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
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
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
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
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
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
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
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
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
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
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<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. \(\frac{sin(M-\frac{1}{2})ω}{sin(\frac{ω}{2})}\)
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\(\frac{sin(M+\frac{1}{2})ω}{sin(\frac{ω}{2})}\).