M. J. Roberts - 10/15/06

Solutions 11-1

Chapter 11 - The Discrete-Time FourierTransform

Solutions

DTFT Direct from Definition

1. From the definition, find the DTFT of

x n = 10rect

4n .

and compare with the Fourier transform table in Appendix E.

X F( ) = x n ej2 Fn

n=

= 10rect4

n ej2 Fn

n=

= 10 ej2 Fn

n= 4

4

X F( ) = 10 ej2 F m 4( )

m=0

8

= 10ej8 F

ej2 Fm

m=0

8

= 10ej8 F 1 e

j18 F

1 ej2 F

X F( ) = 10ej8 F e

j9 F

ej F

ej9 F

ej9 F

ej F

ej F

= 10sin 9 F( )sin F( )

= 90drcl F ,9( )

From the table,

rect

Nw

nF

2Nw

+ 1( )drcl F ,2Nw

+ 1( )

rect

4n

F9drcl F ,9( )

10rect

4n

F90drcl F ,9( )

Check

2. From the definition, derive a general expression for the F and forms of theDTFT of functions of the form,

x n = Asin 2 F

0n( ) = Asin

0n( ) .

M. J. Roberts - 10/15/06

Solutions 11-2

(It should remind you of the CTFT of x t( ) = Asin 2 f

0t( ) = Asin

0t( ) .) Compare

with the Fourier transform table in Appendix E.

X F( ) = x n ej2 Fn

n=

= Asin 2 F0n( )e

j2 Fn

n=

= Ae

j2 F0n

ej2 F

0n

j2e

j2 Fn

n=

X F( ) =A

j2e

j2 F0

F( )ne

j2 F0

+ F( )n

n=

Then, using

ej2 xn

n=

=1

x( )

we get

X F( ) = A / j2( ) 1

F0

F( ) 1F

0F( ) = A / j2( ) 1

F0

F( ) +1

F0

F( )

X F( ) = jA / 2( ) 1

F + F0( ) 1

F F0( )

The form can be found by the transformation, F / 2 .

X j( ) = jA / 2( ) 1

/ 2 +0

/ 2( ) 1/ 2

0/ 2( )

X j( ) = j A

2+

0( ) 2 0( )

DTFT Using Tables and Transforms

3. A DT signal is defined by

x n = sinc n / 8( ) .

Sketch the magnitude and phase of the DTFT of x n 2 .

From the table of transform pairs,

sinc n / w( ) F

wrect wF( ) 1F( )

sinc n / 8( ) F

8rect 8F( ) 1F( )

sincn 2

8

F8rect 8F( ) 1

F( ) ej4 F

M. J. Roberts - 10/15/06

Solutions 11-3

sincn 2

8

F8e

j4 Frect 8 F k( )( )

k =

n-32 32

x[n]

1

F -1 1

|X( F )|8

F -1 1

Phase of X( F )|

- p

p

4. A DT signal is defined by

x n = sin n / 6( ) .

Sketch the magnitude and phase of the DTFT of x n 3 and

x n + 12 .

From the table,

sin 2 F

0n( ) F

j / 2( ) 1F + F

0( ) 1F F

0( )

sinn

6

Fj / 2( ) 1

F + 1 / 12( ) 1F 1 / 12( )

sinn 3( )6

Fj / 2( ) 1

F + 1 / 12( ) 1F 1 / 12( ) e

j6 F

sinn 3( )6

Fj / 2( ) e

j6 FF + 1 / 12 k( ) e

j6 FF 1 / 12 k( )

k =

sinn 3( )6

Fj / 2( ) e

j6 k 1/12( )

= j

F + 1 / 12 k( ) ej6 k +1/12( )

= j

F 1 / 12 k( )k =

sinn 3( )6

F1 / 2( ) F + 1 / 12 k( ) + F 1 / 12 k( )

k =

sinn 3( )6

F 1

21

F + 1 / 2( ) +1

F 1 / 2( )

M. J. Roberts - 10/15/06

Solutions 11-4

Similarly,

sinn + 12( )

6

Fj / 2( ) e

+ j24 k 1/12( )F + 1 / 12 k( ) e

+ j24 k +1/12( )F 1 / 12 k( )

k =

sinn + 12( )

6

Fj / 2( ) e

+ j 24 k 2( )F + 1 / 12 k( ) e

+ j 24 k +2( )F 1 / 12 k( )

k =

sinn + 12( )

6

Fj / 2( ) F + 1 / 12 k( ) F 1 / 12 k( )

k =

sinn + 12( )

6

Fj / 2( ) 1

F + 1 / 2( ) 1F 1 / 2( )

This is the same as the transform of the unshifted function because a shift of 12 is ashift over exactly one period.

n-12 12

x[n]

-1

1

F -1 1

|X( F )|0.5

F -1 1

Phase of X( F )|

- p

p

Phase of X( F )|n

-12 12

x[n]

-1

1

F -1 1

|X( F )|0.5

F -1 1- p

p

5. The DTFT of a DT signal is defined by

X j( ) = 8 rect 2 /( ) / 2( )( ) + rect 2 /( ) + / 2( )( ) 2 ( ) .

