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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 .84 An anglemodulated 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 .85 Repeat P rob. 4.84 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 secondorder 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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