exercises solved5

7/28/2019 Exercises Solved5

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DIGITAL SIGNAL PROCESSING

Discrete Fourier Transform

Solved problems:

1. Find the Discrete Fourier Transform of the following signal:

}1,1,1,1{3,2,1,0],[ ==nnf .

Solution:

The Discrete Fourier Transform (DFT) of the discrete signal f[n]

is F[k]:

NjN

n

kn eWNkWnfkF

21

0

,1,...1,0,][][

=

=== wherefor

In our case, N=4 so the last equation for F[k], written in matrix

notation is:

fTF = , where T is matrix of the transform, with elements Tij=Wij,

i, j = 0..N- 1:

jWjW

jeeW

WWW

WWW

WWW

F

F

F

F

jj

===

===

=

322

24

2

963

642

32

;1)(

;

1

1

1

1

1

1

1

1111

]3[

]2[

]1[

]0[

Using the property of periodicity of W ( ZrWWNrpp

=+ , ) with

basic period N=4, we get:

jWWWWWWWWW ========= 89429924660444 ;1;1

Now we can compute F[k], k=0..3 with this matrix equation:

+

=

=

j

j

jj

jj

F

F

F

F

22

0

22

0

1

1

1

1

11

1111

11

1111

]3[

]2[

]1[

]0[

DFT of the signal f[n], n=0..3 is F[k], k=0..3 = {0, 2-2j, 0, 2+2j}

2. Find the inverse DFT of F[k], k=0,1,2,3 = {0, 2-2j, 0, 2+2j}.

Solution:

The Inverse Discrete Fourier Transform (IDFT) of F[k] is thesignal f[n]:


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1,...1,0,][1

][1

0

==

=

NnWkF

Nnf

N

k

kn, where N

j

eW

2

=

The last equation written in matrix notation is:

FTN

1FTf

*1==

, whereTis the matrix of transform.

From previous we have

=

=

jj

jj

jj

jj

11

1111

11

1111

,

11

1111

11

1111

*TT so

=

+

=

1

1

1

1

22

0

22

0

11

1111

11

1111

1

]3[

]2[

]1[

]0[

j

j

jj

jj

N

f

f

f

f

We got that IDFT of F[k], k=0..3 is the signal f[n], n=0..3 = {1, 1, -1, -1}.

3. Find the Z-transform of the following signal:

f[n], n=...-2,-1,0,1,2,3,5,... = {...0, 0, 1, 1, -1, -1, 0, 0,...}

Find its frequency spectrum for the following frequencies:

2

3,,

2

,0

==== TTTT and

Solution:

The Z-transform of the signal f[n] is:

)1()1()1(11][)( 21121321

=

+=++=+== zzzzzzzzznfzF

n

n

The frequency spectrum of the signal f[n] is:

)( TjeF = Tjez

zF =

)(

)1()1()( 2 TjTjTj eeeF +=

The spectrum for the specific frequencies is:

;222)1()1()1()(

;000)1()1()(

;222)1()1()1()(

;0)11()11()(

32

3

2

3

2

22

0

jjeeeF

eeeF

jjeeeF

eF

jjj

jjj

jjj

j

+=+=+=

==+=

==+=

=+=

If we compare the values of the frequency spectrum with thevalues from the previous example (DFT of the same signal) we can

see that by computing N-point DFT of a signal we actually


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compute N equally-spaced samples from the frequency spectrum of

that signal:

1,...1,0),(|)(][

2

2 ===

=

NkeFeFkFk

Nj

kN

T

Tj

4. Find the even and the odd part of the following signals:

f[n], n=0..3 = {1, 1, -1, -1}, and

F[k], k=0..3 = {0, 2-2j, 0, 2+2j}

Solution:

The even part of f[n] is fp[n], where fp[n] is computed by:

)])[(][(2

1][ nNfnfnf

p

+= ,

while odd part of f[n] is fn[n]:

)])[(][(2

1][ nNfnfnfn =

Where (N-n) means NnN mod)( .

For example, if f[n] is the following signal:

]}1[],2[],...2[],1[],0[{1..0],[ == NfNffffNnnf ,

then, the signal )][( nNf is:

]}1[],2[],...2[],1[],0[{1..0)],[( ffNfNffNnnNf==

In our case, f[n]={1, 1, -1, -1}, so f[(N-n)]={1, -1, -1, 1}

The even part of f[n] is:

)])[(][(2

1][ nNfnfnfp += = {1, 0, -1, 0}, while the odd part is:

)])[(][(2

1][ nNfnfnfn = = {0, 1, 0, -1}.

Similarly, for F[k] we get:

F[(N-k)]={0, 2+2j, 0, 2-2j};

The even part is: Fp[k]={0, 2, 0, 2};

The odd part is: Fn[k]={0, -2j, 0, 2j}.

