Chap 9 Solutions

M. J. Roberts - 10/7/06

Solutions 9-1

Chapter 9 - The Discrete-Time Fourier Series

Solutions

Basics

1. Without using a calculator or computer find the dot products of (a) w

1 and

w

1,

(b) w

1 and

w

2 (c)

w

11 and

w

37 , where

wk

=

W4

0

W4

k

W4

2k

W4

3k

and W

NF

= ej2 / N

F

to show that they are orthogonal.

(a)

w1

=

ej / 2( )

0

ej / 2( )

1

ej / 2( )

2

ej / 2( )

3

=

1

j

1

j

w1

=

ej / 2( )

0

ej / 2( )

1

ej / 2( )

2

ej / 2( )

3

=

1

j

1

j

w1

Hw

1= 1 j 1 j

1

j

1

j

= 1 1+ 1 1 = 0

(b)

w1

=

ej / 2( )

0

ej / 2( )

1

ej / 2( )

2

ej / 2( )

3

=

1

j

1

j

w2

=

ej / 2( )

0

ej / 2( )

2

ej / 2( )

4

ej / 2( )

6

=

1

1

1

1

M. J. Roberts - 10/7/06

Solutions 9-2

w1

Hw

1= 1 j 1 j

1

1

1

1

= 1+ j 1 j = 0

(c)

w11

=

ej / 2( )

0

ej / 2( )

11

ej / 2( )

22

ej / 2( )

33

=

1

j

1

j

w37

=

ej / 2( )

0

ej / 2( )

37

ej / 2( )

74

ej / 2( )

111

=

1

j

1

j

w1

Hw

1= 1 j 1 j

1

j

1

j

= 1 1+ 1 1 = 0

2. Find the DTFS harmonic function of a signal x n with period 4 for which

x 0 = 3 ,

x 1 = 1,

x 2 = 5 and

x 3 = 0 using the matrix multiplication

X =W

Hx

NF

.

N

F= 4 ,

x =

3

1

5

0

,

W =

W4

0W

4

0W

4

0W

4

0

W4

0W

4

1W

4

2W

4

4

W4

0W

4

2W

4

4W

4

8

W4

0W

4

3W

4

6W

4

12

=

1 1 1 1

1 j 1 j

1 1 1 1

1 j 1 j

WH

=

1 1 1 1

1 j 1 j

1 1 1 1

1 j 1 j

X =1

4

1 1 1 1

1 j 1 j

1 1 1 1

1 j 1 j

3

1

5

0

=1

4

1

8 j

3

8 + j

=

1 / 4

2 j / 4

3 / 4

2 + j / 4

M. J. Roberts - 10/7/06

Solutions 9-3

3. Using the direct summation formula find and sketch the DTFS harmonic function

of N

0

n with N

F= N

0.

X k =1

N0

N0

n ej2 kF

0( )n

n= N0

=1

N0

N0

n ej2 kF

0( )n

n= N0

/ 2

N0

/ 2 1

=1

N0

k

|X[k]|

2 4 6 8-8 -6 -4 -2

1N0

k2 4 6 8-8 -6 -4 -2

X[k]

4. One period of a periodic DT function with period, 4, is described by

x n = n n 2 , 0 n < 4 .

Using the summation formula for the DTFS harmonic function and not using the

tables or properties, find the harmonic function X k .

X k =1

N0

x n ej2 kn / N

0

n=0

3

=1

N0

n n 2{ }ej2 kn / N

0

n=0

3

X k =

1

41 e

j2 2k / 4( )=

1

41 e

j k=

1 ej k

4

Harmonic Functions Using Tables and Properties

5. Using the DTFS table of transforms and the DTFS properties, find the DTFSharmonic function of each of these periodic signals using the representation time,

N

F, indicated.

