242 C H A P T E R 4: Frequency Analysis: The Fourier Series
FIGURE 4.1
RC circuit and corresponding phasor
circuit.
vs(t)
1 Ω
1F vc(t)
+
− +
−
1
−j
+
−
Vc
+
−
Vs
corresponding tovc(t):
Vc
Vs
=
−j
1 −j
=
−j( 1 +j)
2
=
√
2
2
∠−π/ 4
SinceVs= 4 ∠π/4, then
Vc= 2
√
2 ∠ 0
so that in the steady state,
vc(t)= 2
√
2 cos(t)
n Eigenfunction approach.Considering the output is the voltage across the capacitor and the input
is the voltage source, the transfer function is obtained using voltage division as
H(s)=
Vc(s)
Vs(s)
=
1 /s
1 + 1 /s
=
1
s+ 1
so that the system frequency response at the input frequency 0 =1 is
H(j 1 )=
√
2
2
∠−π/ 4
According to the eigenfunction property the steady-state response of the capacitor is
vc(t)= 4 |H(j 1 )|cos(t+π/ 4 +∠H(j 1 ))
= 2
√
2 cos(t)
which coincides with the solution found using phasors. n
nExample 4.2
An ideal communication system provides as output the input signal with only a possible delay in
the transmission. Such an ideal system does not cause any distortion to the input signal beyond