美国麻省理工大学：《Communication Systems Engineering（通讯系统工程）》Lecture 3: The Sampling theorem

Lecture 3: The Sampling Theorem Eytan Modiano AA Dept. Eytan Modiano

Lecture 3: The Sampling Theorem Eytan Modiano AA Dept. Eytan Modiano Slide 1

Sampling Given a continuous time waveform, can we represent it using discrete samples? How often should we sample? Can we reproduce the original waveform?

Sampling • Given a continuous time waveform, can we represent it using discrete samples? – How often should we sample? – Can we reproduce the original waveform? � � � � � � � � � � Eytan Modiano Slide 2

The fourier transform Frequency representation of signals · Definition: X(= x()e /2midt x(1)=X(0l2d Notation X()=F[X( X(t)=F-1[X(f)] x()分X(f) Eytan Modiano

The Fourier Transform • Frequency representation of signals ∞ () = ∫−∞ x(t)e− j ft • Definition: X f dt 2π ∞ 2 π x t() = ∫−∞ X( f )e j ft df • Notation: X(f) = F[x(t)] X(t) = F-1 [X(f)] x(t) � X(f) Eytan Modiano Slide 3

Unit impulse 8(t) 6(t)=0,Vt≠0 6(t)=1 6()x(t)=x(0) 6(t-c)x()=x(t) F[6()=|60-/2=2=1 6(t) F6(t) 0()台1 Eytan Modiano

Unit impulse δ(t) δ() t = ∀ 0, t ≠ 0 ∞ δ() =1 ∫−∞ t ∞ δ() () = x(0) ∫−∞ t x t ∫ ∞ δ(t − τ )x(τ ) = x(t) −∞ δ ∞ 2π t F[ ( δ t)] F[ (t)] = ∫−∞δ(t)e− j ftdt = e0 = 1 δ() δ() t ⇔ 1 0 Eytan Modiano Slide 4 1

Rectangle pulse tk1/2 I()={1/2|t=1/2 0 otherwise FIn(]= II(oe/2gd-_[/2 e 4g dt ing Sin(f) Sinc(f jtF 1/2 Eytan Modiano

j t j f Rectangle pulse  1 | | t < 1/ 2  Π() t = 12 / | t |= 12 /  0 otherwise 12 Π ∞ 2π / F[ (t)] = ∫−∞Π(t)e− j ftdt = ∫−12 e− j2πftdt / e − π − e π πf = = Sin() = Sinc f − j f 2π πf () Π( )t 1 1/2 1/2 Eytan Modiano Slide 5

Properties of the Fourier transform Linearity x1()X1(f,x2()X2(=>ax1(t+βx2()∞X1(6+BX2(1 Duality X(x(t)=>x(X(-t)andx(1f)X(f) Time-shifting: x(t-) X(fe-12mtt Scaling: FI(X(at)]=1/a x(t/a Convolution: x(t X(f), y(t Y(f then FIx(t*y(t)]= X(Y( Convolution in time corresponds to multiplication in frequency and vIsa versa x(0)*y()=|xt-)rr

α β α β τ π τ Properties of the Fourier transform • Linearity – x1(t) X1(f), x2(t) X2(f) => αx1(t) + βx2(t) αX1(f) + βX2(f) • Duality – X(f) x(t) => x(f) X(-t) and x(-f) X(t) • Time-shifting: x(t-τ) X(f)e-j2πfτ • Scaling: F[(x(at)] = 1/|a| X(f/a) • Convolution: x(t) X(f), y(t) Y(f) then, – F[x(t)*y(t)] = X(f)Y(f) – Convolution in time corresponds to multiplication in frequency and visa versa ∞ x t( ) * y(t) = x t − τ )y(τ )dτ ∫−∞( Eytan Modiano Slide 6

Fourier transform properties Modulation) ()e/2mod+ YU-fo) Now, cos(x)= +e rej2yf I +x()e -2 x(1)cos(2) Hence, x(t)cos(2for)+ X(-f6)+X(+f Example: x(t= sinc(t), F[sinc(t)]= If Y(t)=sinc(t)cos(2Tf t(1(f-fo)+ll(f+fo))/2 /2 Eytan Modiano

Π π Π Π Fourier transform properties (Modulation) 2π x t e j fot ⇔ X( f − fo () ) e jx + e− jx Now, cos(x) = 2 x t e j fot + x t e− j2πfot x t() cos(2πfot) = () 2π () 2 ( Hence,( x t) cos(2πfot) ⇔ X f − fo ) + X( f + fo ) 2 • Example: x(t)= sinc(t), F[sinc(t)] = Π(f) • Y(t) = sinc(t)cos(2πfot) (Π(f-fo)+Π(f+fo))/2 1/2 Eytan Modiano -fo +fo Slide 7

More properties Power content of signal Autocorrelation Rr()= x(O)x(t-t)dt R()分→X0P Sampling x(t0)=x(1)6(1-t0) x(o)28(t-nto)sampled version of x(t) H∑-m)=∑8-, n=-∞ n=-0 Eytan Modiano

|( More properties • Power content of signal • Autocorrelation • Sampling Eytan Modiano Slide 8 ∫ ∞ ∞ x t −∞|( ) |2 dt = ∫−∞ X f ) |2 df ∞ * Rx () = ∫−∞ τ x(t)x (t − τ )dt Rx () ⇔ | X( f ) |2 τ x to ( )δ(t − to () = x t ) ∞ x t() ∑δ(t − nto ) = sampled version of x(t) n=−∞ ∞ ∞ F[ ∑ δ(t − nto )] = 1 ∑ δ( f − n )] t t n=−∞ o n=−∞ o

The Sampling Theorem X() Band-limited signal X(=0, forall f, I f2w Bandwidth W Sampling Theorem: If we sample the signal at intervals Ts where Ts x(n7 )sin c(-n)I x(0)=>x()sinc[2W(t-a)

The Sampling Theorem X f () • Band-limited signal X f () = 0, for all f ,| f | ≥ W – Bandwidth x t s () = ∑ x(n Ts)sin c[( − n)] 2 W T n=−∞ s ∞ n n () = ∑ x t x( )sin c W[2 (t − )] 2 W 2 W n=−∞ Eytan Modiano Slide 9

Proof )=x)28(-m X()=X(*F∑(-m7 F∑8-m)=r∑(f X()=∑X0 n n=-∞ The Fourier transform of the sampled signal is a replication of the Fourier transform of the original separated by 1/Ts intervals 2/TS Slide 10

Proof ∞ x t () ∑δ(t − nTs ) δ () = x t n=−∞ ∞ X f ( ) * F[ ∑δ(t − nTs )] δ () = X f n=−∞ ∞ ∞ F[ ∑δ(t − nTs )] = 1 ∑ δ( f − n ) n=−∞ Ts n=−∞ Ts 1 ∞ n δ X f () = ∑ X f ( − ) Ts n=−∞ Ts • The Fourier transform of the sampled signal is a replication of the Fourier transform of the original separated by 1/Ts intervals -1/Ts -w w 1/Ts 2/Ts Eytan Modiano Slide 10