resistor

Bandwidth of some bandpass filters

October 18, 2025 math and physics play , , , , , , , , , , , , ,

[Click here for a PDF version of this post]

Karl had a couple of fun filter problems that he chatted with me about. I didn’t remember how do find the bandwidth, nor the resonant frequency of such a circuit, and after he solved the problems, I tried them for myself.

The circuits, as seen in fig. 1, were RC and RL respectively, and both ended up having the same transfer function.

fig 1a. RC bandpass filter.

fig 1b. RL bandpass filter.

 

For my attempt to solve these circuits, I used the loop current method that Karl recently taught me. Perhaps I once knew that method, but if I had, I’d forgotten. This loop method can be nicer than a standard nodal analysis, since it can automatically eliminate some current variables. Here’s the equations for the RC circuit
\begin{equation}\label{eqn:RLCbandwidth:20}
\begin{aligned}
V_a – V_s &= I_1 (1) \\
V_a &= (I_1 – I_2) \inv{s} \\
V_o – V_a &= I_2 \inv{s} \\
V_o &= I_2 (1).
\end{aligned}
\end{equation}
The equations for the RL circuit are
\begin{equation}\label{eqn:RLCbandwidth:40}
\begin{aligned}
V_a – V_s &= I_1 s \\
V_a &= (I_1 – I_2) (1) \\
V_o – V_a &= I_2 (1) \\
V_o &= I_2 s.
\end{aligned}
\end{equation}
The transfer functions for both is
\begin{equation}\label{eqn:RLCbandwidth:60}
H(s) = \frac{V_o}{V_s} = \frac{1}{1/s + 3 + s}.
\end{equation}
A plot of \( \Abs{H(j\omega)} \) can be found in fig. 2, and peaks at \( \omega = 1 \).

fig. 2. Transfer function.

 

Observe that the denominator of this transfer function looks just like a series RLC impedance. For example, for the circuit of fig. 3, we have
\begin{equation}\label{eqn:RLCbandwidth:80}
\frac{I}{V} = \inv{Z} = \inv{ R + L s + \inv{s C} }.
\end{equation}

fig. 3. RLC circuit.

 

Written out in the frequency domain, that impedance is
\begin{equation}\label{eqn:RLCbandwidth:100}
Z(\omega) = R + j \lr{ \omega L – \inv{\omega C} }.
\end{equation}
Max power transfer through this circuit will be for the specific frequency where the impedance is purely real. In this case, that is the frequency \(\omega_0\) that satisfies
\begin{equation}\label{eqn:RLCbandwidth:120}
\omega_0^2 = \inv{L C}.
\end{equation}
Karl’s textbook didn’t define bandwidth in any general sense, but did do so for an RLC circuit of this form, stating that the bandwidth was \( \omega_2 – \omega_1 \) where these are the frequencies of the half (average) power points.

We need to remind ourselves what the formula for average power is
\begin{equation}\label{eqn:RLCbandwidth:140}
\begin{aligned}
P
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re} \lr{ V e^{j\omega t} } \Real \lr{ I e^{j\omega t} } dt \\
&= \inv{4 T} \int_0^T
\lr{ V e^{j \omega t} + \bar{V} e^{-j\omega t} }
\lr{ I e^{j \omega t} + \bar{I} e^{-j\omega t} } dt \\
&= \inv{4} \lr{ V \bar{I} + \bar{V} I } \\
&= \inv{2} \textrm{Re} \lr{ V \bar{I} } \\
&= \inv{2} \textrm{Re} \lr{ I Z \bar{I} } \\
&= \inv{2} \Abs{I}^2 R \\
&= \inv{2} \Abs{V}^2 \frac{R}{\Abs{Z}^2}.
\end{aligned}
\end{equation}
The maximum average power is for purely real impedance
\begin{equation}\label{eqn:RLCbandwidth:160}
P = \inv{2 R} \Abs{V}^2,
\end{equation}
so the half power points are when \( \Abs{Z} = \sqrt{2} R \). For the RLC circuit that is when
\begin{equation}\label{eqn:RLCbandwidth:180}
\Abs{R + j \lr{ \omega L – \inv{\omega C} }}^2 = 2 R^2,
\end{equation}
or
\begin{equation}\label{eqn:RLCbandwidth:200}
R^2 = \lr{ \omega L – \inv{\omega C} }^2.
\end{equation}
We seek solutions for
\begin{equation}\label{eqn:RLCbandwidth:220}
\begin{aligned}
R &= \omega L – \inv{\omega C} \\
-R &= \omega L – \inv{\omega C}.
\end{aligned}
\end{equation}
We find the same solutions for either, both leading to
\begin{equation}\label{eqn:RLCbandwidth:240}
B = \Abs{\omega_2 – \omega_1} = \frac{R}{L}.
\end{equation}
This is the (half-power) bandwidth for the RLC circuit. We may now re-express the transfer functions for the filters in terms of the resonant frequency and bandwidth
\begin{equation}\label{eqn:RLCbandwidth:260}
H(s) = \frac{s}{\omega_0^2 + B s + s^2},
\end{equation}
and see by inspection that \( \omega_0 = 1 \) for our circuit (also seen in the plot) and \( B = 3 \,\textrm{rad/s} \).

