Communication Systems Engineering Episode 1 Part 3 docx

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? � � � � � � � � � � Eytan Modiano Slide 2 The Fourier Transform • Frequency representation of signals ∞ () = ∫ −∞ x(t)e − jft dt • Definition: Xf 2 π ∞ 2 π xt() = ∫ −∞ X( f )e jft df • Notation: X(f) = F[x(t)] X(t) = F-1 [X(f)] x(t) � X(f) Eytan Modiano Slide 3 δδ δ Unit impulse δ (t) t 0, δ () =∀t ≠ 0 ∞ δ () =1 ∫ −∞ t ∞ δ () () = x(0) ∫ −∞ txt ∫ ∞ δ (t − τ )x( τ ) = x(t) −∞ δ ∞ 2 π t F[(t)] δ F[(t)] = ∫ −∞ δ (t)e − jft dt = e 0 = 1 δ () t δ () ⇔ 1 0 Eytan Modiano Slide 4 1 jt jf Rectangle pulse t  1 ||< 1/ 2  t / / Π() =  12 | t |= 12   0 otherwise 12 Π ∞ 2 π / F[(t)] = ∫ −∞ Π(t)e − jft dt = ∫ −12 e − j 2 π ft dt / e − π − e π π f = = Sin() = Sinc f − jf π f 2 π () Π()t 1 1/2 1/2 Eytan Modiano Slide 5 αα α ββ β αα α ββ β ττ τ ππ π ττ τ Properties of the Fourier transform • Linearity – x1(t) <=> X1(f), x2(t) <=>X2(f) => α x1(t) + β x2(t) <=> α X1(f) + β X2(f) • Duality – X(f) <=> x(t) => x(f) <=> X(-t) and x(-f)<=> X(t) • Time-shifting: x(t- τ ) <=> X(f)e -j2 π f τ • 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 ∞ xt xt − τ )y( τ )d τ ()* y(t) = ∫ −∞ ( Eytan Modiano Slide 6 ΠΠ Π ππ π ΠΠ Π ΠΠ Π Fourier transform properties (Modulation) 2 π xt e jf o t ⇔ X( f − f o )() e jx + e − jx Now, cos(x) = 2 xt e jf o t + x t e − j 2 π f o t xt() cos( 2 π f o t) = () 2 π () 2 ( x tHence,( ) cos( 2 π f o t) ⇔ Xf − f o ) + X( f + f o ) 2 • Example: x(t)= sinc(t), F[sinc(t)] = Π (f) • Y(t) = sinc(t)cos(2 π f o t) <=> ( Π (f-f o )+ Π (f+f o ))/2 1/2 Eytan Modiano -f o +f o Slide 7 |( More properties • Power content of signal • Autocorrelation • Sampling Eytan Modiano Slide 8 ∫ ∞ ∞ xt −∞ |( ) | 2 dt = ∫ −∞ Xf ) | 2 df ∞ * R x () = ∫ −∞ x(t)x (t − τ )dt τ R x () ⇔ | X( f ) | 2 τ xt o ( ) δ (t − t o )() = xt ∞ xt() ∑ δ (t − nt o ) = sampled version of x(t) n =−∞ ∞ ∞ F[ ∑ δ (t − nt o )] = 1 ∑ δ ( f − n )] t t n =−∞ o n =−∞ o The Sampling Theorem Xf() () = 0, for all f ,| f | ≥ W • Band-limited signal Xf – Bandwidth < W -w w Sampling Theorem: If we sample the signal at intervals Ts where Ts <= 1/ 2W then signal can be completely reconstructed from its samples using the formula ∞ xt() = ∑ 2W © T s x(nT s )sin c[2W © (t − nT s )] n =−∞ Where, W ≤ W © ≤ 1 − W T s ∞ 1 t WithT = => xt s () = ∑ x(nT s )sin c[( − n)] 2 W T n =−∞ s ∞ n n () = ∑ x( )sin [ 2 xt cW(t − )] 2 W 2 W n =−∞ Eytan Modiano Slide 9 Proof ∞ xt () ∑ δ (t − nT s ) δ () = x t n =−∞ ∞ Xf ()* F[ ∑ δ (t − nT s )] δ () = X f n =−∞ ∞ ∞ F[ ∑ δ (t − nT s )] = 1 ∑ δ ( f − n ) n =−∞ T s n =−∞ T s 1 ∞ n δ () = ∑ Xf − )Xf ( Ts n =−∞ T s • 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 [...]... sample at a rate somewhat greater than 2W which makes reconstruction filters that are easier to realize • Given any set of arbitrary sample points that are 1/ 2W apart, can construct a continuous time signal band-limited to W Eytan Modiano Slide 12 ... rectangular pulse • Now the recovered signal after low pass filtering f ) 2W f )] x (t ) = F 1[ Xδ ( f )Ts Π( 2W ∞ t x (t ) = ∑ x (nTs )Sinc( − n) Ts n = −∞ X ( f ) = Xδ ( f )Ts Π( Eytan Modiano Slide 11 f ) 2W Notes about Sampling Theorem • When sampling at rate 2W the reconstruction filter must be a rectangular pulse – Such a filter is not realizable – For perfect reconstruction must look at samples...Proof, continued • If 1/ Ts > 2W then the replicas of X(f) will not overlap and can be recovered • How can we reconstruct the original signal? – Low pass filter the sampled signal H ( f ) = Ts Π( • Ideal low pass filter is a rectangular pulse • Now the recovered signal after low pass filtering f ) 2W f )] x (t ) = F 1[ Xδ ( f )Ts Π( 2W ∞ t x (t ) = ∑ x (nTs )Sinc( − n) . = ∫ −∞ δ (t)e − jft dt = e 0 = 1 δ () t δ () ⇔ 1 0 Eytan Modiano Slide 4 1 jt jf Rectangle pulse t  1 ||< 1/ 2  t / / Π() =  12 | t |= 12   0 otherwise 12 Π ∞ 2 π / F[(t)]. the Fourier transform of the original separated by 1/ Ts intervals -1/ Ts -w w 1/ Ts 2/Ts Eytan Modiano Slide 10 Proof, continued • If 1/ Ts > 2W then the replicas of X(f) will not overlap. αα α ββ β αα α ββ β ττ τ ππ π ττ τ Properties of the Fourier transform • Linearity – x1(t) <=> X1(f), x2(t) <=>X2(f) => α x1(t) + β x2(t) <=> α X1(f) + β X2(f) • Duality – X(f) <=> x(t) =>