discrete-time fourier series concept - unt computer …rakl/class3010/chapter7.pdf · ·...

1

Discrete-Time Fourier Methods

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl

1

Discrete-Time Fourier Series Concept

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

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 2

The relationship between complex and real sinusoids









2









The Discrete-Time Fourier Series

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

x n[ ] = cx k[ ]e j2πkn /N

k= N∑ cx k[ ] = 1

Nx n[ ]e− j2πkn /N

n=n0

n0 +N−1

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

k= N∑ means a summation over any range of

consecutive k’s exactly N in length.


The Discrete Fourier Transform The discrete Fourier transform (DFT) is almost identical to the DTFS.A periodic discrete-time signal can be expressed as

x n[ ] = 1N

X k[ ]e j2πkn /N

k= N∑ X k[ ] = x n[ ]e− j2πkn /N

n=n0

n0 +N−1

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

k= N∑ means a summation over any range of

consecutive k’s exactly N in length. The main difference between the DTFS and the DFT is the location of the 1/N term. So X k[ ] = N cx k[ ].


3

The Discrete Fourier Transform

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




DFT Example

Find the DFT harmonic function for

x n⎡⎣ ⎤⎦ = u n⎡⎣ ⎤⎦ − u n − 3⎡⎣ ⎤⎦( )∗δ5 n⎡⎣ ⎤⎦using its fundamental period as the representation time.

X k⎡⎣ ⎤⎦ = x n⎡⎣ ⎤⎦e− j2πkn/ N

n= N∑

= u n⎡⎣ ⎤⎦ − u n − 3⎡⎣ ⎤⎦( )∗δ5 n⎡⎣ ⎤⎦e− j2πkn/5

n=0

4

∑

= e− j2πkn/5

n=0

2

∑ = 1− e− j6πk /5

1− e− j2πk /5 =e− j3πk /5

e− jπk /5 × e j3πk /5 − e− j3πk /5

e jπk /5 − e− jπk /5

= e− j2πkn/5

n=0

2

∑ = e− j2πk /5sin 3πk / 5( )sin πk / 5( ) = 3e− j2πk /5 drcl k / 5,3( )


The DFT Harmonic Function

We know that x n[ ] = 1N

X k[ ]e j2πkn /N

k= N∑ so we can find x n[ ]

from its harmonic function. But how do we find the harmonicfunction from x n[ ]? We use the principle of orthogonality likewe did with the CTFS except that now the orthogonality is indiscrete time instead of continuous time.






4

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


The DFT Harmonic Function The projection of a real vector x in the direction of anotherreal vector y is

p = xTyyTy

y

If p = 0, x and y are orthogonal. If the vectors are complex-valued

p = xHyyHy

y

where the xH is the complex-conjugate transpose of x. xTyand xHy are the dot product of x and y.










5


The most common definition of the DFT is

X k[ ] = x n[ ]e− j2πkn /N

n=0

N−1

∑ , x n[ ] = 1N

X k[ ]e j2πkn /N

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.


Convergence of the DFT

•  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



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.

x n[ ] D F TN← →⎯⎯ X k[ ]

From x n[ ] = 1N

X k[ ]e j2πkn /N

k= N∑ we see that x n[ ] is formed

by a linear combination of functions of the form e j2πkn /N eachof which has a period N . Therefore x n[ ] must also be periodicwith period (but not necessarily fundamental period) N .


DFT Properties


DFT Properties


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 functionX k[ ] is purely imaginary.


