Discrete-Time Fourier Series Concept

A signal can be represented as a linear combination of sinusoids.

Discrete-Time Fourier Methods

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Discrete-Time Fourier Series Concept

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Discrete-Time Fourier Series Concept

The relationship between complex and real sinusoids

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Discrete-Time Fourier Series Concept

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Discrete-Time Fourier Series Concept

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1

Discrete-Time Fourier Series Concept

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Discrete-Time Fourier Series Concept

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∑

c x [ k ] e j 2 π kn/N

cx [ k ] =

k= N

1 N

∑

∑

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A periodic discrete-time signal can be expressed as x[n] =

x [ n ] e− j 2 π kn/N

1 N

∑ X[ k ]e

k= N

j 2 π kn/N

X[ k ] =

n0 + N −1

∑ x[n]e

− j 2 π kn/N

n=n0

where X [ k ] is the DFT harmonic function, N is any period of x [ n ]

n=n0

and the notation,

means a summation over any range of

∑

means a summation over any range of

k= N

k= N

consecutive k’s exactly N in length. The main difference between the

consecutive k’s exactly N in length.

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Discrete-Time Fourier Series Concept

The discrete Fourier transform (DFT) is almost identical to the DTFS.

where c x [ k ] is the harmonic function, N is any period of x [ n ] and the notation,

8

The Discrete Fourier Transform

The discrete-time Fourier series (DTFS) is similar to the CTFS. A periodic discrete-time signal can be expressed as x[n] =

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The Discrete-Time Fourier Series

n0 + N −1

Discrete-Time Fourier Series Concept

DTFS and the DFT is the location of the 1/N term. So X [ k ] = N c x [ k ].

11

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2

The Discrete Fourier Transform

The Discrete Fourier Transform

Because the DTFS and DFT are so similar, and because the DFT is so widely used in digital signal processing (DSP), we will concentrate on the DFT realizing we can always form the DTFS from c x [ k ] = X [ k ] / N.

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DFT Example

)

x ⎡⎣ n ⎤⎦ = u ⎡⎣ n ⎤⎦ − u ⎡⎣ n − 3⎤⎦ ∗ δ 5 ⎡⎣ n ⎤⎦ using its fundamental period as the representation time. X ⎡⎣ k ⎤⎦ =

∑

n= N 4

x ⎡⎣ n ⎤⎦ e

2

(

= ∑e n=0 2

n=0

∑ X[ k ]e

j 2 π kn/N

so we can find x [ n ]

k= N

function from x [ n ]? We use the principle of orthogonality like

− j 2 π kn/5

we did with the CTFS except that now the orthogonality is in discrete time instead of continuous time.

1− e− j6π k /5 e− j3π k /5 e j3π k /5 − e− j3π k /5 = = × jπ k /5 − jπ k /5 1− e− j 2π k /5 e− jπ k /5 e −e

= ∑ e− j 2π kn/5 = e− j 2π k /5

1 N

from its harmonic function. But how do we find the harmonic

)

− j 2 π kn/5

We know that x [ n ] =

− j 2 π kn/ N

= ∑ u ⎡⎣ n ⎤⎦ − u ⎡⎣ n − 3⎤⎦ ∗ δ 5 ⎡⎣ n ⎤⎦ e n=0

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The DFT Harmonic Function

Find the DFT harmonic function for

(

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( (

) = 3e )

sin 3π k / 5 sin π k / 5

− j 2 π k /5

(

)

drcl k / 5,3

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The DFT Harmonic Function

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The DFT Harmonic Function

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3

The DFT Harmonic Function

The DFT Harmonic Function

Below is a set of complex sinusoids for N = 8. They form a set of basis vectors. Notice that the k = 7 complex sinusoid rotates counterclockwise through 7 cycles but appears to rotate

The projection of a real vector x in the direction of another real vector y is xT y y yT y If p = 0, x and y are orthogonal. If the vectors are complexvalued

clockwise through one cycle. The k = 7 complex sinusoid is exactly the same as the k = −1 complex sinusoid. This must be true because the DFT is periodic with period N.

