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1 Any sinusoidal signal may be mathematically represented in one of two ways: a time-domain form, v(t) = A cos( t + ) and a frequency-domain (or phasor) form, V(j ) = Aej = A Note the j in the notation V(j ), indicating the ej t dependence of the phasor In the remainder of this chapter, bold uppercase quantities will be employed to indicate phasor voltages or currents (Continued)
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(Concluded) 2 A phasor is a complex number, expressed in polar form, consisting of a magnitude equal to the peak amplitude of the sinusoidal signal and a phase angle equal to the phase shift of the sinusoidal signal referenced to a cosine signal 3 When using phasor notation, it is important to make a note of the speci c frequency, , of the sinusoidal signal, since this is not explicitly apparent in the phasor expression
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EXAMPLE 49 Addition of Two Sinusoidal Sources in Phasor Notation
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Compute the phasor voltage resulting from the series connection of two sinusoidal voltage sources (Figure 428)
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v1 (t) = 15 cos(377t + /4) V v2 (t) = 15 cos(377t + /12) V
Find: Equivalent phasor voltage vS (t)
v2(t)
+ ~ + ~
Analysis: Write the two voltages in phasor form:
V1 (j ) = 15 /4 V V2 (j ) = 15ej /12 = 15 /12 V Convert the phasor voltages from polar to rectangular form: V1 (j ) = 1061 + j 1061 V V2 (j ) = 1449 + j 388 Then VS (j ) = V1 (j ) + V2 (j ) = 2510 + j 1449 = 2898ej /6 = 2898 /6 V Now we can convert VS (j ) to its time-domain form: vS (t) = 2898 cos(377t + /6) V
Comments: Note that we could have obtained the same result by adding the two
v1(t)
vS(t)
+ ~
sinusoids in the time domain, using trigonometric identities: v1 (t) = 15 cos(377t + /4) = 15 cos( /4) cos(377t) 15 sin( /4) sin(377t) V v2 (t) = 15 cos(377t + /12) = 15 cos( /12) cos(377t) 15 sin( /12) sin(377t) V Combining like terms, we obtain v1 (t) + v2 (t) = 15[cos( /4) + cos( /12)] cos(377t) 15[sin( /4) + sin( /12)] sin(377t) = 15(1673 cos(377t) 0966 sin(377t))
Part I
Circuits
= 15 (1673)2 + (0966)2 cos 377t + arctan
0966 1673
= 15(1932 cos(377t + /6) = 2898 cos(377t + /6) V The above expression is, of course, identical to the one obtained by using phasor notation, but it required a greater amount of computation In general, phasor analysis greatly simpli es calculations related to sinusoidal voltages and currents
It should be apparent by now that phasor notation can be a very ef cient technique to solve AC circuit problems The following sections will continue developing this new method to build your con dence in using it Superposition of AC Signals Example 49 explored the combined effect of two sinusoidal sources of different phase and amplitude, but of the same frequency It is important to realize that the simple answer obtained there does not apply to the superposition of two (or more) sinusoidal sources that are not at the same frequency In this subsection, the case of two sinusoidal sources oscillating at different frequencies will be used to illustrate how phasor analysis can deal with this more general case The circuit shown in Figure 429 depicts a source excited by two current sources connected in parallel, where i1 (t) = A1 cos( 1 t) i2 (t) = A2 cos( 2 t) The load current is equal to the sum of the two source currents; that is, iL (t) = i1 (t) + i2 (t) or, in phasor form, IL = I1 + I2 (449) (448) (447)
Figure 429 Superposition of AC currents
I1(t)
I2(t)
Load
At this point, you might be tempted to write I1 and I2 in a more explicit phasor form as I1 = A1 ej 0 I2 = A 2 e j 0 (450)
and to add the two phasors using the familiar techniques of complex algebra However, this approach would be incorrect Whenever a sinusoidal signal is expressed in phasor notation, the term ej t is implicitly present, where is the actual radian frequency of the signal In our example, the two frequencies are not the same, as can be veri ed by writing the phasor currents in the form of equation 446: I1 = Re [A1 ej 0 ej 1 t ] I2 = Re [A2 ej 0 ej 2 t ] (451)
Since phasor notation does not explicitly include the ej t factor, this can lead to serious errors if you are not careful! The two phasors of equation 450 cannot be added, but must be kept separate; thus, the only unambiguous expression for the load current in this case is equation 448 In order to complete the analysis of any circuit with multiple sinusoidal sources at different frequencies using phasors, it is
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