And here's another circuit that we want to analyze using AC circuit analysis
method. This is the RL circuit.
So, we've, we've replaced the capacitor with the inductor, and like we did in the
last case, we're going to say we're driving this circuit with some AC.
Voltage and what I want to know, is what is the magnitude of the voltage across
the inductor as a function of frequency. Now, what we have to do then is use
Kirchhoff's voltage law to find the current through the circuit.
And then once I know the current, I can find the voltage across the inductor.
So, to find VL, I first write Kirchhoff's voltage law.
And so starting here at the ground, if we go up through this, voltage, A/C voltage
source, we pick up a voltage V. The nerves of voltage drop of minus IR,
and then a voltage drop of, across the conductor of minus j omega L times I.
And so, all three voltages have to sum up to 0 because we've completed a circuit,
around that loop. So I can solve this for I, it's just V is
V over R plus j omega L, so put these terms on the other side, and then take
the, factor the I out and solve this. And then the voltage across the inductor
is that current times j omega L. So I have j omega L over this term, times
V, and I'm going to take and factor an R, or divide, multiply both the top and the
bottom by 1 over R. So if I do that, I get j omega L over R,
then the 1 over R makes this factor 1 and then I have a j omega L over R.
So, this is just a little bit of algebraic reorganization there.
Now, I want the magnitude of VL like we did for the magnitude of the voltage
across the capacitor. So I have to compute the magnitude of
this, kind of, ugly looking factor, with J's upstairs and downstairs.
So I've multiplied this factor times its complex conjugate.
And the complex conjugate is just the same factor, but everywhere I had a plus
J, I replace it with a minus J. And then square root of the entire
quantity. So if I multiply this out, j times minus
j is plus 1. Then I have omega squared, L squared over
R squared, the, the numerator. And then down in the denominator, I have
the first term squared, 1 plus omega squared, L squared over R squared.
So you see, the cross terms, when I multiply all this out, I get a plus and a
minus term that look the same. They cancel out.
Then I have plus j times minus j, gives me plus 1.
And then omega L over R, whole thing squared.
So here is the expression for the magnitude of the voltage across the
inductor, divided by the dry voltage. Now, if I plot this factor, it looks like
this. This starts off at zero.
When omega is zero, you see, the numerator is zero, and the denominator is
just one. And so I have a zero.
Now, as omega gets large, the, when omega L over R becomes a lot bigger than 1, I
can ignore the 1. And then I have the same factor upstairs
and downstairs in the numerator and the denominator.
And so this whole thing just becomes 1 when omega L over R becomes large.
So this response, then, low frequencies do not pass through high frequencies do.
Now again lets look at the special place when omega L over R is 1.
So that means when omega equals R over L. So when omega L over R is 1.
Then this whole factor becomes 1 over 1 plus 1 square root.
So that's 0.707. So this is the point where the frequency
is R over L. And at that point, the response of this
filter is 1 over the square root of 2.707.
Now this filter is a high pass filter. High frequencies pass through, low
frequencies don't. So this is just the opposite of the
capacitor, so if I replace this component with a capacitor, I get a low pass
filter. If I put an inductor in there, I get a
high pass filter. Okay, now that you've seen those two
examples, I would like you to take some time, and it's really important that you
do this. take some time and find the transfer
functions for these two circuits. So these circuits are, very similar to
the ones that we just analyzed. but I swapped the position of the
inductor and resistor. And so in this circuit we're treating it
like a filter, and I want to know the voltage across the resistor.
So what I'm doing is I'm applying a certain voltage to the input of this
device. And on the output is the voltage measured
at this point to ground. And so, what you need to do is write down
Kirchhoff's Voltage Law. Find the current through the circuit and
then compute the voltage across this resister.
So do that one and then do the same kind of thing where but for the RC circuit
where we swapped the position of the capacitor and the resister.
And compute the output which is in this case the voltage across the resistor.
And I'd like you to compute the expressions for that those voltages in
both cases, compute their magnitudes, and make a plot like the ones that we just
made. So it's really important that you do
this. So take some time and work that out.
Robert ClarkProvost and Senior Vice President for Research and Professor
Mark BockoDistinguished Professor and Chair