Sketch x n .

M. J. Roberts - 10/15/06

Solutions 11-5

From the table,

sinc n / w( ) F

wrect w / 2( ) 2 ( )

sinc n / 4( ) F

4rect 2 /( ) 2 ( )

ej n / 2

sinc n / 4( ) F4rect

2

22 ( )

ej n / 2

sinc n / 4( ) F4rect

2+

22 ( )

sinc n / 4( ) ej n / 2

+ ej n / 2( ) F

4 rect2

2+ rect

2+

22 ( )

2sinc n / 4( )cos n / 2( ) F4 rect

2

2+ rect

2+

22 ( )

Therefore

x n = 2sinc n / 4( )cos n / 2( )

n-16 16

x[n]

-2

2

6. Sketch the magnitude and phase of the DTFT of

x n = rect

4n cos 2 n / 6( ) .

Then sketch x n .

From the table,

M. J. Roberts - 10/15/06

Solutions 11-6

rect

Nw

nF

2Nw

+ 1( )drcl F ,2Nw

+ 1( )and

cos 2 F

0n( ) F

1 / 2( ) 1F F

0( ) +1

F + F0( )

X F( ) = 9drcl F ,9( ) 1 / 2( ) 1

F 1 / 6( ) +1

F + 1 / 6( )

Since both functions are periodic with period, one, at every impulse in the combfunction the value of the Dirichlet function will be the same.

X F( ) = 9 / 2( ) drcl 1 / 6,9( ) 1

F 1 / 6( ) + drcl 1 / 6,9( ) 1F + 1 / 6( )

X F( ) = 9 / 2( )drcl 1 / 6,9( )sin 3 / 2( )9 sin /6( )

1F 1 / 6( ) +

1F + 1 / 6( )

X F( ) =

1F 1 / 6( ) +

1F + 1 / 6( )

Then, using

cos 2 F

0n( ) F

1 / 2( ) 1F F

0( ) +1

F + F0( )

2cos 2 n / 6( ) F

1F 1 / 6( ) +

1F + 1 / 6( )

and, therefore,

x n = 2cos 2 n / 6( )

n-12 12

x[n]

-2

2

F -1 1

|X( F )|1

F -1 1

Phase of X( F )|

- p

p

7. Sketch the inverse DTFT of

X F( ) = 1 / 2( ) rect 4F( ) 1

F( ) 1/ 2F( ) .

From the table,

M. J. Roberts - 10/15/06

Solutions 11-7

sinc n / w( ) F

wrect wF( ) 1F( ) and

N

0

nF 1

N0

1/ N0

F( ) = F0 F

0

F( )

Therefore

sinc n / 4( ) F

4rect 4F( ) 1F( ) and

2

nF 1

21/ 2

F( )

1 / 4( )sinc n / 4( ) F

rect 4F( ) 1F( )

Therefore using multiplication-convolution duality,

x n = 1 / 4( )sinc n / 4( ) 2

n .

n-16 16

x[n]

-0.1

0.25

8. Find the numerical values of the constants.

(a)

ArectW

n ejB n F

10sin 5 F + 1( )( )sin F + 1( )( )

Find A, W and B.

rectW

nF

sin F 2W + 1( )( )sin F( )

10rect2

nF

10sin 5 F( )sin F( )

10rect2

n ej2 n F

10sin 5 F + 1( )( )sin F + 1( )( )

A = 10 , W = 2 , B = -2

(b) 2

15n 3 rect

3n

FAe

jB F Find A and B.

2

15n 3 rect

3n = 2 n 3

2 n 3

F2e

j6 F

M. J. Roberts - 10/15/06

Solutions 11-8

A = 2 , B = -6

(c) 2 / 3( )

n

u n + 2F Ae

jB F

1 ej2 F

Find A, B and .

nu n

F 1

1 ej2 F

, < 1

2 / 3( )n

u nF 1

1 2 / 3( )ej2 F

2 / 3( )n+2

u n + 2F e

j4 F

1 2 / 3( )ej2 F

2 / 3( )2

2 / 3( )n

u n + 2F e

j4 F

1 2 / 3( )ej2 F

2 / 3( )n

u n + 2F

3 / 2( )2 e

j4 F

1 2 / 3( )ej2 F

A = 9 / 4 , B = -4, = 2 / 3

(d) 4sinc n / 10( ) F

Arect BF( ) 1F( ) Find A and B.

sinc n / N

w( ) FN

wrect N

wF( ) 1

F( )

sinc n / 10( ) F

10rect 10F( ) 1F( )

4sincn 1

10

F40rect 10F( ) 1

F( )

A = 40 , B = 10

9. Find the numerical values of these functions.

(a) x n = 4 2 / 3( )

n

u n X j( )

=

X j( ) =4

1 2 / 3( )ej

X j( ) =4

1 2 / 3( )ej

=4

1+ 2 / 3=

12

5= 2.4

(b) x n = 2rect

3n 2 Find the numerical value of

X F( )

F =1/8.

