Signal Processing and Linear Systems-B.P.Lathi copy

# For example we relate b etter t o t he angle 24 t han

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Unformatted text preview: , use t he r adian u nit a nd, above all, b e c onsistent. I n o ther words, in a given problem or an expression do n ot m ix t he two units. I t is convenient t o use t he v ariable Wo ( radian frequency) t o e xpress 211&quot; Fa: (B.20) wo = 211&quot; F a W ith t his n otation, t he s inusoid in Eq. (B. IS) c an b e expressed as f (t) = C cos (wot + 0) i n which t he p eriod To is given by [see E qs. (B.19) a nd (B.20)] (B.16c) IF(w)1 B.2 = LF(w) = t an- 1 (~) _ t an- 1 (4;) (B.17) • Sinusoids C onsider t he sinusoid f (t) = C cos (211&quot;Fat + 0) (B. IS) We know t hat cos 'P = cos ('P + 2n11&quot;) n = 0, ± 1, ±2, ± 3,··· T herefore, cos 'P r epeats i tself for every change of 211&quot; in t he a ngle 'P. For t he s inusoid in Eq. ( B.IS), t he angle 211&quot; Fat + 0 changes by 211&quot; w hen t changes by 1/ Fa. Clearly, t his sinusoid r epeats every 1/ F a seconds. As a result, t here a re Fo r epetitions p er second. T his is t he f requency of t he sinusoid, a nd t he r epetition interval To given by 1 (B.19) T o= Fa (B.21a) 211&quot; Wo = To (B.21b) a nd This is the polar representation of F(w). Observe t hat 4 +w 2 9 + 16w 2 ' 211&quot; 1 To=--=wo/211&quot; Wo I n f uture discussions, we shall often refer t o wo a s t he frequency of t he s ignal cos (wot + 0), b ut i t s hould b e c learly u nderstood t hat t he f requency of t his sinusoid is F a Hz (Fa = wo/211&quot;), a nd wo is actually t he r adian f requency. T he signals C cos wot a nd C s in wot a re illustrated in Figs. B.6a a nd B .6b respectively. A general sinusoid C cos (wot + 0) c an b e readily sketched by shifting t he signal C cos wot i n Fig. B .6a b y t he a ppropriate a mount. Consider, for example, f (t) = C cos (wot - 60°) T his s ignal c an b e o btained by shifting (delaying) t he s ignal C cos wot (Fig. B.6a) t o t he r ight by a phase (angle) of 60°. We know t hat a sinusoid undergoes a 360° c hange of phase (or angle) in one cycle. A quarter-cycle segment corresponds t o a 90° c hange of angle. Therefore, a n angle of 60° c orresponds t o t wo-thirds of a quarter-cycle segment. We therefore shift (delay) t he s ignal in Fig. B.6a by twothirds of a quarter-cycle segment t o o btain C cos (wot - 60°), as shown in Fig. B.6c. O bserve t hat if we delay C cos wot i n Fig. B .6a by a quarter-cycle (angle of 90° o r 11&quot; / 2 r adians), we o btain t he s ignal C s in wot, d epicted in Fig. B.6b. T his verifies t he well-known trigonometric identity C cos (wot - I) = C s in wot (B.22a) Background 16 17 B.2 Sinusoids l' 1m a Re~ -b (a) F ig. B .7 Phasor addition of sinusoids. Therefore, c C= va 2 +b 2 /I = t an-1 ( Irvt- -~ 2 To (b) (B.23b) ~b) (B.23c) Equations (B.23b) a nd (B.23c) show t hat C a nd /I a re t he m agnitude a nd angle, respectively, of a complex number a - jb. I n o ther words, a - jb = C e i8 . Hence, t o find C a nd /I, we convert a - jb t o polar form a nd t he m agnitude a nd t he angle of t he resulting polar number are C a nd /I, respectively. To summarize, a cos wot + b sin wot = C cos (wot (e) F ig. B .6 Sketchin...
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