[MUSIC] Welcome back to linear circuits. This is Dr. Ferri. This lesson is on phasors. Our goal in this module is going to be able to look at a problem like this, where we got a fairly complicated circuit with reactive elements in it, and we've got a. And our goal is to try to find the output to that circuit. Now, we could use differential equations and solve differential equations for that, but it's really messy that way. So our objective here is to find an easier way of solving circuit problems with sinusoid inputs. We're going to start out by introducing phasors to represent sinusoids. Now, the methods that we're going to go through in this module are based on a sound mathematical foundation, but it's very involved to be able to go through that mathematical foundation as to why this method is going to work. So, we're just going to introduce the method and show how to use it in actual applications without going through the rigorous mathematical foundations to show you that it really does work. We're going to be building on some things that you probably had many years ago. And that is complex numbers. Starting out with we're going to define imaginary numbers, square root of minus one. And you've probably seen it as i before, i is equal to the square root of minus one. Well, in electrical engineering we use i for a different reason, we use it for current, so we use j to represent that imaginary number. We also will be looking at complex numbers, where a is the real part and b is the imaginary part. Now, the other thing we have to remind you about is trigonometry. So, if I've got a triangle like this. Then the opposite side is sine and the adjacent is cosine. And that's with a hypotenuse of 1. So if I had a different hypotenuse, then I just scale through. Let me define phasors. The sine wave is giving in this from right there. And I'm going to define the phasor and polar notation as being this. Where V is simply the amplitude and theta is the angle. And I show the phasor, I indicate it to phasor by showing it is a bolded term. Now it is a complex number and I'm going to represent it. I'm going to draw it here as the imaginary versus the real part as a vector, with a magnitude of the vector being V, and the angle here, from here being theta. Now this is polar notation, and I can convert it to rectangular notation where this is the real component of it. That's A and B is the imaginary component. I can convert between rectangular and polar notation by looking at this as a triangle. With a side the opposite side being b and the aj side being a. So I go back to my trigger identities and trick relationships, so converting from Polar to Rectangular, I just use this formula here, that we saw in the previous field and then to go from Rectangular back to Polar I use Pythagorean theorem. And also use the tan relationship between the B and the A. Now, I just showed phasor in terms of voltage signals, but in terms of current signals, it's the same thing. The amplitude of the cosine shows up here and the angle shows up here. Notice that the frequency doesn't show up at all, and that's because it's explicit. Whenever we have a polar notation or rectangular notation in phasors, we have an implicit understanding of what that frequency is, but it doesn't show up explicitly. Here's some examples. This is my voltage signal, so the 10 shows up right here, the angle shows up right there. If I were to plot it, the imaginary versus the real part and looking at the polar notation, I've got an angle of 45, -45 this way with a magnitude of 10. So when I convert that over to rectangular form, I use 10 times the cosine of theta, plus 10 times the sine of theta all times j. Now the next one is a similar signal, it's ten at an angle of 90 degrees. So let's plot that. So 90 degrees this being zero degrees right there. And this being 90 is going to look like that. And in terms of rectangular coordinates, it's the same as saying that's the imaginary axis and that's the real axis. So it's a purely imaginary number. If I look now at a current signal, notice that I have no angle here, so that means an angle of 0. So if I plot, it's going to be straight out this way. So it is exactly along the real axis, and with a magnitude of 10, so it's going to have a value of 10 there. Now notice again that I have different frequencies here, and the frequencies don't show up at all in these phasors because it's implicit. When I look at this value here I have to keep in mind that that frequency was at 500 radions per second. Now in this particular case, I've got a sine rate. In order to represent it as a phasor I have to first convert it to a cosine. And I did it by subtracting off that 90. So this becomes minus 70 degrees. So I get 10 at minus 70, and I use my standard formulas for converting from polar to rectangle, and I get this. Now most calculators will convert from polar to rectangular for you, so you don't have to remember those formulas. Now I want to go back to looking at phasors of real signals. This is back to that oscilloscope trace that we had, and I'm just going to redraw it over here so you can see a little bit better. Again the input. The input is shown in green and the output is shown in blue. So in terms of phasor notation, the amplitude for the input signal is 1. And the angle is 0 because it's a cosine. Right there with no angle, no lag to it. So that's this one. Now the output, the amplitude is .7, and we already looked at this let me remind you. We already looked at this, the theta was equal to minus 45, so that's what this signal looks like in phasor notation. And we'll look at a few other cases that we might see. This is a particular case where we've got the input and then the output. Now let's look at this. Remember that this is one cycle, and that's 360 degrees. The delay here is half that. So, the delay in the output signal is half that. So, it's at 180 degrees. And, because it's to the right, it's -180. And, the amplitude is the same as what we had up here, so it's .7. So, this is .7 and 180 degrees. Now, this one, the delay. Let's see, this is 360, this is 180, this is half that, or 90. So the delay is 90, so I put a minus in there because it's a delay and it lags to the right. And again, the amplitude is 0.7. So, in this particular case, if this is one case, this is 180, that's 90, this is right there. That is half of 90 or 45. Again, because it's to the right, I make it a -45.7. So this is the exact same signal that we're seeing right there. Let me go through a couple more slides really quickly with working with phasors. If I have to add or subtract phasors, I have to use rectangular form. And just showing an example here. If, this is my first voltage and this is my second voltage. I have to have the same frequency in order to add these together. So I want to add these together. Trigonometrically, it would be really hard to do this, it's just messy. But in phasor form it's a little bit easier. I convert this to phasors, 7 at 30, 3 at -60. And then I have to convert them over to rectangular form. And then from rectangular form I just add these together so I add the really parts and I add the imaginary parts to go this now this is a phasor is rectangular form that's a sum of those two, and then I can go back to polar notation right there. So again I have to use rectangular form if I want to add or subtract phasors If I want to multiply phasors, turns out it's easier to use polar form. I don't have to, but it's easier. So for example, if I want to multiply these two phasors together, I multiply their amplitudes, their magnitudes here, and I add their angles. So if I divide two phasors, I divide their magnitudes and subtract Their angles. As an example, let's look at this is v = 5 at 30. I is = 2 at -60. If I want to multiply them, that's 5 times 2 is 10. And that's 30 plus -60 is -30. If I divide them, then it's 5 divided by 2. And then 30- -60 is 90 degrees. To summarize the key concepts that we've gone through in this lesson, we've introduced the idea of phasers. And the phasers are going to help us to solve circuit problems. Now, by convention we represent our sinosoids in this form with the cosine and angle to it. We introduced phasors in both polar notation as well as rectangular notation. We also tried to relate the phasor notation, like right here, to what real signals will look like. Then we showed how to work with phasors by adding and subtracting we said that you have to do in rectangular form. And multiplying/dividing it turns out it's easier to do in polar form. And just as a reminder, whenever we're working with phasors, if you're trying to combine them by adding and subtracting or multiplying/dividing, we have to make sure that we use the same frequency. We can't combine phasors. That come from different frequencies. All right, thank you very much. [MUSIC]