X F( ) = 2sin 7 F( )sin F( )

ej4 F

X 1/ 8( ) = 2sin 7 / 8( )sin / 8( )

ej / 2

= j2

M. J. Roberts - 10/15/06

Solutions 11-9

(c) X F( ) = rect 10F( ) comb F( ) 1 / 2( ) comb F 1 / 4( ) + comb F + 1 / 4( )

Find the numerical value of x 2 .

x n = 1 / 10( )sinc n / 10( )cos 2 n / 4( )

x 2 = 1 / 10( )sinc 2 / 10( )cos( ) = 1 / 10( )

sin / 5( )/ 5

cos( ) = 0.09355

10. Using the differencing property of the DTFT and the transform pair,

tri n / 2( ) F

1+ cos 2 F( ) ,

find the DTFT of 1 / 2( ) n + 1 + n n 1 n 2( )( ) . Compare it with

Fourier transform found using the table in Appendix E.

T h e f i r s t b a c k w a r d d i f f e r e n c e o f tri n / 2( ) i s

1 / 2( ) n + 1 + n n 1 n 2( )( ) . Applying the differencing

property,

trin

2tri

n 1

2

F1 e

j2 F( ) 1+ cos 2 F( )( )

1 / 2( ) n + 1 + n n 1 n 2( ) F

1 ej2 F( ) 1+ cos 2 F( )( )

1 / 2( ) n + 1 + n n 1 n 2( ) F

1 ej2 F

+e

j2 F+ e

j2 F

2e

j2 F ej2 F

+ ej2 F

2

1 / 2( ) n + 1 + n n 1 n 2( ) F

1 ej2 F

+e

j2 F

2+

ej2 F

2

1

2

ej4 F

2

1 / 2( ) n + 1 + n n 1 n 2( ) F

1 / 2( ) ej2 F

+ 1 ej2 F

ej4 F( )

Other route to the DTFT:

M. J. Roberts - 10/15/06

Solutions 11-10

1 / 2( ) n + 1 + n n 1 n 2( ) F

1 / 2( ) ej2 F

+ 1 ej2 F

ej4 F( )

Check.

11. Using Parseval’s theorem, find the signal energy of

x n = sinc n / 10( )sin 2 n / 4( ) .

Ex

= x n2

n=

= X F( )2

dF1

From the table,

sinc n / w( ) F

wrect wF( ) 1F( )

and

sin 2 F

0n( ) F j

21

F + F0( ) 1

F F0( )

Using the multiplication-convolution duality of the DTFT,

sinc n / 10( )sin 2 n / 4( ) F

10rect 10F( ) 1F( ) j / 2( ) 1

F + 1 / 4( ) 1F 1 / 4( )

Periodic convolution is the same as aperiodic convolution with one period.

sinc n / 10( )sin 2 n / 4( ) F

j5rect 10F( ) 1F( ) F + 1 / 4( ) F 1 / 4( )

sinc n / 10( )sin 2 n / 4( ) F

j5rect 10F( ) 1F + 1 / 4( ) 1

F 1 / 4( )

sinc n / 10( )sin 2 n / 4( ) F

j5 rect 10F( ) 1F + 1 / 4( ) rect 10F( ) 1

F 1 / 4( )

E

x= j5 rect 10F( ) 1

F + 1 / 4( ) rect 10F( ) 1F 1 / 4( )

2

dF1

Since we are integrating only over a range of one, only one impulse in each combis significant.

E

x= 25 rect 10F( ) F + 1 / 4( ) rect 10F( ) F 1 / 4( )

2

dF1

M. J. Roberts - 10/15/06

Solutions 11-11

E

x= 25 rect 10 F + 1 / 4( )( ) rect 10 F 1 / 4( )( )

2

dF1

The square of the sum equals the sum of the squares because there is no crossproduct; the two rectangles do not overlap.

E

x= 25 rect 10 F + 1 / 4( )( )dF

1+ rect 10 F 1 / 4( )( )dF

1{ }

Ex

= 25 dF1/ 4 1/ 20

1/ 4+1/ 20

+ dF1/ 4 1/ 20

1/ 4+1/ 20

= 251

10+

1

10= 5

Fourier Method Relations

12. Sketch the magnitude and phase of the CTFT of

x

1t( ) = rect t( )

and of the CTFS of

x

2t( ) = rect t( ) 8

t( ) .

For comparison purposes, sketch X

1f( ) versus f and

T

0X

2k versus

kf

0 on the

same set of axes. ( T

0 is the period of

x

2t( ) and

T

0= 1 / f

0.)

X

1f( ) = sinc f( )

Using the relationship between an the CTFT of an aperiodic signal and the CTFSof a periodic extension of that signal,

X

2k = f

sX

1f

sk( ) = 1 / 8( )sinc k / 8( )

M. J. Roberts - 10/15/06

Solutions 11-12

f -4 4

|X1( f )|

1

f -4 4

X1( f )

-

kf0

-4 4

|T0X

2[k]|

1

kf0

-4 4

T0X

2[k]

-

13. Sketch the magnitude and phase of the CTFT of

x

1t( ) = 4cos 4 t( )

and of the DTFT of

x

2n = x

1nT

s( )where

T

s= 1 / 16 . For comparison purposes sketch

X

1f( ) and

T

sX

2T

sf( ) versus

f on the same set of axes.

