Welcome back to Electronics. This lesson is on First-Order Lowpass Filters. In our previous lesson, we introduced active filters. In this lesson, we want to introduce active lowpass filters. We'll go through the characteristics and then show you how to design them. So, starting with the characteristics. A Lowpass Filter passes low frequency components and filters out, or we say attenuates, high frequency components. They're characterized by a transfer function H of omega. So if I plot on a linear plot, we have certain characteristics that we define for this. Two particular things that we use in design actually. One is the bandwidth. And the bandwidth is the value at which it's 0.707 of what we call the DC value. First is the frequency plot. DC value is at omega equals zero. And so we look at the value of the plot right there, and I'll call that the DC gain. So the DC gain is that, and then the bandwidth is the frequency at which we're at 0.707 of the DC gain. Now this is a linear plot where I'm plotting the magnitude H of h. The other way we often plot this is a Bode plot, where I'm actually plotting 20 times the log of h. So, then, this plot, this value is going to be 20 times the log, and this is base 10 of the DC gain. And then, at the point where I'm at 3 dB below the DC gain, that is my bandwidth. Now, in this particular class, we're really going to be concentrating a little bit more on the linear plots here rather than the Bode plots. But Bode plots is a common way of seeing this. It's the same plot, except on different scales. This being a log scale here. Log in, in terms of omega. So, this is on a log scale. And this is on a linear scale where the units are decibels. Now, let's look at the characteristics of a First-Order Filter. A first-order filter has this generic frequency response. This, this transfer function right here. And we showed the same plot as what we, what we had before. And the bandwidth is this value that is 1 over tau, where tau is a multiplier of j omega. The way we get that is if I look at the magnitude of H. It's equal to K of DC, 1 over, and in the magnitude of the denominator. Well the magnitude of the denominator is the square root of this sum that squares of the real part plus the imaginary part. So, the real part is 1, squared is 1 plus, the imaginary part squared, which is Tau Omega squared. And at 0.707, this is equal to 0.707 when Tau is equal to 1 over Omega. So, that's the bandwidth. So for that value, this whole thing is equal to 0.707. Now, the DC gain is a value of H when I let it equal to zero. That's because, again, I want to emphasize again, that's the DC, is at Omega equals zero. So, that's just a generic First-Order Filter. So, going from a passive to an active filter is probably the easiest way to design your first filter. It is to look at a passive filter, which is an RC circuit. This is a, a lowpass filter where my input is right here, and my output is measured across this capacitor. And so this is the generic form for this circuit. If I want to make this into an active filter and I want to isolate at the input, then all I have to do is add a buffer circuit at the input. So the buffer circuit doesn't change this waveform. So this V in signal is the same here, V in. The waveform looks the same, doesn't change in amplitude, but it increases the power here. And it isolates it from the imput. So, if I'm measuring a, a measurement device and I want to, I don't want to change my measured signal, then I just add in the buffer circuit before this. And then I've got the characteristics of a lowpass filter. If I want isolation at the output, then I take my circuit and I add my buffer circuit at the output. In this case, if I were to add, say a resister right here, I'm not going to change the out. So, I can add a whole circuit on here, cascade something else to it, and I won't be changing the out. So, this is just taking a simple passive filter and creat, and making it in to active by adding an up amp. Now an alternate way of creating a lowpass filter is to put the filter into the feedback loop. So putting it in the feedback loop means that I can do two things at once. I can isolate at the input and at the output both with this one, with this one circuit, and the other thing is because I've got two resistors in there, I can change the gain, the DC gain. With the other two, I had a DC gain of one. I might not want a DC gain of one. I might want something else. So, this particular circuit has this transfer function, this part right here is my H. Let's go ahead and, and derive where we get this, this transfer function from. I want to start with this circuit right here. It's a little bit simpler. And we're going to use this, because later on when we do high pass filters, we'll start