RLC circuits

A very common use of inductors is as part of a tuned circuit. For instance, each slider on a graphic equalizer selects out a range of frequencies from the audio spectrum, and controls how much gain there should be for those frequencies. A later stage recombines the various frequency bands.

RLC (Resistor-Inductor-Capacitor) circuits are a common way of extracting a frequency range. Here's part of a typical circuit:

The middle of the schematic is the part we will be analyzing. It emphasizes one band of frequencies over the others, as we will see, and the potentiometer (refer back if you've forgotten the potentiometer) on the right acts as a volume control for the that band of frequencies (Vout will be a fraction of VR controlled by the setting of the potentiometer wiper). A graphic equalizer contains many such circuits, each tuned to emphasize a different band. The output of all the channels are recombined after another stage of amplification, shown as the triangle on the right. In this way it separates the frequency bands, allows you to adjust the volume of each one individually, and then recombines them and sends the result to your speakers.

A time-varying voltage comes from your CD player, which is the sound waveform.The yellow elements represent stages of amplification, the one on the left being connected at its input end (left side) to youur CD player. We will treat the output of the one on the left as a voltage source measured with respect to ground (which we will name Vin because it is input to the part of the circuit under study). The input of the amplifier on the right we will assume to be an "observer" only: no current flows into it, but it watches the voltage Vout from our circuit and amplifies it. This concept is familiar to you already, because it's what voltmeters do.Thus it does not affect the behavior of the part of the circuit we are studying.

In analyzing circuits you have to be willing to focus on a chunk of it, and say "I can understand this part on its own, and I'll neglect its interaction with other parts". You can't generally write down or solve the equations of motion for a whole circuit at once. Here we don't know how the yellow amplifier blocks work, but we'll take the left one to be a voltage source and the right one to be a zero-current observer (and amplifier) of voltage.

Typically, coming in from the left is an audio signal, which is a mish-mosh of different frequencies with different amplitudes, like this, where the horizontal axis is time (maybe milliseconds) and the vertical axis is volateg, Vin..

Let's consider instead a pure tone; a "sine-wave" test signal, of the form
Vin = sin(wt)
(2)
where w is the frequency, in radians/second. This is related to frequency f in cycles per second (hertz) by w = 2p f. This waveform of course looks (and sounds) quite boring:

Now we'll find the equations of motion of the circuit. We expect to find two coupled diffeqs, becasue there are two "energy storage" elements: the inductor and the capacitor.

The constitutive relations of the elements are:
VR = R iR 

VL = L i'L 

VC = (1/C) qC

(3)
(4)
(5)
The loop relation and node relations are:
Vin = VL + VC + VR 

iL = iC = iR

(6)
(7)

State Variables

We need one state variable for each capacitor and each mass, thus in this system we will have two state variables. We could choose qC and iL, however qC is actually inconvenient and instead we choose the voltage of the capacitor, VC, which is linearly related to qC via (5). Thus, iL and VC are our state variables, and we will try to eliminate all the other dynamic variables except Vin, which is an input. We quickly obtain these state equations:
i'L = (1/L) (Vin - VC - R iL

V'C = (1/C) iL

(8)
(9)
Now is a good time to notice that we care about VR, and it's been eliminated. But VR is simply R iL, so it won't be hard to recover it after we've solved for iL(t) and VC(t).

We could now solve these diffeqs numerically. That will be homework. Here let's solve them analytically.

Noting that Vin is given by a sine in equation (2), I'm going to make these guesses for iL(t) and VC(t):
iL = A sin(wt) + B cos(wt) 

VC = D sin(wt) + E cos(wt)

(10)
(11)
where the coefficients A, B, D, E are to be adjusted in the hope that a solution satisfying (8) & (9) can be found in this generic class. It turns out that the values of A, B, D, E which make a solution possible are
 

(12)
(13)
These are found by substituting (10) and (11) and Vin = sin(wt) into (8) and (9) and setting the sine terms and the cosine terms equal to one another, resulting in four messy algebraic equations to find the four unknowns. You find D & E.

Remember, the input is Vin = sin(wt), and the output (if the potentiometer is set to max) is VR=RiL. We can put different sine-wave test signals in by varying w.

So, how is this working? You will see in more detail in the Homework, but in the next section we will think about plotting VR as a function of time, considering different frequency inputs. What we will find is that the amplitude of the output is a function of frequency, with the peak value occurring at a particular resonant frequency. That resonant frequency is determined by the values of the inductor and capacitor in the circuit. Thus, each channel of our graphic equalizer has a different capacitor and inductor, such that the frequency it controls is a particular range.