6

The Dirichlet Function

The functional form sin πNt( )N sin πt( )

appears often in discrete-time signal analysis and is given the special name Dirichlet function. That is

drcl t,N( ) = sin πNt( )N sin πt( )


DFT Pairs

e j2πn /N D F TmN← →⎯⎯ mNδmN k − m[ ]

cos 2πqn / N( ) D F TmN← →⎯⎯ mN / 2( ) δmN k − mq[ ]+δmN k + mq[ ]( )

sin 2πqn / N( ) D F TmN← →⎯⎯ jmN / 2( ) δmN k + mq[ ]−δmN k − mq[ ]( )

δN n[ ] D F TmN← →⎯⎯ mδmN k[ ]

1 D F TN← →⎯⎯ NδN k[ ]

u n − n0[ ]− u n − n1[ ]( )∗δN n[ ] D F TN← →⎯⎯ e− jπk n1+n0( )/N

e− jπk /N n1 − n0( )drcl k / N ,n1 − n0( ) tri n / Nw( )∗δN n[ ] D F T

N← →⎯⎯ Nw drcl2 k / N ,Nw( ) , Nw an integer

sinc n /w( )∗δN n[ ] D F TN← →⎯⎯ wrect wk / N( )∗δN k[ ]


The Fast Fourier Transform

One could write a MATLAB program to implement the DFT..

% (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


The Fast Fourier Transform

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. Mstands for computer multiplies and A stands for computer additions.γ N = 2γ ADFT MDFT AFFT MFFT ADFT / AFFT MDFT / MFFT

1 2 2 4 2 1 1 42 4 12 16 8 4 1.5 43 8 56 64 24 12 2.33 5.334 16 240 256 64 32 3.75 85 32 992 1024 160 80 6.2 12.86 64 4032 4096 384 192 10.5 21.37 128 16256 16384 896 448 18.1 36.68 256 65280 65536 2048 1024 31.9 649 512 261632 262144 4608 2304 56.8 113.8

10 1024 1047552 1048576 10240 5120 102.3 204.8


Generalizing the DFT for Aperiodic Signals

Pulse Train

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.


DFT of Pulse Train

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



7

Normalized DFT of

Pulse Train

By plotting versus k / N0 instead of k, the period of the normalized DFT stays at one.



The normalized DFT approaches this limit as the period approaches infinity.



Definition of the Discrete-Time Fourier Transform (DTFT)

Forward F Form Inverse

x n[ ] = X F( )e j2πFn dF1∫

F← →⎯ X F( ) = x n[ ]e− j2πFnn=−∞

∞

∑

Forward Ω Form Inverse

x n[ ] = 12π

X e jΩ( )e jΩn dΩ2π∫

F← →⎯ X e jΩ( ) = x n[ ]e− jΩnn=−∞

∞

∑


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 e- jΩn is periodic with fundamental period 2π / n and 2π is also a period of e− jΩn . The forward DTFT is

X e jΩ( ) = x n[ ]e− jΩnn=−∞

∞

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

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

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

The Discrete-Time Fourier Transform


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

δ n[ ] F← →⎯ 1

α n u n[ ] F← →⎯e jΩ

e jΩ −α=

11−αe− jΩ

, α < 1 , −α n u −n −1[ ] F← →⎯e jΩ

e jΩ −α=

11−αe− jΩ

, α > 1

α n sin Ω0n( )u n[ ] F← →⎯e jΩα sin Ω0( )

e j2Ω − 2αe jΩ cos Ω0( ) +α 2 , α < 1 , −α n sin Ω0n( )u −n −1[ ] F← →⎯e jΩα sin Ω0( )

e j2Ω − 2αe jΩ cos Ω0( ) +α 2 , α > 1

α n cos Ω0n( )u n[ ] F← →⎯e jΩ e jΩ −α cos Ω0( )⎡⎣ ⎤⎦

e j2Ω − 2αe jΩ cos Ω0( ) +α 2 , α < 1 , −α n cos Ω0n( )u −n −1[ ] F← →⎯e jΩ e jΩ −α cos Ω0( )⎡⎣ ⎤⎦

e j2Ω − 2αe jΩ cos Ω0( ) +α 2 , α > 1


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

X F( ) = Ae− j2πFnn=−∞

∞

∑ = A e− j2πFnn=−∞

∞

∑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 mustbe a periodic impulse.


8

The Generalized DTFT

To shorten the process, find the inverse DTFT of a periodicimpulse of the form Aδ1 F( ). Using the formula

x n[ ] = Aδ1 F( )e j2πFndF1∫ = A δ F( )e j2πFndF

−1/2

1/2

∫ = A

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

frequency form A F← →⎯ 2πAδ2π Ω( ).