(a) x n = 6cos 2 n / 32( ) , N

F= 32

Using cos 2 n / N

0( ) FS 1

2N

0

k 1 +N

0

k + 1( ) , N

F= N

0= 32

X k = 3

32k 1 +

32k + 1( )

M. J. Roberts - 10/7/06

Solutions 9-4

k-40 40

|X[k]|

3

k-40 40

Phase of X[k]

- π

π

(b)

x n = 10sin2 n 2( )

12 , N

F= N

0= 12

Using sin 2 n / N

0( ) FS j

2N

0

k + 1N

0

k 1( ) , N

F= N

0= 12

10sin 2 n / 12( ) FS

= j512

k + 112

k 1( )

Then, using x n n

0

FSe

j2 kF0( )n

0X k

X k = j5 comb

12k + 1 comb

12k 1( )e

j3

k

k-40 40

|X[k]|

5

k-40 40

Phase of X[k]

- π

π

We can demonstrate that this solution is correct by reconstituting the signal using

M. J. Roberts - 10/7/06

Solutions 9-5

x n = X k ej2 kn / N

0

k = N0

= j512

k + 112

k 1( )ej k /3

ej2 kn / N

0

k = N0

Since the summation extends only over one period, N

0= 12 , choose the simplest

period, 6 k < 6 . In that period the two comb functions are simply twoimpulses at k = ±1.

x n = j5 k + 1 k 1( )ej2 k n /12 1/6( )

k = 6

5

x n = j5 e

j2 k n /12 1/6( )e

j2 k n /12 1/6( )= j5 e

j2 k n /12 1/6( )e

j2 k n /12 1/6( )

x n = j5 j2sin 2 n / 12 1 / 6( )( ) = 10sin 2n 2

12

We could have chosen a different period, for example 4 k < 16 . Then

x n = j5 k 11 k 13( )ej2 k n /12 1/6( )

k =4

15

x n = j5 e

j22 n /12 1/6( )e

j26 n /12 1/6( )= j5 e

j2 11n /12 11/6( )e

j2 13n /12 13/6( )

x n = j5ej2 12n /12 12/6( )

ej2 n /12+1/6( )

ej2 n /12 1/6( )

= j5ej2 n

=1

ej4

=1

ej2 n /12 1/6( )

ej2 n /12 1/6( )

x n = j5 j2sin 2 n / 12 1 / 6( )( ) = 10sin 2n 2

12

which is the same answer as before (with somewhat more effort).

(c)

x n =x

1n / 8 , n / 8 an integer

0 , otherwise

, N

F= 48

where x

1n = sin 2 n / 6( )

Using the time-scaling property of the DTFS,

M. J. Roberts - 10/7/06

Solutions 9-6

X k =

1

8X

1k

X

1k =

j

26

k + 16

k 1( ) , N

F= N

0= 6

X k =

j

166

k + 16

k 1( ) , N

F= N

0= 48

k-48 48

|X[k]|

0.0625

k-48 48

Phase of X[k]

- π

π

(d) x n = e

j2 n , N

F= 6

Notice that this function, although written as x n = e

j2 n could be

more simply written as x n = 1 because for any integer value of

discrete time, n, the result is always the same, 1. Therefore

e

j2 n= 1

FS

6k

k-40 40

|X[k]|

1

k-40 40

Phase of X[k]

- π

π

M. J. Roberts - 10/7/06

Solutions 9-7

(e)

x n = cos2 n

16cos

2 n 1( )16

, N

F= 16

cos2 n

16cos

2 n 1( )16

FS 1

2

16k 1 +

16k + 1

16k 1 e

j k /8

16k + 1 e

j k /8

cos2 n

16cos

2 n 1( )16

FS 1 ej /8

2comb

16k 1 +

1 ej /8

2comb

16k + 1

k-40 40

|X[k]|

0.25

k-40 40

Phase of X[k]

- π

π

(f)

x n = sin33 n

32= sin

33 2 n

64 , N

F= N

0= 64

The period of this signal is 64.

This form does not appear directly in Appendix E. Therefore we must eitheruse what is in Appendix E with some properties to get to this form or simplyapply the definition of the DTFS harmonic function directly.

