Lec 2 - Discrete Time Fourier Transform

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Representation of Sequences by Fourier TransformsMany sequences can be represented by a Fourier integral of the form as

Discrete Time Fourier TransformMaryam Mahsal Khan (Lecturer)

x[n ] =

1 2

X (e

j

)e jn d.

X (e j ) =

n =

x[n]e

jn

Frequency response of a LTI system is simply the Fourier transform of the impulse response.

B.Sc (CSE) UET Peshawar M.Sc(EE) UTP Malaysia

h[n ] =

1 2

H (e

j

)e jn d.

H (e j ) =

n =

h[n]e

jn

Representation of Sequences by Fourier Transforms

Fourier Transform (Convergence)

The frequency response of discrete-time LTI system is always a periodic function with period 2.j ( + 2 )

Determining the class of signals that can be represented Fourier transform is equivalent to considering the convergence of the infinite sum of the Fourier transform. A sufficient condition for convergence can be found as

H (e

)=

n =

h[n]e

j ( + 2 ) n

=

n =

h[n]e

jn

= H (e )

j

X ( e j ) =

n =

x[n]e

jn

n =

x[n ] e

jn

n =

x[n] <

More generally, H ( e j ( + 2r ) ) = H ( e j ), for r an integer.

Thus, if a sequence is absolutely summable, then its Fourier transform exists. The series can be shown to converge uniformly to a continuous function of . Since a stable sequence is, by definition, absolutely summable, all stable sequences have Fourier transforms.

1

Fourier Transform (Interpretation)Signals: The Fourier Transform X ( e j ) of a signal x[n] describes the frequency content of the signal.j At each frequency 0 , the magnitude spectrum X ( e 0 ) describes the amount of that frequency contained in the signal.

Example 2.19 Ideal Frequency-Selective Filters: Ideal Lowpass filter

At each frequency 0 , the phase spectrum X (e j0 ) describes the location (relative shift) of that frequency component of the signal.

Systems: The frequency response H ( e j ) of a linear system describes how frequencies input to the system are modified:An input frequency component 0 is amplified or attenuated j by a factor H (e 0 ) . An input frequency component 0 is shifted by an amount H (e j0 ).

DTFT of Ideal Low-Pass Filter

Contd..Convergence of the Fourier Transform The oscillatory behaviour Gibbs Phenomena

H M e j =

( ) sin ne nM c n= M

j

2

Frequency-Domain Representation of Discrete-Time SystemsEigenfunctions for LTI systemConsider an input sequence

Frequency-Domain Representation of Discrete-Time Systems (Contd)Frequency response of the system is defined as

x[ n ] = e jn for < n < ,

H ( e j ) =

The corresponding output of a LTI discrete-time system with impulse response h[n ] is

k =

h[k ]e

jk

,

y[n ] = x[n ] * h[n ] =If we define Then we havejn

k =

h[k ]e k =

j (nk )

H ( e j ) =

h[k ]e

= e jn h[k ]e jk . k = ,

Real and imaginary representation

jk

H ( e j ) = H R ( e j ) + jH I ( e j )Magnitude and phase polar representation

y[n ] = H ( e j )e jn .

We define e as an eigenfunction of the system, and the associated eigenvalue is H ( e j ) .

H ( e j ) = H ( e j ) e jH ( e ) .

j

Example 2.17 Frequency Response of the Ideal DelayConsider the ideal delay system defined byy[n ] = x[n nd ], where nd is a fixed integer.If we consider x[n ] = e as input to this system. Then we have the outputjn

Example 2.20 Frequency Response of the Moving-Average SystemThe impulse response of a moving-average system is1 M1 n M 2 , h[n ] = M 1 + M 2 + 1 0 otherwise. The frequency response is

y[n ] = e j ( n nd ) = e jnd e jnTherefore, the frequency response of the ideal delay is

H ( e j ) ===

H ( e j ) = e jnd .Real and imaginary representation H ( e j ) = cos(nd ) + j sin(nd ). Magnitude and phase representationH ( e j ) = 1 and H ( e j ) = nd .

