51. 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. \(\frac{-asinω}{1-2acosω+a^2}\)
Show Explanation Answer: D
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) = \(\frac{-asinω}{1-2acosω+a^2}\).
52. 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. \(\frac{1}{1+2acosω+a^2}\)
Show Explanation Answer: A
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)| = \(\frac{1}{\sqrt{1-2acosω+a^2}}\).
53. 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. \(\frac{sin(M-\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
Show Explanation Answer: C
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\(\frac{sin(M+\frac{1}{2})ω}{sin(\frac{ω}{2})}\).
54. 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
Show Explanation Answer: B
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) = \(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}} = \frac{1-a^2}{1-2acosω+a^2}\).
55. 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)
Show Explanation Answer: D
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)
56. 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}
Show Explanation Answer: A
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}
57. 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. \(\frac{1}{1+2acosω-a^2}\)
Show Explanation Answer: B
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|} = \(\frac{1}{1-2acosω+a^2}\)
58. If x(n) = Aejωn is the input of an LTI system and h(n) is the response of the system, then what is the output y(n) of the system?
A. H(-ω)x(n)
B. -H(ω)x(n)
C. H(ω)x(n)
D. None of the mentioned
Show Explanation Answer: C
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)
59. If the system gives an output y(n) = H(ω)x(n) with x(n) = Aejωnas input signal, then x(n) is said to be the Eigenfunction of the system.
A. True
B. False
Show Explanation Answer: A
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.
60. What is the output sequence of the system with impulse response h(n) = (1/2)nu(n) when the input of the system is the complex exponential sequence x(n) = Aejnπ/2?
A. \(Ae^{j(\frac{nπ}{2}-26.6°)}\)
B. \(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)
C. \(\frac{2}{\sqrt{5}} Ae^{j({nπ}{2}+26.6°)}\)
D. \(Ae^{j(\frac{nπ}{2}+26.6°)}\)
Show Explanation Answer: B
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) = \(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)