0195136047.pdf

3.1 SINUSOIDAL STEADY-STATE PHASOR ANALYSIS 107

whereVmis the peak amplitude andφis the phase angle. This may be expressed in terms of

exponential functions as

v(t)=

Vm

2

[

ej(ωt+φ)+e−j(ωt+φ)

]

=V ̄aejωt+V ̄be−jωt (3.1.18)

whereV ̄a=(Vm/ 2 )ejφandV ̄b=(Vm/ 2 )e−jφ. Note thatV ̄aandV ̄bare complex numbers.

Even though Equation (3.1.17) is a cosine function that is considered here, recall that any

sine, cosine, or combination of sine and cosine waves of the same frequency can be written as a

cosine wave with a phase angle. Some useful trigonometric identities are as follows:

sin(ωt+φ)=cos

(

ωt+φ−

π

2

)

(3.1.19)

Acosωt+Bsinωt=

√

A^2 +B^2 cos(ωt−φ) (3.1.20)

whereφ=tan−^1 (B/A).

By expressing the sinusoidal excitation as the sum of two exponentials, as in Equation

(3.1.18), the method developed to find the response to exponential excitations can easily be

extended with the principle of superposition to obtain the forced response to sinusoidal excitation,

as illustrated in the following example.

EXAMPLE 3.1.3

Consider anRLCseries circuit excited byv(t)=Vmcosωtin the time domain. By using
superposition, solve for the time-domain forced response of the resultant current through the
frequency-domain approach.

Solution

Figures E3.1.3(a) and E3.1.3(b) are equivalent in the time domain. Figure E3.1.3(c) shows the
transformed networks with the use of superposition. It follows then

I ̄ 1 = Vm/^2

R+jωL+( 1 /j ω C)

=

Vm/ 2

R+j[ωL−( 1 /ωC)]

=I 1 ejθ^1

I ̄ 2 = Vm/^2

R−jωL+( 1 /j ω C)

=

Vm/ 2

R−j[ωL−( 1 /ωC)]

=I 2 ejθ^2

where

I 1 =I 2 =

Vm/ 2

R^2 +[ωL−( 1 /ωC)]^2

; θ 1 =−θ 2 =−tan−^1

ωL−( 1 /ωC)

R

The student is encouraged to use a knowledge of complex numbers and check these results.

The corresponding time functions are given by

i 1 (t)=I 1 ejθ^1 ejωt and i 2 (t)=I 2 e−jθ^1 e−jωt

By superposition, the total response is

i(t)=

Vm

[

R^2 +{ωL−( 1 /ωC)}^2

] 1 / 2 cos(ωt+θ^1 )=Imcos(ωt+θ^1 )