The Generalized DTFT

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

x n[ ] = Aδ1 F − F0( )e j2πFn dF1∫ = A δ F − F0( )e j2πFn dF

−1/2

1/2

∫ = Aej2πF0n

Now change x n[ ] to x n[ ] = Acos 2πF0n( ) = A / 2( ) e j2πF0n + e− j2πF0n( ).Then

Acos 2πF0n( ) F← →⎯ A / 2( ) δ1 F − F0( ) +δ1 F + F0( )⎡⎣ ⎤⎦or

Acos Ω0n( ) F← →⎯ πA δ2π Ω −Ω0( ) +δ2π Ω +Ω0( )⎡⎣ ⎤⎦


Forward DTFT Example

Find the forward DTFT of x n[ ] = u n − n0[ ]− u n − n1[ ].

u n − n0[ ]− u n − n1[ ] F← →⎯ u n − n0[ ]− u n − n1[ ]( )e− j2πFnn=−∞

∞

∑ = e− j2πFnn=n0

n1 −1

∑Let m = n − n0 . Then

u n − n0[ ]− u n − n1[ ] F← →⎯ e− j2πF m+n0( )

m=0

n1 −n0 −1

∑ = e− j2πFn0 e− j2πFmm=0

n1 −n0 −1

∑Summing this geometric series

u n − n0[ ]− u n − n1[ ] F← →⎯ e− j2πFn01− e− j2πF n1 −n0( )

1− e− j2πF



u n − n0[ ]− u n − n1[ ] F← →⎯ e− j2πFn01− e− j2πF n1 −n0( )

1− e− j2πF

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

u n − n0[ ]− u n − n1[ ] F← →⎯ e− j2πFn0e− jπF n1 −n0( )

e− jπFe jπF n1 −n0( ) − e− jπF n1 −n0( )

e jπF − e− jπF

By the definition of the sine function in terms of complex exponentials

u n − n0[ ]− u n − n1[ ] F← →⎯e− jπF n0 +n1( )

e− jπFsin πF n1 − n0( )( )

sin πF( )



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

u n − n0[ ]− u n − n1[ ] F← →⎯sin πF n1 − n0( )( )

sin πF( ) , n0 + n1 = 1

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

rect t /w( ) F← →⎯ wsinc wf( ) = sin πwf( )π f

The DTFT is a periodically-repeated sinc function.


More DTFT Pairs

δ n[ ] F← →⎯ 1

u n[ ] F← →⎯1

1− e− j2πF+ 1 / 2( )δ1 F( ) , u n[ ] F← →⎯

11− e− jΩ

+ πδ1 Ω( )

sinc n / w( ) F← →⎯ w rect wF( )∗δ1 F( ) , sinc n / w( ) F← →⎯ w rect wΩ / 2π( )∗δ2π Ω( ) tri n / w( ) F← →⎯ wdrcl2 F,w( ) , tri n / w( ) F← →⎯ wdrcl2 Ω / 2π ,w( ) 1 F← →⎯ δ1 F( ) , 1 F← →⎯ 2πδ2π Ω( ) δN0

n[ ] F← →⎯ 1 / N0( )δ1/N0F( ) , δN0

n[ ] F← →⎯ 2π / N0( )δ2π /N0Ω( )

cos 2πF0n( ) F← →⎯ 1 / 2( ) δ1 F − F0( ) + δ1 F + F0( )⎡⎣ ⎤⎦ , cos Ω0n( ) F← →⎯ π δ2π Ω − Ω0( ) + δ2π Ω +Ω0( )⎡⎣ ⎤⎦ sin 2πF0n( ) F← →⎯ j / 2( ) δ1 F + F0( ) − δ1 F − F0( )⎡⎣ ⎤⎦ , sin Ω0n( ) F← →⎯ jπ δ2π Ω +Ω0( ) − δ2π Ω − Ω0( )⎡⎣ ⎤⎦

u n − n0[ ]− u n − n1[ ] F← →⎯e j2πF

e j2πF −1e− j2πn0F − e− j2πn1F( ) = e− jπF n0 +n1( )

e− jπFn1 − n0( )drcl F,n1 − n0( )

u n − n0[ ]− u n − n1[ ] F← →⎯e jΩ

e jΩ −1e− jn0Ω − e− jn1Ω( ) = e− jΩ n0 +n1( )/2

e− jΩ /2 n1 − n0( )drcl Ω / 2π ,n1 − n0( )

We can now extend the table of DTFT pairs.