p=

p=

xH y y yH y

where the x H is the complex-conjugate transpose of x. xT y and x H y are the dot product of x and y. M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

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The DFT Harmonic Function

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The DFT Harmonic Function

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The DFT Harmonic Function

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The DFT Harmonic Function

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4

The DFT Harmonic Function

Convergence of the DFT

The most common definition of the DFT is N −1

X [ k ] = ∑ x [ n ] e− j 2 π kn/N ,

x[n] =

n=0

1 N

∑ X[ k ]e

j 2 π kn/N

•  The DFT converges exactly with a finite number of terms. It does not have a Gibbs phenomenon in the same sense that the CTFS does

k= N

Here the beginning point for x [ n ] is taken as n0 = 0 . This is the form of the DFT that is implemented in practically all computer languages.

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The Discrete Fourier Transform

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DFT Properties

X [ k ] is called the DFT harmonic function of x [ n ] and k is the

harmonic number just as we have seen in the CTFS. x [ n ] and X [ k ] form a DFT pair based on N points. D FT x [ n ] ←⎯⎯ → X[ k ] N

1 From x [ n ] = N

∑ X[ k ]e

j 2 π kn/N

we see that x [ n ] is formed

k= N

by a linear combination of functions of the form e j 2 π kn/N each of which has a period N. Therefore x [ n ] must also be periodic with period (but not necessarily fundamental period) N.

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DFT Properties

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DFT Properties

It can be shown (and is in the text) that if x [ n ] is an even

function, X [ k ] is purely real and if x [ n ] is an odd function X [ k ] is purely imaginary.

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5

The Dirichlet Function

The functional form

DFT Pairs

D FT e j 2 π n/N ←⎯⎯ → mNδ mN [ k − m ] mN

D FT cos ( 2π qn / N ) ←⎯⎯ → ( mN / 2 ) (δ mN [ k − mq ] + δ mN [ k + mq ]) mN

sin (π Nt ) N sin (π t )

D FT sin ( 2π qn / N ) ←⎯⎯ → ( jmN / 2 ) (δ mN [ k + mq ] − δ mN [ k − mq ]) mN

appears often in discrete-time

D FT δ N [ n ] ←⎯⎯ → mδ mN [ k ] mN

signal analysis and is given the special name Dirichlet function. That is drcl ( t, N ) =

D FT 1←⎯⎯ → Nδ N [ k ] N

( u [ n − n ] − u [ n − n ]) ∗ δ [ n ]←⎯⎯→ e 0

sin (π Nt ) N sin (π t )

31

γ N = 2γ 1 2 2 4 3 8 4 16 5 32 6 64 7 128 8 256 9 512 10 1024

(Acquire the input data in an array x with N elements.)

. Initialize the DFT array to a column vector of zeros.

X = zeros(N,1) ; Compute the Xn’s in a nested, double for loop.

for k = 0 :N - 1 for n = 0 :N - 1 X(k + 1) = X(k + 1) + x(n + 1) * exp(-j * 2 * pi * n *k / N) ; end end

ADFT 2

M DFT 4

AFFT 2

M FFT 1

ADFT / AFFT 1

M DFT / M FFT 4

12 56 240

16 64 256

8 24 64

4 12 32

1.5 2.33 3.75

4 5.33 8

992 4032 16256 65280 261632

1024 4096 16384 65536 262144

160 384 896 2048 4608

80 192 448 1024 2304

6.2 10.5 18.1 31.9 56.8

12.8 21.3 36.6 64 113.8

1047552 1048576 10240

5120

102.3

204.8

33

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34

Generalizing the DFT for Aperiodic Signals

Generalizing the DFT for Aperiodic Signals

Pulse Train

DFT of

Pulse Train

As the period of the rectangular wave increases, the period of the DFT increases

This periodic rectangular-wave signal is analogous to the

continuous-time periodic rectangular-wave signal used to

illustrate the transition from the CTFS to the CTFT.

