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Okay, we're going to change topics now and talk about Band Pass Filters and then
graphic equalizers. So, this is an example of filters you can
build with R, Ls and Cs together. So, it's the next most complicated thing
beyond the simple low-pass and high-pass filters that we looked at with RC and RL
combinations. So, here is a simple Band Pass Filter.
here's the input voltage side and the output voltage side, and I have a
resistor in series with the parallel combination of a capacitor and an
inductor. And what this thing is, is really nothing
more than a voltage divider. So, the input voltage is applied to the
series combination of R. And so, that's the Z1 impedance, and then
the parallel combination of C and L is Z2.
So, Vn Goes through Z1 and then Z2 to ground.
So, that's this, and now the output voltage is just tapped off the center
point here. And so, this is nothing more than the
familiar old voltage divider. And so, the output voltage is the input
voltage times this impedance Z2, divided by the sum of Z1 and Z2.
So, that's the familiar old voltage divider equation.
Now, if we take, the first thing we have to do is figure out, what's the parallel
combination of L and C. So, L in parallel with C is, now
remember, the formula for figuring out the impedance of a parallel combination
of two impedances is the first impedance times the second impedance, divided by
their sum. So, here's the impedance of the inductor,
impedance of the capacitor, and then the sum.
Now, I just, if I multiply the J omega C through the denominator I get a one from
the second term. And then, the, the J omega, J times J,
gives me minus 1. Then, I have an omega squared, and an LC.
Now again, I really encourage you to follow these, go through these slowly and
work this out with paper and pencil yourself as we go along.
So, here is the impedance, Z2, L in parallel with C.
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Now, the output voltage going back to here is that Z2 impedance, so here it is
right here. Divided by the sum of R plus that Z2.
So, here is the output voltage. Now, we can rewrite that.
I want to take and clean things up a little bit by multiplying this through
the denominator. And this is a, a little exercise for you
to start with this formula and reduce it to this format.
Now, we're going to, to clean things up a little bit.
We're going to define two quantities here.
One of them, omega 0, is the resonant frequency of this LC tank circuit here.
And that's 1 over the square root of L times C.
And then, the other thing that we defined we talked about this a couple of weeks
ago, is the Q factor. The Q factor is omega 0.
The resonant frequency times L over R. And so, if you take, and you, you work on
this, work on the algebra of this a little bit, you should be able to reduce
it to this expression. now, what I did also is there is an extra
step here. You work out.
This is the full complex equation here. I want to compute the magnitude of this.
And so, after I have written this all out, I then take the the real part
squared plus the imaginary part squared, square root and that's where this comes
from. So this is a a little bit of a challenge
for you. But I want you to work this out.
this is the output as a function of the input.
And so, it has a Q up here, there's a Q down there and it has this interesting
resonance response. Now, if we take then and, and plot what
this looks like. So, this is the magnitude, remember.
And you can do this in Excel. I, I did this in Excel myself.
And so, I'll plot this as V out over V in as a function of the frequency.
But I'm going to normalize the frequency to the resonant frequency.
So, it's omega over omega nought is the variable that I'm going to use because it
always appears in that combination. Omega over omega nought.
So, my frequencies are always expressed just as a fraction of the resonant
frequency. So, when omega equals omega nought,
you're on resonance for that LC combination.
Now, this what I did here was just showed this curve for different values of omega
nought. one, a half, and two.
And I assumed that the cube was one over the square root of two in each case.
And so just by changing the value of omega nought, plotting this out versus
frequency, you see this kind of response. When omega equals omega nought.
So, let's look at the green curve. And, and the when omega is, when you're
on resonance here. What happens, omega over omega nought is
1. And then, this factor is going to go
away. I've got 1 minus 1, that's gone.
This is 1. There's a Q squared down here.
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And I, when I take the square root of this factor, that's just this same factor
here. And so, it's this over the same factor,
so I get 1. And so, on resonance, on, right at 1.
And then, as I go to 0 frequency, omega, when omega goes to 0, then this term goes
away. This term goes away.
I just have a one in the denominator and upstairs, omega goes to 0, and so the
whole thing goes to 0. And then, when you go to high frequency.
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When omega becomes becomes much bigger than omega nought.
Well then, you can ignore this term and you can actually ignore this term too,
beyond a certain point. And when omega, yeah, because you see
omega, this whole thing is squared. So, I have omega to the 4th over omega
nought to the 4th and that's going to dominate this term.
When omega over omega nought is say 5 5 to the 4th is a lot bigger than 5
squared. And so, you can ignore this term.
And I just get this over, over this factor and this whole thing drops down as
omega squared. So, there's a resonant frequency when
omega equals omega nought. And then, it goes down as 1 over omega
squared for large omega, then it goes to 0 for small omega.
Now, I, this graph is a little misleading because I put in different values of
omega nought, but I plotted them on the point, on, on the same axis.
But I just wanted to show what happens if you move the resonant frequency around.
Keeping the queue fixed.
Robert ClarkProvost and Senior Vice President for Research and Professor
Mark BockoDistinguished Professor and Chair
