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˜ n
2. We find V , the capacitor response phasor associated with the V ˜ n excita-
c s
tion. This can be found by noting that the voltage across the capacitor is equal
to the capacitor impedance multiplied by the current phasor, giving:
n
ZV ˜ n
V = Z I = c s (7.77)
n
˜ n
˜ n
c c Z + Z + Z n
n
n
c R L
where from the results of Section 6.8, particularly Eqs. (6.83) through (6.85),
we have:
Z = 1 (7.78)
n
c j2π nC
Z = j2π nL (7.79)
n
L
Z = R (7.80)
n
R
3. Finally, we use the linearity of the ODE system and write the solution as
the linear superposition of the solutions corresponding to the response to
each of the basis functions; that is,


 ZV n nt 
n ˜

Vt() = Re  ∑ c s e j2π  (7.81)
c n n n
 n Z + Z + Z L 


R
c
leading to the expression:

 V n ˜ nt 

Vt() = Re  ∑ s e j2π  (7.82)
c 2
 n 1 − (2π nLC + j(2π n RC 
)
)


Homework Problem
Pb. 7.27 Consider the RLC circuit. Assuming the same notation as in Section
6.5.3, but now assume that the source potential is given by:
V = V cos (ω t)
6
s 0
a. Find analytically the potential difference across the capacitance.
(Hint: Write the power of the trigonometric function as function of
the different multiples of the angle.)

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