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32

The Fast Fourier Transform

One could write a MATLAB program to implement the DFT.

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There is a function in MATLAB fft which accomplishes the same goal and is typically much faster. This table compares the speeds of the two methods. M stands for computer multiplies and A stands for computer additions.

.

%

D FT N

D FT sinc ( n / w ) ∗ δ N [ n ] ←⎯⎯ → wrect ( wk / N ) ∗ δ N [ k ] N

The Fast Fourier Transform

%

N

D FT N

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%

− jπ k ( n1 +n0 ) /N

( n1 − n0 ) drcl ( k / N, n1 − n0 ) e− jπ k /N 2 tri ( n / N w ) ∗ δ N [ n ] ←⎯⎯→ N w drcl ( k / N, N w ) , N w an integer 1

35

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6

Generalizing the DFT for Aperiodic Signals

Generalizing the DFT for Aperiodic Signals

Normalized

DFT of

Pulse Train

The normalized DFT approaches this limit as the

period approaches infinity.

By plotting versus k / N 0 instead of k, the period of the normalized DFT stays at one.

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37

Definition of the Discrete-Time Fourier Transform (DTFT)

Forward x[n] = ∫ X( F ) e

F Form j 2 π Fn

1

1 2π

∫

2π

The function e− jΩ appears in the forward DTFT raised to the nth power. It is periodic in Ω with fundamental period 2π . n is an integer. Therefore

∞

∑ x[n]e

e- jΩn is periodic with fundamental period 2π / n and 2π is also a period

− j 2 π Fn

of e− jΩn . The forward DTFT is

n=−∞

Ω Form

Forward x[n] =

dF ←⎯→ X ( F ) =

( )

38

The Discrete-Time Fourier Transform

Inverse

F

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∞

( ) ∑ x[n]e

X e jΩ =

Inverse

− jΩn

n=−∞

∞

( ) ∑ x[n]e

F X e jΩ e jΩn dΩ ←⎯ → X e jΩ =

a weighted summation of functions of the form e− jΩn , all of which repeat with

− jΩn

( )

every 2π change in Ω. Therefore X e jΩ is always periodic in Ω with period

n=−∞

2π . This also implies that X ( F ) is always periodic in F with period 1.

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39

DTFT Pairs

By generalizing the CTFT to include transform that have impulses we were able to find CTFT's of some important practical functions. The same is true of the DTFT. The DTFT of a constant

F δ [ n ] ←⎯ →1

e jΩ 1 e jΩ 1 F = , α 1 e − α 1 − α e− jΩ e − α 1 − α e− jΩ e jΩα sin ( Ω0 ) e jΩα sin ( Ω0 ) F F n n α sin ( Ω0 n ) u [ n ] ←⎯→ j 2Ω , α < 1 , − α sin ( Ω0 n ) u [ −n − 1] ←⎯→ j 2Ω , α >1 jΩ 2 jΩ e − 2α e cos ( Ω0 ) + α e − 2α e cos ( Ω0 ) + α 2 F α n cos ( Ω0 n ) u [ n ] ←⎯ →

jΩ

e jΩ ⎡⎣ e jΩ − α cos ( Ω0 ) ⎤⎦

e

j 2Ω

− 2α e cos ( Ω0 ) + α jΩ

2

F , α < 1 , − α n cos ( Ω0 n ) u [ −n − 1] ←⎯ →

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The Generalized DTFT

We can begin a table of DTFT pairs directly from the definition. (There is a more extensive table in the text.)

F α n u [ n ] ←⎯ →

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e jΩ ⎡⎣ e jΩ − α cos ( Ω0 ) ⎤⎦

e

j 2Ω

− 2α e cos ( Ω0 ) + α jΩ

2

, α >1

41

X(F ) =

∞

∑ Ae

n=−∞

− j 2 π Fn

∞

= A ∑ e− j 2 π Fn n=−∞

does not converge. The CTFT of a constant turned out to be an impulse. Since the DTFT must be periodic that cannot the the transform of a constant in discrete time. Instead the transform must be a periodic impulse.