X

1f( ) = 2 f 2( ) + f + 2( )

x

2n = 4cos 4 nT

s( ) = 4cos n / 4( )

X

2F( ) = 2

1F 1 / 8( ) +

1F + 1 / 8( )

X2

F( ) = 2 F 1 / 8 k( ) + F + 1 / 8 k( )k =

X2

Tsf( ) = 2 f / 16 1 / 8 k( ) + f / 16 + 1 / 8 k( )

k =

X2

Tsf( ) = 32 f 2 16k( ) + f + 2 16k( )

k =

TsX

2T

sf( ) = 2 f 2 16k( ) + f + 2 16k( )

k =

M. J. Roberts - 10/15/06

Solutions 11-13

f -16 16

|X1( f )|

2

f -16 16

Phase of X1( f )

- p

p

f

f

-16 16

|TsX

2(T

sf )|

2

-16 16

Phase of TsX

2(T

sf )

- p

p

......

......

14. Sketch the magnitude and phase of the DTFT of

x

1n =

sinc n / 16( )4

and of the DTFS of

x

2n =

sinc n / 16( )4

32n .

For comparison purposes sketch X

1F( ) versus F and

N

0X

2k versus

kF

0 on

the same set of axes.

From the table,

sinc n / w( ) F

wrect wF( ) 1F( )

sinc n / 16( )4

F4rect 16F( ) 1

F( )

M. J. Roberts - 10/15/06

Solutions 11-14

therefore

X1

F( ) = 4rect 16F( ) 1F( ) = 4 rect 16 F q( )( )

q=

.

The fundamental frequency of the periodic signal is the reciprocal of its period,

F

0= 1 / N

0= 1 / 32 .

Using the results of the analysis of periodic extensions of aperiodic DT signals,

X

2k = F

0X

1kF

0( )

X2

k = 1 / 32( )X1

k / 32( ) =1

8rect 16 k / 32 q( )( )

q=

= 1 / 8( ) rect k / 2 16q( )q=

F -1 1

|X1(F )|

4

F -1 1

Phase of X1(F )

- p

p

kF0

-1 1

N0|X

2[k]|

4

kF0

-32 32

Phase of X2[k]

- p

p

DTFT

M. J. Roberts - 10/15/06

Solutions 11-15

15. Find the DTFT of each of these signals:

(a) x n = 1 / 3( )

n

u n 1

X j( ) = x n ej n

n=

= 1 / 3( )n

u n 1 ej n

n=

= 1 / 3( )n

ej n

n=1

X j( ) = 1 / 3( )m+1

ej m+1( )

m=0

=e

j

3

m+1

m=0

X j( ) =e

j

3

ej

3

m

m=0

=e

j

3

1

1 ej

/ 3=

ej

3 ej

Alternate Solution:

x n = 1 / 3( )

n

u n 1 = 1 / 3( )n

u n n

Using

nu n

F 1

1 ej

and

nF

1

x n = 1 / 3( )n

u n 1 / 3( ) n =1

1e

j

3

1

x n =

1 1e

j

3

1 ej

/ 3=

ej

3

1 ej

/ 3=

ej

3 ej

Second Alternate Solution:

x n = 1 / 3( ) 1 / 3( )

n 1

u n 1

X j( ) =

1

3

1

1 ej

/ 3e

j=

ej

3 ej

(b) x n = sin n / 4( ) 1 / 4( )

n

u n 2

sin n / 4( ) = sin 2 n / 8( ) = cos 2 n 2( ) / 8( )

M. J. Roberts - 10/15/06

Solutions 11-16

x n = 1 / 4( )

2

cos 2 n 2( ) / 8( ) 1 / 4( )n 2

u n 2

ncos

0n( )u n

F1 cos

0( )ej

1 2 cos0( )e

j+

2e

j2, < 1

X j( ) = 1 / 4( )2

ej2

1 1 / 4( )cos / 4( )ej

1 1 / 2( )cos / 4( )ej

+ 1 / 16( )ej2

=e

j2

16

1 2 / 8( )ej

1 2 / 4( )ej

+ 1 / 16( )ej2

Alternate Solution:

x n =e

j n / 4e

j n / 4

j21 / 4( )

n

u n 2 =1

j2e

j / 4/ 4( )

n

ej / 4

/ 4( )n

( )u n 2

x n =1

j2e

j / 4/ 4( )

2

ej / 4

/ 4( )n 2

ej / 4

/ 4( )2

ej / 4

/ 4( )n 2

( )u n 2

ej / 4

4

n

u nF 1

1 ej / 4

/ 4( )ej

ej / 4

4

n 2

u n 2F e

j2

1 ej / 4

/ 4( )ej

X j( ) =1

j2

ej / 4

4

2

ej2

1 ej / 4

/ 4( )ej

ej / 4

4

2

ej2

1 ej / 4

/ 4( )ej

X j( ) =e

j2

j2

ej / 4

/ 4( )2

1 ej / 4

/ 4( )ej( ) e

j / 4/ 4( )

