[SOUND] Welcome back to Linear Circuits, this is Dr Ferri. This lesson is on the frequency response for linear plots. Our objective is to introduce the frequency response as a way of showing how a system or circuit processes signals of different frequencies. We're going to be building upon the frequency spectrum that we've already covered, and then also transfer functions. This is something that we covered in the last module. This, for example, is a transfer function of an RC circuit. The frequency response is defined here. In this particular course, we're defining it as the plot, so it's the plot of the magnitude and angle of the transfer function. So I find a transfer function, which is complex valued, and I compute the magnitude and plot it as well as the angle, and plot that versus frequency. Now going back to the RC circuit that I mentioned before, we had computed the transfer function in the last module, and we found it to be equal to this quantity here. If I find the magnitude of this, it's really equal to the magnitude of the denominator, which is the real part squared plus the imaginary part square root of that. And then the angle is equal to this, which is the angle of the numerator is 0. And this is the angle of the denominator with the minus sign, because that's in the denominator. If I plot those two, so I plot the magnitude. So this is H of omega and then this is the angle of H of omega, and we oftentimes just write that as a magnitude and angle versus omega. What I find in this particular plot is it starts out at the value of one at omega equals 0. So at omega equals 0 the magnitude's one and the angles can be 0. And as it falls off, as omega increases, now this is plotted for a particular value of RC, but the general shape is going to be the same. Just how quickly it falls off will depend on RC. Now let's recall what we know about transfer functions since this is a slide that we covered in the last module. We have an input in terms of a phasor into our circuit, which has this transfer function, and this is our phasor output. So in the time domain, it looks like this, a cosine in, a cosine out. Now this output amplitude is found from this function right here, it's the input amplitude times the magnitude of the transfer function at that frequency is equal to the output. And the phase angle out is given from this relationship right here. The input angle plus the angle of the transfer function at that frequency is equal to the output angle. Now let's apply this. Let's take a look at this example, where we're given this frequency response of a circuit, and we want to find the steady state response to this input. So we want to use those relationship between the input and output amplitudes. So in this particular case, we have a DC component, and then we've got a component with omega equals 200. So that's right here and right here. So I follow this up and I get this value, is about, say, 0.4, and then this value is 1. So this is equal to H at 200, the magnitude, and if I look at the angle here, the angle is about, H at 200, is about, say, minus 65 degrees. And up here I've got the value of 1, that's H(0). In particular, H(0), it has no angle, so it's just real valued. So then I go back over here and I look at my output for this case, and I'll write it as V out. Okay, I look at the DC component, I take this DC component and multiply it by this value. So that's 2 times 1 is 2. Plus I look at this component right here, the amplitude is 1, I multiply it by the amplitude here, it's 0.4. Cosine of 200t and then this has a 0 angle, so I just have to worry about this one. So it's minus 65 degrees. And that's the output corresponding to this input to a circuit that has this frequency response. Now let's look at this in a little bit more general view, looking at plots of what the input and output look like. So in this case I've got an input that looks like this. I've got a high frequency superimposed on a low frequency, so I can actually kind of see what the two parts to it too are. That's the low frequency component in the red dashed, and then the high frequency is super imposed on it. So I've got two frequencies going in, and the output of this circuit gives something like this. So in the frequency domain, I've got two frequencies, a low frequency which is this one in red, and then the high frequency. The low frequency is at 50 radians per second, high frequency at 800, both have an amplitude equal to 1. Now I put it through this transfer function and at 50, the value is, say, about 0.9, and at 800 the value is about 0.2. So the frequency spectra of my output is just taking this input frequency spectra multiplying it term by term times corresponding value of H. So this is H of omega, and so this is at 0.9 And this is at 0.2. And so what you can see is the input signal, these two frequencies are pretty much matched in amplitude. But the output, the low frequency dominates, and you could see that the high frequency signal is much smaller, and that corresponds right there to the small value. To summarize the key concepts, we've introduced the idea of a frequency response as being a plot of the transfer function. Because the transfer function is complex valued, we plot the magnitude versus the frequency, as well as the angle versus the frequency. And then we also showed how the transfer function or the frequency response can be used to determine the steady state sinusoidal response of a circuit, two signals at different frequencies. And the key thing here is that we use this relationship. How the input and output amplitudes are related through that transfer function and how the input and output phases are related through that transfer function as well. All right, thank you very much. [SOUND]