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

To summarize a cos wot b sin wot c cos wot e f ig

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Unformatted text preview: g a sinusoid. Alternatively, if w e advance C sin wot by a quarter-cycle, we o btain C cos wot. Therefore, C sin (wot + ~) = C cos wot (B.22b) This observation means sin wot lags cos wot by 90°(7':/2 radians), or cos wot leads sin wot by 90°. B.2-1 Addition o f Sinusoids T wo sinusoids having t he same frequency b ut different phases a dd t o form a single sinusoid of t he same frequency. This fact is readily seen from t he well-known trigonometric i dentity C cos (wot i n which + /I) = a cos wot a =Ccos/l, + bsin wot b = - Csin /I • E xample B .6 In the following cases, express f (t) as a single sinusoid: (a) f (t) = cos wot - v'3 sin wot ( b) f (t) = - 3cos wot + 4sin wot ( a) In this case, a = 1,b = - v'3, and from Eqs. (B.23) C =V12 +(v'3)2=2 = C cos /I cos wot - C sin /I sin wot (B.23a) + /I) in which C a nd /I a re given by Eqs. (B.23b) a nd (B.23c), respectively. These h appen t o b e t he m agnitude and angle, respectively, of a - jb. T he process of adding two sinusoids with t he same frequency can be clarified by using p hasors t o r epresent sinusoids. We represent t he sinusoid C cos (wot + /I) by a phasor of length C a t a n angle /I w ith t he horizontal axis. Clearly, t he sinusoid a cos wot is r epresented by a horizontal phasor o flength a (/I = 0), while b sin wot = b cos (wot - ~) is r epresented by a vertical phasor o flength b a t a n angle -7': / 2 w ith t he horizontal (Fig. B.7). Adding these two phasors results in a phasor of length C a t a n angle /I, as depicted in Fig. B.7. From this figure, we verify t he values of C a nd /I found in Eqs. (B.23b) a nd (B.23c), respectively. P roper care should be exercised in computing /I. Recall t hat t an - 1 ( ~b) i tan-1C~a)' Similarly, tan-1C::~) i - tan-1(~). E lectronic calculators cannot make this distinction. When calculating such a n angle, i t is advisable t o n ote t he q uadrant where t he angle lies and n ot t o rely exclusively on a n electronic calculator. A foolproof m ethod is t o convert t he complex number a - jb t o p olar form. T he m agnitude of t he resulting polar number is C a nd t he angle is /I. T he following examples clarify this point. ( )=tan- 1 (4) =60° B ackground 18 B .3 1£ S ketching S ignals i Therefore, f (t) = 2 cos (wot 1m + 60°) Re ..... -126.9' We can verify t his result by drawing phasors corresponding to t he two sinusoids. The sinusoid cos wot i s represented by a phasor of unit length a t a zero angle with the horizontal. T he phasor sin Wo t is represented by a unit phasor a t an angle of - 90° with t he horizontal. Therefore, - v3 s in wot is represented by a phasor of length v 3 a t 90° with the horizontal, as depicted in F ig. B.8a. The two phasors added yield a phasor of length 2 a t 60° with t he horizontal (also shown in Fig. B.8a). Therefore, f (t) = 2 cos (wot ( b) (a) -4 Re ..... + 60°) F ig. B .8 P hasor a ddition of sinusoids in Example B.6. Alternately, we n ote t hat a - jb = 1 + j v3 = Hence, C = 2 a nd e = 7r/3. Observe t hat...
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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.

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