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7

The Generalized DTFT

The Generalized DTFT

Now consider the function X ( F ) = Aδ1 ( F − F0 ) , − 1 / 2 < F0 < 1 / 2. Its inverse DTFT is

To shorten the process, find the inverse DTFT of a periodic

x [ n ] = ∫ Aδ1 ( F − F0 ) e j 2 π Fn dF = A

impulse of the form Aδ1 ( F ) . Using the formula x [ n ] = ∫ Aδ1 ( F ) e j 2 π Fn dF = A 1

1/2

∫ δ (F)e

1

j 2 π Fn

1/2

∫ δ (F − F )e

j 2 π Fn

0

−1/2

dF = Ae j 2 π F0 n

(

)

Now change x [ n ] to x [ n ] = A cos ( 2π F0 n ) = ( A / 2 ) e j 2 π F0 n + e− j 2 π F0 n .

dF = A

−1/2

Then

proving that the DTFT of a constant A is Aδ1 ( F ) or, in radian-

F A cos ( 2π F0 n ) ←⎯ → ( A / 2 ) ⎡⎣δ1 ( F − F0 ) + δ1 ( F + F0 ) ⎤⎦

F frequency form A ←⎯ → 2π Aδ 2 π ( Ω ) .

or F A cos ( Ω0 n ) ←⎯ → π A ⎡⎣δ 2 π ( Ω − Ω0 ) + δ 2 π ( Ω + Ω0 ) ⎤⎦

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43

Forward DTFT Example

∞

∑(

n = −∞

)

u [ n − n0 ] − u [ n − n1 ] e− j 2 π Fn =

F u [ n − n0 ] − u [ n − n1 ] ←⎯ → e− j 2 π Fn0

n1 −1

∑ e− j 2π Fn

u [ n − n0 ] − u [ n − n1 ] ←⎯→

n1 − n0 −1

∑

e

− j 2 π F ( m + n0 )

=e

− j 2 π Fn0

m=0

n1 − n0 −1

∑

e

e− jπ F ( n1 − n0 ) e jπ F ( n1 − n0 ) − e− jπ F ( n1 − n0 ) e − jπ F e jπ F − e − jπ F By the definition of the sine function in terms of complex exponentials F u [ n − n0 ] − u [ n − n1 ] ←⎯ → e− j 2 π Fn0

− j 2 π Fm

m=0

Summing this geometric series

F u [ n − n0 ] − u [ n − n1 ] ←⎯ →

1 − e− j 2 π F ( n1 − n0 ) u [ n − n0 ] − u [ n − n1 ] ←⎯→ e− j 2 π Fn0 1 − e− j 2 π F F

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u [ n − n0 ] − u [ n − n1 ] ←⎯→

(

sin π F ( n1 − n0 ) sin (π F )

)

)

46

More DTFT Pairs

We can now extend the table of DTFT pairs. F δ [ n ] ←⎯ →1

1 + (1 / 2 )δ1 ( F ) 1 − e− j 2 π F F sinc ( n / w ) ←⎯ → w rect ( wF ) ∗ δ1 ( F ) F u [ n ] ←⎯ →

, n0 + n1 = 1

, ,

F tri ( n / w ) ←⎯ → w drcl2 ( F, w )

F tri ( n / w ) ←⎯ → w drcl2 ( Ω / 2π , w ) F 1←⎯ → 2πδ 2 π ( Ω )

,

F δ N0 [ n ] ←⎯ → (1 / N 0 )δ1/ N0 ( F )

F δ N0 [ n ] ←⎯ → ( 2π / N 0 )δ 2 π / N0 ( Ω )

,

F cos ( 2π F0 n ) ←⎯ → (1 / 2 ) ⎡⎣δ1 ( F − F0 ) + δ1 ( F + F0 ) ⎤⎦

sin (π wf ) πf The DTFT is a periodically-repeated sinc function.

