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Lecture 41 Practical sampling and reconstruction
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Lecture 4 2 Outline F Practical sampling –Aperture effect –Non ideal filters –Non-band limited input signals F Practical reconstruction F Practical digital systems F Discrete time Fourier transform
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Lecture 4 3 Practical sampling F Practical sampling differs from ideal in the following respects –The sample (impulse) train actually consists of pulses of duration –Real signal are time limited, therefore cannot be band limited (the uncertainty principle of Fourier transform) –Reconstruction filters are not ideal
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Lecture 4 4 xa(t)xa(t) n= s(t) = (t nT) x s (t) = x a (nT) (t nT) n= sample and hold filter 1 h(t)h(t)
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Lecture 4 5 Xa(j)Xa(j) 1 s >2 1/T /T |H s (j )| // // H (j ) = sin ( /2) e -j /2
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Lecture 4 6 s >2 |X s (j )| /T/T /T X s (j ) = H(j ). 1 T X a (j jk s ) k= - If / there is no significant distortion over signal band (otherwise equalization can compensate distortion) Aperture effect
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Lecture 4 7 Non ideal filters F The problem of non-ideal filtering can be combatted by increasing 1/T F Effectively, we are only using the middle portion of the filter, where it is closer to perfect
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Lecture 4 8 Non-band limited signals F We may be only interested in a low frequency portion of a wide-band signal, e.g. speech only needs upto 3-4 Khz F There may be high frequency, wideband additive noise in the input signal F So we prefilter | c 1 0, | H aa (j ) =
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Lecture 4 9 Anti-aliasing filter F Filter prior to sampling removes higher frequency components which could have been moved into the lower frequency range by aliasing F Of course, we are distorting the signal, but this is in a frequency range in which we are not interested F If we were, we would need to use a higher sampling frequency
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Lecture 4 10 Practical D/A conversion F We don’t have perfect interpolation F Sample and hold F The impulse response of a sample and hold filter is h(t) 1 h(t)h(t) T
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Lecture 4 11 Frequency domain F We need to compensate by adding a compensated reconstruction filter after the sample and hold process |H s (j )| // // H o (j ) = sin ( /2) e -j /2
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Lecture 4 12 |H r (j )| // e j /2, | sin ( /2 ) ~ Compensated reconstruction filter 0, | ~ H r (j ) =
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Lecture 4 13 Ideal system A/DD/A h(n)
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Lecture 4 14 Practical system h(n) A/D T Antialiasing pre-filter Sample and Hold T Sample and Hold Compensated Reconstruction Filter H aa (j ) D/A
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Lecture 4 15 Effective frequency response H eff (j ) = H r (j ) H 0 (j ) H(e j T ) H aa (j )
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Lecture 4 16 xa(t)xa(t) n= S(t) = (t nT) x s (t) = x a (nT) (t nT) n= convert to discrete sequence x[n] = x a (nT)
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Lecture 4 17 Back to sampling Let x a (t) aperiodic x s (t) = x a (nT) (t nT) aperiodic X s (j ) = x a (nT) (t nT) e - j t dt n=
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Lecture 4 18 X s (j ) = x a (nT) (t nT) e - j t dt x a (nT) e -j nT = x[n] e -j n where = T X( e j ) = x[n] e -j n is defined as the discrete time Fourier transform of x[n] n=
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Lecture 4 19 F From the sampling theorem, X s (j ) = X a (j kj s ) F thus, X( e j ) = X a (j j ) In other words, on going from the continuous domain to the discrete domain, we undergo a scaling or normalization in frequency from = s to = 2 F There is also a corresponding time normalization from t=T to n=1 n= 1T1T k= 1T1T TT 2 k T
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Lecture 4 20 Discrete time Fourier transform X(e j ) = x[n] e -j n –continuous and periodic with period 2 x[n] = X(e j ) e j n d –discrete and aperiodic n= 1
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Lecture 4 21 Example F Let x[n] = u[n]-u[n-M] = [n-k] Then X(e j ) = x[n] e -j n = e -j n = = e -j M-1)/2 k=0 M-1 n=0 M-1 n= 1-e -j M 1-e -j sin( M/2) sin( /2)
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Lecture 4 22 sin( M/2) sin( /2) |X(e j )| = / // 0...
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Lecture 4 23 Reading F Discrete time signals: Sections F Sampling: Sections 8.2 F Reconstruction, quantization, coding: Sections 8.2 - p.363 F Practical sampling and reconstruction: Sections 8.2 - p.355 F Fourier transform: Chapter 4
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