From Appendix E,

e

j2 n / N0

FS

N0

k 1

Using the frequency-shifting property,

e

j2 mn / N0 e

j2 n / N0

FS

N0

k m 1

M. J. Roberts - 10/7/06

Solutions 9-8

e

j2 n m+1( )/ N0 FS

N0

k m + 1( )or

e

j2 mn / N0

FS

N0

k m

Then

ej2 mn / N

0 + ej2 mn / N

0

2= cos 2 mn / N

0( ) FS 1

2N

0

k m +N

0

k + m( )and

ej2 mn / N

0 ej2 mn / N

0

j2= sin 2 mn / N

0( ) FS j

2N

0

k + mN

0

k m( )

X k =

j

264

k + 3364

k 33( )

k-64 64

|X[k]|

0.5

k-64 64

Phase of X[k]

- π

π

We can demonstrate that this solution is correct by reconstituting the signalusing

x n = X k ej2 kn / N

0

k = N0

=j

264

k + 3364

k 33( )ej2 kn / N

0

k = N0

Since the summation extends only over one period, N

0= 64 , choose the simplest

period, 32 k < 32 .

x n =j

264

k + 3364

k 33( )ej2 kn / N

0

k = 32

31

We must now determine for which values of k, 32 k < 32 , the periodicimpulse functions are not zero. Take the first periodic impulse function,

M. J. Roberts - 10/7/06

Solutions 9-9

64k + 33 .

Its impulses occur whenever k + 33 is an integer multiple of 64. The only valueof k in the range, 32 k < 32 , for which that is true is k = 31. Similarly, for

64k 33

the only value for which it is non-zero is k = 31. Therefore we can write thesummation as

x n =j

2k 31 k + 31( )e

j2 kn / N0

k = 32

31

=j

2e

j2 31n /64( )e

j2 31n /64( )( )

x n =j

2e

j2 31n /64( )e

j2 31n /64( )( ) =j

2j2sin 2 31n / 64( )( )( )

= sin 2 31n / 64( )( ) = sin 31 n / 32( )

But this can also be written as

x n =j

2e

j2 64n /64( )

=1

ej2 33n /64( )

ej2 64n /64( )

=1

ej2 33n /64( )

=j

2e

j2 33n /64( )e

j2 33n /64( )( )

x n =

j

2j2sin 2 33n / 64( )( )( ) = sin 33 n / 32( )

If these two results are to both be correct

sin 31 n / 32( ) = sin 33 n / 32( )

for any integer value of n. We can write

sin 31 n / 32( ) = sin 31n / 32 2 n( ) = sin31 n 64 n

32

= sin 33 n / 32( ) = sin 33 n / 32( )

proving that the two expressions are indeed equivalent for integer values of n.

(g) x n = rect

5n

11n

M. J. Roberts - 10/7/06

Solutions 9-10

Using

rectW

nN

0

nFS 2W + 1

N0

drcl k / N0,2W + 1( )

and

drcln

2m + 1,2m + 1 =

2m+1n and the fact that

2W + 1 = N

0

X k =

11k

Agrees with 1

FS

N0

k because x n = rect

5n

11n = 1

and in 1

FS

N0

k , N

0 can be arbitrarily chosen. The meaning

of the result is the same regardless of the choice of N

0.

k-40 40

|X[k]|

1

k-40 40

Phase of X[k]

- π

π

This result was obtained for a period of N

0= 11. But when the function is a

constant, we can choose any period we like and get an equivalent result(because a constant repeats exactly in any “period” you choose). For

example, if we let N

0= 4 the transform pair,

1

FS

N0

k , yields

X k =

4k . Then, reconstituting the signal from its DTFS,

x n = X k ej2 kn / N

0

k = N0

=4

k ej2 kn / N

0

k = N0

Summing over the period, 2 k < 2 , yields

x n =4

k ej2 kn / 4

k = 2

1

= ej2 0( )n / e

= 1 .