1 F u [ n ] ←⎯ → + πδ1 ( Ω ) 1 − e− jΩ F sinc ( n / w ) ←⎯ → w rect ( wΩ / 2π ) ∗ δ 2 π ( Ω )

,

F 1←⎯ → δ1 ( F )

Compare this to the CTFT of a rectangular pulse of width w centered at t = 0.

F cos ( Ω0 n ) ←⎯ → π ⎡⎣δ 2 π ( Ω − Ω0 ) + δ 2 π ( Ω + Ω0 ) ⎤⎦

,

F sin ( 2π F0 n ) ←⎯ → ( j / 2 ) ⎡⎣δ1 ( F + F0 ) − δ1 ( F − F0 ) ⎤⎦ ,

F rect ( t / w ) ←⎯ → w sinc ( wf ) =

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(

e− jπ F ( n0 + n1 ) sin π F ( n1 − n0 ) e − jπ F sin (π F )

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Forward DTFT Example

Consider the special case of n0 + n1 = 1 ⇒ n0 = 1 − n1 (making the function a discrete-time rectangular pulse of width 2n0 +1 centered at n = 0. F

1 − e− j 2 π F ( n1 − n0 ) 1 − e− j 2 π F

Factor out e− jπ F ( n1 − n0 ) from the numerator and e− jπ F from the denominator

n = n0

Let m = n − n0 . Then F

44

Forward DTFT Example

Find the forward DTFT of x [ n ] = u [ n − n0 ] − u [ n − n1 ]. F u [ n − n0 ] − u [ n − n1 ] ←⎯ →

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F u [ n − n0 ] − u [ n − n1 ] ←⎯ → F u [ n − n0 ] − u [ n − n1 ] ←⎯ →

47

F sin ( Ω0 n ) ←⎯ → jπ ⎡⎣δ 2 π ( Ω + Ω0 ) − δ 2 π ( Ω − Ω0 ) ⎤⎦

e j 2π F e− jπ F ( n0 + n1 ) e− j 2 π n0 F − e− j 2 π n1 F = ( n1 − n0 ) drcl ( F, n1 − n0 ) e −1 e − jπ F

(

j 2π F

)

e jΩ e− jΩ( n0 + n1 ) /2 e− jn0 Ω − e− jn1Ω = ( n1 − n0 ) drcl (Ω / 2π , n1 − n0 ) e jΩ − 1 e− jΩ /2

(

)

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48

8

DTFT Properties

DTFT Properties

n

F x [ n − n0 ] ←⎯ → e− j 2 π Fn0 X ( F ) ,

(

F e j 2 π F0 n x [ n ] ←⎯ → X ( F − F0 ) ,

F e jΩ0 n x [ n ] ←⎯ → X e j (Ω−Ω0 )

(

F x* [ n ] ←⎯ → X* ( −F ) ,

(

)

F x [ n ] − x [ n − 1] ←⎯ → 1 − e− j 2 π F X ( F ) ,

∞

∑e

) x [ n ] − x [ n − 1] ←⎯→ (1 − e ) X ( e )

∑ x[n]

n=−∞

(Note:

49

DTFT Properties

F

j0

− jΩ

2π

− jΩ

F

,

jΩ

jΩ

F

∑e

= δ1 ( F ) ,

jΩ see note

jΩ

∞

n=−∞ ∞

jΩ

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j 2 π Fn

X e jΩ

∑ x [ m ]←⎯→ 1 − e

m=−∞

F

see note

)

( ) + π X ( e )δ ( Ω ) x [ −n ] ←⎯→ X ( e ) x [ n ] ∗ y [ n ] ←⎯→ X ( e ) Y ( e ) x [ n ] y [ n ] ←⎯→ (1 / 2π ) X ( e )  Y ( e ) n

1 X ( 0 )δ1 ( F ) , 2

F x [ n ] y [ n ] ←⎯ → X ( F )  Y( F )

(

− jΩ

+

F x [ n ] ∗ y [ n ] ←⎯ → X ( F ) Y( F ) ,

F x* [ n ] ←⎯ → X* e− jΩ F

− j 2π F

F x [ −n ] ←⎯ → X ( −F ) ,

)

