tutorial no 2 solution

Department of Electronics and Telecommunication Engineering Yeshwantrao Chavan College of Engineering Nagpur

ET317: Digital signal Processing

Tutorial No: 2: Discrete Time Fourier Transform

©Course Instructors: Prof. V. R. Gupta, E.T. Department, YCCE, Nagpur.

Q. 1) Compute the DFT of each the following finite length sequences considered to be of length N

(Where N is even).

a) [ ] [ ]x n n

b) 0 0[ ] [ ] 0 1x n n n n N

c) 1, 0 1

[ ]0, 0 1

n even n Nx n

n odd n N

d) 1, 0 ( / 2) 1

[ ]0, ( / 2) 1

n Nx n

N n N

e) , 0 1

[ ]0,

na n Nx n

otherwise

Solution: Q. 1a)

We know that the N-point DFT of a Discrete Time Sequence ( )x n is given by the formula

1 2

0

( ) ( ) 0,1,2,........., 1N j kn

N

n

X k x n e for k N

Given: 1 0

( ) ( )0

nx n n

elsewhere

Therefore

1 2

0

( ) ( )

( ) 1 0,1,2,.........., 1

N j knN

n

X k n e

X k for k N

Thus we got the important DFT pair as given below

( )n 1 DFT

N-point





Solution: Q. 1b)


1 2

0

( ) ( ) 0,1,2,........., 1N j kn

N

n

X k x n e for k N

Given: 0

0

1( ) ( )

0

n nx n n n

elsewhere

Therefore

0

1 2

0

2

( ) ( )

( ) 0,1,2,.........., 1

N j knN

n

j knN

X k n e

X k e for k N

Thus we got the important DFT pair as given below

Solution: Q. 1c)

Given: 1, 0 1

[ ]0, 0 1

n even n Nx n

n odd n N


1 2

0

( ) ( ) 0,1,2,........., 1N j kn

N

n

X k x n e for k N

( )n

02j knNe

DFT

N-point





In this question ( )x n is only present for the even values of ‘n’ and the sequence ( )x n is zero for odd values of

‘n’. Hence the length of the sequence is2

N. Thus the formula for computing the DFT of this sequence will be as

follows:

2

122

0

( ) (2 ) 0,1,2,.........., 1N

N

j kn

n

X k x n e for k N

As it is given that, the value of the sequence is 1 for even values of ‘n’ thus the above equation can be

written as

12 4

0

2

( )/

( ) 0,1,2,.........., 1

1( )

1

N

j knN

n

j k

j k N

X k e for k N

eX k

e

Thus, 2 2

0,( )

0

NN kX k

elsewhere

Solution: Q. 1d)

Given: 1, 0 ( / 2) 1

[ ]0, ( / 2) 1

n Nx n

N n N


1 2

0

( ) ( ) 0,1,2,........., 1N j kn

N

n

X k x n e for k N





Thus

12 2

0

( ) 0,1,2,........., 1

N

j knN

n

X k e for k N

(2 )/

1( ) 0,1,2,........., 1

1

j k

j k N

eX k for k N

e

2

(2 )/

0

2( )

1

0 , 0 ( 1)

N

j k N

k

X k k odde

k even k N

Solution: Q. 1e)

Given: , 0 1

[ ]0,

na n Nx n

otherwise


1 2

0

( ) ( ) 0,1,2,........., 1N j kn

N

n

X k x n e for k N

Thus,

1 2

0

( ) 0,1,2,........., 1N j kn

n N

n

X k a e for k N

1 2

0

( ) 0,1,2,........., 1

nN j kN

n

X k ae for k N

2

(2 )/

1( ) 0,1,2,........., 1

1

N j k

j k N

a eX k for k N

ae

(2 )/

1( ) 0,1,2,........., 1

1

N

j k N

aX k for k N

ae





Q. 2) Consider the complex sequence

0 , 0 1[ ]

0,

j ne n N

x notherwise

a) Find the Fourier transform X(ejw

) of x[n].

b) Find the N-point DFT X[k] of the finite-length sequence x[n].

c) Find the DFT of x[n] for the case 00

2 k

N

, where k0 is an integer.

Solution: Q. 2a)

The Fourier transform X(ejw

) of x[n] is given by the formula.

( ) ( )j j n

n

X e x n e

0

1

0

( )N

j nj j n

n

X e e e

0

1( )

0

( )N

j nj

n

X e e

0

0

( )

( )

1( )

1

j Nj

j

eX e

e

0

00

00

0

( )2

( ) ( )2 2

( )/2( )/2

( )/2

1

( )1

Nj

N Nj j

j

jj

j

e

e eX ee

ee

0 0

00

0 0

( ) ( )2 2( )

( )/22( )/2 ( )/2

( )

N Nj j

Nj

jj

j j

e eX e e e

e e





0 0

0

0 0

( ) ( )12 2( )

2

( )/2 ( )/2( )

N Nj jN

jj

j j

e eX e e

e e

0

1 0( )2

0

2sin ( )2

( )( )

2sin2

Nj

j

N

X e e

0

1 0( )2

0

sin ( )2

( )( )

sin2

Nj

j

N

X e e

Solution: Q. 2b)

Given: 0 , 0 1

[ ]0,

j ne n N

x notherwise


1 2

0

( ) ( ) 0 1N j kn

N

n

X k x n e for k N

0

1 2

0

( ) 0 1N j kn

j n N

n

X k e e for k N

0

1 2

0

( ) 0 1

nN j kj N

n

X k e e for k N

0

21

0

( ) 0 1

nkN j

N

n

X k e for k N





0

0

2

2

1( )