If we chose the period, 2 k < 6 we would get

M. J. Roberts - 10/7/06

Solutions 9-11

x n =4

k ej2 kn / 4

k =2

5

= ej2 4( )n / 4

= 1 ,

which is exactly the same result. In general, for any choice of period and anyrange of k covering one period,

x n =N

0

k ej2 kn / N

0

k = k0

k0

+ N0

1

= ej2 qN

0n / N

0 = 1

where the integer q lies in the range, k

0q < k

0+ N

0.

(h) x n = rect

2n

21n 3 ,

N

F= 21

Using

rectN

w

nN

0

nFS

2Nw

+ 1

N0

drcl k / N0,2N

w+ 1( )

and x n n

0

FSe

j2 kn0

/ N0 X k ,

X k = 5 / 21( )drcl k / 21,5( )e

j2 k /7

k-40 40

|X[k]|

0.25

k-40 40

Phase of X[k]

- π

π

6. Find the DTFS harmonic function of

x n =3

m3

m 1

m=

n

with N

F= N

0= 3 .

M. J. Roberts - 10/7/06

Solutions 9-12

3m

3m 1

FS1 / 3 1 / 3( )e

j2 k /3 , N

F= N

0= 3

Using

x mm=

nFS

X k

1 ej2 k / N

0

, k 0

3

m3

m 1

m=

nFS 1

3

1 ej2 k /3

1 ej2 k /3

=1

3 , k 0 ,

N

F= N

0= 3

This is correct because, since the representation time is 3, the corresponding time-

domain function is 3

m which (except for an unknown constant) is the

accumulation of the backward difference of itself,

x n =3

m3

m 1

m=

n

=3

m + C .

7. A periodic DT signal, x n , is exactly described for all discrete time by its

DTFS,

X k =

8k 1 +

8k + 1 + j2

8k + 2 j2

8k 2( )e

j k / 4 .

(a) Write a correct analytical expression for x n in which 1 , j, does not

appear.

cos 2 n / 8( ) FS 1

28

k 1 +8

k + 1( )

2cos 2n / 8( ) FS

8k 1 +

8k + 1

sin 4 n / 8( ) = sin 2n / 4( ) FS j

28

k + 28

k 2( )

4sin 2 n / 4( ) FS

j28

k + 28

k 2( )

2cos2 n 1( )

8+ 4sin

2 n 1( )4

FS 8k 1 +

8k + 1

+ j28

k + 2 j28

k 2

ej2 k /8

Therefore

x n = 2cos2 n 1( )

8+ 4sin

2 n 1( )4

(b) What is the value of x n at n = 10 ?

M. J. Roberts - 10/7/06

Solutions 9-13

x 10 = 2cos2 10 1( )

8+ 4sin

2 10 1( )4

= 2cos 11 / 4( ) + 4sin 11 / 2( )

x 10 = 2cos 11 / 4( ) 4sin 11 / 2( ) = 2cos 3 / 4( ) 4sin 3 / 2( ) = 2 1 / 2( ) 4 1( )

x 10 = 4 2 / 2 = 4 2 2( ) / 2 = 3.657 / 1.414 = 2.586

8. Find the average signal power of

x n = rect

4n

20n

directly in the DT domain and then find its harmonic function X k and find the

signal power in the k domain and show that they are the same.

In the time domain:

Px

=1

N0

x n2

n= N0

=1

20rect

4n

20n

2

n= 20

=1

20rect

4n

20n

2

n= 10

9

Px

=1

201

n= 4

4

=9

20

In the k domain:

rect

4n

20n

FS 9

20drcl k / 20,9( )

Px

= X k2

k = N0

= 9 / 20( )drcl k / 20,9( )2

k = 20

= 9 / 20( )2

drcl k / 20,9( )2

k = 20

Doing the summation numerically in MATLAB, P

x= 9 / 20 . Check.

9. Using the frequency shifting property of the DTFS find the time-domain signal

x n corresponding to the harmonic function,

X k = 7 / 32( )drclk 16

32,7 .

