In this video, I'm going to solve for the transfer function for a filter formed by cascading a first order low pass filter, and a first order high pass filter. We're going to put the transfer function in standard form, and look at the limitations of a filter formed in this way. Let's say we form a filter, by cascading. Two first-order filters. So we have a first-order, low-pass filter that has some transfer function, Hlp of omega. We cascade that with another first-order filter. A first-order high-pass filter, that has a transfer. Function H, HP of omega and this is the output voltage. So the overall filter is formed by cascading these two first water filters. And the overall function of this cascade is the ratio of this VO to this VI. A first order low pass filter has a transfer function. Standard form for the denominator would be s over omega u plus 1. It's a low pass filter, so we have a 1 in the numerator. And I'll call the gain constant k1. This high pass filter has, again. Same form of denominator. But I'll call its minus 3 db frequency omega l. It's a high pass filter so we have the highest order term in the numerator s over omega l, and it'll have some gain constant, say k2. So the overall transfer function V O over V I would be equal to the product of the two transfer functions, H H P of omega time H L P of omega is equal to. K1 K2 1 over s over omega u plus 1 times s over omega l over s over omega l plus 1. And remember that in writing a transfer function like this s is equal to j omega. Now what I want to do is write the transfer function for this cascade in standard form of a second-order filter, and you can see that when we multiply these two transfer functions out, we're going to have a second-order denominator, an S squared term in the denominator. So by mani, manipulating this and putting it into standard form for a second-order filter, remember. For a second-order filter, we can write the denominator like this. S over omega naught squared, plus one over q, s over omega naught plus one. Some gang constant K out front, and then the numerator is composed of one of the terms of the denominator, or some combination of these terms [NOISE] So, let, let me just multiply out this transfer function. So, I can write that Vo over Vi is equal to K1. K 2 [NOISE] and the numerator we still have our S over omega L. And I'm just going to distribute these terms so I have S squared, omega U omega O plus S times one over omega U. Plus 1 over omega L plus 1. And I can write this, this 1 over omega U plus 1 over omega L, I could write that as the sum over the product. So I can write this as K1 K2 s over omega l, here we have an s squared, omega u, omega l, plus s times omega u plus omega l, over omega u omega l plus one. Now remember we want this to be in standard form given by this. So we want the numerator term, this is an S to the first power term, we want this term to be the same as the S to the first power term in the denominator. Which means then I want, instead of a one over omega l here, I want it to be equal to this. So I can re-write the transfer function like this. K one, K two. I'm going to bring the one over omega l out front, and I'm going to replace it by. And S times omega U plus omega L over omega U omega L, and then in the denominator I can write this as S over the squa.re root of omega U. Omega L all squared plus s omega u plus omega L over omega u omega L plus 1. Now, I have this in standard form because these two terms are the same. The problem here is, of course, is that these two transfer functions are not the same. To make it the same I have to cancel out this term by multiplying this by an omega u omega L over omega u plus omega L. So now this transfer function is the same as this transfer function. But it's now written in standard form where we can identify k, q, and omega naught. So let me make just one more adjustment to this I'm just going to bring this term to this side. So I'm going to have, write this as K1. K2, omega U over omega U plus omega L, and then we have our transfer function in standard form. S, omega U plus omega L over omega U. Omega l, s over the square root of omega u and omega l, squared, plus s omega u omega l plus, over omega u times omega l plus 1. And now that it's in standard form, we can identify by inspection a mega nought. Remember a mega nought is what is, what is sitting right here in this position So I can write that a mega nought is equal to the square root, or the geometric mean of. Omega U, the upper cutoff frequency, and the lower cutoff frequency. And then we know that the term sitting here, must be equal to one over Q omega knot. So I can write that one over Q omega know is equal to mega U, plus omega L over omega U omega L. And then I can substitute this value for omega naught into here, so I have that Q is equal to, it would be omega U omega L. Over omega u plus omega L times 1 over the square root of omega u, omega L, is equal to. So one of these square roots would cancel one of these square roots so I can write it as the square root. I'm going to make a U, I make it O, over, I make it U plus I make it O. SO, here is one of our equations, I make a note in terms of I make a U and I make a O, here is another equation, the quality factors in terms of the tape, and then by inspection we can see that the game constant k must be equal to k one, k two times omega u over omega u plus omega l. [NOISE] And the form of this transfer function it's a second order denominator and the numerator consists of the middle term, the s to the first power term, which indicates that this is a bat, band pass filter. So by cascading a first order low pass filter and a first order high pass filter we obtain a second order band pass filter. But there's some restrictions to. Creating a filter in this way. And we can determine that with these equations. Now remember, the overall magnitude of the transfer function would, would look something like this. A low pass filter and a low pass filter cascaded. Where this is omega L, and this is omega U. Now we can see that, if omega U and omega L are far apart, in other words, omega U is much much greater than omega L, then because omega U dominates omega L here, we can neglect this, this becomes 1, and this overall gain here, k, is equal to k1 k2. So if we pick K one and K two equal to each other, we could have a flat band here equal to the product of K one and K two. But, as omega U gets closer to omega L, you can see that the gain decreases. So as omega L moves in this direction, and omega U moves in this direction, we'll start to get filters that look like. Like this, and then as we, the two frequencies get closer together, you can see that the gain, from my picture here, the gain K, overall K drops. Because as, say omega, you were equal to omega at all, then this would give us a factor of one half. K1, k2. Now look at how the quality factor changes as a mega u and mega l become closer together. So if a mega u were exactly equal to a mega l here, so say a mega u equals a mega l, then q would be equal to. Omega U over two omega U is equal to a quality factor of one half. And remember the quality factor for ban pass filter Q is equal to F not over the band width. So if I keep our F not here a constant as I decrease the band width I move omega U and omega L closer and closer together. You can see that the bandwidth decreases, so the quality factor increases as, as omega u moves in this direction and omega l moves in this direction. So it turns out that this is the maximum quality factor that you can get from forming a second order bandpass in this way. So the maximum value of Q is one half. And, as a mega U and a mega L move further apart, the Q decreases from this value, approaching a Q of zero. Right, a center frequency with an infinite bandwidth. So you can form a band pass filter in this way but there are some limitations on the transfer functions that you can implement
