Signal Processing and Linear Systems-B.P.Lathi copy

H int s ee exercise e45b 1 7 1 for each of t he

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Unformatted text preview: xactly ( without a ny e rror) from i ts s amples t aken u niformly a t a r ate F . > 2 B s amples p er s econd. I n o ther w ords, t he m inimum s ampling f requency is F . = 2 B Hz. t T o p rove t he s ampling t heorem, c onsider a signal f (t) ( Fig. 5 .la) w hose spectrum is b andlimited t o B Hz ( Fig. 5 .lb).t F or c onvenience, s pectra a re shown a s f unctions o f W as well a s o f F (Hz). S ampling f (t) a t a r ate o f F . Hz ( F. s amples p er s econd) c an b e a ccomplished by m ultiplying f (t) b y a n i mpulse t rain 8 r(t)(Fig. 5 .lc), c onsisting o f u nit i mpulses r epeating p eriodically e very T s econds, w here T = l /F• . T he r esult is t he s ampled s ignal7(t) r esented in Fig. 5 .ld. T he s ampled s ignal consists o f i mpulses s paced e very T s econds ( the s ampling i nterval). T he n th i mpulse, l ocated a t t = n T, h as a s trength f (nT), t he v alue o f f (t) a t t = n T. [ F(w) = 0 for t -.. = 1000". J (t) = f (t)8r(t) = L f(nT)8(t - nT) (5.1) n 4 .8-4 An angle-modulated signal is described by the equation 'PEM(t) = lOcos (wet + 0.1 sin 2000".t) ( a) F ind t he frequency deviation t:.F ( b) E stimate the bandwidth of 'PEM(t). 4 .8-5 Repeat P rob. 4.8-4 if 'PEM(t) = 5 cos (Wet + 20 sin 1000".t + 10 sin 2000".t) tThe theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose spectrum exists over a frequency band :Fe - ~ < I:FI < :Fe + ~ has a bandwidth of B Hz. Such a signal is uniquely determined by 2 B samples per second. In general, the sampling scheme is a bit more complex in this case. It uses two interlaced sampling trains, each at a rate of B samples per second (known as second-order sampling). See, for example, the references.1,2 tThe spectrum F(w) in Fig. 5.1b is shown as real, for convenience. However, our arguments are valid for complex F(w) as well. 319 5 3 20 S ampling 5.1 321 T he S ampling T heorem I f we a re t o r econstruct f (t) f rom f (t), we s hould b e a ble t o r ecover F (w) f rom F(w). T his r ecovery is possible i f t here is no o verlap b etween s uccessive cycles o f F(w). F igure 5 .le i ndicates t hat t his r equires F (Ol) Fs?:: 2 B Also, t he s ampling i nterval T = 1 / F •. (5.5) T herefore 1 T<- 2B ( b) I) T T hus, a s l ong as t he s ampling f requency F s is g reater t han t wice t he s ignal b andwidth B ( in h ertz), F (w) will c onsist o f n onoverlapping r epetitions o f F (w). I n s uch a c ase, Fig. 5.1e i ndicates t hat f (t) c an b e r ecovered from i ts s amples f (t) b y p assing t he s ampled s ignalf(t) t hrough a n i deal lowpass filter o f b andwidth B Hz. T he m inimum s ampling r ate F . = 2 B r equired t o r ecover f (t) f rom i ts s amples l et) is called t he N yquist r ate for f (t), a nd t he c orresponding s ampling i nterval T = 1 /2B is called t he N yquist i nterval for f (t).t ( I) ,tttttttt --I T I-- (c) (5.6) t- E xample 5 .1 I n t his example, we examine t he effects of sampling a signal a t t he Nyquist rate, below t he Nyquist r ate (undersampling) a nd above t he Nyquist r ate (oversampling). Consider a signal f (t) = sinc 2(511't) (Fig. 5.2a) whose spectrum is F(w) = 0.2.6...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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