Spectrum Representation

Transcription

1 Spectrum Representatin Lecture #4 Chapter 3 99

2 What Is this Curse All Abut? T Gain an Appreciatin f the Varius Types f Signals and Systems T Analyze The Varius Types f Systems T Learn the Sills and Tls needed t Perfrm These Analyses. T Understand Hw Cmputers Prcess Signals and Systems 100

3 What is a Spectrum? A signal is a functin f time which can be represented by a series f sinusidal functins r sinusidal cmpnents. These sinusidal cmpnents have different frequencies, different amplitudes, and different phases. Therefre, the plts f frequency versus amplitude and phase fr the sinusidal cmpnents which cmprise the signal are called the Frequency Spectrum r Spectrum f the signal. 101

4 Decmpsitin f a signal We can express the fllwing representatin f a functin: xt () A Acs( ft ) 1 j ft { } 1 Where X A is a real cnstant (DC cmpnent) and X e X e j X A e is the phasr fr frequency f Here we see that there are +1 frequency cmpnents fr x(t), 0 and with each frequency there is a phasr. 10

5 Decmpsitin f a signal Fr example, the th frequency has a phasr X with amplitude, A, and phase, θ. j ft () cs( ) { } 1 1 j xt A A ft X e Xe Where X DC cmpnent and X A e is the phasr fr frequency f X 0 X 1 X X X 0 f 1 f f f This is really nly half f the spectrum. frequency 103

6 Alternative frm f the Spectrum Using Euler s frmula, let rewrite x(t) : A A xt A A ft X e e e e j j ft j j ft () cs( ) { } 1 1 X j f * t X j ft X { e e } 1 Using this apprach, we see that there are +1 frequency cmpnents Or we can say that fr each where 1, there is a psitive frequency f with phasr X / and a negative frequency -f with phasr X * /. Therefre, we say that the spectrum is tw-sided. X X / * / X 1* / X 1 / X X / * / X * / X / X 0 -f -f -f -f 1 0 f 1 f f f frequency 104

7 An Example xt ( ) cs(00 t 3) 8cs(500t ) 10 7e e 7e e j 3 j00t j 3 j00t 4e e 4e e j j500t j j500t 4e -jπ/ 7e jπ/3 10 7e -jπ/3 4e jπ/ f 105

8 Fr the 3 rd terms in this sum Let s define: A Furthermre, define tatin Change f Then, A f X X * xt X e e j j j X * ( A e )* A e A e X A fr 0 a j ft j ft () { } 1 1 Ae j fr 0 106

9 Using the new ntatin anther (simpler) frm appears X X * X X * xt () A { e e } A e e j ft j ft j ft j ft A A A A A e e e e A e e e e j j ft j j ft j j ft j j ft ( )* Using the new ntatin, we nw have A A A e e e e j j f t j j f t 1 1 In the secnd sum, we replace with - A A A e e e e j j f t j j f t 1 1 j ft j ft j ft 1 1 a a e a e a e The Spectrum is nw assciated with the 1 's a 107

10 Multiplicatin f Sinusids A beat nte (r frequency) is the result f multiplying sinusids We see this in many applicatins: Music and bradcasting, What is the spectrum f the prduct f sinusids? 108

11 Prduct f Sinusids xt ( ) cs( ft) cs( ft); let's assume f f 1 1 j f1t j f1t j ft j ft e e e e ( )( ) j ( f1 f) t j ( f1 f) t j ( f1f) t j ( f1f) t 1 e e e e ( ) 1 (cs ( f ) cs ( ) ) 1 f t f1 f t (f 1 + f ) -(f 1 - f ) 0 f 1 - f f 1 + f f 109

12 Prduct f Sinusids The prduct f sinusids prduce a signal which is the sum f sinusids whse frequencies are the sum and difference f the frequencies f the riginal sinusids. In fact, if we define f 1 = f c as a center frequency and f c >> f, such that f = Δf (a small value cmpared t f C ), then the prduct f the tw sinusids will yet frequencies ± Δf arund f c 1 x( t) cs( fct) cs( ft) (cs ( fc f) tcs ( fc f) t) (f c + Δf) -f c -(f c Δf) f c 0 f c - Δf f c + Δf f 110

13 Examples Tw signals 0 Hz and 00 Hz Resultant prduct signal at 0 Hz and 180 Hz An envelp effect results (Beat Signal)

14 Examples Tw signals 180 Hz and 00 Hz Resultant prduct signal at 0 Hz and 380 Hz tice the 0 Hz cmpnent (A single 0 Hz signal is added t emphasis the 0 Hz cmpnent)

15 Amplitude Mdulatin Amplitude Mdulatin r AM is the technique used t bradcast AM radi. The infrmatin (e.g., the newscaster s vice) is mdulates a carrier signal. Usually, the carrier signal is at a frequency (the carrier frequency) which is much higher than the infrmatin t be bradcast (the AM band is 660 Hz thrugh 1600 Hz while the infrmatin is a vice signal which is between 150 Hz and 4 Hz) The frm f AM is: x(t)=v(t)cs(πf c t) 113

16 Spectrum f an AM Signal Let v(t) = A+Bcs(πf i t) Then the AM signal becmes: x(t)=v(t)cs(πf c t)= (A+Bcs(πf i t)) cs(πf c t) = A cs(πf c t) +Bcs(πf i t) cs(πf c t) = A cs(πf c t)+b/(cs(π[f c +f i ]t)+cs(π[f c -f i ]t) 114

17 Spectrum f an AM Signal x(t)= A cs(πf c t)+b/(cs(π[f c +f i ]t)+cs(π[f c -f i ]t) The spectrum has 3 cmpnents: at f c, f c +f i and f c -f i where the latter are called the sidebands f the AM signal. f c -f i is the lwer sideband and f c +f i is the upper sideband B/4 A/ B/4 A/ B/4 B/4 -(f c + Δf) -f c -(f c Δf) 0 f c - Δf f c + Δf f c 115

18 The wavefrm f an AM Signal x(t)= (1+0.5 cs(π0t)) cs(π00t)

19 Hmewr Exercises: Prblems: 3.1, Instead use xt ( ) sin (3 t) 117