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Unformatted text preview: eying ( FSK), because t he i nformation digits are t ransmitted by shifting t he c arrier
frequency. = 99.9 MHz
100.1 MHz F or P M Because m (t) increases a nd decreases linearly with time, t he i nstantaneous frequency increases linearly from 99.9 t o 100.1 MHz over a halfcycle a nd decreases linearly from 100.1
t o 99.9 M Hz over t he r emaining halfcycle of t he m odulating signal (Fig. 4.43b). :Fi = Fe + ~; m (t) = 108 + 5 m (t) = 108  51[m(t)]minl = 108 _ 10 5 = 99.9 MHz
8
(Fi)max = 10 + 5 [m(t)]max = 100.1 MHz
(F;)min 'PPM (t) = A cos [wet .w j I I
99.9 MHz I (a) t , n(t) 1 l IJ'FM(t) IJ'PM(t)
t ..... • (c) l n ~~~ ~~~ ~~ ~ ~ ~~~ ~~ ~ ~ ~~ ~ ~ ~ ~ ~ ~ + ~m(t)] {A sin wet
 A sin wet t ...... ~~~~~~~~~~~~~ ~~~~~~~~~~~~~~ t ...... when m (t) =  1
when m (t) = 1 T his P M wave, illustrated in Fig. 4.44c, has t he s ame frequency Fe = 100 MHz everywhere.
However, t here a re phase discontinuities of 11" r adians a t t he i nstants where impulses of m (t)
are located. At these instants, t he c arrier phase shifts by rr i nstantaneously. A finite phase
shift in zero time implies infinite instantaneous frequency (dB / dt = 00) a t t hese instants.
This conclusion agrees w ith o ur observation a bout m (t).
T his scheme of carrier phase modulation by a digital signal is called p haseshifting
k eying ( PSK), b ecause information digits are t ransmitted by shifting t he c arrier phase.
Note t hat P SK may also be viewed as a DSBSC m odulation by m (t).
T he a mount of phase discontinuity in 'PPM (t) a t t he i nstant where m (t) is discontinuous is kpmd, where md is t he a mount o f discontinuity in m (t) a t t hat i nstant. In t he
p resent example, t he a mplitude of m (t) changes b y 2 (from  1 t o 1) a t t he discontinuity.
Hence, t he p hase discontinuity in 'PPM (t) is kpmd = ~ (2) = rr r adians, which confirms our
earlier result.
When m (t) is a d igital signal (as in Fig. 4.44a), ' PPM (t) shows a phase discontinuity
where m (t) h as a j ump discontinuity. In such a case t he p hase deviation k pm(t) m ust be
restricted t o a range (11", 11") in order to avoid ambiguity in demodulation. For example,
if kp were 311"/2 in t he p resent example, t hen (d) (b) F ig. 4 .44 1 = 100 MHz 100.1 MHz + kpm(t)] = A cos [wet Because m et) switches back a nd f orth from a value of  20,000 t o 20,000, t he c arrier
frequency switches back a nd f orth from 99.9 t o 100.1 MHz every halfcycle of m (t), as
illustrated i n Fig. 4.43d.
T his i ndirect m ethod of sketching P M (using m (t) t o frequencymodulate a carrier)
works as l ong as m (t) is a c ontinuous signal. I f m (t) is discontinuous, m (t) contains
impulses, a nd t his m ethod is n ot so convenient. I n such a case, a direct approach should
b e used. T his is d emonstrated in t he n ext example.
•
1 1.
= :F.e + 2k f. (t ) = 108 + 4m (t )
11" m T he derivative m (t) = 0 everywhere except for impulses of s trength ± 2 a t t he p oints of
discontinuities of m (t) (Fig. 4.44c). This fact means t he c arrier frequency is Fe = 100
MHz everywhere except a t...
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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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