Unformatted text preview: g a sinusoid. Alternatively, if w e advance C sin wot by a quartercycle, 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.21 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 wellknown
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 tan1C~a)' Similarly, tan1C::~) i  tan1(~). 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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