6.5 Addition of Sinusoidsof the SameFrequency
Adding two sinusoidsof the same f'requencybut ditl'erentamplitudesand phasesresultsin another
sinusoid(sin or cos) of same fiequency. The resulting amplitude and phaseare different from the
amplitude,and phaseof the two original sinusoids,as illustratedwith the examplebelow.
Example 6-2; Consider an electrical circuit with two elementsR and /- connectedin seriesas
shownin Fig. 6. 15.
i(t)
'R
,
T
=6 sin2 t v
t, =vR+vL
P L= 8cos2tv
=6sin2t A
Figure6, l5: Addition of sinusoidsin an RL circuit'
In Fig. 6.15, the cunent i(r):6r;n12t) amp flowing throughthe circuit producestwo voltages:
yn :6sin(2t) V acrossthe resistor and yL : 8cos(2t) V acrossthe inductor. The total voltage
voltagev(t) acrossthe current sourcecan be obtainedusing KVL as
v (t):Yn 1 1 ;+ Y r(r)'
Sinusoids in Engineering
rEt
v.
+ 8cos(2r)
v(r): 6rin12r)
(6.s)
(sine or cosine) of frequency
The total voltage given by equation (6.5) can be written as one sinusoid
the resulting sinusoid' In terms
2radls.The objective is io find the amplitude and the phaseangle of
of a sine function,
: u sin(2t* 0):
+ Scos(2r)
v(r): 6s1t12r)
6.6
identit
Whjre&-sbjective is to find M and@.Usingthetrigonometric
as
written
(6.6)
be
can
equation
side,
right
(D\n
the
cos(A)sin
6sin(2r)*8cos(2r) : (Mcosf)sin(2r)+ (Msin{)cos(2t)'
ofsin(2r) andcos(2t)on bothsidesofequation(6'7)gives
Equatingthecoefficients
Mcos(S| : 6
sines:
cosines: Msin(@) : 8'
(6.7)
(6.8)
(6.e)
to polar form as
To determine the magnitude M and phase@,equations(6.8) and (6'9) are converted
shownin Fig. 6.16.
in an RL circuit'
Figure 6.l6: Determinationof magnitudeand phaseof the resultingsinusoid
Therefore,
M:
:
o H:i::'''
-
Therefore,v(r) : 6sin(2r) a Scos(2r): l0sin(2r + 53'13") V' The amplitudeof the^voltage
0r2l
sinusoidis lO V the angularfrequencyis Q:2 rad/s,the linearfrequen? tt "f-: Zn: 2n: i
:O'927rad'and
isT=n:3'l4lsec,thephaseangle=53'I3":(53'I3')(##)
Hz,theperiod
thetime shift can be calculatedas
T:
a
a)
r53
6.5 Addition of Sinusoidsof the SameFrequency
o.927
L
:
0'464 s.
The plot of the voltage and current waveformsis shown in Fig. 6.17, ltcan be seenfrom Fig. 6. 17
rhar the volrage waveform is shifted to the left by 0.464 s (time shift). In other words, the voltage in
the RL circuit leads the current by 53.3'. It will be shown later that it is oppositein an RC circuit,
where the voltage waveform lags the cuffent.
Figure 6. l7: Voltageand current relationshipfor an RL circuit.
Note: The voltageu(t) can also be representedas one sinusoidusing the cosinef'unctionas
* Qr).
v(r): 6s;n12r)
+ 8cos(2t): tt cos(2r
(6.10)
The amplitude"M andthe phaseangle @1can be determinedusing a proceduresimilar to that outlined
above. Using rhe trigonometric identity cos(A + B) : cos(A)cos (B) - sin (A) sin (B) on the right
sideof equation(6.10)gives
6 s i n (2 r) * 8 c o s (2 r): (-Ms i n fl ) si n(2t) * (Mcos@ 1)cos(2t).