M. J. Roberts - 10/7/06

Solutions 9-14

rect

3n

32n

FS7 / 32( )drcl k / 32,7( )

rect3

n32

n( )ej2 16n /32( ) FS 7

32drcl

k 16

32,7

rect3

n32

n( )ej n FS 7

32drcl

k 16

32,7

rect3

n32

n( ) 1( )n

FS 7

32drcl

k 16

32,7

This frequency shifting causes the sign of the time-domain function to alternate.

10. Find the DTFS harmonic function for

x n = rect

3n

8n

with the representation time, N

F= 8 . Then, using MATLAB, plot the DTFS

representation,

xF

n = X k ej2 kn /8

k =0

7

over the DT range 8 n < 8 . For comparison, plot the function,

xF 2

n = X k ej2 kn /8

k =13

20

over the same range. The plots should be identical.

rect

3n

8n

FS7 / 8( )drcl k / 8,7( )

M. J. Roberts - 10/7/06

Solutions 9-15

n-8 8

xF

[n]

1

k = 0:7

n-8 8

xF

[n]

1

k = 13:20

System Response to Periodic Excitation

11. Graph the response of the DT system in Figure E-11 to the periodic excitation

x n =

4n

4n 2 over the time range 0 n < 8 .

x[n] + -y[n]

D

0.8 10Figure E-11

The difference equation is

y n 0.8y n 1 = 10x n 1 .

The excitation can be written as

x n = 1 / 4( )1

k 1 ej k( )

X k

ej2 kn / 4

k = 4

. The

response is in the form

y n = Y k ej2 kn / 4

k = 4

.

The difference equation for a single harmonic becomes

Y k e

j2 kn / 40.8Y k e

j2 k n 1( )/ 4

= 10 1 / 4( )1

k 1 ej k( )e

j2 k n 1( )/ 4 .

Simplifying,

M. J. Roberts - 10/7/06

Solutions 9-16

Y k =10 1 / 4( )

1k 1 e

j k( )ej k / 2

1 0.8ej k / 2

and, since 1

k is just the DT constant 1,

y n = 2.5

1 ej k( )e

j k / 2

1 0.8ej k / 2

ej kn / 2

k = 4

.

n7

y[n]

-8

8

12. A DT system described by the difference equation

2 y n + y n 1 = 4x n

is excited by x n = 1

6n . Graph the

x n and

y n over the time period

0 n < 12 .

From the difference equation the frequency response is

H e

j( ) =4

2 + ej

.

Then the ratio of the harmonic function of the response to the harmonic functionof the excitation is

Y k

X k=

4

2 + ej2 k / N

F

=4

2 + ej k /3

.

Using 1

FS

NF

k and

N0

nFS 1

N0

mk , N

F= mN

0 we get the

DTFS pair

x n = 1

6n

FSX k =

6k 1 / 6( )

1k

So the harmonic function of the response is

M. J. Roberts - 10/7/06

Solutions 9-17

Y k =

4

2 + ej k /3 6

k 1 / 6( )1

k( )and the response is

y n =4

2 + ej k /3 6

k 1 / 6( )1

k( )ej kn /3

k = 6

y n =4

2 + ej k /3 6

k 1 / 6( )1

k( )ej kn /3

k =0

5

=4

3

1

6

4ej kn /3

2 + ej k /3

k =0

5

n12

x[n]1

n12

y[n]2

DTFS Basics

13. Based on a representation time N

F= 4 the DTFS harmonic function

X k of a

signal x n has the following values.

X 1 = 2 j2 , X 0 = 4 , X 1 = 2 + j2 , X 2 = 3

(a) What is the numerical value of X 3 ?

Because of the periodicity of X, X 3 = X 1 = 2 j2

(b) What is the numerical value of X 22 ?

X 22 = X 22 4 5 = X 2 = 3

(c) What is the numerical average value of x n ?

