This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Properties of Fourier Transform for Discrete Time Signals”.

1. If x(n)=x_{R}(n)+jx_{I}(n) is a complex sequence whose Fourier transform is given as X(ω)=X_{R}(ω)+jX_{I}(ω), then what is the value of X_{R}(ω)?

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Explanation: We know that

By substituting e-jω = cosω – jsinω in the above equation and separating the real and imaginary parts we get

2. If x(n)=x_{R}(n)+jx_{I}(n) is a complex sequence whose Fourier transform is given as X(ω)=X_{R}(ω)+jX_{I}(ω), then what is the value of x_{I}(n) ?

d) None of the mentioned

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Explanation: We know that the inverse transform or the synthesis equation of a signal x(n) is given as

By substituting ejω = cosω + jsinω in the above equation and separating the real and imaginary parts we get

3. If x(n) is a real sequence, then what is the value of X_{I}(ω)?

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Explanation: If the signal x(n) is real, then x

_{I}(n)=0

We know that,

Now substitute xI(n)=0 in the above equation=>xR(n)=x(n)

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)

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Explanation: We know that, if x(n) is a real sequence

If we combine the above two equations, we get

X*(ω)=X(-ω)

a) True

b) False

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Explanation: We know that if x(n) is a real signal, then xI(n)=0 and xR(n)=x(n)

We know that,

Since both XR(ω) cosωn and XI(ω) sinωn are even, x(n) is also even

=>

6. If x(n) is a real and odd sequence, then what is the expression for x(n)?

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Explanation: If x(n) is real and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently

XR(ω)=0

7. What is the value of X_{R}(ω) given X(ω)=1/(1-ae^{-jω} ) ,|a|<1?
a) asinω/(1-2acosω+a^{2} )

b) (1+acosω)/(1-2acosω+a^{2} )

c) (1-acosω)/(1-2acosω+a^{2} )

d) (-asinω)/(1-2acosω+a^{2} )

View Answer

Explanation: Given, X(ω)= 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)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a^2 ) This expression can be subdivided into real and imaginary parts, thus we obtain X

_{R}(ω)= (1-acosω)/(1-2acosω+a

^{2}).

8. What is the value of X_{I}(ω) given X(ω)=1/(1-ae^{-jω} ) ,|a|<1?
a) asinω/(1-2acosω+a^{2} )

b) (1+acosω)/(1-2acosω+a^{2} )

c) (1-acosω)/(1-2acosω+a^{2} )

d) (-asinω)/(1-2acosω+a^{2} )

View Answer

Explanation: Given, X(ω)= 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)= (1-ae^jω)/((1-ae^(-jω) )(1-ae^jω)) = (1-acosω-jasinω)/(1-2acosω+a^2 ) This expression can be subdivided into real and imaginary parts, thus we obtain X

_{I}(ω)= (-asinω)/(1-2acosω+a

^{2}).

9. What is the value of |X(ω)| given X(ω)=1/(1-ae^{-jω} ) ,|a|<1?
a) 1/√(1-2acosω+a^{2} )

b) 1/√(1+2acosω+a^{2})

c) 1/(1-2acosω+a^{2} )

d) 1/(1+2acosω+a^{2} )

View Answer

Explanation: For the given X(ω)=1/(1-ae

^{-jω}) ,|a|<1 we obtain X

_{I}(ω)= (-asinω)/(1-2acosω+a

^{2}) and X

_{R}(ω)= (1-acosω)/(1-2acosω+a

^{2})

We know that |X(ω)|=√(〖X_R (ω)〗^2+〖X_I (ω)〗^2 )

Thus on calculating, we obtain

|X(ω)|= 1/√(1-2acosω+a

^{2})

10. If x(n)=A, -M

c) Asin[(M+1/2)ω]/sin[(ω/2)]
d) sin[(M+1/2)ω]/sin(ω/2)

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Explanation: Clearly, x(n)=x(-n). Thus the signal x(n) is real and even signal. So, we know that

11. What is the Fourier transform of the signal x(n)=a^{|n|}, |a|<1?
a) (1+a^{2})/(1-2acosω+a^{2})

b) (1-a^{2})/(1-2acosω+a^{2})

c) 2a/(1-2acosω+a^{2} )

d) None of the mentioned

View Answer

Explanation: First we observe x(n) can be expressed as

x(n)=x1(n)+x2(n)

where x1(n)= a

^{n}, 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)= 1/(1-ae^(-jω) ) and X2(ω)= (ae^jω)/(1-ae^jω )
Now, X(ω)= X1(ω)+ X2(ω)= 1/(1-ae^(-jω) )+(ae^jω)/(1-ae^jω ) = (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) e^{jωk}. X(-ω)

b) e^{jωk}. X(ω)

c) e^{-jωk}. X(-ω)

d) e^{-jωk}. X(ω)

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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}

View Answer

Explanation: 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)=a^{n}u(n), |a|<1?
a) 1/(1+2acosω+a^{2} )

b) 1/(1-2acosω+a^{2} )

c) 1/(1-2acosω-a^{2} )

d) 1/(1+2acosω-a^{2} )

View Answer

Explanation: Given x(n)= a

^{n}u(n), |a|<1 The auto correlation of the above signal is rxx(l)=1/(1-a

^{2}) a

^{|l|}, -∞< l <∞ According to Wiener-Khintchine Theorem, S

_{xx}(ω)=F{ r

_{xx}(l)}= [1/(1-a

^{2})].F{a

^{|l|}} = 1/(1-2acosω+a

^{2})

**Sanfoundry Global Education & Learning Series – Digital Signal Processing.**

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