Fourier Transform Properties for Discrete Time Signal MCQ

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)=\(-2\sum_{n=1}^∞ x(n) sin⁡ωn\)

=>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

\(X(ω)=X_R(ω)=A(1+2∑_{n=1}^∞ cos⁡ωn)\)

On simplifying the above equation, we obtain

X(ω)=A\(\frac{sin⁡(M+\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}\).

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