Lecture 15:
Recall: RLC circuit equation
LQ00 + RQ0 + (1/C)Q = V (t)
Initial conditions: Q(0) = q0, I(0) = Q0(0) = 0.
Last time: considered homogeneous case V (t) ≡ 0.
Example: L = 0.25, R = 10, C = 10−3.
0.25Q00 + 10Q0 + 103Q = 0 or
Q00 + 40Q0 + 4000Q = 0.
√
r = −20 ± (1/2) −14400 = −20 ± 60i
General solution:
Q(t) = c1e−20t cos(60t) + c2e−20t sin(60t)
Use initial conditions:
Q(0) = q0 = c1, I(0) = 0 = −20c1 + 60c2, and so c2 = q0/3
Solution to IVP:
Q(t) = q0e−20t cos(60t) + (q0/3)e−20t sin(60t)
p
√
2
2
Let A = c1 + c2 = ( 10/3)q0 and φ = arctan(c2/c1) = arctan(1/3) ≈
0.3218. Then as in the case of LC circuit (i.e. pure periodic motion),
q
−20t
−20t
Q(t) = e A cos(60t − φ) = e
c21 + c22 cos(60t − arctan(1/3))
Non-homogeneous case: forcing term V (t) 6= 0
Let’s simplify to LC circuit.
1
LQ00 + (1/C)Q = V (t).
IVP: Q(0) = 0, Q0(0) = I(0) = 0.
Recall general solution to homogeneous equation was:
Q(t) = c1 cos(ω0t) + c2 sin(ω0t)
√
where ω0 := 1/ LC, the natural frequency.
Let V (t) = V cos(ωt), with ω 6= ω0; here ω is the forcing frequency.
(V is a constant, the amplitude of the forcing term)
Use method of undetermined coeffs.: Guess particular solution
Qp(t) = A cos(ωt) + B sin(ωt)
Plug Qp(t) into LC ODE:
LQ00p + (1/C)Qp = L(−Aω 2 cos(ωt) − Bω 2 sin(ωt))
+(1/C)(A cos(ωt) + B sin(ωt)) = V cos(ωt)
So,
(−ω 2L + (1/C))A cos(ωt) + (−ω 2L + (1/C))B sin(ωt) = V cos(ωt).
Since this holds for all t,
A=
V
,
(−ω 2L + (1/C))
√
Using ω0 = 1/ LC, we can re-write
A=
V
L(ω02 − ω 2)
2
B=0
So, the general solution to the non-homogeneous equation is:
Q(t) = c1 cos(ω0t) + c2 sin(ω0t) +
V
cos(ωt)
L(ω02 − ω 2)
Determine c1, c2 from initial conditions:
0 = Q(0) = c1 +
V
,
L(ω02 − ω 2)
so c1 = − L(ω2V−ω2) .
0
0 = I(0) = Q0(0) = c2ω0
so c2 = 0. So, solution to IVP is:
Q(t) =
V
(cos(ωt) − cos(ω0t))
L(ω02 − ω 2)
and by a trig identity, this equals
=
2V
ω0 − ω
ω0 + ω
sin(
t)
sin(
t)
L(ω02 − ω 2)
2
2
View this as the product of
1):
sin( ω02+ω t) times
2):
2V
L(ω02 −ω 2 )
sin( ω02−ω t)
If ω is very close to ω0, then 1) is rapidly oscillating sin wave and
2) is a slowly oscillating sin wave with large period. Get beats.
3
Lecture 16:
Last time, considered LC circuit
LQ00 + (1/C)Q = V cos(ωt)
IVP: Q(0) = 0, Q0(0) = I(0) = 0.
Recall general solution to homogeneous equation was:
Q(t) = c1 cos(ω0t) + c2 sin(ω0t)
√
where ω0 := 1/ LC, the natural frequency.
We assumed that ω 6= ω0.
Guess for a particular solution:
Qp(t) = A cos(ωt) + B sin(ωt)
If ω ≈ ω0, we get beats.
Now, let’s consider case: ω = ω0. So, forcing frequency and
natural frequency agree.
Then the guess for particular solution is different:
Qp(t) = t(A cos(ω0t) + B sin(ω0t))
Plugging Qp(t) into the non-homogeneous ODE, we get
A = 0, B =
V
2Lω0
(by some hairy calculation).
So, general solution of non-homogeneous ODE is:
Q(t) = c1 cos(ω0t) + c2 sin(ω0t) + (
4
V
)t sin(ω0t)
2Lω0
V
)(sin(ω0t)+ω0t cos(ω0t)
2Lω0
Initial conditions (Q(0) = 0, Q0(0) = I(0) = 0) yield: c1 = 0, c2 =
Q0(t) = −c1ω0 sin(ω0t)+c2ω0 cos(ω0t)+(
0.
So, we get a wave with fixed frequency ω0 and linearly increasing
V
amplitude: ( 2Lω
)t.
0
Laplace Transform
3 steps:
1. Transform IVP for 2nd order linear ODE with constant coeffs.
to an algebraic equation
2. Solve algebraic equation.
3. Take inverse transform to get solution to ODE.
It when the non-homogeneous part is discontinuous (but with only
a few discontinuities). Forcing terms might be impulses of different
amplitudes over different time intervals.
Laplace transform involves an improper integral:
The type of improper integral we use is:
Z A
Z ∞
f (t)dt := lim
f (t)dt
A→∞
a
a
If the limit exists we say that the improper integral is convergent;
otherwise divergent.
There is a lot of theory in the text (section 6.1), but we will spare
you most of it.
Definition: The Laplace transform of a function f (t) is:
Z ∞
L(f ) =
e−stf (t)dt
0
5
Note that we only need f (t) to be defined on t > 0.
Note that L(f ) is a function of a new variable s.
Note that L(f ) may diverge for some s.
However, if L(f ) converges for some value s0, then it converges
for all s > s0.
Example 4:
Z ∞
Z A
L(1) =
e−stf (t)dt = lim
e−stdt
A→∞
0
0
1
e−st A
e−sA 1
|0 = − lim (
− )= .
= − lim
A→∞ s
A→∞
s
s
s
provided s > 0 (otherwise the integral is divergent – the area under
the curve is infinite).
Example 5:
at
∞
Z
−st at
L(e ) =
e
A
Z
e(a−s)tdt
e dt = lim
A→∞
0
0
e(a−s)A
1
1
e(a−s)t A
|0 = lim (
−
)=
.
= lim
A→∞ a − s
A→∞ a − s
a−s
s−a
provided that s > a (otherwise the integral is divergent).
Note that when a = 0, Example 5 becomes Example 1.
Example 6’: Let
f (t) =
Then
Z
L(f ) =
1
−st
e
0
1 0≤t≤1
0 t>1
e−st 1 1 − e−s
dt = −
| =
s 0
s
provided s > 0.
6
Example 6: Let
1 0≤t<1
f (t) = k
t=1
0
t>1
where k is any constant. Then
1 − e−s
L(f ) =
.
s
The value of the constant k does not matter.
7