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ECE 4710: Lecture #7 1 Overview  Chapter 3: Baseband Pulse & Digital Signaling  Encode analog waveforms into baseband digital signals »Digital signaling.

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Presentation on theme: "ECE 4710: Lecture #7 1 Overview  Chapter 3: Baseband Pulse & Digital Signaling  Encode analog waveforms into baseband digital signals »Digital signaling."— Presentation transcript:

1 ECE 4710: Lecture #7 1 Overview  Chapter 3: Baseband Pulse & Digital Signaling  Encode analog waveforms into baseband digital signals »Digital signaling is low cost, flexible, and has many other benefits  Compute spectrum for the digital signals »What is the bandwidth required for effective transmission of digital signal?  Filter baseband signals to minimize bandwidth »Required to ensure efficient utilization of available spectrum  Examine affect of filtering on ability to recover digital information »Filtering can introduce Inter Symbol Interference (ISI) leading to imperfect recovery of digital data

2 ECE 4710: Lecture #7 2 Pulse Amplitude Modulation  Pulse Amplitude Modulation = PAM  Basic method for converting analog signal to a pulse-type signal where amplitude of pulse represents analog information  First step required for converting an analog waveform to a PCM (digital) signal »PCM = Pulse Code Modulation (studied next)  PAM is used by itself in a few limited applications  Two classes of PAM signals: »Natural sampling or gating  easy to generate »Instantaneous “flat-top” sampling  PCM conversion

3 ECE 4710: Lecture #7 3 Natural Sampling PAM  Natural Sampling PAM  Let w(t) be analog waveform bandlimited to B Hz  f s = 1 / T s  2 B (Nyquist Sampling)   = duration of sampling pulse < T s  d = duty cycle =  / T s » % time the pulse is “on” in a each period

4 ECE 4710: Lecture #7 4 Natural Sampling PAM w(t) s(t) w s (t) = w(t) s(t)

5 ECE 4710: Lecture #7 5 PAM Generation  Generation of natural sampling PAM is easily done with existing CMOS hardware  bi-lateral switch + clock

6 ECE 4710: Lecture #7 6 PAM Spectrum  What is frequency spectrum of natural sampling PAM?  Recall impulse sampling results:  Multiplication in time is convolution in frequency:  Spectrum of impulse sampled signal is the spectrum of the unsampled signal that is repeated every f s Hz t TsTs f  fsfs 2f s -f s -2f s 0... s(t)

7 ECE 4710: Lecture #7 7 PAM Spectrum  In the limit as   0 then PAM s(t)  impulse sampling  Natural sampling PAM is a real-world way of impulse sampling  It should not be surprising that spectrum of the unsampled signal is repeated every f s Hz in the PAM signal spectrum  What is effect of non-ideal (   0) PAM sampling?  Series of impulse functions whose amplitudes vary as a function of sin x / x

8 ECE 4710: Lecture #7 8 PAM Spectrum  Ideal Impulse Sampling  Non-ideal natural sampling PAM t TsTs f  fsfs 2f s -f s -2f s 0... s(t) f t TsTs  fsfs 2f s -f s -2f s 0... s(t) 

9 ECE 4710: Lecture #7 9 PAM Spectrum  PAM spectrum is convolution of baseband analog signal spectrum, W(f), with non-ideal PAM sampling function  Assume rectangular baseband spectrum: fsfs 2f s -f s -2f s 0...

10 ECE 4710: Lecture #7 10  Baseband spectrum repeated @ harmonics of sampling frequency  PAM BW is many times larger than original signal BW  For d = 1/3 and f s = 4 B the FNBW is 3 f s = 12 B for PAM signal  factor of 12 increase !! PAM Spectrum

11 ECE 4710: Lecture #7 11 PAM Recovery  Original spectrum can be recovered using LPF on n = 0 harmonic  Harmonics have identical replicas of shape (not amplitude) of W(f) assuming Nyquist sampling f s  2 B (no aliasing or spectral folding)  Why choose f s = 4 B (example) when Nyquist rate is f s = 2 B ??  Oversampling is often done since ideal “brick wall” LPF is not physically realizable  Real LPF filter rolloff can allow signal energy from adjacent spectral copies to distort baseband signal spectrum  Previous example assumes bandlimited baseband signals  Pre-filtering of large bandwidth baseband analog signals is done to limit bandwidth and eliminate aliasing

12 ECE 4710: Lecture #7 12  Product detection can also be used to recover baseband spectrum from other harmonics, e.g. n = 1, 2, 3, etc.  Multiply PAM signal with cos (2  n f s t ) and then LPF »Shifts frequency band @ n f s back to baseband ( f = 0)  Why? »Baseband PAM signal may be corrupted by low frequency noise »Higher bands are usually free of most noise »Use more complicated product detection to get better signal PAM Recovery

13 ECE 4710: Lecture #7 13 Flat-Top PAM  Instantaneous sample and hold of analog signal produces “Flat-Top PAM”  Instantaneously sampled w(t) determines amplitude of constant rectangular pulse  Compare to natural sampling PAM  Amplitude variation of w(t) preserved within pulse width

14 ECE 4710: Lecture #7 14 Flat-Top PAM w(t) s(t) w s (t) = w(t) s(t)

15 ECE 4710: Lecture #7 15 Flat-Top PAM Spectrum  Spectrum for flat-top PAM is given by  The amplitude of original baseband spectrum W(f) is distorted by H(f) !!  Compare to natural sampling PAM  c n is constant and does not distort spectral shape of W(f)

16 ECE 4710: Lecture #7 16 Flat-Top PAM Spectrum  Assume bandlimited baseband spectrum for W(f)

17 ECE 4710: Lecture #7 17  Harmonic ( n  0) replicas of W(f) significantly distorted by H(f)  Baseband ( n = 0) W(f) replica has small distortion at higher frequencies Flat-Top PAM Spectrum

18 ECE 4710: Lecture #7 18 Flat-Top PAM Recovery  LPF baseband spectrum  Can compensate for high-frequency distortion by adding additional gain to higher frequencies of low pass filter response »Equalization filter with transfer function  1/ H(f)  Product detection can also be used  Add pre-filter before product multiplier to make spectrum flat

19 ECE 4710: Lecture #7 19 PAM Signal Transmission  Large BW of PAM signals is undesirable for wireless and/or long-distance transmission  PAM BW is many times larger than original baseband signal BW  Magnitude and phase response of channel will likely distort PAM signal considerably  S/N performance of PAM is worse than straight transmission of analog signal  Primary use of PAM is for conversion to PCM  Digital PCM signal has many performance advantages compared to analog or PAM signal transmission


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