Welcome back to electronics. This is Dr. Ferri. In this lesson, we will do a review of transfer functions. In our previous lesson, we did a preview of impedances. Now, impedances is a basic type of component that we're going to need when we do transfer functions. So in this lesson, we will review transfer functions and we show how they are used to characterize a circuit. And this will be leading into our next lesson which will be to look at frequency response curves. Let's define a transfer function in terms of two-port networks. By two-port networks, that means I've got an input and an output. So my input is a voltage here and my output is a voltage there. And in this case I'm looking at the input to be a sinusoidal input. And I'm looking at the output in steady state, and it's also sinusoidal. And notice that it's at the same frequency, but it has a different amplitude and a different phase. For shorthand notation, we draw what we call, or define what we call a phaser, which is just a shorthand notation to show the input amplitude, or we can call it the magnitude, and the angle. And the output phaser is the output amplitude and the corresponding phase. And the transfer function just shows what happens in the circuit. So, how does the circuit handle the input to give you that output? And we define the transfer function as being whatever impedance function that we have. In our last lesson, we showed impedance functions that, when we came up with some sort of function of omega, multiplying by my input to give me my output. That's what we did when we did circuit analysis in our last lesson. And what we're saying here is whatever multiplied by that input to give me the output That's the transfer function. So in terms of phasers, I just substitute in for the phaser here, this whole thing is a phaser. And over here, the output, everything here is the phaser form of the output. Now if I do that, I can take the magnitude of both sides, and the angle of both sides. So, if I take the magnitude of both sides of this. I would get the magnitude of H times A is equal to, that's A in is equal to A out. So, I've just written it this way in terms of a function A out is equals to something. I'm trying to solve for A out. The output amplitude and that equal to the input amplitude times the magnitude that transfers function at that frequency omega. So it's important this is a given omega so at that frequency omega. And the phase angle right here on the output is equal to the angle of H, plus at that same frequency, plus the angle of the input. And that was gotten just by taking the angle of both sides, that's angle of H plus the angle of theta is equal to the angle of the output. Summary of Simple Circuits. We analyzed two of these circuits in our last lesson on impedances. And so the two circuits that we looked at was this one right here and this one right here. And what we had was our output what we solved using voltage divider laws. Our output was equal to 1 over 1+RC j omega times V in. And we define the transfer function as anything that multiplies my input to give me my output. So that's how we came up with this transfer function. And if instead I took the voltage across a resistor rather than the capacitor I can do the same sort of analysis and get this transfer function. And in the last lesson we actually did solve this transfer function right here. So there is a summary of simple circuits and their transfer functions. So in summary, we defined a transfer function for two-port networks and we showed some simple transfer functions of simple circuits. In our next lesson we will use the transfer function representation to show how to come up with frequency responses. Thank you.