Average power for circuit elements

February 9, 2016 ece1236 , , , , , , ,

[Click here for a PDF of this post with nicer formatting] or [Click here for my notes compilation for this class]

In [2] section 2.2 is a comparison of field energy expressions with their circuit equivalents. It’s clearly been too long since I’ve worked with circuits, because I’d forgotten all the circuit energy expressions:

\begin{equation}\label{eqn:averagePowerCircuitElements:20}
\begin{aligned}
W_{\textrm{R}} &= \frac{R}{2} \Abs{I}^2 \\
W_{\textrm{C}} &= \frac{C}{4} \Abs{V}^2 \\
W_{\textrm{L}} &= \frac{L}{4} \Abs{I}^2 \\
W_{\textrm{G}} &= \frac{G}{2} \Abs{V}^2 \\
\end{aligned}
\end{equation}

Here’s a recap of where these come from

Energy lost to resistance

Given
\begin{equation}\label{eqn:averagePowerCircuitElements:40}
v(t) = R i(t)
\end{equation}

the average power lost to a resistor is

\begin{equation}\label{eqn:averagePowerCircuitElements:60}
\begin{aligned}
p_{\textrm{R}}
&= \inv{T} \int_0^T v(t) i(t) dt \\
&= \inv{T} \int_0^T \textrm{Re}( V e^{j \omega t} ) \Real( I e^{j \omega t} ) dt \\
&= \inv{4 T} \int_0^T
\lr{V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{I e^{j \omega t} + I^\conj e^{-j \omega t} }
dt \\
&= \inv{4 T} \int_0^T
\lr{
V I e^{2 j \omega t} + V^\conj I^\conj e^{-2 j \omega t}
+ V I^\conj + V^\conj I
}
dt \\
&= \inv{2} \textrm{Re}( V I^\conj ) \\
&= \inv{2} \textrm{Re}( I R I^\conj ) \\
&= \frac{R}{2} \Abs{I}^2.
\end{aligned}
\end{equation}

Here it is assumed that the averaging is done over some integer multiple of the period, which kills off all the exponentials.

Energy stored in a capacitor

I tried the same sort of analysis for a capacitor in phasor form, but everything cancelled out. Referring to [1], the approach used to figure this out is to operate first strictly in the time domain. Specifically, for the capacitor where \( i = C dv/dt \) the power supplied up to a time \( t \) is

\begin{equation}\label{eqn:averagePowerCircuitElements:80}
\begin{aligned}
p_{\textrm{C}}(t)
&= \int_{-\infty}^t C \frac{dv}{dt} v(t) dt \\
&= \inv{2} C v^2(t).
\end{aligned}
\end{equation}

The \( v^2(t) \) term can now be expanded in terms of phasors and averaged for

\begin{equation}\label{eqn:averagePowerCircuitElements:100}
\begin{aligned}
\overline{{p}}_{\textrm{C}}
&= \frac{C}{2T} \int_0^T \inv{4}
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} }
\lr{ V e^{j \omega t} + V^\conj e^{-j \omega t} } dt \\
&= \frac{C}{2T} \int_0^T \inv{4}
2 \Abs{V}^2 dt \\
&= \frac{C}{4} \Abs{V}^2.
\end{aligned}
\end{equation}

Energy stored in an inductor

The inductor energy is found the same way, with

\begin{equation}\label{eqn:averagePowerCircuitElements:120}
\begin{aligned}
p_{\textrm{L}}(t)
&= \int_{-\infty}^t L \frac{di}{dt} i(t) dt \\
&= \inv{2} L i^2(t),
\end{aligned}
\end{equation}

which leads to

\begin{equation}\label{eqn:averagePowerCircuitElements:140}
\overline{{p}}_{\textrm{L}}
= \frac{L}{4} \Abs{I}^2.
\end{equation}

Energy lost due to conductance

Finally, we have conductance. In phasor space that is defined by

\begin{equation}\label{eqn:averagePowerCircuitElements:160}
G = \frac{I}{V} = \inv{R},
\end{equation}

so power lost due to conductance follows from power lost due to resistance. In the average we have

\begin{equation}\label{eqn:averagePowerCircuitElements:180}
\begin{aligned}
p_{\textrm{G}}
&= \inv{2 G} \Abs{I}^2 \\
&= \inv{2 G} \Abs{V G}^2 \\
&= \frac{G}{2} \Abs{V}^2
\end{aligned}
\end{equation}

References

[1] J.D. Irwin. Basic Engineering Circuit Analysis. MacMillian, 1993.

[2] David M Pozar. Microwave engineering. John Wiley & Sons, 2009.