EE314HW3 - Homework 3 Chapter4: 4.2-1 [iii]. 4.2-3. 4.2-4....

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Unformatted text preview: Homework 3 Chapter4: 4.2-1 [iii]. 4.2-3. 4.2-4. 4.2-2. 4.2-8. 4.3-1. 4.3-3. 4.3-8. 4.4-2. 4.4-3. 4.4-4. Due: Feb. 12 [Friday] 4.2-1 (iii) m(t) = coswmylt-coswmgt : Cos 10001rt-cos 30007rt = % (cos 27rfm_1i + cos 27rfmlgt) : % (cos 2000M + cos40007rt)——i fmJ : IDODHz fm'2 = 2000Hz See Fig. S4.2—la(iii) for the graphical results. Modulated signal spectrum Mm (iii) 1/2 (kHz) me) f -2000 —lO00 1000 2000 -12-ll -9 -8 8 9 ll 12 4.2-3 (3) go (t) 3 m (t) ‘ c053 (wet) = m (t) {% cos (wet) + icos(3wct)] The term %m(t)cos (wet) is the desired modulated signal, whose spectrum is centered at iwc. The remaining term %m (t) cos (3wct) is the unwanted term, which represents the modulated signal with carrier frequency 3 fc with spectrum centered at i312 as shown in Figure 84.2-3. A bandpass filter centered at ifc allows the passageof the desired term gm (t) cos (wet). but suppresses the unwanted term §m (t) cos {Swat}. Hence1 this system works as desired with the output %m (t) cos (wet). (b) Figure 84.2—3 shows the two Signalspectra at points b and c. (c) The minimum usable value of f.C to avoid-spectral folding overlap after modulation is B where B is the bandwidth of the lowpass signal m(t). - (d) Yes. 9,. (t) e m(t} . sins (Luci) = mU) sin (Luci) + 41m (3wct)] 2 mm cos (we: —— + icos (3wct — The term gm (6) cos (wet i g) is the desired term as in part (a) in the form k - m (t) cos (tact + 6) with a = fig. (e) Readers may verify that the expansion of cos (nwct) always contains a term cos (tact) when n is odd. This is not true when n is even. Hence, the system works for a carrier cos” (wet) only when n is odd. Signal @ (b) c —fC + B Signal @ (c) fckB fc 3f: f(Hz) Fig, 84.2—3 4.2-4 (a) We use the ring modulator of Figure 4.6a with the carrier frequency f1 = lUOVkHz (wl : 200w x 103) and the output bandpass filter centered at fc = 5f: 2 500 kHz. The output vfit) is; wilt)? — [17102) cos(w1£) ~ émfi) COS (3w1t) + gm“) cos (5w1t)-+ . . The output bandpass filter suppresses all the terms except the one centered at 500 kHz (corresponding to the carrier fC 2 5f1). Hence, the filter output is DSB—SC AM signal: y{t) : sigma) cos (Stalt) (b) This is the desired output km(t) cos (tact) with k : 4/51r. 4.2—7 (a) Figure 842-? shows the signals at. points a,b and c spectrum @ poian spectrum @ point 8 kHz MHz) MA( ) 45 40 -5 0 5 :0 15 45 _IO _5 0 5 IO 15 spectrum. @ point c ' fad-i7.) v10 -5 5 10 IS ‘20 25 30 35 .35 do -25 -20 -15 cos 20007rt cos 100007rt Fig. 842—? (b) Fiom the spectrum at. point c, it is clear that the channel bandwidth must be at least 30000 Hz (from 5000 Hz to 35000 Hz), (0) Figure 34.2-7 shows the re ’ ' C- ceiver diagram to recover both m1 (t) and m2 (t)from the modulated signal at point 4.2-8 (a) Figure 84.2-8 shows the output signal spectrum Y(f} after the scrambler processing. - N f ) suppresaed suppressed (b) Observe that Y(w) is the same as M(w) with the frequency spectrum inverted, that is, the high frequencies are shifted to lower frequencies and vice verse. Thus, the scrambler in Fig. P4.2-8 inverts the frequency spectrum. To get back to the original spectrum M(w), we need to invert the spectrum Y(w) once again. This can be done by passing the scrambled signal 1; (t) through the same scrambler in Fig 842—8. 4.3-1 SAM (15) = 2 [b + %m (0] cos (wct) _ = (25 + m (5)}005 (tact) With WC 2 20000” (a) PC 2 33—2 2: 262 Because T 0.1 __._. 2 3' 2 T 4 m2t=~f 2tdt:-—/ 2 :_ f) T ml) OJ mmdt 3 we have P5 = émzfi) = g and PC + P3 = 2!;2 + (bl b :1 D 2 n=fififi= '-100%=25% 2+5 (c) Assume b = 1 See Fig. 84.3-1 (d) Ifb=0‘5,Ps=§,Pa+P§=252+ a - . .c .7 _ V _ I _ switl=2ib+95mmlwsuct Fig. 84.3—1 . 