3.2 Pulse Amplitude Modulation(Nature sampling or gating 显示该图片 Definition: if o(t) is an analog waveform bandlimited to B hertz, the PAM signal that uses natural sampling is o(t=o(ts(t) Where s(t) is a rectangular wave switching waveform and f=I/T>=2B s(t)=∑Ⅱ(t-k7,]/r) k=-oo ∑ k→)=∑Ⅱ(-k, k=-oo k
6 3.2 Pulse Amplitude Modulation(Nature sampling or gating) • Definition: if ω(t) is an analog waveform bandlimited to B hertz, the PAM signal that uses natural sampling is ωs (t)=ω(t)s(t) Where s(t) is a rectangular wave switching waveform and fs=1/Ts>=2B ( ) ∑∏ ) 1 ∑∏( ) ( s(t) ∑∏ [ - ]/ ∞ -∞ ∞ -∞ ∞ -∞ = = = = − = − = k k s k s d k T t k t t k T
3.2 Pulse Amplitude Modulation(PAM signal with natural sampling W(t) Ws(t) (a)Baseband Analog Waveform s(t) t (b)Switching waveform with Duty Cycle d=T/T=1/3 Figure 3-1 PAM Signal with natural sampling
7 3.2 Pulse Amplitude Modulation(PAM signal with natural sampling) (a) Baseband Analog Waveform S(t) t (b) Switching waveform with Duty Cycle d=τ /T =1/3 t W(t) W (t) s Figure 3-1 PAM Signal with natural sampling ≈ T τ
3.2 Pulse Amplitude Modulation( Generation of PAM with natural sampling) Analog bilateral switch Ws(t=W(t*s(t) s(o) clock Figure 3-2 Generation of PAM with natural sampling(gating) 8
8 3.2 Pulse Amplitude Modulation(Generation of PAM with natural sampling) clock S(f) Analog bilateral switch Ws(t)=w(t)*s(t) Figure 3-2 Generation of PAM with natural sampling (gating)
3.2 Pulse Amplitude Modulation(spectrum of a naturally sampled PAM) Theorem: the spectrum for a naturall sampled PAM signal is W()=Flo(O]=d 2 Sin na w(f-nys) =OO gn Where fs=1/Ts, Os=ifs, the duty cycle of s(t) is d=t/Ts and W(=Flo(t)I is the spectrum of the original unsampled waveform s()=(r-k7;1/r ∑Ⅱ(--k-)=∑∏(-kΣcnem0 k=-∞ k= 9
9 3.2 Pulse Amplitude Modulation(spectrum of a naturally sampled PAM) • Theorem: the spectrum for a naturally sampled PAM signal is : Where fs=1/Ts , ωs=2πfs , the duty cycle of s(t) is d=τ/Ts , and W(f)=F[ω(t)] is the spectrum of the original unsampled waveform. ∑ ∞ =-∞ ( - n ) πnd sin πnd ( ) = [ω ( )] = n s s s W f F t d W f f ( ) ∑∏ ∑∏ ∑ ∑∏ ∞ =-∞ ω ∞ =-∞ ∞ =-∞ ∞ =-∞ ) 1 - τ ) = ( τ - τ = ( s(t) = [ - ]/ τ k j n t n k k s k s s c e d k T t k t t k T ＝
3.2 Pulse Amplitude Modulation(spectrum of a naturally sampled PAM) wU) The spectrum of the PAM with natural sampling is a function of the spectrum of the analog input waveform The spectrum of the input analog waveform is repeated at harmonics of the sampling frequency wAn-2 l)w-d the PAM spectrum is zero for +3f, +6f and so on Because of d=1/3 and the spectrum in these harmonic bands is nulled out by the(sin(x)/x)function ( The bandwidth of the pam signal is smpingwitd= L3 and/=4B much larger than the bandwidth of Figure 33 Spectrum of a PAM waveform the original ananlog signal. the null bandwidth for the envelope of the Fig 3-3 Spectrum of a PAMPAM signalis 3f=12B, that is, the nt waveform with Natural bandwidth of this PAM signal is tell sampling(p132) times the bandwidth of the analog 10 Signa
10 3.2 Pulse Amplitude Modulation(spectrum of a naturally sampled PAM) • The spectrum of the PAM with natural sampling is a function of the spectrum of the analog input waveform • The spectrum of the input analog waveform is repeated at harmonics of the sampling frequency • the PAM spectrum is zero for ±3fs , ±6fs , and so on,Because of d=1/3, and the spectrum in these harmonic bands is nulled out by the (sin(x)/x) function • The bandwidth of the PAM signal is much larger than the bandwidth of the original ananlog signal. The null bandwidth for the envelope of the PAM signal is 3fs=12B, that is, the null bandwidth of this PAM signal is 12 times the bandwidth of the analog signal. • Fig. 3-3 Spectrum of a PAM waveform with Natural sampling(p132)