[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:04.80,Default,,0000,0000,0000,,>> In this video, we're going to show\Nhow phasors can be used to perform
Dialogue: 0,0:00:04.80,0:00:11.72,Default,,0000,0000,0000,,a certain subset of\Ntrigonometric calculations.
Dialogue: 0,0:00:11.72,0:00:13.89,Default,,0000,0000,0000,,The phasors don't always lend themselves,
Dialogue: 0,0:00:13.89,0:00:17.01,Default,,0000,0000,0000,,we can't always replace trig with phasors.
Dialogue: 0,0:00:17.01,0:00:19.23,Default,,0000,0000,0000,,But under certain circumstances we can,
Dialogue: 0,0:00:19.23,0:00:22.17,Default,,0000,0000,0000,,and it turns out that\Nthose circumstances are very common
Dialogue: 0,0:00:22.17,0:00:25.96,Default,,0000,0000,0000,,in dealing with sinusoidal\Nsteady state circuits.
Dialogue: 0,0:00:25.96,0:00:29.07,Default,,0000,0000,0000,,So I have defined two different voltages,
Dialogue: 0,0:00:29.07,0:00:35.85,Default,,0000,0000,0000,,V1 is equal to 5 times the\Ncosine of 300t plus 50 degrees.
Dialogue: 0,0:00:35.85,0:00:40.96,Default,,0000,0000,0000,,A second one V2, three cosine\Nof 300t plus 25 degrees.
Dialogue: 0,0:00:40.96,0:00:45.23,Default,,0000,0000,0000,,The first thing we note is\Nthat the frequencies of
Dialogue: 0,0:00:45.23,0:00:47.21,Default,,0000,0000,0000,,the sinusoids have got to\Nbe the same in order for
Dialogue: 0,0:00:47.21,0:00:50.47,Default,,0000,0000,0000,,the phasor approach to work.
Dialogue: 0,0:00:50.47,0:00:53.50,Default,,0000,0000,0000,,But given in, that in RLC circuits,
Dialogue: 0,0:00:53.50,0:00:55.57,Default,,0000,0000,0000,,driven by a single frequency,
Dialogue: 0,0:00:55.57,0:00:57.35,Default,,0000,0000,0000,,all the voltages and currents in
Dialogue: 0,0:00:57.35,0:01:00.22,Default,,0000,0000,0000,,that circuit will be oscillating\Nat that same frequency.
Dialogue: 0,0:01:00.22,0:01:03.88,Default,,0000,0000,0000,,So this limitation that\Nthe frequencies have to be the same,
Dialogue: 0,0:01:03.88,0:01:07.14,Default,,0000,0000,0000,,isn't a big deal for us in\Nwhat we're going to be doing.
Dialogue: 0,0:01:07.14,0:01:10.55,Default,,0000,0000,0000,,Each of these is oscillating\Nat different frequencies,
Dialogue: 0,0:01:10.55,0:01:13.59,Default,,0000,0000,0000,,at the same frequency but\Nwith different phase offsets.
Dialogue: 0,0:01:13.59,0:01:15.44,Default,,0000,0000,0000,,What I wanted to do here,\Nis show you how we can
Dialogue: 0,0:01:15.44,0:01:17.24,Default,,0000,0000,0000,,use phasors to add and subtract,
Dialogue: 0,0:01:17.24,0:01:20.58,Default,,0000,0000,0000,,and multiply and divide\Nthese trigonometric functions.
Dialogue: 0,0:01:20.58,0:01:25.34,Default,,0000,0000,0000,,To do so, we're going to be using\Nthe phasor representations of these.
Dialogue: 0,0:01:25.34,0:01:29.21,Default,,0000,0000,0000,,So I've gone ahead and define\Nthose first, define those here.
Dialogue: 0,0:01:29.21,0:01:34.74,Default,,0000,0000,0000,,The phasor representation\Nof V1 is in polar form,
Dialogue: 0,0:01:34.74,0:01:37.66,Default,,0000,0000,0000,,is equal to 5e to the j50,
Dialogue: 0,0:01:37.66,0:01:39.76,Default,,0000,0000,0000,,five is the amplitude,
Dialogue: 0,0:01:39.76,0:01:42.50,Default,,0000,0000,0000,,50 is the angle.
Dialogue: 0,0:01:42.50,0:01:44.34,Default,,0000,0000,0000,,To get it to rectangular coordinates,
Dialogue: 0,0:01:44.34,0:01:49.08,Default,,0000,0000,0000,,it's just 5 times the cosine of\N50 plus 5 times j sine of 50.
Dialogue: 0,0:01:49.08,0:01:53.06,Default,,0000,0000,0000,,Or 3.21 plus j3.82.
