{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 5

# Chapter 5 - Chapter 5 H mte 2000'9‘“ t"-sooo t"...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 5 H mte) - 2000 '9‘“) t" -sooo t" t... t—a IO-IM q.qM '10.“ 10M 4.," IOM :on Fig.55.1-1 5.1-1 in this case fc =10 MHz. In, = 1 and m; = 8000. For FIVI : A} = twp/27. = 2W x 105/21: = 105 Hz, Also j. = 107. Hence. (f.)....,. = 10' +105 = 10.1 MHz. and (Li-n... = 107 - 105 = 9.9 MHz. The carrier frequency increases linearly from 9.9 M Hz to 10.1 MHz over a quarter (rising) cycle of duration (1 seconds. For the next (1 seconds. when m(t) = 1. the carrier frequency remains at 10.1 MHz. Over the next quarter (the falling) cycle of duration a. the carrier frequency decreases linearly from 10.1 MHz to 9.9 MHz.. and over the last quarter cycle, when ml!) = —-l. the carrier frequency remains at 9.9 MHz. This cycles repeats periodically with the period 4a seconds as shown in Fig. 85.1-ia. For PM : A] = kpmm: = 50: x 3000/27: = 2 x 10‘ Hz. Also I. = 10". Hence. (1.)...“ = 10’ + 2 x 103 = 10.2 MHz. and (12).“... = 107 — 2 x 10” = 9.8 MHz. Figure S5.1-1b shows mu). We conclude that the frequency remains at 10.2 MHz over the (rising) quarter cycle. where 1ii(l) = 8000. For the next a seconds. vh(t) = 0. and the carrier frequency remains at 10 MHz. Over the next a seconds. where n’t(t) = -8000. the carrier frequency remains at 9.8 MHz. Over the last quarter cycle 1h(t) = 0 again, and the carrier frequency remains at 10 MHz. This cycles repeats periodically with the period 4n seconds as shown in Fig. 55.1-1. 5.1-2 In this case ft = 1 MHz. 111,. = l and "1;: 2000. For PM : A] = mop/2n = 20.0007/21r = 10‘ Hz. Also 1. = 1 MHz. Hence. (1.)...” = 10° + 10‘ = 101 MHz. and (I.)min = 106 — 10" = 0.99 MHz. The carrier frequency rises linearly from 0.99 MHz to 1.01 MHz over the cycle (over the interval - 1°- 3 < f < 395:). Then instantaneously. the carrier frequency falls to 0.99 MHz and starts rising linearly to 10.01 MHz over the next cycle. The cycle repeats periodically with period 10'3 as shown in Fig. 55.1-2a. For PM : Here. because mU) has jump discontinuities. we shall use a direct approach. For convenience. we select the origin for m(t) as shown in Fig. 85.1-2. Over the interval -3 to we can express the message signal as mU) = 2000?. Hence. emu) = cos [21r(10)6t + 321nm] = cos [21r(10)°t + 220001] =- cos [21r(10)6! + moon] = cos [2x (106 + 500) 1] At the discontinuity. the amount of jump is 1nd = 2. Hence. the phase discontinuity is kpmd = 1r. Thereibre, the. carrier frequency is constant throughout at 106 + 500 Hz. But at the points of discontinuities. there is a 45 5.1-3 5.2-1 5.2-2 5.2-3 5.2-4 5.2-5 Fig. 55.1.2 phase discontinuity of 11' radians as shown in Fig. S5.1-2b. In this case, we (3) Wm“) = Acos [wet + k,1n(f)] = 10 cos[10,000 We are given that ¢p,\1(t) = 10 cos (13.000!) with k, = 1000. Clearly. m(t (b) t I Therefore k, / m(o)da =1000/ t Hence St = / m(a) do = m(f.) = In this case k! = 10001: and k, = 1. For m(?) = 2 cos 1001+ 18 cos 200011! and must maintain k, < 1r because there is a discontinuity of the amount 2. For k, > 1r, the phase discontinuity will be higher than 21: giving rise to ambiguity in demodulation. t+ k,m(r)] ) = 3!. over the interval |t| S 1. I l inﬁll = .4 COS [~21 + k1] m(a)da] = 10 cos [10.0001 + hf 112(c)(la] "1(a) do = 3000! 3 111(1) 2 ~200 sin lOOt — 36.00011' sin 2000M Therefore 17:, = 20 and "1;, = 36. 0001! + 200. Also the baseband signal bandwidth 8 = 20001r/21' = 1 kHz. For FM : : A] = k,m,,/2n = 10.000. and em = 2(Af + B) = 2(20.ooo +1000)= 42 kHz. For PM : : Af = k,m}/21r = 18.000 + I? Hz. and Bpm = 2(Af + B) = 2(18.031.83 +1000) == 38.06366 kHz. 95531“) = 10 cos(w¢t + 0.1 sin 2000171). Here. the baseband signal bandwidth B = 20001r/21r = 1000 Hz. Also: 01.0) = w; + 2001 cos 20001” Therefore. A.» = 2001: and A f = 100 Hz and em = 2(A f + B) = 2000 + 1000) = 2.2 kHz. ,9“. (f) = 5 cos(.act + 20 sin 10001" + 10 sin 2000“). Here. the baseband signal bandwidth B = 20001r/21r = 1000 Hz. Also. w,(1) = we + 20. 