Signals and Systems - Electrical Engineering

28 C H A P T E R 0: From the Ground Up!

Assume that the steady-state response of this circuit (i.e.,vc(t)ast→∞) is also a sinusoid

vc(t)=Ccos( 0 t+ψ)

of the same frequency as the input, with amplitudeCand phaseψto be determined. This response

must satisfy the differential equation, or

vi(t)=

dvc(t)

dt

+vc(t)

Acos( 0 t)=−C 0 sin( 0 t+ψ)+Ccos( 0 t+ψ)

=C 0 cos( 0 t+ψ+π/ 2 )+Ccos( 0 t+ψ)

=C

√

1 +^20 cos( 0 t+ψ+tan−^1 (C 0 /C))

Comparing the two sides of the above equation gives

C=

A

√

1 +^20

ψ=−tan−^1 ( 0 )

for a steady-state response

vc(t)=

A

√

1 +^20

cos( 0 t−tan−^1 ( 0 )).

Comparing the steady-state responsevc(t)with the input sinusoidvi(t), we see that they both have the

same frequency 0 , but the amplitude and phase of the input are changed by the circuit depending

on the frequency 0. Since at each frequency the circuit responds differently, obtaining the frequency

response of the circuit will be useful not only in analysis but in the design of circuits.

The sinusoidal steady-state is obtained using phasors. Expressing the steady-state response of the

circuit as

vc(t)=Re

[

Vcej^0 t

]

whereVc=Cejψis the corresponding phasor forvc(t), we find that

dvc(t)

dt

=

dRe[Vcej^0 t]

dt

=Re

[

Vc

dej^0 t

dt

]

=Re

[

j 0 Vcej^0 t

]

By replacingvc(t),dvc(t)/dt, obtained above, and

vi(t)=Re

[

Viej^0 t

]

where Vi=Aej^0