# 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

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

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. –$$\sum_{n=-∞}^∞ x(n)cos⁡(ωn)$$

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)

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

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

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. $$\frac{-asinω}{1-2acosω+a^2}$$

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. $$\frac{-asinω}{1-2acosω+a^2}$$

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. $$\frac{1}{1+2acosω+a^2}$$

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. $$\frac{sin⁡(M-\frac{1}{2})ω}{sin⁡(\frac{ω}{2})}$$

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