0001r cos 10001rt + 20,0001 cos 2000M Therefore. A.» = 20.0001r+20.0001r = 40.0001r and AI = 20 kHz and 85” kHz. The baseband signal bandwidth 8 = 3 x 1000 = 3000 Hz. For FM : A] = 2;? = 1293‘ =15.951kHz and am = 2(Af + 13) = For PM: A; = “'2' = = 31.831 kHz and 13m = 2(Af+ B) The baseband signal bandwidth 8 = 5 x 1000 = 5000 Hz. For FM : 46 = 2(AI+B) = 2(20.000+1000) = .22 37.831 kHz. = 66.662 kHz. Aj=ilg£=mg§ﬂ=1knzend Bp,\1=2(Af+B)=2(2+5)=14 kHz. .1 For PM : To ﬁnd BPM. we observe from Fig. 85.1-2 that thM (t) is essentially a sequence of sinusoidal pulses of width T = 10‘“ seconds and of frequency fr, = 1 MHz. Such a pulse and its spectrum are depicted in Figs. 3.22c and d. respectively. The bandwidth of the pulse, as seen from Fig. 3.22d, is tin/T rad/s or 2/1" Hz. Hence. BPM = 2 kHz. 5-2-6 (al FOP FM = A! = k-‘k? = = 100 kHz and the baseband signal bandwidth B = 33:79: = 1 kHz. Therefore BFM = 2(A/ + B) = 202 kHz For PM: A} = = 10—42395! =10 kHz and BPM = 2(Af + B) = 2(10+l) = 22 kHz. (b) m(I) = 2 sin 200011.. and B = 20001r/21r = 1 kHz. Also m, = 2 and m; = 4000a. For-FM: Af=3%‘2=l°£%f"£=2oomz.snd 3m = 2(AI + B) = 2(200 + 1) = 402 kHz For PM: Af = “'2' = mam = 20 kHz and 13m = 2(Af + B) = 2(20 + 1) = 42 kHz. (c) m(t) = sin 40001rt. and B = 40001r/21r = 2 kHz. Also m, = 1 and m.’, = 40007t. For-FM: Aj=%z=m°°#n=ioomz,and BFM = 2(Af + B) = 2(100 + ‘2) = 204 kHz For PM: a; = = 394515991 =20 kHz and an, = 2(A1+ B) =2(2o+2)=44 kHz. (d) Doubling the amplitude of m (f) roughly doubles the bandwidth of both FM and PM. Doubling the frequency of mm {expanding the spectrum 51(w) by a factor 2] has hardly any effect on the FM bandwidth. However. it roughly doubles the bandwidth of PM. indicating that PM spectrum is sensitive to the shape of the baseband sport rum. FM spectrum is relatively insensitive to the nature of the spectrum M(..;). 5.2-? From pair 22(Table 3.1). we obtain r", <==9 1r NJ“. The spectrum MM) = ﬁn' “1" is a Gaussian pulse. which decays rapidly. lts 3 dB bandwidth is 1.178 i'ad/s=0.]87 Hz. This is an extremely small bandwidth compared to A f . Also vhf!) = —2fr"‘/2. The spectrum of rii(t) is M'(.u) = jwlllw) = jﬁwe‘”2/‘. This spectrum also decays rapidly away from the origin. and its bandwidth can also be assumed to be negligible compared to Af. For FM: Aj=£lgz =2°%r—‘-1.—.3kuzand Br31z2Af=2x3=6kHL For PM : To find "1;. we set the derivative of rh(r) = —2t¢!"2/2 equal to zero. This yields iii(t) = —2a":/2 + “an-'2"; = 0‘ =s z = .1— J5 and 111;, = riﬂvlil = 0.858. and A] = = = 3.432 kHz and am e 2(Af) = 2(3.432) = 6.864 kHz. 5.3-1 The block diagram of the design is shown in Fig. 55.3-1. 5.3-2 5.4-1 5.4-2 5.4-3 and Fig. sea-2 The block diagram of the design is shown in Fig. 85.3-2. (a) tppuﬂ) = A cos [(4.ch + krm(!)] When this ppm!) is passed through an ideal PM demodulator, the output is kpriiu) This signal. when passed through an ideal integrator. yields kpmh). Hence. FM demodulator followed by an ideal integrator acts as a PM demodulator. However. if m(!) has a discontinuity. 1M!) = 00 at the point(s) of discontinuitY- and the 5y5tem will fail. (b) ¢m(t)=Acos [uci+k,/ m(o)dol Wth this signal ppm!) is passed through an ideal PM demodulator, the output is It; It 1n(o)da. When this signal is passed through an ideal diﬂ’erentiator. the output is kjm(t). Hence. PM demodulator. followed by an ideal differentiator. acts as PM demodulator regardless of whether m(t) has jump discontinuities or not. Figure 55.4-2 shows the waveforms at points b. c. d. and e. The ﬁgure is self explanatory. From Eq. (5.30). the Laplare transform of the phase error 9,(t) is given by S eds) = s + e.(s) For 91"!) = kfz, 9‘“) = and 2k 920*) " hats + AKIN-0] The steady—state phase error {Be}. (5 33)] is 2k ' t = ' .‘ s = = 33310“) l’l‘l‘e“ ) s(s+AK) Hence. the incoming signal cannot be tracked. If s + a 2k ’ x t the“ 69(5) " 32 {g + AKS:+¢)] and 21: 2k 3319"” = We“) = = m Hence. the incoming signal can be tracked within a constant phase 2k/Aka radians. Now. if .2 - - I '1.- H(s) = \$ + a; T I then 6.(s) = z .0: 52 [s + 4Kgﬁ+u+n] r ﬁ(t)-l‘ '9(~-lim——gL—-——O ' - all-I2)“ e ')— 1-O #3 + AK(.«2 + M +15) ln this case. the incoming signal can be tracked with zero phase error. 48 Fiﬁ—+454 HlWlHH/IHHHH .f HHWHHHHHIHHH ** W ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

Chapter 5 - Chapter 5 H mte 2000'9‘“ t"-sooo t"...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online