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L23 315 f11


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Signals and Systems Lecture Slides

Signals and Systems Lecture Slides

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  • 1. L23 315 F11EE 315Fall 2011 SAMPLING THEOREM PROOFIntroduction: The sampling theorem for sequences is valid for sequences that areideally band limited to 1/(2n); the Fourier transform has a cycle that isidentically zero for frequencies larger than 1/(2n). Such band-limitedsequences can be recovered from equally spaced samples that areseparated by the integer n. Thus the amount of storage required to storesuch a sequence can be reduced by a factor of n. The sampling theorem is represented in the block diagram below.The input is to an ideal low pass filter that eliminates any aliasingfrequency components in the input signal. If the input sequence is ideallyband limited, the output of this filter, fin[*] is equal to the input sequence, in[*] . The second system in the cascade merely outputs every nthcomponent of the filtered input sequence: sfin[*]= fin[n*]. The thirdsystem in the cascade, the up sampler, merely replaces the deletedcomponents of the filtered input, sfin[ ], with zeroes. This zero paddedsequence is then input to the initial filter. The output of this cascade of foursystems is equal to the filtered input, fin[*].Block Diagram:Analysis of the System: This system is best understood in the frequency domain; and iseasiest to understand with visual plots of the Fourier transforms of thesequences involved. The first Fourier transform is plotted below; since Fourier transformsare complex valued in general, a two dimensional “face” represents thisthree dimensional nature of the Fourier transform; the shading of the eyesat small frequency magnitudes represents that the Fourier transform isdifferent at positive and negative frequencies. When the sequence is real,the Fourier transform at a negativefrequency equals the conjugate of thecorresponding positive frequency. 1
  • 2. The input is drawn to represent the potential presence of frequencycomponents at aliasing frequencies. For this visualization, the downsampling factor is two, n=2. In the cycle of the periodic Fourier transform,the aliasing components are at frequencies with magnitudes between ¼and ½. For a general down sampling factor of n, the aliasing componentsare at frequency magnitudes greater than 1/(2n).The Anti-Aliasing Filter: Since the transfer function of the anti-aliasing filter is, in general,given as below: the transfer function of a cycle of the anti-aliasing filter iszero when the magnitude of the frequency is greater than1/(2n). Theoutput of the initial anti-aliasing filter is ideally band limited to a centeredband of width 1/n. ì ì ï 1, ·- * < 1 ï IN(·), ·-* < 1 ï  L(·) = í   2n , FIN(·) = í  2n , ï 0, otherwise ï î ï 0, î otherwise Since the cycle of the transfer function is identically zero atfrequencies above 1/(2n), the cycle of the Fourier transform is also zero atfrequencies with magnitudes above 1/(2n). The sequence at the outputof the anti aliasing filter is ideally band limited to 1/(2n). The visualization ofthe output of the anti aliasing filter follows. The unit pulse response of this ideal low pass filter can be easilyevaluated with symbolic Matlab; such a Matlab script follows.Symbolic Matlab Script:>>syms f k n real;l=int(exp(i*2*pi*f*k),f,-1/n/2,1/n/2) <l=sin((pi*k)/n)/(pi*k)>>>zero=simplify(l-(sinc(k/n))/n) <zero=0>The unit pulse response of this ideal low pass filter is upr[*]= (1/ n)sinc(*/ n) . 2
  • 3. The Down Sampler: The second system in the cascade of four systems is the downsampler where the sampling rate is reduced by a factor of n; the downsampler output only contains every nth sample of the input sequence,sfin[*]=fin[n*]. The relation between the input and output Fourier transformsof a down sampler was shown earlier to be: 1 n-1 æ ·- j ö SFIN(·) = å FIN ç    ÷. n j=0 è n øThe Fourier transform at the output of the down sampler, by a factor of n,is scaled in amplitude by (1/n), stretched in frequency by a factor of n,and is repeated periodically with period unity. The visualization of the Fourier transform of the output of the downsampler has been scaled in amplitude by 1/n and stretched by a factor ofn to fill the entire band of frequencies. The periodic repetition of theFourier transform does nothing to the shape of the Fourier transform sinceit is periodic with unity period. The amplitude of the Fourier transform ismerely multiplied by n; however, it had already been divided by n. Thenet effect on the amplitude of the Fourier transform is to leave theamplitude unchanged at A. 3
  • 4. The up sampler is defined in the frequency domain; the output Fouriertransform is merely the input Fourier transform compressed by a factor ofn: SFINS(·) = SFIN(n·) . The down sampler output Fourier transform is thus   periodic with period ½; the figure that follows visualizes the down sampleroutput. The low frequency portion of this Fourier transform is identical tothe output of the anti aliasing filter; if an anti aliasing filter is used to filterthe down sampler output the filter output will be equal to the output ofthe anti aliasing filter. In some sense, this anti aliasing filter serves to interpolate for the zero ì ï SFINS(·), ·-* < 1 ï  OUT(·) = í  2n = FIN(·)  ï ï 0, î otherwiseA visualization of the sampling system output Fourier transform follows; It isidentical to the output of the anti aliasing filter; when the system input isideally band limited the input to the sampling system is then equal to theoutput of the sampling system. 4
  • 5. The numeric Matlab script below illustrates the manner in which the finalfilter interpolates the up sampled signal. The unit pulse response of thefinal filter; the down sampling factor is n=3. The unit pulse response of thefilter is upr[*]= sinc(*/ n) ; It is assumed there are only three non zerocomponents in the up sampler: ì ï 1, * = 0 ï 3, * = 3 ï sfins [*] = í 3 ï 2, * = 6 ï ï 0, otherwise îNumeric Matlab Script:>>n=3;sfins=[1 3 2];>>dt=.01;t=-10:dt:10;s1=sfins(1)*sinc(t./n);s2=sfins(2)*sinc((t-3)./n); s3=sfins(3)*sinc((t-6)./n);>>plot(t,s1,:,t,s2,:,t,s3,:,t,s1+s2+s3),hold on,plot(3*(0:2),sfins,*), xlabel(*,t),ylabel(sfins(*),sfins(t)),title(n+3) 5
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