ω)=0
XI(ω)=\(-2\sum_{n=1}^∞ x(n) sinωn\)
=>x(n)=-\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω
3. What is the value of XR(ω) given X(ω)=\(\frac{1}{1-ae^{-jω}}\),|a|<1?
a) \(\frac{asinω}{1-2acosω+a^2}\)
b) \(\frac{1+acosω}{1-2acosω+a^2}\)
c) \(\frac{1-acosω}{1-2acosω+a^2}\)
d) \(\frac{-asinω}{1-2acosω+a^2}\)
Explanation: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)
This expression can be subdivided into real and imaginary parts, thus we obtain
XR(ω)=\(\frac{1-acosω}{1-2acosω+a^2}\).
4. What is the value of XI(ω) given \(\frac{1}{1-ae^{-jω}}\), |a|<1?
a) \(\frac{asinω}{1-2acosω+a^2}\)
b) \(\frac{1+acosω}{1-2acosω+a^2}\)
c) \(\frac{1-acosω}{1-2acosω+a^2}\)
d) \(\frac{-asinω}{1-2acosω+a^2}\)
Explanation: Given, X(ω)=\(\frac{1}{1-ae^{-jω}}\), |a|<1
By multiplying both the numerator and denominator of the above equation by the complex conjugate of the denominator, we obtain
X(ω)=\(\frac{1-ae^{jω}}{(1-ae^{-jω})(1-ae^{jω})} = \frac{1-acosω-jasinω}{1-2acosω+a^2}\)
This expression can be subdivided into real and imaginary parts, thus we obtain
XI(ω)=\(\frac{-asinω}{1-2acosω+a^2}\).
5. What is the value of |X(ω)| given X(ω)=1/(1-ae-jω), |a|<1?
a) \(\frac{1}{\sqrt{1-2acosω+a^2}}\)
b) \(\frac{1}{\sqrt{1+2acosω+a^2}}\)
c) \(\frac{1}{1-2acosω+a^2}\)
d) \(\frac{1}{1+2acosω+a^2}\)
Explanation: For the given X(ω)=1/(1-ae-jω), |a|<1 we obtain
XI(ω)=(-asinω)/(1-2acosω+a2) and XR(ω)=(1-acosω)/(1-2acosω+a2)
We know that |X(ω)|=\(\sqrt{X_R (ω)^2+X_I (ω)^2}\)
Thus on calculating, we obtain
|X(ω)| = \(\frac{1}{\sqrt{1-2acosω+a^2}}\).
6. If x(n)=A, -M<n<M,; x(n)=0, elsewhere. Then what is the Fourier transform of the signal?
a) A\(\frac{sin(M-\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
b) A2\(\frac{sin(M+\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
c) A\(\frac{sin(M+\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
d) \(\frac{sin(M-\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
Explanation: Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that
\(X(ω)=X_R(ω)=A(1+2∑_{n=1}^∞ cosωn)\)
On simplifying the above equation, we obtain
X(ω)=A\(\frac{sin(M+\frac{1}{2})ω}{sin(\frac{ω}{2})}\).
7. What is the Fourier transform of the signal x(n)=a|n|, |a|<1?
a) \(\frac{1+a^2}{1-2acosω+a^2}\)
b) \(\frac{1-a^2}{1-2acosω+a^2}\)
c) \(\frac{2a}{1-2acosω+a^2}\)
d) None of the mentioned
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)=\(\frac{1}{1-ae^{-jω}}\) and X2(ω)=\(\frac{ae^{jω}}{1-ae^{jω}}\)
Now, X(ω)=X1(ω)+ X2(ω)=\(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}}=\frac{1-a^2}{1-2acosω+a^2}\).
8. If X(ω) is the Fourier transform of the signal x(n), then what is the Fourier transform of the signal x(n-k)?
a) ejωk. X(-ω)
b) ejωk. X(ω)
c) e-jωk. X(-ω)
d) e-jωk. X(ω)
Explanation: 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)
9. What is the convolution of the sequences of x1(n)=x2(n)={1,1,1}?
a) {1,2,3,2,1}
b) {1,2,3,2,1}
c) {1,1,1,1,1}
d) {1,1,1,1,1}
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}
10. What is the energy density spectrum of the signal x(n)=anu(n), |a|<1?
a) \(\frac{1}{1+2acosω+a^2}\)
b) \(\frac{1}{1-2acosω+a^2}\)
c) \(\frac{1}{1-2acosω-a^2}\)
d) \(\frac{1}{1+2acosω-a^2}\)
Explanation: 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|} = \(\frac{1}{1-2acosω+a^2}\)