Physics 120 - David Kleinfeld - Spring 2016
1
Notes on time domain circuit analysis
1.1
Background
The most common first-order differential equation in circuit analysis, such as for a
RC high-pass or low-pass filter, has constant coefficients and is of the form
τ
dV (t)
+ V (t) = F (t).
dt
(1.1)
The full solution is the solution to the homogeneous equation, denoted VH (t) plus the
solution to the inhomogeneous solution, denoted VI (t). The homogeneous equation
is the part with F (t) = 0, i.e., the equation for the decay from V (0) for which the
homogeneous solution is, to within a constant,
VH (t) = V (0)Φ(t)
(1.2)
Φ(t) = e−t/τ .
(1.3)
with
.
The inhomogeneous solution satisfies the full equation
τ
dVI (t)
+ VI (t) = F (t)
dt
(1.4)
and we take as an ansatz that the inhomogeneous solution is proportion to the
homogeneous solution times a yet to be detemined function of time α(t), i.e.,
VI (t)
=
=
Φ(t)(t)α(t)
e−t/τ α(t)
(1.5)
and
dVI (t)
dα(t)
1
= τ e−t/τ
− e−t/τ α(t).
dt
dt
τ
If we plug these back into the original equation, we get
τ
dα(t)
= et/τ F (t)
dt
or
α(t) =
Zt
0
dt0 t0 /τ
e
F (t0 ) + Constant
τ
1
(1.6)
(1.7)
(1.8)
and, substituting back for α(t) gives
−t/τ
VI (t)
= e
Zt
0
Zt
=
0
dt0 t0 /τ
e
F (t0 ) + Constant
τ
(1.9)
dt0 −(t−t0 )/τ
e
F (t0 ) + Constant.
τ
Thus
V (t)
= VI (t) + VH (t)
=
Zt
0
(1.10)
dt0 −(t−t0 )/τ
e
F (t0 ) + V (0− )e−t/τ
τ
The integral for the driven response is call the ”convolution” integral. The above
relation is a special case, for constant coefficients, between the input and the driven
response for a linear system. In general, the response of such a linear system to a
pulse (”delta function”) is given by Φ(t) and
VI (t) =
Zt
0
1.2
dt0
Φ(t − t0 ) F (t0 ).
τ
(1.11)
Example of step input
Here we have a signal that is F (t) = 0 for t < 0 and F (t) = V0 for t ≥ 0.
V (t)
=
Zt
0
dt0 −(t−t0 )/τ
e
V0 + V (0− )e−t/τ
τ
(1.12)
= V0 (1 − e−t/τ ) + V (0− )e−t/τ
1.3
Example of sine input
Here we have a signal that is F (t) = 0 for t < 0 and F (t) = V0 sin(ω0 t) for t ≥ 0.
For focus only on the driven part and take V (0− ) = 0.
V (t)
=
Zt
0
dt0 −(t−t0 )/τ
e
V0 sin(ω0 t0 )
τ
= V0 e
= V0
−t/τ
e−t/τ
2i
Zt
(1.13)
dt0 t0 /τ eiω0 t − e−iω0 t
e
τ
2i
0
Zt/τ
dx ex eiω0 τ x − e−iω0 τ x
0
2
t/τ
e−t/τ Z
= V0
dx e(1+iω0 τ )x − e(1−iω0 τ )x
2i
0
et/τ eiω0 t − 1 et/τ e−iω0 t − 1
−
= V0
2i
1 + iω0 τ
1 − iω0 τ
"
!
#
1
1
eiω0 t
e−iω0 t
1
−t/τ
= V0
−
−e
−
2i
1 + iω0 τ
1 − iω0 τ
1 + iω0 τ
1 − iω0 τ
−t/τ
!
e
(1 − iω0 τ )eiω0 t − (1 + iω0 τ )e−iω0 t + 2iω0 τ e−1/τ
1 + (ω0 τ )2
#
"
eiω0 t + eiω0 t
1
eiω0 t − eiω0 t
−t/τ
− ω0 τ
+ ω0 τ e
= V0
1 + (ω0 τ )2
2i
2
"
#
1
ω0 τ
ω0 τ
−t/τ
= V0
sin(ω0 t) −
cos(ω0 t) +
e
1 + (ω0 τ )2
1 + (ω0 τ )2
1 + (ω0 τ )2
1
= V0
2i
"
#
The sine term represents a faithful transmission of the input and the cosine terms
represents a a phase suited version, while the exponential represents the transient
from turning the signal on at t=0. With a bit more algebra, recalling that sin(a − b)
= sin(a)cos(b) - cos(a)sin(b), we can simply this.
V (t)
V0

ω0 τ
1
ω0 τ

q
sin(ω0 t) − q
cos(ω0 t) + q
e−t/τ 
= q
2
2
2
2
1 + (ω0 τ )
1 + (ω0 τ )
1 + (ω0 τ )
1 + (ω0 τ )
V0

h
i
ω0 τ

sin ω0 t − atan(ω0 τ ) + q
= q
e−t/τ 
2
2
1 + (ω0 τ )
1 + (ω0 τ )
(1.14)
At short times, i.e., t << τ ,
V (t) ≈
ω0 τ
t
2
1 + (ω0 τ ) τ
(1.15)
and in steady state, , i.e., t >> τ ,
h
i
V0
V (t) = q
sin ω0 t − atan(ω0 τ )
1 + (ω0 τ )2
3
(1.16)