9

DTFT Properties

α x n[ ]+ β y n[ ] F← →⎯ α X F( ) + β Y F( ) , α x n[ ]+ β y n[ ] F← →⎯ α X e jΩ( ) + β Y e jΩ( ) x n − n0[ ] F← →⎯ e− j2πFn0 X F( ) , x n − n0[ ] F← →⎯ e− jΩn0 X e jΩ( ) e j2πF0n x n[ ] F← →⎯ X F − F0( ) , e jΩ0n x n[ ] F← →⎯ X e j Ω−Ω0( )( )If y n[ ] = x n /m[ ] , n /m an integer

0 , otherwise⎧⎨⎩

⎫⎬⎭

then y n[ ] F← →⎯ X mF( ) or y n[ ] F← →⎯ X e jmΩ( )x* n[ ] F← →⎯ X* −F( ) , x* n[ ] F← →⎯ X* e− jΩ( )x n[ ]− x n −1[ ] F← →⎯ 1− e− j2πF( )X F( ) , x n[ ]− x n −1[ ] F← →⎯ 1− e− jΩ( )X e jΩ( )


DTFT Properties

x m[ ]m=−∞

n

∑ F← →⎯X F( )

1− e− j2πF+ 1

2X 0( )δ1 F( ) , x m[ ]

m=−∞

n

∑ F← →⎯X e jΩ( )1− e− jΩ

+π X e j0( )δ2π Ω( )

x −n[ ] F← →⎯ X −F( ) , x −n[ ] F← →⎯ X e− jΩ( ) x n[ ]∗y n[ ] F← →⎯ X F( )Y F( ) , x n[ ]∗y n[ ] F← →⎯ X e jΩ( )Y e jΩ( ) x n[ ]y n[ ] F← →⎯ X F( )Y F( )see note , x n[ ]y n[ ] F← →⎯ 1 / 2π( )X e jΩ( )Y e jΩ( )see note

e j2πFnn=−∞

∞

∑ = δ1 F( ) , e jΩnn=−∞

∞

∑ = 2πδ2π Ω( )

x n[ ] 2

n=−∞

∞

∑ = X F( ) 2 dF1∫ , x n[ ] 2

n=−∞

∞

∑ = 1 / 2π( ) X e jΩ( ) 2dΩ

2π∫Note: x t( )y t( ) = x τ( )y t −τ( )dτ

T∫ where T is a period of both x and y( )


DTFT Properties

Find and graph the inverse DTFT of

X F( ) = rect 50 F −1 / 4( )( ) + rect 50 F +1 / 4( )( )⎡⎣ ⎤⎦ ∗δ1 F( )Start with

sinc n /w( ) F← →⎯ wrect wF( )∗δ1 F( )In this case w = 50.

1 / 50( )sinc n / 50( ) F← →⎯ rect 50F( )∗δ1 F( )Then, using the frequency-shifting property

e j2πF0n x n[ ] F← →⎯ X F − F0( ) e jπn /2 1 / 50( )sinc n / 50( ) F← →⎯ rect 50 F −1 / 4( )( )∗δ1 F( )


DTFT Properties

Frequency shifting the other direction

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

Combining the last two results and using cos x( ) = ejx + e− jx

2 1 / 25( )sinc n / 50( )cos πn / 2( ) F← →⎯

rect 50 F −1 / 4( )( ) + rect 50 F +1 / 4( )( )⎡⎣ ⎤⎦ ∗δ1 F( )


DTFT Properties Time scaling in discrete time is quite different from timescaling in continuous-time. Let z n[ ] = x an[ ]. If a is not an integer some values of z n[ ] are undefined and a DTFT cannot be found for it. If a is an integer greater than one, some values of x n[ ] will not appear in z n[ ] because of decimation and there cannot be a unique relationship between their DTFT’s


DTFT Properties

Time scaling does not work for time compression because ofdecimation. But it does work for a special type of time expansion.