(6.1l )
Equatingthe coetficients of sin(2 r) and cos(2r) on both sidesof equation(6. I I ) gives
sines:
Msin(@r) :
-6
(6.t2)
Sinusoidsin Engineering
t54
c o s i n e s:
M c o s(@ 1)
8,
(6.r3)
To determinethe magnitudeM andphase@1
, equations(6.12) and (6.13) are convertedto polar form
as shownin Fig. 6.18.
Figure 6. l8: Determinationof magnitudeand phasefor a cosinefunction.
Therefore,
t PI :
J6, +8,
:10
h : 'atan2(-6'8)
:
-36.87"
v(r) : 6s1t12r)+ Scos(2t): lQcos(2r- 36.87')V'
Therefore,
canalsobe obtaineddirectlyfrom the sinefunctionusingthe trig identitysin0 :
This expression
cos(O- 90"). Therefore,
l0sin(2t+ 53.13")
l0cos((2r+ 53.I 3") - 90")
l0cos(2r* 36.87').
In general, the results of this example can be expressedas follows:
A c o s o t * Bs i n a tl
1/ez +& cos(orr*atan2(B,A))
A cos crrt* Bsin all
\/ Az + 82 sin( arr * atan2(A,B)) .
r-7------:
Example 6-3: ConsidertheRC circuitshownin Fig. 6.19.
-t
I
6.5 Addition of Sinusoids of the Same
+
l0cos l20nr V
'=h+'c
5 sin l20nr V
= cos(I20 7u)mA
t20n
Figure6.l9: Additionof sinusoids
in anRC circuit.
In Fig. 6.19,thecurrentl(r) : sesl120rr) mA flowingthroughthecircuitproduces
rwo volrages:
vn : lOcos(l20at) acrosstheresistorantluc.:5sin(l20zrr) acrossthecapacitor.
Thetotalvoltage
voltagev(r) canbe obtainedusingKVL as
v(r)
v6(t) + v6.(r)
l Oc o s (l 2Oxt) + _5si n(l 20zr).
(6.r4)
The total voltage given by equation (6. 14) can bc written as a single sinusoid
of frequency l20z
rad/s. The objective is to find the arnplitudeand the phaseangle ofthe sinusoid.
The total voltage
can be written as a cosine function as
l0cos(I 20zr) + Ssin(I ZOttt):M cos(t}Ont+ tz).
:(McosS2)cos(t2on)
+ (* M sing2)sin( I 20ar),
(6.l5)
wherethetrigonometric,*".,.{isemp|oyedontherighthandsideoftheequation.Equatinos$2rjntjonbothsidesof
equation
(6.l-5)gives
cosrnes:
M cos(Q2)
IO
sines:
M sin(Q2)
-5.
(6.l6)
(6.17)
To determinethe rnagnitudeM andphaseangle
@2,eqirations(6. l6) and (6.17) areconvertedto the
polar fbrm as shown in Fig. 6,20. Therefbre,
M
:
v/W+s,
w
:
:
I l.18
atan2(--5,
l0)
-26.57",
:
whichgivesu(r): 1l.lScos(120nt - 26'57")v'
v
x
sin$2 = -1
Figure 6.20: Determinationof magnitudeand phaseof the resultingsinusoidin an RC circuit
:
the linear
The amplitudeof the voltagesinusoid is I 1.8 V, the angularfrequencyis o) l}On radls,
period is T : 16.7ms, the phaseangleis -26'57", and the time
frequencyit f :
#:6oHz,the
- -shift is 1.23ms.
figure
The plot of the voltage and current waveforms is shown in Fig. 6.21' It can be sei:n from the
voltage
words,
the
other
that ihe voltage waveform is shifted to the right by 1.23 ms (time shift). In
in this RC circuit lags rhe current by 26.57".
<--
Voltagc in V
l.o
o
- t.0
fbr anRC circuit'
6.21:Voltageandcurrentrelationship
F.lUure