hw 4 solutions - EE 3 1 HWF Solo-Hons 4.3-8 The signal at...

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Unformatted text preview: EE 3+1 HW‘F Solo-Hons 4.3-8 The signal at point a is [A + m{t)] cos (wet). The signal at point b is: :r (r) = [A + m [t)]2co52 (wet) = W (1 + cos (2am The lowpass filter suppresses the term containing cos (Email. Hence. the signal at point e is: ‘2 m 2 2 2 IMF/=1 +2A ét)+m (t) =A?[1+2n:4(t)+(m7(t))] Usually, in (t) [A < 1 for most of the time. This condition is violated only when m (t) is near its peak, Hence, the output at point d is: A2 y(t)ss—§-+Am(t) A blocking capacitor will suppress the dc term A2/2, yielding the output mm. From the signal in (t), we see that the distortion component is m2 (t) /2. 4.4-1 In Fig. 4.14, when the carrier is cos [{Aw) t + 6] or sin [(Aw) t + 5}. we have: at; (t) = 2 [m1 (t) cos (wet) + m; (t) sin (wct)] cos [(Luc + Aw) t 4i- 6] = 2m; (1:) cos (wet) cos [(wc + Aw) t + 6] + 21112 {t} sin (tact) cos {(wc + Aw): + 6] = on (t) {cos[(Aw) t + 6] + cos [(2% + Aw) t + 6]} + ring (t) {sin [(2.1.2 + Aw}: + 61 — sin [(Aw) t + 5]} Similarly 322 (t) = m1 (t) {sin [(2% + Aw)t + 5] + sin [(Aw) t + 5]} + m; (t) {cos [[Aw) t + 6] — cos [(ch + Awlt + 5}} After 11 (t) and .122 (t) are passed through low-pass filter. the outputs are urn"1 (t) = nil (t) cos [{Aw) t + 6] — m: (t) sin [(Aw) t + 6] mg (i) = mg (t) sin [(Aw) t + 61+ mg (t) cos t + 5] 4-4-2 To generate a DSB-SC signal from m(t), we multiply m(t) by cos (wet). However, to generate the 383 Signals of the same relative magnitude, it is convenient to multiply m (t) by 2cos (tact). This also avoids the nuisance of the fractions 1/2, and yields the DSB-SC spectrum M(tc—wc)+M(w+wc) r We suppress the USB spectrum (above we and below —wc) to obtain the LSB spectrum. Similarly. to obtain the USB spectrum, we suppress the LSB spectrum (between —w., and we) from the USB-SC spectrum. Figures-84.4-2a 311d 54.4-2b show the three cases Mm ¢ out f1 ‘ T i' 1‘ I 550 650 f .150 .50 I 50 [m f 4550 550 450 4500 350 450 ___s£L hm Sis 45° “350 0 350 450 f .aso-sso .450 350 o 350 450 550 650 . ' f Fig. S4.4-2a (a) From Fig. S4A—2a. we can exoress 41515336) = 2 cos {700110 + cos (900st) and ¢U53 (t) = cos {11007rt)+ ' 2cos (13001rt). (b) From Fig. 54.4-2b. we can express: . Qing (t) = % [cos(4001rt) + cos (6001rt}] and $333 (t) = % [cos(1400‘rrt) + cos(16001rt)]. I I I I" I I [In 400 400 u m In I mono: sun-1m 0 mm mail! I was“) 0 I I ll”I [ I [H4 .300 -2oo a 200 300 )' IAIU) .IKIJ JD“ 0 7m m f Fig. 84.4-2b 4.4-5 Because Mi. (f) = —jM (flsgn (f), the transfer function of a Hilbert transformer is: HU) = —j SEHU) If we apply 111;; (t) at. the input of the Hilbert transformer; Y (f), the spectrum of the output. signal y (t) is YU) = Ma {fiHU} = [-iMU} 5gn(f}1[—j SSUUH = “Mm This shows that the Hilbert transform of min. (t) is —m(t). To show that the energies of mm and ma (t) are equal, we have: on +00 Em=f+ Mom] [M(f)|2df +m an +Cfl +00 Em =f mitt) dt=f+ 1Mh(r)I2dI=/_ 1M(f)1215gn(f)l2df=/_m thfJIRdf=Em -m _ 4.5-1 From Eq. (4.25) 1 H:- (I + fa) + H.- (I — fa) Figure 54.5—1 shows H.- (f — fa) + H.- (f + fa) and the reciprocal, which is Ho (f). Hoff)= If! 5213 Fig. 54.5-1 ...
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