Let z n[ ] = x n /m[ ] , n /m an integer0 , otherwise

⎧⎨⎩

. Then Z F( ) = X mF( ).

So the time-scaling property of the DTFT is

z n[ ] = x n /m[ ] , n /m an integer0 , otherwise

⎧⎨⎩

, z n[ ] F← →⎯ X mF( )

or z n[ ] F← →⎯ X e jmΩ( )


10

DTFT Properties

In the time domain, the response of a system is the convolutionof the excitation with the impulse response y n[ ] = x n[ ]∗h n[ ]In the frequency domain the response of a system is the productof the excitation and the frequency response of the system

Y e jΩ( ) = X e jΩ( )H e jΩ( )


DTFT Properties

Find the signal energy of x n[ ] = 1 / 5( )sinc n /100( ). Thestraightforward way of finding signal energy is directly from

the definition Ex = x n[ ] 2

n=−∞

∞

∑ .

Ex = 1 / 5( )sinc n /100( ) 2

n=−∞

∞

∑ = 1 / 25( ) sinc2 n /100( )n=−∞

∞

∑In this case we run into difficulty because we don't know howto sum this series.


DTFT Properties

We can use Parseval's theorem to find the signal energy fromthe DTFT of the signal.

1 / 5( )sinc n /100( ) F← →⎯ 20rect 100F( )∗δ1 F( )Parseval's theorem is

x n[ ] 2

n=−∞

∞

∑ = X F( ) 2 dF1∫

For this case

Ex = 20rect 100F( )∗δ1 F( ) 2 dF1∫ = 20rect 100F( ) 2 dF

−1/2

1/2

∫

Ex = 400 dF−1/200

1/200

∫ = 4


Transform Method Comparisons






11

Transform Method Comparisons Using the equivalence property of the impulse and fact that both e j2πF and δ1 F( ) have a fundamental period of one,

Y F( ) = 1 / 2( ) e jπ /6 δ1 F −1 /12( )e jπ /6 − 0.9

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

⎡⎣⎢

⎤⎦⎥

Finding a common denominator and simplifying,

Y F( ) = 1 / 2( )δ1 F −1 /12( ) 1− 0.9e jπ /6( ) +δ1 F +1 /12( ) 1− 0.9e− jπ /6( )1.81−1.8cos π / 6( )

Y F( ) = 0.4391 δ1 F −1 /12( ) +δ1 F +1 /12( )⎡⎣ ⎤⎦ + j0.8957 δ1 F +1 /12( )−δ1 F −1 /12( )⎡⎣ ⎤⎦ y n[ ] = 0.8782cos 2πn /12( ) +1.7914sin 2πn /12( ) y n[ ] = 1.995cos 2πn /12 −1.115( )


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

DTFT is defined by X F( ) = x n[ ]e− j2πFnn=−∞

∞

∑ and the DFT

is defined by X k[ ] = x n[ ]e− j2πkn /N

n=0

N−1

∑ . If the signal x n[ ] is causal

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

range. Then X F( ) = x n[ ]e− j2πFnn=0

N−1

∑ . Now let F→ k / N yielding

X k / N( ) = x n[ ]e− j2πkn /N = X k[ ]n=0

N−1

∑



The result

X k / N( ) = x n[ ]e− j2πkn /N = X k[ ]n=0

N−1

∑is the DTFT of x n[ ] at a discrete set of frequencies F = k / N orΩ = 2πk / N . If that resolution in frequency is not sufficient, Ncan be made larger by augmenting the previous set of x n[ ] valueswith zeros. That reduces the space between frequency pointsthereby increasing the resolution. This technique is called zeropadding.


