[<< wikibooks] Circuit Theory/Convolution Integral/Examples/example49/VL
Given that the source voltage is (2t-3t2), find voltage across the resistor.
This is the VL solution.
Outline:

=== Transfer Function ===

H
(
s
)
=

V

L

V

S

=

s

4
+
s
+

1

0.25
s

{\displaystyle H(s)={\frac {V_{L}}{V_{S}}}={\frac {s}{4+s+{\frac {1}{0.25s}}}}}
simplify(s/(4 + s + 1/(0.25*s)))

H
(
s
)
=

s

2

s

2

+
4
s
+
4

{\displaystyle H(s)={\frac {s^{2}}{s^{2}+4s+4}}}

=== Homogeneous Solution ===
solve(s^2 + 4.0*s + 4.0,s)
There are two equal roots at s = -2, so the solution has the form:

V

L

h

(
t
)
=
A

e

−
2
t

+
B
t

e

−
2
t

+

C

1

{\displaystyle V_{L_{h}}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}}

=== Particular Solution ===
After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor.

V

L

p

=
0

{\displaystyle V_{L_{p}}=0}
This also means that C1 has to be zero.

=== Initial Conditions ===
So far the full equation is:

V

L

(
t
)
=
A

e

−
2
t

+
B
t

e

−
2
t

{\displaystyle V_{L}(t)=Ae^{-2t}+Bte^{-2t}}
Initial voltage is all across the inductor.

V

L

(
0
)
=
1
=
A

{\displaystyle V_{L}(0)=1=A}

A
=
1

{\displaystyle A=1}
At this point will have to do integral .. to get to the current. There is no other way to use the known initial conditions: current (initially zero), and VC (initially zero). Will have to introduce integration constant and then evaluate that. More chance of mistakes, more complex, so start over with something else.

i
(
t
)
=

1
L

∫

0

t

V

L

(
t
)
d
t
=

∫

0

t

(

e

−
2
x

−
B
(
x
)

e

−
2
x

)
d
x

{\displaystyle i(t)={\frac {1}{L}}\int _{0}^{t}V_{L}(t)dt=\int _{0}^{t}(e^{-2x}-B(x)e^{-2x})dx}
f := (exp(-2*x) - B*x*exp(-2*x));
S :=int(f,x=0..t)

i
(
t
)
=
B
∗
(

e

−
2
t

(
2
t
+
1
)

4

−

1
4

)
−

e

−
2
t

2

+

1
2

+

C

1

{\displaystyle i(t)=B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}+C_{1}}

i
(
0
)
=
0
=

C

1

{\displaystyle i(0)=0=C_{1}}
Ok so C1 is zero. Now need to find B.
Find B by doing another integral to get VC:

V

C

(
t
)
=

1
C

∫

0

t

i
(
t
)
d
t
=
4
∗

∫

0

t

(
B
∗
(

e

−
2
t

(
2
t
+
1
)

4

−

1
4

)
−

e

−
2
t

2

+

1
2

)
d
x

{\displaystyle V_{C}(t)={\frac {1}{C}}\int _{0}^{t}i(t)dt=4*\int _{0}^{t}(B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}})dx}
f := (4*(B*(exp(-2*x)*(2*x+1)/4 -1/4) - exp(-2*x)/2 + 1/2));
S :=int(f,x=0..t)

V

C

(
t
)
=
B
+
2
t
+

e

−
2
t

−
B
t
−
B

e

−
2
t

−
B
t

e

−
2
t

−
1
+

C

1

{\displaystyle V_{C}(t)=B+2t+e^{-2t}-Bt-Be^{-2t}-Bte^{-2t}-1+C_{1}}

V

C

(
0
)
=
0
=
B
+
1
−
B
−
1
+

C

1

{\displaystyle V_{C}(0)=0=B+1-B-1+C_{1}}

C

1

=
0

{\displaystyle C_{1}=0}
Still doesn't help us find B. Guess B = 2 since (2t-Bt) has to equal zero if VC is going to converge on 1. Then see if VC(∞) = 1:

V

C

(
t
)
=
2
+

e

−
2
t

−
2

e

−
2
t

−
2
t

e

−
2
t

−
1

{\displaystyle V_{C}(t)=2+e^{-2t}-2e^{-2t}-2te^{-2t}-1}
At t = ∞ what is the 2te-2t term's value?

limit(B*t*exp(-t),t = infinity)

V

C

(
∞
)
=
2
−
1
=
1

{\displaystyle V_{C}(\infty )=2-1=1}
Yes! B = 2 works ... looks like the only thing that works .... So:

i
(
t
)
=
2
∗
(

e

−
2
t

(
2
t
+
1
)

4

−

1
4

)
−

e

−
2
t

2

+

1
2

=
t

e

−
2
t

{\displaystyle i(t)=2*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}=te^{-2t}}
simplify(2*(exp(-2*t)(2*t+1)/4-1/4) - exp(-2*t)/2 + 1/2)
This means that VR is:

V

R

(
t
)
=
4
∗
i
(
t
)
=
4
t

e

−
2
t

{\displaystyle V_{R}(t)=4*i(t)=4te^{-2t}}

=== Impulse Solution ===
Taking the derivative of the above get:

V

R

δ
(
t
)
=
4

e

−
2
t

−
8
t

e

−
2
t

{\displaystyle V_{R}\delta (t)=4e^{-2t}-8te^{-2t}}

=== Convolution Integral ===

V

R

(
t
)
=

∫

0

t

(
4

e

−
2
(
t
−
x
)

−
8
(
t
−
x
)

e

−
2
(
t
−
x
)

)
(
2
x
−
3

x

2

)
d
x

{\displaystyle V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx}
f := (4*exp(-2*(t-x)) - 8*(t-x)exp(-2*(t-x)))*(2*x-3*x^2);
S :=int(f,x=0..t)

V

R

(
t
)
=
8
−
8

e

−
2
t

−
10
t

e

−
2
t

−
6
t

{\displaystyle V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t}
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.