with this configuration, as well. I'm going to just derive this, and then we'll substitute back in for these impedances up here. So, if I look at this. I want to do a derivation for this. I want to get V out in terms of V in. Okay. So I can do a KVL across here. Looking at that as plus. That's equivalent to having a V in here with respect to ground. And remember that the voltage drop here is zero. So I can get an equation that is V in is equal to Z1 times I. Now, the I that's going up here is the same I that's going in here. Because, remember that there's zero current going into here. Zero current into there. So all the I goes through this way. So that is the same I. So Z1i plus zero, if I do around here, because that's a, a zero full drop there. And the other KVL that I can do is across this way. Over to here, and then down to ground. So, doing that one, I get V in is equal to Z1i plus Zfi plus the out. Well, these two are going to cancel. And what I'm left with is V out is equal to minus Z sub f i. And I can actually solve for i from this equation. I get i is equal to V in over Z1. So I plug that into here, I get V out is equal to minus Zf over Z1 times V in. And so this is, this is simple. I've, I've been able to come up with the input output equation for this circuit right here. Now I want to use that to go up to this circuit. Well, what's ZF in this circuit? ZF is equal to the resistor and parallel with the impedance of this capacitor. So if I write that, I get 1 over, 1 over Rf, and 1 over the impedance of the capacitor is j omega c. So if that's Zf, and then Z1 is equal to R1. So if I make these substitutions into this equation, I derive the transfer function that we needed. Let's look at the frequency characteristics of a lowpass filter. This is a transfer function right here. And if I calculate the magnitude and the angle, I get this. So the magnitude is, I just take the, in the denominator I'm going to take the sum of the squares of the real part and the imaginary part and then take the square root of that. And then the angle, notice that I've got a minus sign right there, so the angle starts out a hun, at 180 degrees. And then I subtract off the angle of the denominator. The DC gain is found by setting the frequency equal to zero. So, H of zero is that DC gain. The bandwidth is the frequency at which I have 0.707 of the DC value. So, this plot right here is the magnitude, and I've got 0.707 of the DC value. Now as the DC value. The magnitude of the DC value. So, this plot, because this is magnitude, this has gotta be equal to, positive value. So I've stripped off the negative sign. And then the frequency at which we have .707 of that DC value. The magnitude of that DC value is the bandwidth. So this range, right here, we call the pass band and this range, the stopband. The passband is frequencies at which I pass through a signal with very little attenuation, and the stopband is when I start attenuating those signals. So, low frequencies are passed through. And high frequencies are attenuated. The angle looks like this, the angle plot, so remember it starts out at 180, and then at high frequencies, this will, will go to 90 degrees. So, what we want to do is an example here. We want to design an inverting lowpass filter to have a DC gain of -2, and a bandwidth of of 500 radians per second. So, let's go back to these formulas. The DC gain is H of 0, which is minus Rf over R1 and that is equal to -2. And the bandwidth, Omega sub B is equal to 1 over RfC. So the bandwidth is equal to 1 over RfC. And that is equal to 500 radians per second. So now I have two equations and I actually have three param, design parameters, Rf, R1 and C. So I can just pick one. So I'll say pick R1 is equal to 1000 ohms. Very common resistor. Let's go ahead and pick that one. So if I solve for that, then Rf would be equal to 2000 ohms, and C would be equal to one microfarad. And that's the design of my circuit, so I would build this, and I would get a filter with this bandwidth and that DC gain. So in summary, a lowpass filter passes low frequency signals and attenuates high frequency signals. We looked at three first-order lowpass configurations. Two of them were made up by just taking a passive lowpass filter and putting a buffer circuit on it, either at the input or at the output to provide is, isolation wherever we want. Now these particular circuits, this one right here, and this one right here, provided a DC gain equal to one. And they were noninverting. So, in this particular case we have an inverting amplifier. It provides isolation at both the input and the output and it also has a DC gain that is not one, so it can actually amplify our signal. Minus Rf over R1. So, in our next lesson we will look at high pass filters. Thank you.