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 j2πFn dF1∫

and the inverse DFT is defined by x n[ ] = 1N

X k[ ]e j2πkn /N

k=0

N−1

∑ .

We can approximate the inverse DTFT by

x n[ ] ≅ X k / N( )e j2πFn dFk /N

k+1( )/N

∫k=0

N−1

∑ = X k / N( ) e j2πFn dFk /N

k+1( )/N

∫k=0

N−1

∑

x n[ ] ≅ X k / N( ) ej2π k+1( )n /N − e j2πk /N

j2πnk=0

N−1

∑ = ej2πn /N −1j2πn

X k / N( )e j2πkn /N

k=0

N−1

∑

x n[ ] ≅ e jπn /N sinc n / N( ) 1N


k=0

N−1

∑



For n << N ,

x n[ ] ≅ 1N


k=0

N−1

∑This is the inverse DFT with X k[ ] = X k / N( ).

Use this result to find the inverse DTFT of

X F( ) = rect 50 F −1 / 4( )( ) + rect 50 F +1 / 4( )( )⎡⎣ ⎤⎦ ∗δ1 F( )with the inverse DFT.



N = 512 ; % Number of pts to approximate X(F)

k = [0:N -1]' ; % Harmonic numbers

% Compute samples from X(F) between 0 and 1 assuming

% periodic repetition with period 1

X = rect(50 *(k / N - 1 / 4)) + rect(50 *(k / N - 3 / 4)) ;

% Compute the approximate inverse DTFT and

% center the function on n = 0

xa = real(fftshift(ifft(X))) ;

n = [-N / 2:N / 2 - 1]' ; % Vector of discrete times for plotting

% Compute exact x[n] from exact inverse DTFT

xe = sinc(n / 50).*cos(pi*n / 2) / 25 ;


12



The Four Fourier Methods


Relations Among Fourier Methods

Discrete Frequency Continuous Frequency

Continuous Time x t( )y t( ) F S← →⎯⎯ X k[ ]∗Y k[ ] x t( )y t( ) F← →⎯ X f( )∗Y f( )Discrete Time x n[ ]y n[ ] F S← →⎯⎯ Y k[ ]X k[ ] x n[ ]y n[ ] F← →⎯ X F( )Y F( )


Continuous Time x t( )y t( ) F S← →⎯⎯ T0 X k[ ]Y k[ ] x t( )∗y t( ) F← →⎯ X f( )Y f( )Discrete Time x n[ ]y n[ ] F S← →⎯⎯ N0 Y k[ ]X k[ ] x n[ ]∗y n[ ] F← →⎯ X F( )Y F( )

Multiplication-Convolution Duality




Continuous Time 1T0

x t( ) 2 dtT0∫ = X k[ ] 2

k=−∞

∞

∑ x t( ) 2 dt−∞

∞

∫ = X f( ) 2 df−∞

∞

∫

Discrete Time x n[ ] 2

n= N∑ = 1

NX k[ ] 2

k= N∑ x n[ ] 2

n=−∞

∞

∑ = X F( ) 2 dF1∫

Parseval’s Theorem




Continuous Time x t − t0( ) F S← →⎯⎯ cx k[ ]e− j2πkt0 /N x t − t0( ) F← →⎯ X jω( )e− jωt0

Discrete Time x n − n0[ ] F S← →⎯⎯ X k[ ]e− j2πkn0 /N x n − n0[ ] F← →⎯ X e jΩ( )e− jΩn0


Continuous Time x t( )e+ j2πkt /T F ST← →⎯⎯ cx k − k0[ ] x t( )e+ jω0t F← →⎯ X j ω −ω0( )( )

Discrete Time x n[ ]e+ j2πkn /N F SN← →⎯⎯ X k − k0[ ] x n[ ]e+ jΩ0n F← →⎯ X e j Ω−Ω0( )( )

Time and Frequency Shifting




discrete-time fourier series concept - unt computer …rakl/class3010/chapter7.pdf · ·...

Documents