2

1 ej / 4

/ 4( )ej( )

1 ej / 4

/ 4( )ej( ) 1 e

j / 4/ 4( )e

j( )

X j( ) =e

j2

j32

ej / 4( )

2

ej / 4( )

2

ej / 4

/ 4( )ej

ej / 4( )

2

+ ej / 4( )

2

ej / 4

/ 4( )ej

1 ej / 4

/ 4 + ej / 4

/ 4( )ej

+ ej2

/ 16

X j( ) =e

j2

j32

ej / 2

ej / 2

ej / 4

/ 4( )ej

+ ej / 4

/ 4( )ej

1 1 / 2( )cos / 4( )ej

+ 1 / 16( )ej2

M. J. Roberts - 10/15/06

Solutions 11-17

X j( ) =e

j2

j32

j2 1 / 4( ) ej / 4

ej / 4( )e

j

1 1 / 2( )cos / 4( )ej

+ 1 / 16( )ej2

X j( ) =e

j2

16

1 1 / 4( )sin / 4( )ej

1 1 / 2( )cos / 4( )ej

+ 1 / 16( )ej2

=e

j2

16

1 2 / 8( )ej

1 2 / 4( )ej

+ 1 / 16( )ej2

(c) x n = sinc 2 n / 8( ) sinc 2 n 4( ) / 8( )

Using

sinc n / w( ) F

wrect wF( ) 1F( )

X F( ) = 8 / 2( ) rect 8F / 2( ) 1

F( ) 8 / 2( ) rect 8F / 2( ) 1F( ) e

j8 F

X F( ) = 4 /( )

2

rect 4 f /( ) 1F( ) e

j8 F

x n = 4 /( )sinc 2 n 4( ) / 8( )

(d) x n = sinc

22 n / 8( )

Using

sinc n / w( ) F

wrect wF( ) 1F( )

X F( ) = 8 / 2( )rect 8F / 2( ) 1

F( ) 8 / 2( )rect 8F / 2( ) 1F( )

X F( ) = 8 / 2( )

2

rect 8F / 2( ) 1F( ) rect 8F / 2( ) 1

F( )

X F( ) = 8 / 2( )

2

rect 8F / 2( ) rect 8F / 2( ) 1F( )

X F( ) = 8 / 2( )

2

2 / 8( ) tri 8F / 2( ) 1F( ) = 8 / 2( ) tri 8F / 2( ) 1

F( )

16. Sketch the magnitudes and phases of the DTFT’s of the following functions:

(a) rect

2n

M. J. Roberts - 10/15/06

Solutions 11-18

Using rect

Nw

nF

2Nw

+ 1( )drcl F ,2Nw

+ 1( ) ,

rect

2n

F5drcl F ,5( )

F -1 1

|X( F )|

5

F -1 1

Phase of X( F )

- p

p

(b) rect

2n 5 n( )

Using

nF

1 and x n y n

FX F( )Y F( )

rect

2n 5 n( ) F

5drcl F ,5( ) 5( ) = 25drcl F ,5( )

F -1 1

|X( F )|

25

F -1 1

Phase of X( F )

- p

p

(c) rect

2n 3 n + 3

Using x n n

0

Fe

j2 Fn0 X F( )

M. J. Roberts - 10/15/06

Solutions 11-19

rect

2n 3 n + 3

F15drcl F ,5( )e

j6 F

F -1 1

|X( F )|

15

F -1 1

Phase of X( F )

- p

p

(d) rect

2n 5 4n( ) = rect

2n 5 n( )

rect

2n 5 4n( ) F

25drcl F ,5( )

F -1 1

|X( F )|

25

F -1 1

Phase of X( F )

- p

p

(e) rect

2n

8n

Using

N0

nF 1

N0

1/ N0

F( ) = F0 F

0

F( ) ,

rect

2n

8n

F5 / 8( )drcl F ,5( ) 1/8

F( )

M. J. Roberts - 10/15/06

Solutions 11-20

rect2

n8

nF

5 / 8( ) drcl F ,5( ) F k / 8( )k =

rect2

n8

nF 5

8drcl k / 8,5( ) F k / 8( )

k =

F -1 1

|X( F )|

0.625

F -1 1

Phase of X( F )

- p

p

(f) rect

2n

8n 3

Using the result of (e) and x n n

0

Fe

j2 Fn0 X F( ) ,

rect2

n8

n 3F

5 / 8( )ej6 F

drcl m / 8,5( ) F m / 8( )m=

F -1 1

|X( F )|

0.625

F -1 1

Phase of X( F )

- p

p

(g)

rect2

n8

2n = rect2

n 2n 8mm=

= rect2

n 2 n 4m( )m=

M. J. Roberts - 10/15/06

Solutions 11-21

rect2

n8

2n = rect2

n n 4mm=

= rect2

n4

n

Therefore

rect

2n

82n

F5 / 4( )drcl F ,5( ) 1/ 4

F( )

rect2

n8

2nF

5 / 4( ) drcl F ,5( ) F m / 4( )m=

rect2

n8

2nF

5 / 4( ) drcl m / 4,5( ) F m / 4( )m=

F -1 1

|X( F )|

1.25

F -1 1

Phase of X( F )

