Fourier series and an RC circuit
Circuit
Resistor R in series with capacitor C, input x(t) is voltage across combination, output y(t) is
voltage across capacitor.
Resistor:
vr (t) = Ri(t)
Capacitor:
i(t) = C
d
vc (t)
dt
and
v(t) = vc (t) + vr (t)
Eliminating i(t) and letting v(t) = x(t) and y(t) = vc (t) gives DE
x(t) = RC
dy(t)
+ y(t)
dt
System interpretation
The system is linear (linear constant coefficient DE), and therefore has an impulse response h(t).
(Don’t yet know how to find it.)
The input/output relationship can therefore be written in the form
y(t) = h(t) ∗ x(t)
Input complex exponential x(t) = c1 ejω0 t :
Z ∞
Z ∞
h(λ)c1 ejω0 (t−λ) dλ
h(λ)x(t − λ)dλ =
y(t) =
−∞
−∞
Z ∞
= c1
h(λ)e−jω0 λ dλ ejω0 t = d1 ejω0 t
−∞
Finding output coefficients
Return to DE:
dy(t)
+ y(t)
dt
Consider input x(t) = c1 ejω0 t . Know the output is of the form y(t) = d1 ejω0 t . Coefficient c1 is
known: want to find d1 .
x(t) = RC
Substitute into DE and solve:
d jω0 t + d1 ejω0 t
d1 e
dt
=⇒ c1 ejω0 t = jω0 RCd1 ejω0 t + d1 ejω0 t
=⇒ c1 = (jω0 RC + 1)d1
c1 ejω0 t = RC
So
d1 =
1
c1
1 + jω0 RC
Multiple component signal
Input complex exponential x(t) = c1 ejω0 t + c2 ej2ω0 t :
Z ∞
h(λ)x(t − λ)dλ
y(t) =
−∞
Z ∞
Z ∞
jω0 (t−λ)
=
h(λ)c1 e
dλ +
h(λ)c2 ej2ω0 (t−λ) dλ
−∞
−∞
Z ∞
−jω0 λ
= c1
h(λ)e
dλ ejω0 t
−∞
Z ∞
−j2ω0 λ
+ c2
h(λ)e
dλ ej2ω0 t
−∞
= d1 ejω0 t + d2 ej2ω0 t
Complex exponentials in, complex exponentials out at same frequencies. Only need to find the
coefficients d1 and d2 given c1 and c2
Coefficients for multi-component signal
You can (and should) do same for two-component signal. Let
x(t) = c1 ejω0 t + c2 ej2ω0 t ,
y(t) must be of form
y(t) = d1 ejω0 t + d2 ej2ω0 t ,
substitute into the DE and solve (algebraically!) for d1 and d2 .
In general, if input and output are
x(t) =
∞
X
ck ejkω0 t
and
y(t) =
k=−∞
∞
X
dk ejkω0 t ,
k=−∞
then
dk =
1
ck
1 + jkω0 RC
In general
If there is a LCCDE linking inputs and outputs, the FS coefficients for input and output will be
found to obey
dk = H(kω0 )ck
where for the RC circuit we have
H(ω) =
1
1 + jωRC
To work out how a signal is modified by the circuit, only need to know the values of H(ω) at the
frequencies present
H(ω) is just a complex number for each value of frequency ω. The plot of the magnitude and
phase of H as a function of ω is called the Bode plot of the system.
Bode plot for RC circuit
2
1
0.9
0.8
0.7
|H(ω)|
0.6
0.5
0.4
0.3
0.2
0.1
0
-2*pi
-pi
0
pi
2*pi
-pi
0
ω
pi
2*pi
2
1.5
1
∠ H(ω)
0.5
0
-0.5
-1
-1.5
-2
-2*pi
Observe
Time domain description of signals x(t) and y(t) −→ differential equation linking input and output
of system. Can solve for y(t) given x(t), but no intuition.
Instead, think about signals as (weighted linear) combinations of complex exponentials (or combinations of frequencies)
∞
X
x(t) =
ck ejkω0 t
and
y(t) =
k=−∞
∞
X
dk ejkω0 t
k=−∞
−→ algebraic equation linking input and output:
dk = H(kω0 )ck
Much easier to understand.
One component approximation to square wave
Input
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0.5
1
1.5
2
Output
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
-0.25
-2
-1.5
-1
-0.5
0
3
Two component approximation to square wave
Input
1
0.5
0
-0.5
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Input component 1
Output component 1
1
0.3
0.2
0.5
0.1
0
0
-0.1
-0.5
-0.2
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
Input component 2
-0.5
0
0.5
1
1.5
2
0.5
1
1.5
2
0.5
1
1.5
2
Output component 2
1
0.3
0.2
0.5
0.1
0
0
-0.1
-0.5
-0.2
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2
-1.5
-1
-0.5
0
Output
0.3
0.2
0.1
0
-0.1
-0.2
-2
-1.5
-1
-0.5
0
Many component approximation to square wave
Input (N=30)
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
0.5
1
1.5
2
Output
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
-0.15
-0.2
-0.25
-2
-1.5
-1
-0.5
0
4