Fourier Transform Chapter 10 Sampling and Series

Transcription

1 Fourier Transform Chapter 0 Sampling and Series

2 Sampling Theorem Sampling Theorem states that, under a certain condition, it is in fact possible to recover with full accuracy the values intervening between regularly spaced samples >> The sample set can be fully equivalent to the complete set of function. Condition >> The function should be band limited, that is, it should have a Fourier transform that is nonzero over a finite range of the transform variable and zero elsewhere. DSP(digital signal processing) can be applied to the sample set at the same time

3 The interval between samples is crucial >> The samples had to be very close together, not much would be gained. In order to reconstruction original signal, by Nyquist-Shanon sampling theorem, fs(sampling frequency) >= fc(cut-off frequency)

4 Illustration of fineness of sampling (a) Sinc(x) > spectrum is flat where s< l/l and zero beyond. This sample interval is wide because it is. (b) sinc (x/) > spectrum cuts off at the same points ( s= l/l) (c),(d) are some samples for experiment.

5 Band-limited function (a) f(x) : band-limited function (b) Fourier Transform F(s) is zero when s>s c - S c (cut-off frequency) - cut-off transform (c) Samples (interval values are equal to each interval not exceeding /S c - (d) Detail representation of (c)

6 Band-limited function In general a cut-off transform is of the form Inverse transform is Where g(x) is arbitrary

7 Shah function Shah symbol is convenient, because multiplication by symbol is equivalent to sampling Definition : n III ( x) ( x n) Property III ( x) f ( x) n III ( x) * f ( x) III ( x) III ( s) f ( n) ( x n) n f ( x n)

8 Prove the sampling theorem - Information about f(x) is conserved only at the sampling points therefore, if f(x) can be reconstructed from f(x)iii(x/t), the theorem is proved. x f ( x) III( ) f ( n ) ( x n ) n - The transform of III(x/) is III(s) - The transform f ( x) III ( x / ) III ( s)* F ( s)

9 s c Over sampling s c s c Under sampling aliasing (island overlap) - multiplication of the original function by II(x/T) has the effect of replicating the spectrum F(s) at intervals T-

10 - If F(s c ) is not zero, the islands have cliffs - a) on the point of happening Multiplication by (s/s c ) permits exact recovery of F(s) - b) F(s) behave impulsively at s=s c If F(s) is even, (s/s c ) / If F(s) is odd, both cancel Odd harmonic component disappears in the sampling process Critical sampling A function whose Fourier transform is zero for s >s c is fully specified by values spaced at equal intervals not exceeding /S c - save for harmonic term with zeros at the sampling points

11 Interpolation - The numerical process of calculating intermediate points from samples - Process of recovery was to multiply a transform by (s/s c ) Convolution with s c sincs c x will yield f(x) directly from III(x/)f(x) in the function domain Fourier transform of s c sincs c x is (s/s c )

12 Rectangular filtering - Rectangular filtering : remove from a function spectral components - A summation can be approximated whose frequencies exceed a certain limit

13 Approximate rectangular filtering with a cutoff at S c is carried out by reading off f(x) at half the critical sampling interval and taking the convolution with s c sincs c x. Numerical procedure For achieving the desired Filter characteristic (s)

14 Undersampling Band Limited function g(x) is defined by sampled values. Effects g(x) has a cutoff spectrum of the desired extent and may at first sight appear to be a product of rectangular filtering (not the same as product of rectangular filtering but good approximation) Too-coarse sampling contains contributions from high freq. Components of f(x) : Aliasing Reinforce the even part of the spectrum, odd part is diminished

15 Fourier Series - The Fourier series will be exhibited as an extreme situation of the Fourier transform - The Fourier series associated with the periodic function g(x), with frequency f and period T ) sin cos ( 0 nfx b nfx a a n n T T T T T T dx x g T a nfxdx x g T n b nfxdx x g T n a ) ( 0 )sin (, )cos (

16 Let T=, then a n -j b n is related to the one-period segment g(x)(x) by Fourier transform a ib Periodic Function: n n g( x) e inx inx dx g( x) ( x) e dx Let p( x) III( x)* f the period of p( x) is ( x) unity f ( x n) : p( x ) p( x) There will be no regular Fourier transform for p(x) Hence we multiply by a factor (x) that die out to zero p ( x) dx cannot converge

17

18 If f(x) = g(x) (x): The Fourier Transform of the one-period segment III( s) F( s) e isx ds n0 n0 i nx { F( n) e } i nx F(0) F ( n) e conjugate F(0) Re F cos nx Im F sin nx a 0 { F( s) e an cos nx bn sin nx i sx } ds This is precisely the value that was quoted earlier for the complex coefficient a n -i b n.

19 Deriving the Fourier integral from the Fourier series

20 Gibbs Phenomenon By omitting terms beyond a certain limiting frequency, the phenomenon of overshoot associated with discontinuities, or sharp changes, in the periodic function. If the fundamental frequency is s 0, and frequencies up to ns 0 are retained The spectrum had been multiplied by a rectangle function (s/s c ), ns 0 <s c <(n+)s 0, we can choose (n+/)s 0 and (s/(n+)s c ) = convolution with (n+)s 0 sinc[(n+)s 0 x] When the series is summed for terms up to frequencies ns 0,the sum is ( x)*(n ) s sinc(n s p 0 ) 0 x

21 Gibbs Phenomenon What happens at a discontinuity When x=0, the result will be (n+)s 0 sinc[(n+)s 0 x]*sgn(x) sincx * sgnx x 0 sinct dt Si( x ) And if we change the scale factors of the sinc function, compressing it by a factor N=(n+)s 0, and strengthening it by a factor N, so as to retain its unit area N sincnx * sgnx x 0 sinct dt Si( Nx )

22 Gibbs Phenomenon Overshoot The overshoot, amounting to 9% of the amount of the discontinuity, remains at 9%, but the maximum is reached nearer to discontinuity. There is 9% overshoot on both sides of discontinuity regardless of the inclusion of more and more terms

23 Fourier coefficients The series coefficients a n and b n for a periodic function p(x) of unit period. a n ib n inx p( x) e dx Note that the integral has the form of a finite Fourier transform. The finite transform can be expressed as a standard transform as follows: p( x) ( x) e inx dx

24 Fourier coefficients As an example n [0(x n)] By using shah Func. Its transform Note that form sinc 0 (0x) e sinc 0 isx s III ( s) 0 dx ( 0x)* III( x) s III ( s) 0 came The general form x f ( x)* III( ) F( s) III( Ts) T T for a period T

25 The Shah symbol is its own Fourier transform Consider function f ( x) e x n ( x n) e For a given small value of, the function f(x) represents a row of narrow Gaussian spikes of width As ->0 each spike narrows in on an integral value of x and increases in height. The function f(x) possesses a regular Fourier transform, since f(x) is integrable, and there are no discontinuities. m F( s) e e m ( s m)

26 The Fourier series is Therefore r e n ( x n) m is periodic (T=) cos mx The function F(s) is a row of Gaussian spikes of width with maxima lying on a broad Gaussian curve of - As ->0 a suitable defining sequence for the shah symbol results e ( m) f x e m e r x ( ) cos mx m e m e x e imx m