Average value of x n = 4

M. J. Roberts - 10/7/06

Solutions 9-18

Average value of x n is

1 / N

F( ) x nn= N

F

. From the DTFS harmonic

func t ion de f in i t i on , X k = 1 / N

F( ) x n ej2 kn / N

F

n= NF

a n d

X 0 = 1 / N

F( ) x nn= N

F

. Therefore

Average value of x n = X 0 = 4

14. Each signal in Figure E-14 is graphed over a range of exactly one fundamental

period. Which of the signals have harmonic functions X k that have a purely

real value for every value of k? Which have a purely imaginary value for everyvalue of k?

(a) (b)

-5 0 5-5

0

5

n-5 0 5

-5

0

5

10

n

x[n]

x[n]

Figure E-14

(a) Neither even nor odd. DTFS harmonic function is then neither purely realnor purely imaginary

(b) Even function. Therefore the DTFS harmonic function is purely real.(Even though the symmetry does not look quite right this really is exactlyone period of an even periodic function.)

Harmonic Functions Using Tables and Properties

15. A DTFS harmonic function X k is found for a signal, using a representation,

time, N

F= 16 . If

X 2 = 1 j , what is the numerical value of

X 18 ?

The DTFS is periodic with period 16. Therefore X 18 is exactly one period

removed from X 2 and they are, therefore, equal.

X 18 =

X 2 = 1 j

What is the numerical value of X 2 ?

X 2 = 1+ j

M. J. Roberts - 10/7/06

Solutions 9-19

X k =1

NF

x n ej2 kn / N

F

n= NF

X k =1

NF

x n ej2 kn / N

F

n= NF

When the sign of k is changed every term in the sum is changed to its complexconjugate. Therefore the overall harmonic function is also changed to its complexconjugate.

16. Using the DTFS table of transforms and the DTFS properties, find the harmonicfunction of each of these periodic signals using the representation time

N

F,

indicated.

(a) x n = e

j2 n /16

24n ,

N

F= 48

From the table,

e

j2 n /16 FS

16k 1 ,

N

F= 16

Therefore, using the change of period property,

ej2 n

16 FS 16k / 3 1 , k / 3 an integer

0 , otherwise

, N

F= 48

Also, from the table,

NF

= mN0

ej2 n / N

0FS

NF

k m

48 = 3 16

ej2 n /16 FS

48k 3

The function, 16

k / 3 1 , is a sequence of impulses. The impulses occur

wherever k / 3 1 = 16m , m any integer. Rearranging, the impulses occur where

k 3 = 48m . This describes a periodic impulse of the form, 48

k 3 .

Therefore the two answers agree.

From the table,

N0

nFS

1 / N0 ,

N

F= 24

Therefore, using the time-scaling property,

M. J. Roberts - 10/7/06

Solutions 9-20

N

0

nFS 1 / 24 , k / 2 an integer

0 , otherwise ,

N

F= 48

Then, since the two DTFS’s are both done with reference to the samerepresentation time, using the multiplication-convolution duality property,

X k = 248

k 3 , k / 3 an integer

0 , otherwise

1 / 24 , k / 2 an integer

0 , otherwise

The non-zero impulses in the first harmonic function occur at values of k forwhich k + 3 is an integer multiple of 48. Therefore all these k’s must be odd.The values of k for which the second harmonic function is non-zero are all even.Therefore

X k = 0 , for all k.

This result implies that the original DT function, x n = e

j2 n /16

24n , is

zero. A periodic convolution of two periodic signals using a common period (48in this case) is equivalent to an aperiodic convolution either of the two functionswith one period of the other function. Therefore

x n = ej2 n /16

n + n 24( )one period of the periodic impulse

function using the common period of 48

= ej2 n /16

+ ej2 n 24( )/16

x n = ej2 n /16

+ ej2 n /16

ej48 /16

= ej2 n /16

+ ej2 n /16

ej3

= 1

= ej2 n /16

ej2 n /16

= 0

Check.