M2 1 e jn M 1 + M 2 + 1 n = M1

1 e jM1 e j ( M 2 +1) M1 + M 2 + 1 1 e j1 sin( ( M 1 + M 2 + 1) / 2) j ( M 2 M1 ) / 2 e . M1 + M 2 + 1 sin( / 2)

3

Example 2.20 Frequency Response of the Moving-Average System (Contd)

Application Example - Signal SmoothingA common DSP application is the removal of noise from a signal corrupted by additive noise. A simple 3-point moving average algorithm is given by: % Signal Smoothing by Averaging clf; % Generate random noise, d[n] R = 51; d = 0.8*(rand(R,1) - 0.5); % Generate uncorrupted signal, s[n] m = 0:R-1; s = 2*m.*(0.9.^m); % Generate noise corrupted signal, x[n] x = s + d'; subplot(2,1,1); plot(m,d','r-',m,s,'g--',m,x,'b-.'); xlabel('Time index n');ylabel('Amplitude'); legend('d[n] ','s[n] ','x[n] '); % do smoothing x1 = [0 0 x];x2 = [0 x 0];x3 = [x 0 0]; y = (x1 + x2 + x3)/3; subplot(2,1,2); plot(m,y(2:R+1),'r-',m,s,'g--'); legend( 'y[n] ','s[n] '); xlabel('Time index n');ylabel('Amplitude');

1 y[n] = ( x[n 1] + x[n] + x[ n + 1]) 38 6 Amplitude 4 2 0 -2 0 5 10 15 20 25 30 Time index n 35 40 45 50 d[n] s[n] x[n]

H ( e j0 ) = H * ( e j0 )?

8 6 4 2 0 y[n] s[n]

Amplitude

0

5

10

15

20 25 30 Time index n

35

40

45

50

Symmetry Properties of the Fourier TransformSymmetry properties of the Fourier transform are often very useful for simplifying the solution of problems. Some basic definitionsConjugate-symmetric sequence Conjugate-antisymmetric sequence* x e [n ] = x e [ n ]* o

Symmetry Properties of the Fourier Transform (Contd)Similarly, a Fourier transform X ( e j ) can be decomposed into a sum of conjugate-symmetric and conjugate-antisymmetric functions.X ( e j ) = X e ( e j ) + X o ( e j )X e ( e j ) = 1 * X ( e j ) + X * ( e j ) = X e ( e j ) 2

xo [n ] = x [ n ] Any sequence can be expressed as a sum of a conjugatesymmetric and conjugate-antisymmetric sequence1 * x[ n ] + x * [ n ] = x e [ n ] 2 1 * * x o [ n ] = x[ n ] x [ n ] = x o [ n ] 2 x e [n ] =

[

]

X o ( e j ) =

1 * X ( e j ) X * ( e j ) = X o ( e j ) 2

[

]

[ [

] ]

x[n ] = x e [n ] + x0 [n ]

Re{ x[n ]} =

1 ( x[n ] + x * [n ]) 2

j Im{x[n ]} =

1 ( x[n ] x * [n ]) 2

Even sequence (real): Odd sequence (real):

x e [n ] = x e [ n ]

x o [ n ] = xo [ n ]

For real sequences, the real part of the Fourier transform is an even function, and the imaginary part is an odd function.

4

Fourier Transform Theorems

Frequency Response of LCCDE

Example: Determining an h[n] from Difference Equationy[n] 1 1 y[n 1] = x[n] x[n 1] 2 4 1 1 j j H (e ) = 1 e ) e 4 2 1 1 e j 4 ) = 1 1 e j 2 1 j e 1 ) = 4 1 j 1 j 1 e 1 e 2 2 Transform Tablesn n 1

H (e

j

j

H (e

j

H (e From

j

1 1 1 h [ n ] = u [n ] 4 2 2

u [n 1 ]

5

Example 2.22 Determining the Impulse Response from the Frequency ResponseThe frequency response of a high-pass filter with delay

e jnd H ( e j ) = 0

c < < . < c

H ( e j ) = e jnd (1 H lp ( e j )) = e jnd e jnd H lp ( e j ),1, < c H lp ( e j ) = 0, c < .

hlp [n ] =

sin c n n

h[n ] = [n nd ]

sin c ( n nd ) (n nd )

6

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