⎧x [ n / m ] , n / m an integer ⎫ F F jmΩ If y [ n ] = ⎨ ⎬ then y [ n ] ←⎯→ X ( mF ) or y [ n ] ←⎯→ X e , otherwise ⎩0 ⎭

F

m=−∞

( )

F x [ n − n0 ] ←⎯ → e− jΩn0 X e jΩ

X(F )

∑ x [ m ]←⎯→ 1 − e

F F α x [ n ] + β y [ n ] ←⎯ → α X ( F ) + β Y ( F ) , α x [ n ] + β y [ n ] ←⎯ → α X ( e jΩ ) + β Y ( e jΩ )

jΩn

= 2πδ 2 π ( Ω )

n=−∞ 2

= ∫ X ( F ) dF , 2

1

∞

∑ x[n]

2

n=−∞

= (1 / 2π ) ∫

2π

( )

X e jΩ

x ( t )  y ( t ) = ∫ x (τ ) y ( t − τ ) dτ where T is a period of both x and y T

2

dΩ

)

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50

DTFT Properties

Find and graph the inverse DTFT of X ( F ) = ⎡⎣ rect ( 50 ( F − 1 / 4 )) + rect ( 50 ( F + 1 / 4 )) ⎤⎦ ∗ δ1 ( F )

Frequency shifting the other direction

F e− jπ n/2 (1 / 50 ) sinc ( n / 50 ) ←⎯ → rect ( 50 ( F + 1 / 4 )) ∗ δ1 ( F )

Start with F sinc ( n / w ) ←⎯ → wrect ( wF ) ∗ δ1 ( F )

Combining the last two results and using cos ( x ) =

In this case w = 50.

F → (1 / 25 ) sinc ( n / 50 ) cos (π n / 2 ) ←⎯ ⎡⎣ rect ( 50 ( F − 1 / 4 )) + rect ( 50 ( F + 1 / 4 )) ⎤⎦ ∗ δ1 ( F )

F → rect ( 50F ) ∗ δ1 ( F ) (1 / 50 ) sinc ( n / 50 ) ←⎯

Then, using the frequency-shifting property F e j 2 π F0 n x [ n ] ←⎯ → X ( F − F0 )

e

jπ n/2

e jx + e− jx 2

F → rect ( 50 ( F − 1 / 4 )) ∗ δ1 ( F ) (1 / 50 ) sinc ( n / 50 ) ←⎯

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51

DTFT Properties

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52

DTFT Properties

Time scaling in discrete time is quite different from time scaling in continuous-time. Let z [ n ] = x [ an ]. If a is not an

Time scaling does not work for time compression because of

integer some values of z [ n ] are undefined and a DTFT cannot

decimation. But it does work for a special type of time expansion. ⎧x [ n / m ] , n / m an integer Let z [ n ] = ⎨ . Then Z ( F ) = X ( mF ) . , otherwise ⎩0 So the time-scaling property of the DTFT is

be found for it. If a is an integer greater than one, some values of x [ n ] will not appear in z [ n ]

⎧x [ n / m ] , n / m an integer z[n] = ⎨ , , otherwise ⎩0

because of decimation and there cannot be a unique relationship between their DTFT’s M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

53

F z [ n ] ←⎯ → X ( mF )

(

F or z [ n ] ←⎯ → X e jmΩ

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)

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9

DTFT Properties

DTFT Properties

Find the signal energy of x [ n ] = (1 / 5 ) sinc ( n /100 ) . The

In the time domain, the response of a system is the convolution

straightforward way of finding signal energy is directly from

of the excitation with the impulse response

the definition E x =

y[n] = x[n] ∗ h[n] In the frequency domain the response of a system is the product of the excitation and the frequency response of the system

( )

∞

∑ x[n]

2

.

n=−∞

Ex =

∞

∑ (1 / 5 ) sinc ( n /100 )

n=−∞

( ) ( )

2

∞

= (1 / 25 ) ∑ sinc 2 ( n /100 ) n=−∞

In this case we run into difficulty because we don't know how to sum this series.