(b) x n = rect

5n

24n( )sin 2 n / 6( ) ,

N

F= 24

X k = 11 / 24( )drcl k / 24,11( ) j / 2( )

24k + 4

24k 4( )

X k = 11 / 24( )drcl k / 24,11( ) j / 2( ) k + 4 k 4( )

X k = j11

48drcl

k + 4

24,11 drcl

k 4

24,11

M. J. Roberts - 10/7/06

Solutions 9-21

(c) x n = x

1n x

1n 1 where

x

1n = tri n / 8( )

20n ,

N

F= 20

X

1k = 8 / 20( )sinc

28k / 20( )

20k = 2 / 5( )sinc

22k / 5( )

20k

X1

k =2

5sinc

22 k 20q( )

5q=

Using x n x n 1

FS1 e

j2 k / NF( )X k

X k =2

51 e

j k /10( ) sinc2

k 20q

5q=

17. Find the signal power of

x n = 5sin 14 n / 15( ) 8cos 26 n / 30( ) .

The fundamental period of this signal is 30.

x n = 5sin2 14 n

308cos

2 13 n

30

X k = j5 / 2( )

30k + 14

30k 14( ) 4

30k 13 +

30k + 13( )

Px

= X k2

k = 30

= j5 / 2( )30

k + 1430

k 14( ) 430

k 13 +30

k + 13( )2

k = 15

14

P

x= 5 / 2( )

2

+ 5 / 2( )2

+ 42

+ 42

= 25 / 2 + 32 = 89 / 2 = 44.5

18. F i n d t h e D T F S h a r m o n i c f u n c t i o n X k o f

x n = rect

1n 1 rect

1n 4( ) 6

n using a representation time N

F= 6 .

Then plot the partial sum x

Nn = X k e

j nk /3

k = N

N

for N = 0,1,2 and then

the total sum x n = X k e

j nk /3

k =0

5

.

x n = rect

1n 1

6n rect

1n 4

6n

M. J. Roberts - 10/7/06

Solutions 9-22

Using

NF = N0

rectNwn[ ] N0

n[ ] FS 2Nw +1

N0

drcl k / N0 ,2Nw +1( )

X k =1

6

sin k / 2( )sin k / 6( )

ej k /3

1

6

sin k / 2( )sin k / 6( )

ej4 k /3

=1

6

sin k / 2( )sin k / 6( )

ej k /3

ej4 k /3( )

X k =1

6e

j5 k /6sin k / 2( )sin k / 6( )

ej k / 2

ej k / 2( ) =

j

3e

j5 k /6sin

2k / 2( )

sin k / 6( )

xN

n =j

3

sin2

k / 2( )sin k / 6( )

ej5 k /6

ej nk /3

k = N

N

=j

3

sin2

k / 2( )sin k / 6( )

ej k n 5/ 2( )/3

k = N

N

n-12 12

x[n]

-2

2

N = 0

n-12 12

x[n]

-2

2

N = 1

n-12 12

x[n]

-2

2

N = 2

n-12 12

x[n]

-2

2

Total Sum from 0 to 5

19. Find and sketch the magnitude and phase of the DTFS harmonic function of

x n = 4cos 2 n / 7( ) + 3sin 2 n / 3( )

which is valid for all discrete-time.

The least common period of these two signals is N

0= 21. Using the tables and

the change-of-period property of the DTFS, the DTFS harmonic function is

M. J. Roberts - 10/7/06

Solutions 9-23

X k = 2

21k 3 +

21k + 3( ) + j3 / 2( )

21k + 7

21k 7( )

k-21 21

|X[k]|

2

k-21 21

Phase of X[k]

- π

π

20. Match each DT function in Figure E-20 to the magnitude of its DTFS harmonicfunction (if there is a match) using a period of 16 for both in every case.

