0:46

What we found is that at low frequency, we had a magnitude of zero decibels, and

Â our corner frequency was defined as 1 over RC.

Â And that's the frequency at which we're three decibels below zero,

Â so we're at minus 3 right there, that's the corner frequency.

Â 1:05

Now let's compare this to what we now know about lowpass filters.

Â So a lowpass filter has these characteristics,

Â it's got the passband region and the passband gain is right here.

Â And the bandwidth is when we're three decibels below the passband region.

Â So if we compare these two, what we find that for a lowpass RC filter

Â the bandwidth is the same as the corner frequency, which is 1 over RC.

Â So if I have to design a lowpass filter, what I can say is that if I want to pass

Â through frequencies below maybe 100 radians per second then solve for RC.

Â The other thing to note is that we have no choice over the pass ban gain,

Â it's going to be 0.

Â 2:09

Now notice I take my RC filter, instead of putting my output over the capacitor I

Â know take it over the resistor, and I now get this frequency response right here.

Â The magnitude in the high frequency is zero decibels.

Â So this is our, we call it.

Â 3:09

And then this is the pass band region.

Â And now the gain, what we see is that I don't have a choice here.

Â The passband gain g is

Â going to be equal to 0 d b,

Â and that's in the passband gain in terms of decibels,

Â which is g c d b right here.

Â 3:44

Let me go through an example.

Â Design an RC filter that attenuates frequencies above 200 Hz.

Â Okay, let's stop and think about what sort of filter do we need?

Â We want to attenuate above 200 hertz.

Â So in other words, I want a low pass filter.

Â 4:05

I want to attenuate above something, and

Â the definitions that we've gone through have been in terms of mega.

Â This is f in hertz, I have to convert that to a mega before I use any of my formulas.

Â So what I'm going to want to do as a bandwidth is equal to

Â 200(2 pi), and

Â that is 1256 radians per second.

Â So I use that in my formulas.

Â 4:39

So this particular case for

Â a lowpass filter, I have to take the voltage across a capacitor.

Â Omega c is equal to this, for this particular circuit.

Â I equate that to Omega B for lowpass filter, and in this particular case,

Â I know what this is, this is 1256.

Â I have a choice.

Â I can pick R or C.

Â I tend to pick C first,

Â because I have a more limited choice over C typically than I do R, so

Â let me just say let C equal something simple, 1 microfarad.

Â And then I want to solve for R from this formula here.

Â I have that over 2.

Â One over the bandwidth, times C.

Â And if I solve that, I get 796 ohms, and if I'm going to pick one,

Â I would just pick an 800 ohm resistor, is close enough to this.

Â Because remember, resistors always have some tolerances in them, anyway.

Â So this is the design of a filter, the thing was to know.

Â What sort of filter am I going to use.

Â And then to use the formulas in order to get the correct values.

Â 5:51

Let's look at another example.

Â This is design an RC filter that attenuates below this.

Â Below 50 Hz,

Â that means that I want to attenuate below, I want something that looks like this.

Â So this is a high path filter, this is the circuit that I would build for

Â a high path filter taking the voltage across that resistor.

Â The corner frequency is this so I would convert 50 Hz

Â 6:28

I use that in this formula so I end up with R = 1 over

Â omega c times c and let's again pick my c.

Â Let C equals to 1 microferad, and

Â that means that I can solve for R equal to 3184 Ohms,

Â and then this would be the high pass filter.

Â 6:56

We've also looked at the Bode plots of RLC circuits and

Â we see that we can Define them as being filters as well.

Â So for example, this particular filter, we looked at the frequency response.

Â And it depends on whether it's overdamped or underdamped what it looks like.

Â But in both cases, they look like lowpass filters.

Â 7:24

What is the bandwidth of this thing?

Â Well, I use the same definition that I always do, which is three decibels

Â below the low frequency point.

Â So that's a little bit different here.

Â So it's going to be out here.

Â 7:45

And this is the band width here, and what you can see is this particular circuit is

Â one that the only difference between them is the value or R, so the value of R

Â will depend on will effect what the band width is because of this resonance area.

Â So I don't really have a good formula to give you in terms of what that

Â bandwidth is.

Â I know the corner frequency is one over the square root of LC, so

Â it's somewhere close in the range of what this bandwidth is.

Â But the bandwidth do we really have to go and

Â measure because these plots differ based on what ours is.

Â To summarize, we've looked at some specific circuits.

Â For low pass circuits, we looked at an RC circuit that looks like this where we took

Â the voltage across the capacitor.

Â And we found that the bandwidth was 1 over RC.

Â And for the RLC circuit, again we took the voltage across the capacitor.

Â 8:37

For a high pass circuit, we looked at the RC circuit over here.

Â Where we took the voltage across the resistor and

Â we found that the quarter frequency was 1/RC.

Â Now this particular circuit's not one that we have examined before, but

Â it is also a highpass circuit.

Â So I took, in this case, C voltage across C inductor.

Â Now one thing that I wanted to point out here is that I showed the input a little

Â bit different that I've shown it elsewhere.

Â Here, its shown as a source, but

Â a lot of times when we implement a filter like this, a circuit as a filter.

Â We connect it to something say a sensor input or something else.

Â It's not specifically a source, so I want to be just a little bit more generic

Â here and say, whatever this voltage is across here, that's going to be my input.

Â So it's the same here.

Â This is a voltage across here.

Â That's my input.

Â Now in all of these cases, the passband gain is 1,

Â and that's something that you'll get with these sorts of circuits.

Â It's not something that is always desirable and partly for that reason,

Â when people implement circuits as filters, they'll use op amps in there.

Â An op amp is something that we haven't introduced so far, but it is what we call

Â an active filter, and it gives us a lot of characteristics, a lot of benefits.

Â For example a passband gain that is not one, that we might want to implement.

Â So if ever you have to build a circuit, you might want to search

Â on the internet for active circuits, active filters, or opand filters.

Â All right, thank you.

Â [SOUND]

Â