4.313 \ (a) According to Eq. (4.101), the carrier amplitude is A : 1:13 : 0—}qu = 13.34. The carrier power is PC : i3 = 88.89 (b) The sideband power is m2(t)/2. Because of symmetry of amplitude values every quarter cycle, the power of m (it) may be computed by averaging the signal energy over a; quarter cycle only. Over a quarter cycle m (t) can be represented as m (t) = (Wt/Tu (see Fig. 84.3-3). Note that To = 104 Hence, 1 W4 40: 2 2 = _ = m (t) TOM/D [TO] dt 33.34 flzwm 2 The Sideband power is = 16.67 The efficiency is PS 16.67 PC + R. 88.89 +16.67 x 100% = 15.79% 4.38 The signal at point a is [A + m(t)}cos (wet). The signal at point b is: A2 + 2Am (a) + m2 (t) a: (t) = [A + m (it)]2 cos2 (wait) 2 2 (1 + cos (2wct}) The lowpass filter suppresses the term containing cos (ZwCE) . Hence, the signal at point c is: ' 2 mm 2 A2+2Am§tJ+m2(t) : A72 [1+ 21120:) + ] Usually, in (t) /A < l for most of the time. This condition is violated only when m (t) is near its peak, Hence,,the output at point cl is: A2 y[£)m—2+~+Am(t) A blocking capacitor will suppress the dc term 212/2, yielding the output m(t). Fiom the signal in (t), we see that the distortion component is m2 (t) /2, 4.4- sign of t 2 To generate a DSB-SC signal from mm, we multiply m (t) by cos (wet). However, to generate the 3313 MS of the same relative magnitude, it is convenient to multiply m (t) by 2 cos (wet). This also avoids the nuisance he fractions 1/2, and yields the DSB-SC spectrum M(w—wc)+M(w+w,) we SuDDress the USB spectrum (above we USE Spectrum, we suppress the LSB spectr and 34.4-21) show the three cases. and below —wC) to obtain the LSB spectrum. Similarly, to obtain the urn {between —wc and sec) from the DSB‘SC spectrum. Figures 844-251 Mm T waif} I ' T i 5 j; i I I 0.5 .150 ,50 50 [00 f 650 450 7350(1) 350 450 550 650 f mm mm”) 7 I l l L “'5 i l ' i i "5 i “‘50 '350 0 35” ‘50 f 450550 «:50 .350 o 350 450 550 650 i | f Fig, 844-23, (a) From Fig. S4_4-2a, we can express tbng (t) : 2 cos (700nt) + cos (900m) and (15935- (t) = cos (1100“) + Zoos (13007rt), ‘ (b) From Fig. S4.4-2b, we can express: ¢L33 (t) = flees (4000i) + cos (600m)} and (buss (t) : flees (140001) + cos (1600025)]. 03 ) Mm maU [ 1 U4 [ I [H4 .300 .200 o 200 300 I 7200—700 7300200 0 200 300 700 :00 f 0355“) (D0556) IM {E 1/4 0 700 000 f 000 000 o_ 200 3.00 f 7300 7700 Fig. 344—219 4.4-3 qfiusg (t) = m (t) cos (wet) — mh (t) sin (was) and 051,33 (t) : m (t) cos (wet) + mh (t) sin (wet) (a) m(t) = cos (100w) + 2003 (30010) and mh ('t) = sin (1007rt) + 25111 (300m). The carrier is cos 10007:. Hence. (:5ng (t) : {cos (100m) + 2 cos (300m)] cos (100015) + [sin (100015) + 2 sin (300711)] sin (10001?) 2 cos (1000 — 1000)t + 2005 (1000 — 300:0); (,‘bUSB (t) = [cos(1001rt) + 2003 (3000t)] cos (1000t) — (sin (1000;) + 251:) (300m)) sin (1000:) = cos (1000 + 1000)]5 + 2003 (1000 + 30077)t cos(1007rt) 005(5001rt) : %c05(4001r£} + %cos(60071t) and mh (t) = ésin(4007rt) + %sin(6001rt). (b) m0 — Hence, (MW 0) : g 005 (400m) + écos (600m)] cos(1000:) + [%51n(4001rt) + gsin (600m)] sin (1000:) : JZ—cos (1000 i 4000)]: + écos (1000 — 6000)t (15,153 (i) = cos (4000i) + %COS(6007¥1)] 005 (1000:) w Sin (4000:) + % Sin (600m)] Sin (1000:) 1 écos (1000 + 40071113 + écos (1000 + 6007023 4.4-4 m (i) = TrBsinc2 (27rBt) with B = 2 kHz and fC = 10 kHz. (3) M (f) = $13 (511:? = %A See Figure 544-43.. MU) Spectrum of 2m{t)cos(cmt) I ! 1/2 1/2 //\\. / r , \\ V J m" _\_ .B 0 B Jt-B -fc -fc+B 0 {DB ,2 fi+B Fig. S4.4—4a (b) The solution is shown in the Fig. 84.4-4b. 'I’LsisU) 1/2 -fc JHB 0 va fv Fig. 34.4-41) (C) ‘PLSB (i) = f:f€”+2"B 77,17; (f + fc - 2TH?) aflmdf + LEE—2n; fi (f * fc + QarB)ei2”fidf : figflflgfi [cos (21rfct) — COS (2w (fc * 27713) t)] + fi sin (27rfct) (PUSBU) = ’f“ _L ‘ 7r 7f : fjffljzflB 41rB (f+ fc+2WB)eJ'2 ftdf+fic+2 3‘47}? (f_fc_2fl_B)ej21rftdf l | M" (2“)? {005’ (27TH) *' COS (27r (fc + 2m?) 25)] - 2+“ sin (27rfct) ...
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EE314HW3 - Homework 3 Chapter4: 4.2-1 [iii]. 4.2-3. 4.2-4....

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