- p

p

(h) rect

2n

5n

rect

2n

5n

F5drcl F ,5( ) 1 / 5( ) 1/5

F( )

rect2

n5

nF

drcl F ,5( ) F m / 5( )m=

rect2

n5

nF

drcl m / 5,5( ) F m / 5( )m=

Then, using

drcln

2m + 1,2m + 1 =

2m+1n

M. J. Roberts - 10/15/06

Solutions 11-22

rect2

n5

nF

5m F m / 5( )

m=

5

m F m / 5( )m=

has an impulse at every 5th integer value of m and

therefore has an impulse at every integer value of F. Therefore

rect

2n

5n

F

1F( )

and, since 1

F

1F( ) , that implies that

rect

2n

5n = 1 .

F -1 1

|X( F )|

1

F -1 1

Phase of X( F )

- p

p

17. Sketch the inverse DTFT’s of these functions.

(a) X F( ) =

1F( ) 1

F 1 / 2( )

Using 1

F

1F( ) and

e

j2 F0n

x nF

X F F0( )

1 e

j n F

1F( ) 1

F 1 / 2( )

e

j n / 2e

j n / 2e

+ j n / 2( ) F

1F( ) 1

F 1 / 2( )

j2e

j n / 2sin n / 2( ) F

1F( ) 1

F 1 / 2( )

2e

j n+1( )/ 2sin n / 2( ) F

1F( ) 1

F 1 / 2( )

2 cos n + 1( ) / 2( ) + j sin n + 1( ) / 2( ) sin n / 2( ) F

1F( ) 1

F 1 / 2( )

M. J. Roberts - 10/15/06

Solutions 11-23

2 cos n + 1( ) / 2( )sin n / 2( )

sin n / 2( ) + j sin n + 1( ) / 2( )sin n '2( )=0

F

1F( ) 1

F 1 / 2( )

2sin

2n / 2( ) F

1F( ) 1

F 1 / 2( )

n-12 12

x[n]

2

(b) X F( ) = j

1F + 1 / 8( ) j

1F 1 / 8( )

Using 1

F

1F( ) and

e

j2 F0n

x nF

X F F0( )

je

j n / 4je

j n / 4 Fj

1F + 1 / 8( ) j

1F 1 / 8( )

j j2sin n / 4( )( ) F

j1

F + 1 / 8( ) j1

F 1 / 8( )

2sin n / 4( ) F

j1

F + 1 / 8( ) j1

F 1 / 8( )

n-12 12

x[n]

-2

2

(c) X F( ) = sinc 10 F 1 / 4( )( ) + sinc 10 F + 1 / 4( )( ) 1

F( )

Using

sinc t / w( ) T

0

t( ) = wf0

cos t / w( ) + T0

/ w 1( )drcl f0t,T

0/ w 1( )

from Appendix A (because T

0/ 2w is an integer),

M. J. Roberts - 10/15/06

Solutions 11-24

X F( ) = 1 / 10( )cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )

+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )

cos 10 F( )e

j2 FndF

1

Fcos 10 F( )

1

2e

j10 F+ e

j10 F( )ej2 Fn

dF1

Fcos 10 F( )

1

2e

j2 F n+5( )dF

1+ e

j2 F n 5( )dF

1

Fcos 10 F( )

1

2

cos 2 F n + 5( )( ) + j sin 2 F n + 5( )( ) dF1

+ cos 2 F n 5( )( ) + j sin 2 F n 5( )( ) dF1

Fcos 10 F( )

These integrals are zero unless n = ±5 . Therefore

1 / 2( ) n + 5 + n 5( ) F

cos 10 F( ) .

Then using rect

Nw

nF

2Nw

+ 1( )drcl F ,2Nw

+ 1( ) ,

rect

4n

F9drcl F ,9( )

Combining inverse transforms,

1 / 2( ) n + 5 + n 5( ) + rect

4n

Fcos 10 F( ) + 9drcl F ,9( ) .

Then, using e

j2 F0n

x nF

X F F0( )

ej n / 2

1 / 2( ) n + 5 + n 5( ) + rect4

n{ }

Fcos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )

and

ej n / 2

1 / 2( ) n + 5 + n 5( ) + rect4

n{ }

Fcos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )

Then, finally

M. J. Roberts - 10/15/06

Solutions 11-25

1

10

ej n / 2

1 / 2( ) n + 5 + n 5( ) + rect4

n{ }+e

j n / 21 / 2( ) n + 5 + n 5( ) + rect

4n{ }

F 1

10

cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )

+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )

The impulses on the left side cancel and we get

1 / 5( )rect4

n cos n / 2( ) F 1

10

cos 10 F 1 / 4( )( ) + 9drcl F 1 / 4,9( )

+cos 10 F + 1 / 4( )( ) + 9drcl F + 1 / 4,9( )

n-12 12

x[n]