1

kj N

N

kj

N

eX k

e

0

00

0 0

0

2

22 2

2 2

2 2

2 2

2

2

1

( )1

k Nj

Nk N k Nj jN N

j k j k

N N

j k

N

e

e eX k

e e

e

0 0

0 0

0 0

2 22 2 2 2

2 2

2 2

2 2

( )

k N k Nj j

k N j k N NjN N

j k j k

N N

e eX k e e

e e

0

2 1 0

2

0

22sin

2( )

1 22sin

2

k Nj

N

k N

NX k e

k

N

0

2 1 0

2

0

2sin

2( )

1 2sin

2

k Nj

N

k N

NX k e

k

N

Note that 2( ) ( )jk

N

X k X e

Solution: Q. 2c)

Here we need to compute the DFT of x[n] for the case 00

2 k

N

, where k0 is an integer. To find this we

need to replace 00

2 k

N

in the solution of Q. 2b)





1 2

0

( ) ( ) 0 1N j kn

N

n

X k x n e for k N

01 2 2

0

( ) 0 1N j k n j kn

N N

n

X k e e for k N

01 2 ( )

0

( ) 0 1N j k k n

N

n

X k e for k N

0

0

2 ( )

2 ( )

1( )

1

j k k

k kj

N

eX k

e

0

2 1( )

02

0

sin ( )( )

sin ( ) /

Nj k k

Nk k

X k ek k N

Q. 3) Find the N-point DFT of the following finite duration sequence of length L (N>=L)

, 0 1[ ]

0,

A n Lx n

otherwise

Solution: Q. 3)

Given: , 0 1

[ ]0,

A n Lx n

otherwise


1 2

0

( ) ( ) 0 1N j kn

N

n

X k x n e for k N

1 2

0

( ) 0 1L j kn

N

n

X k Ae for k N





1 2

0

( )L j kn

N

n

X k A e

2 /

2

1( )

1

j kL N

kj

N

eX k A

e

( 1)/ sin( / )( )

sin( / )

j L N kL NX k Ae

k N

Q. 4) Compute the DFT of the following sequence x(n)= [0, 1, 2, 1] and check the validity of your

answer by calculating its IDFT.

Try to solve this question by your own.

Q. 5) Let ( )jX e denote the Fourier transform of the sequence1

[ ] ( )2

n

x n u n

. Let [ ]y n denote

a finite duration sequence of length 10; i.e., [ ] 0, 0, [ ] 0, 10.y n n and y n n The 10-point

DFT of [ ]y n , denoted by [ ]Y k , corresponds to 10 equally spaced samples of ( )jX e i.e.,

2 /10[ ] ( )j kY k X e . Determine [ ]y n .

Solution: Q. 5)

We know that, Fourier transform of the sequence1

[ ] ( )2

n

x n u n

is given by

( ) ( )j j n

n

X e x n e

0

1( )

2

n

j j n

n

X e e

0

1( )

2

n

j j

n

X e e





1( )

11

2

j

j

X e

e

To find [ ]Y k , will take 10 equally spaced samples of ( )jX e

2 /10( ) ( ) , 0 9j

kY k X e k

Thus we have the 10-point DFT of [ ]y n

(2 /10)

1( ) , 0 9 (1)

11

2

j k

Y k k

e

But here we are asked to find [ ]y n . So we need to take IDFT of ( )Y k . But here I will use the formula of

DFT which will give us the relationship between [ ]y n and ( )Y k

210

9

0

( ) [ ] 0 9 (2)j kn

n

Y k y n e k

Recall that

point DFT

(2 / )

1(3)

1

NNn

j k N

aa

ae

So, by comparing equation number 1, 2 and 3, we may infer that

10

1

2[ ] , 0 9

11

2

n

y n n





Solve the following questions

Q. 1) Determine the 8-point DFT of the sequence 1 0 3

[ ]0

nx n

otherwise

Q. 2) Determine the 8-point DFT of the sequence [ ] {1,2,3,4}x n

Q. 3) Determine the 6-point DFT of the sequence [ ] cos5

nx n

Q. 4) Determine the 4-point DFT of the sequence

1[ ] 2 ( ) 4 ( 1) 2 ( 2) 4 ( 3)x n n n n n

2[ ] ( ) 6 ( 1) 3 ( 3)x n n n n

Unsolved Problems from the book: “Digital signal processing: Principles, Algorithms, and

Applications” by John G. Proakis & D. G. Manolakis, Fourth Edition.

Chapter No. 7

Problem Nos.: 7.1, 7.5, 7.6, 7.12, 7.16, 7.17, 7.23





Some Basic Formulas are required for solving these numerical which are as follows:

2

2

1 2

0

1

0

1 2

0

1

0

: ( ) ( ) 0 1

( ) ( ) 0 1; ,

1: ( ) ( ) 0 1

1( ) ( ) 0 1; ,

jN

jN

N j knN

n

Nkn

N N

n

N j knN

n

Nkn

N N

n

DFT X k x n e for k N

OR

X k x n W for k N where W e

IDFT x n X k e for n NN

OR

x n X k W for k N where W eN

2

1 2

1

1 2

1

1 1

11

Nn N N

N

N N for a

a a afor a

a

1

0

1

11

1

Nn N

N for a

a afor a

a

cos sinie i

2 cos(2 ) sin(2 ) cos(2 ) 1 for all the valuesof

sin(2 ) 0 for all the valuesof

j ke k i k k k

k k

1cos( ) sin( ) cos( )

1

sin( ) 0 for all the valuesof

j kfor k even

e k i k kfor k odd

k k

tutorial no 2 solution

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