5. Find the DFT of the signals fp[n] and fn[n], using the resultsfrom the previous example.

Solution:


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On figure 14 are shown the correspondences between the real, the

imaginary, the even and the odd parts of signal and his Discrete

Fourier Transform. The black arrow connects the signals

that form transform pair.

Figure 14

According to this property of symmetry of the Discrete Fourier

Transform, if DFT of f[n]={1, 1, -1, -1} is F[k]={0, 2-2j, 0, 2+2j}, then DFT of

the even part of f[n], fp[n], which is real, is the real even part

of F[k], that is Fp[k]={0, 2, 0, 2} (computed in the previous example).Similarly, DFT of the odd part of f[n], fn[n], which is also real,

according to this property is the imaginary odd part of F[k], that

is Fn[k]={0, -2j, 0, 2j}.

6. If the signal f[n], n=0..N-1 is real and even, show that the

signal F[k]=DFT{f[n]}, k=0..N-1 is also real and even.

Solution:We know that the signal f[n] is real i.e. ][][ * nfnf = , and that is

also even, i.e. )][(][ nNfnf = .

We need to show that:

][][ * kFkF = and )][(][ kNFkF = .

If we start from the definition of DFT ][* kF we get:

=

=

=

===

1

0

1

0

*1

0

** ][][]][[][N

n

knN

n

knN

n

knWnfWnfWnfkF ;

Now, if we use the relation )][(][ nNfnf = :

=

=

1

0

* )][(][N

n

kNknWWnNfkF , 1

0==

WW

kNdue to periodicity of W

=

=

1

0

)(* )][(][N

n

nNkWnNfkF

The value of the congruence (N-n) is equal to the difference N-n

for all 0n . For n=0, (N-n)=0, while N-n=N. But, the value of)( nNk

W is same as))(( nNk

W for n=0, so we can make this

replacement:


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][][)][(][1

0

1

0

))((*kFWmfWnNfkF mk

N

m

N

n

nNk===

=

=

We got that ][][* kFkF = which means that F[k] is real.

In order to show that F[k] is even we start from )][( kNF :

=

=

1

0

))((][)][(

N

n

kNnWnfkNF , and same as previously, the value of the

congruence (N-k) is different from N-k only for k=0. But for k=0,))(( kNn

W

is same as)( kNn

W, so again we can make the replacement

))(( kNnW

=

)( kNnW

:

=

=

==

1

0

1

0

)(][][)][(

N

n

knNnN

n

kNnWWnfWnfkNF , 1

0==

WW

Nn

][][)][()][(][)][(*

1

0

*1

0

**1

0kFkFWnfWnfWnfkNF

N

n

knN

n

knN

n

kn

=====

=

=

=

We got that ][)][( kFkNF = which means that F[k] is even.

7. If the length of x[n] is N=4, and if its 8-point DFT is:

}23,3,21,1,21,3,23,5{7..0],[8 jjjjkkX ++== , find the 4-

point DFT of the signal x[n].

Solution:The samples of ][8 kX are eight equally-spaced samples from the

frequency spectrum of the signal x[n]:

kT

TjeXkX

82)(][8

=

= , k=0..7

More precisely, they are samples from the spectrum for the

following frequencies: }4

7,

2

3,

4

5,,

4

3,

2,

4,0{

=T .

With 4-point DFT of x[n] we get 4 samples from the spectrum of x[n]:

kT

TjeXkX

42)(][4

=

= , k=0..3

These samples are for the following frequencies:

}2

3,,

2,0{

=T

If we compare the two sets of frequencies we can easily see

that:


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3]6[)(]3[

1]4[)(]2[

3]2[)(]1[

5]0[)(]0[

82

3

4

84

82

4

80

4

===

===

===

===

XeXX

XeXX

XeXX

XeXX

j

j

j

j

So, 4-point DFT of x[n] is: }3,1,3,5{][4 =kX

8. Two N-length signals are given, N is even number:

1..0],[1..0],[ == NnngNnnf i

The relation between f[n] and g[n] is: ][)1(][ nfng n = .

Find the relation between ]}[{][ ngDFTkG = and ]}[{][ nfDFTkF = .

Solution:

=

=

==

1

0

1

0

][)1(][][N

n

N

n

knnknWnfWngkG

if we replacen)1( by

jne :

=

=

==

1

0

2

221

0

2

][][][N

n

nN

Njkn

NjN

n

jnkn

Nj

eenfeenfkG

)]2

[(][][

1

0

)

2

(2

NkFenfkG

N

n

Nkn

N

j

+==

=

+

, k=0..N-1

We got that G[k] is circular shifted (for2

Nsamples) version of

F[k].

