Lecture 12 Modulation and Sampling

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EE 2 spring 2-22 Handou #25 Lecure 2 Modulaion and Sampling The Fourier ransform of he produc of wo signals Modulaion of a signal wih a sinusoid Sampling wih an impulse rain The sampling heorem 2

Convoluion and he Fourier ransform suppose f(), g() have Fourier ransforms F (), G() he convoluion y = f g of f and g is given by y() = f(τ)g( τ) dτ (we inegrae from o because f() and g() are no necessarily zero for negaive ) from he able of Fourier ransform properies: Y () =F ()G() i.e., convoluion in he ime domain corresponds o muliplicaion in he frequency domain Modulaion and Sampling 2 2

Muliplicaion and he Fourier ransform he Fourier ransform of he produc is given by Y () = 2π y() =f()g() Y = (F G) 2π F (λ)g( λ)dλ i.e., muliplicaion in he ime domain corresponds o convoluion in he frequency domain Modulaion and Sampling 2 3

example: f() =e, F() = 2 + 2 g() = cos2, G() = πδ( 2) + πδ( + 2) he Fourier ransform of y() =e cos 2 is given by Y () = 2π = 2 F (λ)g( λ) dλ F (λ)δ( λ 2) dλ + 2 = 2 F ( 2) + F ( + 2) 2 F (λ)δ( λ + 2) dλ = +( 2) 2 + +( + 2) 2 Modulaion and Sampling 2 4

2.8.5 f().6.4 F ().2.5 3 2 2 3 3 2 2 3.5.8 y() Y ().6.4.5.2 3 2 2 3 3 2 2 3 Modulaion and Sampling 2 5

Sinusoidal ampliude modulaion (AM) u() y() =u()cos Fourier ransform of y cos Y () = 2π U() (πδ( )+πδ( + )) = 2 U( )+ 2 U( + ) cos is he carrier signal y() is he modulaed signal he Fourier ransform of he modulaed signal is he Fourier ransform of he inpu signal, shifed by ± Modulaion and Sampling 2 6

Sinusoidal ampliude modulaion Baseband Signal U() Carrier πδ(+ ) * πδ( ) = Modulaed Signal (/2)U(+ ) (/2)U( ) Modulaion and Sampling 2 7

example: u() =2+cos, =2 3 u() 2.5 2 U() =4πδ()+πδ( )+πδ(+) 4π.5 2 4 6 8 π π 3 2 y() Y () = 2 U( 2) + U( + 2) 2 2π 2π 2 3 2 4 6 8 π/2 π/2 π/2 π/2 2 9 9 2 Modulaion and Sampling 2 8

demodulaion y() =u()cos z() lowpass filer u() cos Fourier ransform of y and z: Y () = 2 U( )+ 2 U( + ) Z() = 2π Y () (πδ( )+πδ( + )) = 2 Y ( )+ 2 Y ( + ) = 4 U( 2 )+ 2 U()+ 4 U( +2 ) if U is bandlimied, we can eliminae he s and 3rd erm by lowpass filering Modulaion and Sampling 2 9

Sinusoidal ampliude demodulaion Modulaed Signal (/2)U(+ ) (/2)U( ) Demodulaion Cosine πδ(+ ) * πδ( ) Demodulaed Signal (/4)U(+2 ) = (/2)U() Lowpass Filer (/4)U( 2 ) Modulaion and Sampling 2

Applicaion Suppose for example ha u() is an audio signal (frequency range Hz 2kHz) We raher no ransmiu direcly using elecromagneic waves: he wavelengh is several km, so we d need very large anennas we d be able o ransmi only one signal a a ime he Navy communicaes wih submerged submarines in his band Modulaing he signal wih a carrier signal wih frequency 5 khz o 5 GHz: allows us o ransmi and receive he signal allows us o ransmi many signals simulaneously (frequency division muliplexing) Modulaion and Sampling 2

Sampling wih an impulse rain Muliply a signal x() wih a uni impulse rain wih period T p() = δ( kt ) k= X x() y() Sampled signal: y() =x() δ( kt) = x(kt)δ( kt) x() k= k= y() T 2T (a rain of impulses wih magniude..., x( T ), x(), x(t ), x(2t ),...) Modulaion and Sampling 2 2

The Fourier ransform of an impulse rain rain of uni impulses wih period T : p() = k= δ( kt) T T 2T 3T 4T 5T 6T Fourier ransform (from able): P () = 2π T 2π/T k= δ( 2πk T ) 2π T 2π T 4π T 6π T 8π T π T 2π T Modulaion and Sampling 2 3

Consequences of Sampling << 2π/T 3Τ 4Τ Τ 2Τ = 2π/T Τ 2Τ 3Τ 4Τ = 4π/T Τ 2Τ 3Τ 4Τ Frequencies well below he sampling rae ( <<2π/T) are sampled in he sense we expec. Frequencies a muliples of he sampling rae ( =2πn/T) looklike hey are consan. We can ell hem from DC. These frequencies alias as DC. Modulaion and Sampling 2 4

Frequency domain inerpreaion of sampling The Fourier ransform of he sampled signal is Y = (X P ), 2π i.e., heconvoluion of X wih P () = 2π T k= δ( 2πk T ) Y () = 2π = T = T = T k= k= X(λ)P ( λ) dλ X(λ) k= δ( λ 2πk T ) dλ X(λ)δ( λ 2πk T ) dλ X( 2πk T ) Modulaion and Sampling 2 5

example: sample x() =e a differen raes 2.8.5 x().6.4 X().2.5 3 2 2 3 3 2 2 3 x sampled wih T =(2π/T =6.3) 2.8.5 y().6.4 Y ().2.5 3 2 2 3 3 2 2 3 Modulaion and Sampling 2 6

x sampled wih T =.5 (2π/T =2.6) 4 3.5.8 3 y().6.4 Y () 2.5 2.5.2.5 3 2 2 3 3 2 2 3 x sampled wih T =.2 (2π/T =3.4).8 8 y().6.4 Y () 6 4.2 2 3 2 2 3 5 5 Modulaion and Sampling 2 7

The sampling heorem can we recover he original signal x from he sampled signal y? example: a band-limied signal x (wih bandwidh W ) X() W W Fourier ransform of y() = k Y () /T x(kt)δ( kt): W W 2π/T 4π/T Modulaion and Sampling 2 8

suppose we filery hrough an ideal lowpass filer wih cuoff frequency c, i.e., we muliply Y () wih H() = { c c c if W c 2π/T W,henhe resul is H()Y () =X()/T, i.e., we recover X exacly Y () /T H() W W 2π/T 4π/T Y ()H() /T W W Modulaion and Sampling 2 9

same signal, sampled wih T = π/w Y () 2π/T 4π/T we can sill recover X() perfecly by lowpass filering wih c = W sample wih T>π/W Y () 2π/T 4π/T X() canno be recovered from Y () by lowpass filering Modulaion and Sampling 2 2

he sampling heorem suppose x is a band-limied signal wih bandwidh W, i.e., X() =for >W and we sample a a rae /T y() = k= x(kt)δ( kt) hen we can recover x from y if T π/w he sampling rae musbealeas/t = W/π samples per second (W/π is called he Nyquis rae) he disorion inroduced by sampling below he Nyquis rae is called aliasing Modulaion and Sampling 2 2