[SOUND] Welcome back to Linear Circuits. This is Dr. Ferri. This lesson is on lowpass, highpass filter design. Our objective is to design RC filters. We're going to be building upon the concept of filters that we've introduced in a previous lesson, as well as the building plots that we looked at when we were examining RC and RLC circuits. Looking back at the Bode plot of an RC circuit. This particular circuit we had this frequency response. This is H of omega. In decibels, so it's 20 times the log of that. What we found is that at low frequency, we had a magnitude of zero decibels, and our corner frequency was defined as 1 over RC. And that's the frequency at which we're three decibels below zero, so we're at minus 3 right there, that's the corner frequency. Now let's compare this to what we now know about lowpass filters. So a lowpass filter has these characteristics, it's got the passband region and the passband gain is right here. And the bandwidth is when we're three decibels below the passband region. So if we compare these two, what we find that for a lowpass RC filter the bandwidth is the same as the corner frequency, which is 1 over RC. So if I have to design a lowpass filter, what I can say is that if I want to pass through frequencies below maybe 100 radians per second then solve for RC. The other thing to note is that we have no choice over the pass ban gain, it's going to be 0. Where this is Gdb it's going to be zero decibels. Let's do a similar thing for a high pass filter. Now notice I take my RC filter, instead of putting my output over the capacitor I know take it over the resistor, and I now get this frequency response right here. The magnitude in the high frequency is zero decibels. So this is our, we call it. Maybe our high frequency gain. Our corner frequency is the frequency at which we're three decibels below that high frequency gain. So that's about right here. That's omega sub-c right there. Comparing that to what we know about highpass filters, when we're at three decibels below the highpass gain, that's a corner frequency. So we've got the corner frequency matches, our definition matches between The high pass filter and the RC Bode plot. So it's 1 over RC, is that value right there. And then this is the pass band region. And now the gain, what we see is that I don't have a choice here. The passband gain g is going to be equal to 0 d b, and that's in the passband gain in terms of decibels, which is g c d b right here. Okay, so we have enough now to design a filter. Let me go through an example. Design an RC filter that attenuates frequencies above 200 Hz. Okay, let's stop and think about what sort of filter do we need? We want to attenuate above 200 hertz. So in other words, I want a low pass filter. I want to attenuate above something, and the definitions that we've gone through have been in terms of mega. This is f in hertz, I have to convert that to a mega before I use any of my formulas. So what I'm going to want to do as a bandwidth is equal to 200(2 pi), and that is 1256 radians per second. So I use that in my formulas. So this particular case for a lowpass filter, I have to take the voltage across a capacitor. Omega c is equal to this, for this particular circuit. I equate that to Omega B for lowpass filter, and in this particular case, I know what this is, this is 1256. I have a choice. I can pick R or C. I tend to pick C first, because I have a more limited choice over C typically than I do R, so let me just say let C equal something simple, 1 microfarad. And then I want to solve for R from this formula here. I have that over 2. One over the bandwidth, times C. And if I solve that, I get 796 ohms, and if I'm going to pick one, I would just pick an 800 ohm resistor, is close enough to this. Because remember, resistors always have some tolerances in them, anyway. So this is the design of a filter, the thing was to know. What sort of filter am I going to use. And then to use the formulas in order to get the correct values. Let's look at another example. This is design an RC filter that attenuates below this. Below 50 Hz, that means that I want to attenuate below, I want something that looks like this. So this is a high path filter, this is the circuit that I would build for a high path filter taking the voltage across that resistor. The corner frequency is this so I would convert 50 Hz to radients per second and I get 314, radians per second. I use that in this formula so I end up with R = 1 over omega c times c and let's again pick my c. Let C equals to 1 microferad, and that means that I can solve for R equal to 3184 Ohms, and then this would be the high pass filter. We've also looked at the Bode plots of RLC circuits and we see that we can Define them as being filters as well. So for example, this particular filter, we looked at the frequency response. And it depends on whether it's overdamped or underdamped what it looks like. But in both cases, they look like lowpass filters. So this is a lowpass filter. What is the bandwidth of this thing? Well, I use the same definition that I always do, which is three decibels below the low frequency point. So that's a little bit different here. So it's going to be out here. So this is the bandwidth here. And this is the band width here, and what you can see is this particular circuit is one that the only difference between them is the value or R, so the value of R will depend on will effect what the band width is because of this resonance area. So I don't really have a good formula to give you in terms of what that bandwidth is. I know the corner frequency is one over the square root of LC, so it's somewhere close in the range of what this bandwidth is. But the bandwidth do we really have to go and measure because these plots differ based on what ours is. To summarize, we've looked at some specific circuits. For low pass circuits, we looked at an RC circuit that looks like this where we took the voltage across the capacitor. And we found that the bandwidth was 1 over RC. And for the RLC circuit, again we took the voltage across the capacitor. For a high pass circuit, we looked at the RC circuit over here. Where we took the voltage across the resistor and we found that the quarter frequency was 1/RC. Now this particular circuit's not one that we have examined before, but it is also a highpass circuit. So I took, in this case, C voltage across C inductor. Now one thing that I wanted to point out here is that I showed the input a little bit different that I've shown it elsewhere. Here, its shown as a source, but a lot of times when we implement a filter like this, a circuit as a filter. We connect it to something say a sensor input or something else. It's not specifically a source, so I want to be just a little bit more generic here and say, whatever this voltage is across here, that's going to be my input. So it's the same here. This is a voltage across here. That's my input. Now in all of these cases, the passband gain is 1, and that's something that you'll get with these sorts of circuits. It's not something that is always desirable and partly for that reason, when people implement circuits as filters, they'll use op amps in there. An op amp is something that we haven't introduced so far, but it is what we call an active filter, and it gives us a lot of characteristics, a lot of benefits. For example a passband gain that is not one, that we might want to implement. So if ever you have to build a circuit, you might want to search on the internet for active circuits, active filters, or opand filters. All right, thank you. [SOUND]