Y e jΩ = X e jΩ H e jΩ

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55

DTFT Properties

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56

Transform Method Comparisons

We can use Parseval's theorem to find the signal energy from the DTFT of the signal. F → 20rect (100F ) ∗ δ1 ( F ) (1 / 5 ) sinc ( n /100 ) ←⎯

Parseval's theorem is ∞

∑ x[n]

n=−∞

2

= ∫ X ( F ) dF 2

1

For this case E x = ∫ 20rect (100F ) ∗ δ1 ( F ) dF = 2

1

1/2

∫

20rect (100F ) dF 2

−1/2

1/200

E x = 400

∫

dF = 4

−1/200 M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

57

Transform Method Comparisons

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58

Transform Method Comparisons

59

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60

10

Transform Method Comparisons

Transform Method Comparisons

Using the equivalence property of the impulse and fact that both

The DFT can often be used to find the DTFT of a signal. The

e j 2 π F and δ1 ( F ) have a fundamental period of one,

DTFT is defined by X ( F ) =

δ ( F − 1 /12 ) − jπ /6 δ1 ( F + 1 /12 ) ⎤ ⎡ Y ( F ) = (1 / 2 ) ⎢ e jπ /6 1 jπ /6 +e e − 0.9 e− jπ /6 − 0.9 ⎥⎦ ⎣ Finding a common denominator and simplifying, Y ( F ) = (1 / 2 )

− j 2 π Fn

and the DFT

N −1

is defined by X [ k ] = ∑ x [ n ] e− j 2 π kn/N . If the signal x [ n ] is causal n=0

δ1 ( F − 1 /12 ) (1 − 0.9e jπ /6 ) + δ1 ( F + 1 /12 ) (1 − 0.9e− jπ /6 )

and time limited, the summation in the DTFT is over a finite range of n values beginning with 0 and we can set the value of N by letting N − 1 be the last value of n needed to cover that finite

1.81 − 1.8 cos (π / 6 )

Y ( F ) = 0.4391 ⎡⎣δ1 ( F − 1 /12 ) + δ1 ( F + 1 /12 ) ⎤⎦

N −1

range. Then X ( F ) = ∑ x [ n ] e− j 2 π Fn . Now let F → k / N yielding

+ j0.8957 ⎡⎣δ1 ( F + 1 /12 ) − δ1 ( F − 1 /12 ) ⎤⎦

n=0

y [ n ] = 0.8782 cos ( 2π n /12 ) + 1.7914 sin ( 2π n /12 )

N −1

X ( k / N ) = ∑ x [ n ] e− j 2 π kn/N = X [ k ]

y [ n ] = 1.995 cos ( 2π n /12 − 1.115 )

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∞

∑ x[n]e

n=−∞

n=0

61

Transform Method Comparisons

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Transform Method Comparisons

We can also use the DFT to approximate the inverse DTFT. The inverse DTFT is defined by x [ n ] = ∫ X ( F ) e j 2 π Fn dF

The result N −1

X( k / N ) = ∑ x[n]e

1

− j 2 π kn/N

= X[ k ]

and the inverse DFT is defined by x [ n ] =

n=0

is the DTFT of x [ n ] at a discrete set of frequencies F = k / N or

We can approximate the inverse DTFT by

Ω = 2π k / N. If that resolution in frequency is not sufficient, N

N −1 ( k+1) /N

can be made larger by augmenting the previous set of x [ n ] values

x[n] ≅ ∑

with zeros. That reduces the space between frequency points thereby increasing the resolution. This technique is called zero

x[n] ≅ ∑ X( k / N )

k=0

∫

k /N

N −1

padding.

63

k=0

k /N

∫

e j 2 π Fn dF

1 N −1 ∑ X ( k / N ) e j 2π kn/N N k=0

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64

Transform Method Comparisons

For n