RLC Circuits as Filters
Samantha R. Summerson
11 September, 2009
1
Circuits with Resistors and Capacitors (continued)
Recall from the last lecture (9/9) that we found the impedance of the capacitor to be
𝑍𝐶 =
1
.
2𝜋𝑓 𝐶
If 𝑓 tends to zero, the impedance tends to infinity; thus, the impedance looks like an open circuit. If 𝑓 tends
to infinity, the impedance tends to zero; in this case, the impedance looks like a short circuit. In general, a
circuit acts as a filter on the input signals (voltages or currents). There are different typs of filters.
𝑅1
+
±
𝑅2
𝐶
𝑣𝑜𝑢𝑡
−
If we plot the magnitude of the ratio of the output and input voltages, we see that
∣𝑣𝑜𝑢𝑡 ∣
∣𝑣𝑖𝑛 ∣
𝑓
Figure 1: Magnitude of the transfer function. This is a low-pass filter.
From the drawing, it is clear that the circuit represents a low-pass filter (LPF). How could we create a
high-pass filter (filter out low frequencies, pass high frequencies). Consider the use of an inductor.
The impedance of an inductor is
𝑍𝐿 = 𝑗2𝜋𝑓 𝐿.
1
𝑅
𝑣𝑖𝑛
±
𝐿
Figure 2: Circuit corresponding to a high-pass filter.
As 𝑓 tends to zero, the impedance also tends to zero and appears to be a short circuit. As 𝑓 tends to infinity,
the impedance also tends to infinity and appears to be an open circuit. Plotting the magnitude of the ratio
of output and input voltages for this circuit, we see that it is a high-pass filter (HPF).
∣𝑣𝑜𝑢𝑡 ∣
∣𝑣𝑖𝑛 ∣
𝑓
Figure 3: Magnitude of the transfer function. This is a high-pass filter.
If we combine resistors, capacitors, and inductions, we can create a band-pass filter (BPF).
𝑅
𝑣𝑖𝑛
±
𝐿
Figure 4: Circuit correspond to a band-pass filter.
The last circuit we analyzed (voltage source in series with a resistor and capacitor), we found that for
𝑣𝑖𝑛 (𝑡)
= 𝐴𝑐𝑜𝑠(2𝜋𝑓 𝑡),
{
}
= 𝑅𝑒 𝐴𝑒𝑗2𝜋𝑓 𝑡 ,
2
∣𝑣𝑜𝑢𝑡 ∣
∣𝑣𝑖𝑛 ∣
𝑓
Figure 5: Magnitude of the transfer function. This is a band-pass filter.
the output voltage was a cosine function of the same frequency:
{
}
1
𝑗2𝜋𝑓 𝑡
𝑣𝑜𝑢𝑡 (𝑡) = 𝑅𝑒
𝐴𝑒
𝑗2𝜋𝑓 𝑅𝐶 + 1
1
𝐴𝑐𝑜𝑠(2𝜋𝑓 𝑡 − 𝑡𝑎𝑛−1 (2𝜋𝑓 𝑅𝐶)).
= √
(2𝜋𝑓 𝑅𝐶)2 + 1
Definition 1. We define the transfer function for a circuit as
𝑣𝑜𝑢𝑡
𝐻(𝑓 ) =
.
𝑣𝑖𝑛
The transfer function is only real when the circuit consists of sources and resistors. Otherwise, it is a
complex function.
𝐻(𝑓 ) = ∣𝐻(𝑓 )∣𝑒𝑗∠𝐻(𝑓 )
Since the range of a transfer function, 𝐻(𝑓 ), is contained in the field of complex numbers, we must plot both
the magnitude and phase of 𝐻(𝑓 ). Note that this magnitude of 𝐻 is an even function and the phase of 𝐻
is an odd function.