-0.2

0.2

(d) X F( ) = F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1

2F( )

X F( ) = F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1 / 2( ) 1

F( ) +1

F 1 / 2( )

X F( ) =1

2

F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1F( )

+ F 1 / 4( ) + F 3 / 16( ) + F 5 / 16( ) 1F 1 / 2( )

X F( ) =1

2

1F 1 / 4( ) +

1F 3 / 16( ) +

1F 5 / 16( )

+1

F 1 / 4 1 / 2( ) +1

F 3 / 16 1 / 2( ) +1

F 5 / 16 1 / 2( )

Then, using 1

F

1F( ) and

e

j2 F0n

x nF

X F F0( ) we get

1

2

ej n / 2

+ ej3 n /8

+ ej5 n /8

+ej3 n / 2

+ ej11 n /8

+ ej13 n /8

F 1

2

1F 1 / 4( ) +

1F 3 / 16( ) +

1F 5 / 16( )

+1

F 3 / 4( ) +1

F 11 / 16( ) +1

F 13 / 16( )

or

M. J. Roberts - 10/15/06

Solutions 11-26

cos n / 2( ) + cos 3 n / 8( ) + cos 5 n / 8( )

F 1

2

1F 1 / 4( ) +

1F 3 / 16( ) +

1F 5 / 16( )

+1

F 3 / 4( ) +1

F 11 / 16( ) +1

F 13 / 16( )

n-16 16

x[n]

-3

3

18. A signal, x n , has a DTFT ,

X F( ) = 10sinc 5F( ) 1

F( ) . What is its signal

energy?

Using rect

Nw

nF

2Nw

+ 1( )sinc 2Nw

+ 1( ) F( ) 1F( ) ,

x n = 2rect

2n .

Ex

= 2rect2

n2

n=

= 4 1n= 2

2

= 20

19. A signal, x n , has a DTFT,

X F( ) =

1F 1 / 4( ) +

1F + 1 / 4( ) + j

1F + 1 / 3( ) j

1F 1 / 3( ) .

What is the fundamental period, N

0, of

x n ?

Using cos 2 F

0n( ) F

1 / 2( ) 1F F

0( ) +1

F + F0( ) and

sin 2 F

0n( ) F

j / 21

F + F0( ) 1

F F0( ) ,

x n = 2cos2 n

4+ 2sin

2 n

3 .

The least common multiple of the two periods of the two sinusoids, 4 and 3, is 12.Therefore

N

0= 12 .

20. The DTFT of x n = 2 n + 3 3 n 3 can be expressed in the form,

X F( ) = Asin bF( ) + Ce

dF . Find the numerical values of A, b, C and d.

X F( ) = 2e

+ j6 F3e

j6 F= 2e

+ j6 F2e

j6 Fe

j6 F= j4sin 6 F( ) e

j6 F

M. J. Roberts - 10/15/06

Solutions 11-27

A = j4 , b = 6 , C = -1 , d = j6

21. Let x n be a DT signal and let

y n = x m

m=

n

. If Y j( ) = cos 2( ) ,

x n consists of exactly four DT impulses. What are their numerical strengths

and locations?

Y j( ) = cos 2( ) =

X j( )1 e

jX j( ) = cos 2( ) 1 e

j( )

X j( ) =

1

2e

j2+ e

j2( ) 1 ej( ) =

1

2e

j2+ e

j2e

je

j3( )

x n =

1

2n + 2 + n 2 n + 1 n 3( )

Impulse #1. Strength = 1/2 Located at n = -2

Impulse #2. Strength = 1/2 Located at n = 2

Impulse #3. Strength = -1/2 Located at n = 3

Impulse #4. Strength = -1/2 Located at n = -1

22. A signal, x n = 4cos 2 n / 15( ) + 2cos 2 n / 9( ) is the excitation of a system

whose impulse response is h n = rect

Nw

n ( N

w is an integer). When

N

w= 22 the response,

y n , of the system is zero. The response of the system is

also zero for some larger values of N

w. Find the next smallest positive numerical

integer value of N

w greater than 22 that makes the response zero. (Hint: The

zeros of drcl F , N( ) occur where F is an integer multiple of 1/N, except when F

itself is an integer.)

X F( ) = 2

1F 1 / 15( ) +

1F + 1 / 15( ) + j

1F + 1 / 9( ) 1

F 1 / 9( )

H F( ) = 2N

w+ 1( )drcl F ,2N

W+ 1( )

The response is the product of X and H. Therefore, for the response to be zero,the impulses in X must occur at zeros of H. The zeros of H occur at integer

multiples of F

0= 2N

w+ 1( )

1

. To have zeros at the impulse locations, this F

0

M. J. Roberts - 10/15/06

Solutions 11-28

must be a common divisor of 1/15 and 1/9. The greatest common divisor of thosetwo numbers is 1/45. That yields

2N

w+ 1 = 45 and

N

W= 22 . The next greatest

common divisor of 1/15 and 1/9 is 1/90 but that does not yield an integer value of

N

W. The next greatest common divisor of 1/15 and 1/9 is 1/135 yielding

N

W= 67 .