(a)-A (b)-F (c)-none (d)-B (e)-E

0 5 10 15-1

0

1

x[n]

n 0 5 10 15

-2

0

2

x[n]

n0 5 10 15

0

0.5

1

x[n]

n0 5 10 15

-1

0

1

x[n]

n0 5 10 15

0

0.5

1

x[n]

n

A B C D E F

0 5 10 150

0.5

|X[k

]|

k0 5 10 15

0

0.5

|X[k

]|

k0 5 10 15

0

0.05

0.1

|X[k

]|

k0 5 10 15

0

0.1

0.2

|X[k

]|

k0 5 10 15

0

0.5

1

|X[k

]|

k0 5 10 15

0

0.5

1

|X[k

]|k

Figure E-20

System Response to Periodic Excitation

21. Find and plot versus N

F= 2n

0 the signal power of the response

y n to the

periodic excitation x n = u n u n n

0( ) 2n0

n in the system in Figure E-

21 for 0 n

0< 10 .

x[n] + -y[n]

D0.9

Figure E-21

y n = x n 0.9 y n 1

or

M. J. Roberts - 10/7/06

Solutions 9-24

y n + 0.9 y n 1 = x n .

From the difference equation, the frequency response is

H ej( ) =

1

1+ 0.9ej

.

So the ratio of the response harmonic function to the excitation harmonic functionis

Y k

X k=

1

1+ 0.9ej2 k / N

F

Using

u n n0

u n n1( ) N

0

nFS e

j k n1+ n

0( )/ N0

ej k / N

0

n1

n0

N0

drclk

N0

,n1

n0

, NF

= N0

we get the DTFS pair

x n = u n u n n0( ) 2n

0

nFS

X k =e

j k / 2

ej k / 2n

0

1

2drcl

k

2n0

,n0

, NF

= 2n0

.

Then

Y k =1

1+ 0.9ej k / n

0

ej k / 2

ej k / 2n

0

1

2drcl

k

2n0

,n0

The average signal power of the response is, from Parseval’s theorem,

Y k2

k = N0

=1

2

1

1+ 0.9ej k / n

0

ej k / 2

ej k / 2n

0

drclk

2n0

,n0

2

k =0

2n0

1

.

Y k2

k = N0

=1

2

drcl2 k

2n0

,n0

1+ 0.9ej k / n

0

2

k =0

2n0

1

M. J. Roberts - 10/7/06

Solutions 9-25

n0

10

Py

60

22. Find and graph the response of the system in Figure E-22 to the periodic

excitation x n = 2

8n

8n 2 over the range 0 n < 16 .

x[n]

+

+ -

+

y[n]

D+

+

D

0.3

0.9

0.7

Figure E-22

Relating this to the standard Direct-Form-2 realization, the difference equation is

y n / 0.9 + 0.3y n 1 + 0.7 y n 2 = x n + x n 2

and the frequency response is

H ej( ) =

1+ ej2

1.111+ 0.3ej

+ 0.7ej2

= 0.91+ e

j2

1+ 0.27ej

+ 0.63ej2

.

Therefore the harmonic response is

H k =Y k

X k= 0.9

1+ ej k / 2

1+ 0.27ej k / 4

+ 0.63ej k / 2

Using

N0

nFS 1

N0

mk , N

F= mN

0, and the time-shifting property of

the DTFS we get the DTFS pair

M. J. Roberts - 10/7/06

Solutions 9-26

x n = 2

8n

8n 2

FSX k = 1 / 8( )

1k 2 e

j k / 2( ) , NF

= 8

Therefore

Y k = 0.91+ e

j k / 2

1+ 0.27ej k / 4

+ 0.63ej k / 2

1

81

k 2 ej k / 2( )

Y k = 0.1125

1+ ej k / 2( ) 2 e

j k / 2( )1+ 0.27e

j k / 4+ 0.63e

j k / 2

and

y n = 0.1125

1+ ej k / 2( ) 2 e

j k / 2( )1+ 0.27e

j k / 4+ 0.63e

j k / 2e

j2 kn /8

k = 8

n16

x[n]

1

2

n16

y[n]

1

2