Find io (t) if Vs (t) = 1 + cos(3t).
Series RLC circuit with two initial conditions Example 33 for wikibook circuit theory Choose Starting Point
edit
Because of the initial conditions, going to start with Vc (t) and then work our way through the initial conditions to io .
H
(
s
)
=
V
c
V
s
=
1
s
C
1
s
C
+
1
1
R
1
+
1
s
L
+
R
2
{\displaystyle H(s)={\frac {V_{c}}{V_{s}}}={\frac {\frac {1}{sC}}{{\frac {1}{sC}}+{\frac {1}{{\frac {1}{R_{1}}}+{\frac {1}{sL}}}}+R_{2}}}}
The MuPad commands are going to be:
L :=1; R1:=.5; R2:=1.5; C:=.5;
simplify((1/(s*C))/(1/(s*C) + 1/(1/R1 + 1/(s*L)) + R2))
Which results in:
H
(
s
)
=
8
s
+
4
8
s
2
+
11
s
+
4
{\displaystyle H(s)={\frac {8s+4}{8s^{2}+11s+4}}}
Homogeneous Solution
edit
Set the denominator of the transfer function to 0 and solve for s:
solve(8*s^2 + 11*s + 4)
Imaginary roots:
s
1
,
2
=
−
11
±
7
i
16
{\displaystyle s_{1,2}={\frac {-11\pm {\sqrt {7}}i}{16}}}
So the solution has the form:
V
c
h
=
e
−
11
t
16
(
A
cos
7
t
16
+
B
sin
7
t
16
)
+
C
{\displaystyle V_{c_{h}}=e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C}
After a very long time the capacitor opens, no current flows, so all the source drop is across the capacitor. The source is a unit step function thus:
V
c
p
=
1
{\displaystyle V_{c_{p}}=1}
mupad screen shot leading up to computing constant B Adding the particular and homogenous solutions, get:
V
c
(
t
)
=
1
+
e
−
11
t
16
(
A
cos
7
t
16
+
B
sin
7
t
16
)
+
C
{\displaystyle V_{c}(t)=1+e^{-{\frac {11t}{16}}}(A\cos {\frac {7t}{16}}+B\sin {\frac {7t}{16}})+C}
Doing the final condition again, get:
V
c
(
∞
)
=
1
=
1
+
C
⇒
C
=
0
{\displaystyle V_{c}(\infty )=1=1+C\Rightarrow C=0}
Which implies that C is zero.
From the given initial conditions, know that Vc (0+ ) = 0.5 so can find A:
V
c
(
0
+
)
=
0.5
=
1
+
A
⇒
A
=
−
0.5
{\displaystyle V_{c}(0_{+})=0.5=1+A\Rightarrow A=-0.5}
Finding B is more difficult. From capacitor terminal relation:
VC := 1 + exp(-11*t/16)*(-.5*cos(7*t/16) + B*sin(7*t/16))
IT := diff(VC,t)
The total current is:
i
T
(
t
)
=
C
d
V
c
d
t
=
11
e
−
11
t
16
16
(
0.5
cos
7
t
16
−
B
sin
7
t
16
)
+
7
e
−
11
t
16
16
(
0.5
sin
7
t
16
+
B
cos
7
t
16
)
{\displaystyle i_{T}(t)=C{dV_{c} \over dt}={\frac {11e^{-{\frac {11t}{16}}}}{16}}(0.5\cos {\frac {7t}{16}}-B\sin {\frac {7t}{16}})+{\frac {7e^{-{\frac {11t}{16}}}}{16}}(0.5\sin {\frac {7t}{16}}+B\cos {\frac {7t}{16}})}
The loop equation can be solved for the voltage across the LR parallel combination:
V
C
+
V
L
R
+
R
2
C
d
V
c
d
t
−
V
s
=
0
{\displaystyle V_{C}+V_{LR}+R_{2}C{dV_{c} \over dt}-V_{s}=0}
V
L
R
=
V
s
−
V
C
−
R
2
i
t
=
1
−
V
c
−
1.5
∗
i
t
{\displaystyle V_{LR}=V_{s}-V_{C}-R_{2}i_{t}=1-V_{c}-1.5*i_{t}}
VLR := 1 - VC - 1.5*IT
We know from the inductor terminal relation that:
i
L
=
1
L
∫
V
L
R
d
t
+
C
1
{\displaystyle i_{L}={\frac {1}{L}}\int V_{LR}dt+C_{1}}
IL := 1/.5 * int(VLR,t)
At this point mupad gave up and went numeric. In any case, it is clear from t = ∞ where the inductor current has to be zero that the integration constant is zero. This enables us to compute B from the inductor initial condition.
t :=0
Set the time to zero, set IL equal to the initial condition of .2 amps and solve for B:
solve(IL=0.2, B)
And get that B is -0.2008928571 ...
mupad screen shot finding the desired output io The desired answer is io which is just VLR /R_1. To calculate need to start new mupad session because t is zero now. Start with:
B := -0.2008928571;
R1 :=0.5;
Repeat the above commands up to VLR and then add:
io = VLR/R1
i
o
=
3
e
−
11
t
16
(
0.0879
cos
7
t
16
−
0.219
sin
7
t
16
)
−
0.0625
e
−
11
t
16
(
0.5
cos
7
t
16
+
0.201
sin
7
t
16
)
{\displaystyle i_{o}=3e^{-{\frac {11t}{16}}}(0.0879\cos {\frac {7t}{16}}-0.219\sin {\frac {7t}{16}})-0.0625e^{-{\frac {11t}{16}}}(0.5\cos {\frac {7t}{16}}+0.201\sin {\frac {7t}{16}})}
