Digital Signal Processing Multiple Choice Questions on “Properties of Fourier Transform for Discrete Time Signals”.
1. If x(n)=xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω)=XR(ω)+jXI(ω), then what is the value of XR(ω)?
A. (sum_{n=0}^∞)xR (n)cosωn-xI (n)sinωn
B. (sum_{n=0}^∞)xR (n)cosωn+xI (n)sinωn
C. (sum_{n=-∞}^∞)xR (n)cosωn+xI (n)sinωn
D. (sum_{n=-∞}^∞)xR (n)cosωn-xI (n)sinωn
Answer: C
Clarification: We know that X(ω)=(sum_{n=-∞}^∞) x(n)e-jωn
By substituting e-jω = cosω – jsinω in the above equation and separating the real and imaginary parts we get
XR(ω)=(sum_{n=-∞}^∞)xR (n)cosωn+xI (n)sinωn
2. If x(n)=xR(n)+jxI(n) is a complex sequence whose Fourier transform is given as X(ω)=XR(ω)+jXI(ω), then what is the value of xI(n)?
A. (frac{1}{2π} int_0^{2π})[XR(ω) sinωn+ XI(ω) cosωn] dω
B. (int_0^{2π})[XR(ω) sinωn+ XI(ω) cosωn] dω
C. (frac{1}{2π} int_0^{2π})[XR(ω) sinωn – XI(ω) cosωn] dω
D. None of the mentioned
Answer: A
Clarification: We know that the inverse transform or the synthesis equation of a signal x(n) is given as
x(n)=(frac{1}{2π} int_0^{2π}) X(ω)ejωn dω
By substituting ejω = cosω + jsinω in the above equation and separating the real and imaginary parts we get
xI(n)=(frac{1}{2π} int_0^{2π})[XR(ω) sinωn+ XI(ω) cosωn] dω
3. If x(n) is a real sequence, then what is the value of XI(ω)?
A. (sum_{n=-∞}^∞ x(n)sin(ωn))
B. –(sum_{n=-∞}^∞ x(n)sin(ωn))
C. (sum_{n=-∞}^∞ x(n)cos(ωn))
D. –(sum_{n=-∞}^∞ x(n)cos(ωn))
Answer: B
Clarification: If the signal x(n) is real, then xI(n)=0
We know that,
XI(ω)=(sum_{n=-∞}^∞ x_R (n)sinωn-x_I (n)cosωn)
Now substitute xI(n)=0 in the above equation=>xR(n)=x(n)
=> XI(ω)=-(sum_{n=-∞}^∞ x(n)sin(ωn)).
4. Which of the following relations are true if x(n) is real?
A. X(ω)=X(-ω)
B. X(ω)=-X(-ω)
C. X*(ω)=X(ω)
D. X*(ω)=X(-ω)
Answer: D
Clarification: We know that, if x(n) is a real sequence
XR(ω)=(sum_{n=-∞}^∞) x(n)cosωn=>XR(-ω)= XR(ω)
XI(ω)=-(sum_{n=-∞}^∞) x(n)sin(ωn)=>XI(-ω)=-XI(ω)
If we combine the above two equations, we get
X*(ω)=X(-ω)
5. 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
Answer: A
Clarification: 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
6. 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
Answer: B
Clarification: 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)=(-2sum_{n=1}^∞ x(n) sinωn)
=>x(n)=-(frac{1}{π} int_0^π)[XI(ω) sinωn] dω
7. 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})
Answer: C
Clarification: 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}).
8. 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})
Answer: D
Clarification: 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}).
9. 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})
Answer: A
Clarification: 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}}).
10. If x(n)=A, -M
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})})
Answer: C
Clarification: 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})}).
11. 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
Answer: B
Clarification: 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, n1(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}).
12. 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)
Answer: D
Clarification: 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)
13. 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}
Answer: A
Clarification: 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}
14. 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})
Answer: B
Clarification: 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})