2. The relation between 12

..0],[1..0],[ ==N

nngNnnf and , where N

is even number, is:

12

..0],2

[][][ =++= NnNnfnfng

Find the relation between ]}[{][ ngDFTkG = , 12

..0 =N

k and the

frequency spectrum of the signal f[n], )( TjeF .

Solution:

=

=

++==

12

0

12

0

])

2

[][(][][

N

n

kn

N

n

knW

NnfnfWngkG


7/9

=

=

++=

12

0

12

0

]2

[][][

N

n

kn

N

n

knW

NnfWnfkG

We multiply the second sum byk

N

W

2 (its value is 1):

=

=

++=

12

0

2

12

0

]2

[][][

N

n

kN

kn

N

n

knWW

NnfWnfkG

=

+

=

++=

12

0

)2

(1

2

0

]2

[][][

N

n

kN

n

N

n

knW

NnfWnfkG

=

=

+=

1

2

12

0

][][][N

N

m

km

N

n

knWmfWnfkG

12

..0],2[][][][1

0

221

0

====

=

=

NkkFenfWnfkGN

n

knN

jN

n

kn

12

..0,)(]2[][2

2 ====

NkeFkFkG

kN

T

Tj

We got that the samples of G[k] are2

Nequally-spaced samples

from the frequency spectrum of f[n].

9. From the real signal 1..0],[ = Nnnf , where N is even number,

the following signals gi[n], i=1..4 are constructed:

1..0],1[][1 == NnnNfng

=

==

12..],[

1..0],[][2

NNnNnf

Nnnfng

=

==

12..,0

1..0],[][3

NNn

Nnnfng

=

120,0

120],2

[][4

Nnn

Nnnn

fng

neparno,e

parno,e

Find the DFT of the signals gi[n], i=1..4, and find their relation

with the frequency spectrum )( TjeF of the signal f[n].

Solution:

Its easy to see that the signal ][1 ng is the inverted signal f[n].


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=

=

==

1

0

21

0

11 ]1[][][N

n

knN

jN

n

kn enNfWngkG

If we make replacement nNm = 1 :

=

=

==

1

0

)(2

)1(21

0

)1(2

1 ][][][

N

m

kmN

jNkN

jN

m

mNkN

j

emfeemfkG

=

=

1

0

)(222

1 ][][N

m

kmN

jkN

jNkN

j

emfeekG

Since 1

2

= Nk

Nj

e

we get that:

][][

2

1 kFekGk

Nj

=

And the relation between ][1 kG and )(TjeF e:

kN

T

Tjk

Nj

eFekG

2

2

1 )(][=

= , k=0..N-1

Since the module ofk

Nj

e

2

is 1, it turns out that with DFT of the

signal ][1 ng we get samples from the amplitude spectrum of the

signal f[n].

The signal ][2 ng has length of 2N, and its obtained by periodic

extension of f[n].

=

=

+=

122

21

0

2

2

2 ][][][N

Nn

knN

jN

n

knN

j

eNnfenfkG

If we make replacement m=n-N:

kNN

jN

m

kmN

jN

n

knN

j

eemfenfkG

=

=

+= 221

0

2

21

0

2

2

2 ][][][

kN

m

kmN

jN

n

knN

j

emfenfkG )1(][][][1

0

2

21

0

2

2

2 +=

=

=

))1(1(][][1

0

2

2

2k

N

n

knN

j

enfkG +=

=

We got that the odd-indexed samples of ][2 kG are zero, while the

remaining are equal to the samples of 2 ][kF :

= =

120,0

120,)(2][

2

22

Nkk

NkkeFkG k

NT

Tj

neparno,e

parno,e


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The signal ][3 ng has length of 2N, and its obtained by adding N

zeros to f[n].

kN

T

Tjkn

NjN

n

N

n

knN

j

eFenfengkG

2

22

21

0

12

0

2

2

33 )(][][][

=

=

=

=== , k=0..2N-1

We got that the samples of ][3 kG are 2N equally-spaced samples

from the spectrum of f[n].

The signal ][4 ng also has length of 2N samples, and its obtained

from f[n] by inserting one zero after every sample from f[n]. The

DFT of ][4 ng is:

=

=

==

1

0

22

212

0

2

2

44 ][][][N

m

kmN

jN

n

knN

j

emfengkG

kN

T

TjN

n

knN

j

eFenfkG

2

1

0

2

4 )(][][=

=

== , k=0..2N-1

We got that the samples of ][4 kG are 2N equally-spaced samples

from the spectrum of f[n] for the frequency range from 0 to 4 ,

which means that ][4 kG can be obtained with one periodic

expansion of ][kF .

exercises solved5

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