23. In Figure E-23 are some DT signals, numbered 1-14. Below them are some DTFTmagnitude plots. For each DTFT magnitude plot, identify the corresponding DTsignal by writing its number in the space provided. If there is no match, write“none”.

1. 3sinc n( ) 2.

5sinc n / 4( ) 2sinc n / 4( ) 3.

7cos 2 n / 8( )

4.

n + 1 n 1 5. 3sinc n / 4( ) 6.

4sin 2 n / 8( )

7. 2 / 3( )

n

u n 8. 2rect

2n 3 9.

4

4n

10.

42

n 11.

3sinc2

n / 4( ) 12.

n + 1 + n 1

13. 2rect

3n 3 14.

1 / 3( )

n

u n

6 10 9 8 3 7

F 1 1

|X(F)|

2

F 1 1

|X(F)|

2

F 1 1

|X(F)|

1

F 1 1

|X(F)|

10

F 1 1

|X(F)|

3.5

F 1 1

|X(F)|

3

5 12 14 1 4 2

F 1 1

|X(F)|

12

F 1 1

|X(F)|

2

F 1 1

|X(F)|

1.5

F 1 1

|X(F)|3

F 1 1

|X(F)|

2

F 1 1

|X(F)|

160

13 11

F 1 1

|X(F)|

14

F 1 1

|X(F)|

12

M. J. Roberts - 10/15/06

Solutions 11-29

Figure E-23

Fourier Method Relations

24. Using the relationship between the CTFT of a signal and the CTFS of a periodicextension of that signal, find the CTFS of

x t( ) = rect t / w( ) T

0

t( )and compare it with the table entry.

X f( ) = wsinc wf( )

X k = f

0X kf

0( ) = f0wsinc wkf

0( ) = w / T0( )sinc wk / T

0( )

25. A CT signal, x t( ) , has a CTFT,

X f( ) = 4rect f / 2( ) . A new signal,

x

pt( ) , is

formed by periodically repeating x t( ) with a period of 8. The CTFS harmonic

function of x

pt( ) is

X

pk . Find

X

pk .

X

pk = f

pX kf

p( ) where fp

= 1 / Tp

= 1 / 8

X

pk = 1 / 8( )X k / 8( ) = 1/2( )rect k / 16( )

X

p3 = 1 / 2( )rect 3 / 16( ) = 1 / 2 .

26. An aperiodic signal, x t( ) , has a CTFT,

X f( ) =

0 , f > 4

4 f , 2 < f < 4

2 , f < 2

= 4 tri f / 4( ) 2 tri f / 2( ) .

Let

xp

t( ) = x t( ) 2t( ) = x t 2k( )

k =

and let X

pk be its CTFS harmonic

function.

(a) Sketch the magnitude of X f( ) .

M. J. Roberts - 10/15/06

Solutions 11-30

4

4

-4

-4

f

|X( f )|

(b) Find an expression for X

pk .

Xp

k = f0

X kf0( ) =

1

2

0 , k / 2 > 4

4 k / 2 , 2 < k / 2 < 4

2 , k / 2 < 2

= 2 trik / 2

4tri

k / 2

2.

Xp

k =1

2

0 , k > 8

4 k / 2 , 4 < k < 8

2 , k < 4

= 2 tri k / 8( ) tri k / 4( )

(c) Find the numerical values of X

p3 ,

X

p5 ,

X

p10 .

X

p3 = 1 / 2( ) 2 = 1 or

X

p3 = 2 tri 3 / 8( ) tri 3 / 4( ) = 10 / 8 1 / 4 = 1

X

p5 = 1 / 2( ) 4 5 / 2( ) = 3 / 4 or

X

p5 = 2 tri 5 / 8( ) tri 5 / 4( ) = 6 / 8 0 = 3 / 4

X

p10 = 1 / 2( ) 0 = 0 or

X

p10 = 2 tri 10 / 8( ) tri 10 / 4( ) = 0 0 = 0

27. Using the relationship between the DTFT of a signal and the DTFS of a periodicextension of that signal, find the DTFS of

rect

Nw

nN

0

n

and compare it with the table entry.

M. J. Roberts - 10/15/06

Solutions 11-31

X F( ) = 2N

w+ 1( )drcl F ,2N

w+ 1( )

From the text,

X

pk = 1 / N

p( )X kFp( ) .

Therefore,

Xp

k =2N

w+ 1

N0

drcl k / N0,2N

w+ 1( ) .

28. A DT signal, x n , is formed by sampling a CT signal,

x t( ) = 12sinc 5t( ) , with a

time between samples, T

s= 0.1. The DTFT of

x n is

X

DTFTF( ) . Find the

numerical value of X

DTFT0.2( ) .

XDTFT

F( ) = fs

XCTFT

fs

F k( )( )k =

,

XCTFT

f( ) =12

5rect

f

5

XDTFT

F( ) = 24 rect10 F k( )

5k =

= 24 rect 2 F k( )( )k =

XDTFT

0.2( ) = 24 rect 2 0.2 k( )( )k =

= 24rect 0.4( ) = 24