∣𝐻(𝑓 )∣ = ∣𝐻(−𝑓 )∣
∠𝐻(𝑓 )
= −∠𝐻(−𝑓 )
By our definition of the transfer function,
𝑣𝑜𝑢𝑡 = 𝐻(𝑓 )𝑣𝑖𝑛 .
Consider a real input to a LTI system.
𝑣𝑖𝑛
=
{
}
𝑅𝑒 𝑉𝑖𝑛 𝑒𝑗2𝜋𝑓 𝑡
{
}
𝑅𝑒 ∣𝑉𝑖𝑛 ∣𝑒𝑗𝜃 𝑒𝑗2𝜋𝑓 𝑡
=
∣𝑉𝑖𝑛 ∣𝑐𝑜𝑠(2𝜋𝑓 𝑡 + 𝜃)
=
We can write the output in terms of 𝐻.
𝑣𝑜𝑢𝑡
=
{
}
𝑅𝑒 𝐻(𝑓 )𝑉𝑖𝑛 𝑒𝑗2𝜋𝑓 𝑡
{
}
𝑅𝑒 ∣𝐻(𝑓 )∣𝑒𝑗∠𝐻(𝑓 ) ∣𝑉𝑖𝑛 ∣𝑒𝑗𝜃 𝑒𝑗2𝜋𝑓 𝑡
{
}
𝑅𝑒 ∣𝐻(𝑓 )∣∣𝑉𝑖𝑛 ∣𝑒𝑗2𝜋𝑓 𝑡+∠𝐻(𝑓 )+𝜃
=
∣𝐻(𝑓 )∣∣𝑉𝑖𝑛 ∣𝑐𝑜𝑠(2𝜋𝑓 𝑡 + ∠𝐻(𝑓 ) + 𝜃)
=
=
3
If you put a sinusoid of a certain frequency into an LTI system (circuit), the output will be a sinusoid of
the same frequency, but with a possibly different amplitude and phase. The above analysis is similar for an
input of 𝑠𝑖𝑛.
𝑣𝑖𝑛
=
=
𝑣𝑜𝑢𝑡
=
𝑉𝑖𝑛 𝑠𝑖𝑛(2𝜋𝑓 𝑡)
{
}
𝐼𝑚 ∣𝑉𝑖𝑛 ∣𝑒𝑗𝜃 𝑒𝑗2𝜋𝑓 𝑡
}
{
𝐼𝑚 ∣𝐻(𝑓 )∣∣𝑉𝑖𝑛 ∣𝑒𝑗2𝜋𝑓 𝑡+∠𝐻(𝑓 )+𝜃
=
∣𝐻(𝑓 )∣∣𝑉𝑖𝑛 ∣𝑠𝑖𝑛(2𝜋𝑓 𝑡 + ∠𝐻(𝑓 ) + 𝜃)
2
Equivalent RLC Circuits
+
𝑅
𝑣𝑖𝑛
±
𝑣
𝐶
−
What does this circuit looks like from the terminals? It turns out that we write equivalent circuits for
general RLC circuits just like we did for circuit with only sources and resistors. The above circuit can be
equivalently written as a voltage source in series with an equivalent impedance.
𝑍𝑒𝑞
𝑣𝑖𝑛
+
±
𝑣
−
We find 𝑍𝑒𝑞 by zero-ing out the voltage source and using our rule for impedances in parallel.
𝑍𝑒𝑞
1
1
+
𝑅 𝑍𝐶
(
1
+ 𝑗2𝜋𝑓 𝐶
𝑅
=
(
=
=
)−1
(
=
)−1
1 + 𝑗2𝜋𝑓 𝐶𝑅
𝑅
𝑅
1 + 𝑗2𝜋𝑓 𝐶𝑅
)−1
For circuits with conductors and inductors, we can use the same equivalent circuits techniques as before.
The only difference is that we solve for 𝑍𝑒𝑞 , an equivalent impedance, rather than 𝑅𝑒𝑞 .
4