Dialogue: 0,0:01:53.06,0:01:59.18,Default,,0000,0000,0000,,Similarly, the phasor representation\Nof this time function.
Dialogue: 0,0:01:59.18,0:02:01.34,Default,,0000,0000,0000,,The amplitude is three,
Dialogue: 0,0:02:01.34,0:02:03.51,Default,,0000,0000,0000,,the phase angle is 25,
Dialogue: 0,0:02:03.51,0:02:06.58,Default,,0000,0000,0000,,so we have 3e to the j25 in its polar form,
Dialogue: 0,0:02:06.58,0:02:08.44,Default,,0000,0000,0000,,or in rectangular coordinates,
Dialogue: 0,0:02:08.44,0:02:14.70,Default,,0000,0000,0000,,this phasor would be 2.72 plus j1.27.
Dialogue: 0,0:02:14.70,0:02:19.62,Default,,0000,0000,0000,,So phasors are the complex representations\Nof the trigonometric functions.
Dialogue: 0,0:02:19.62,0:02:22.19,Default,,0000,0000,0000,,Phasors can be written in\Neither their polar form,
Dialogue: 0,0:02:22.19,0:02:23.62,Default,,0000,0000,0000,,or the rectangular form.
Dialogue: 0,0:02:23.62,0:02:29.36,Default,,0000,0000,0000,,It turns out that when we're\Nadding two sinusoids together,
Dialogue: 0,0:02:29.36,0:02:32.10,Default,,0000,0000,0000,,it's easier because we're going\Nto be adding their phasors.
Dialogue: 0,0:02:32.10,0:02:35.27,Default,,0000,0000,0000,,It's easier to add with\Nrectangular coordinates.
Dialogue: 0,0:02:35.27,0:02:39.29,Default,,0000,0000,0000,,Because when you're adding, you're\Nadding two complex numbers,
Dialogue: 0,0:02:39.29,0:02:41.42,Default,,0000,0000,0000,,it's just the real part plus the real part,
Dialogue: 0,0:02:41.42,0:02:43.76,Default,,0000,0000,0000,,and the imaginary part plus imaginary part.
Dialogue: 0,0:02:43.76,0:02:53.19,Default,,0000,0000,0000,,So then we can say\Nphasor V1 plus phasor V2,
Dialogue: 0,0:02:53.19,0:02:58.80,Default,,0000,0000,0000,,we'll add the phasors and then switch\Nthe result back to the time domain.
Dialogue: 0,0:02:58.80,0:03:09.16,Default,,0000,0000,0000,,Phasor V1 plus phasor V2 is\Nequal to 3.21 plus j3.83.
Dialogue: 0,0:03:09.16,0:03:11.85,Default,,0000,0000,0000,,That's phasor V1 in\Nrectangular coordinates,
Dialogue: 0,0:03:11.85,0:03:20.76,Default,,0000,0000,0000,,plus phasor V2 in rectangular\Ncoordinates is 2.72 plus j1.27.
Dialogue: 0,0:03:20.76,0:03:22.64,Default,,0000,0000,0000,,You add those up and you
Dialogue: 0,0:03:22.64,0:03:33.96,Default,,0000,0000,0000,,get 5.93 plus j5.10.
Dialogue: 0,0:03:33.96,0:03:35.68,Default,,0000,0000,0000,,Now, let's just convert those back to
Dialogue: 0,0:03:35.68,0:03:38.53,Default,,0000,0000,0000,,polar coordinates because\Nin polar coordinates,
Dialogue: 0,0:03:38.53,0:03:41.95,Default,,0000,0000,0000,,it is easiest to see what the magnitude and
Dialogue: 0,0:03:41.95,0:03:46.50,Default,,0000,0000,0000,,the phasors that we use then to\Nrewrite it in its time domain form.
Dialogue: 0,0:03:46.50,0:03:49.32,Default,,0000,0000,0000,,So in polar coordinates,
Dialogue: 0,0:03:49.32,0:03:52.46,Default,,0000,0000,0000,,this is then equal to 7.82e
Dialogue: 0,0:03:52.46,0:03:59.91,Default,,0000,0000,0000,,to the j40.67 degrees.
Dialogue: 0,0:03:59.91,0:04:02.46,Default,,0000,0000,0000,,I'd suggest that you stop the video now.
Dialogue: 0,0:04:02.46,0:04:03.97,Default,,0000,0000,0000,,Just go ahead make that conversion from
Dialogue: 0,0:04:03.97,0:04:06.74,Default,,0000,0000,0000,,the rectangle coordinates back to this.
Dialogue: 0,0:04:06.74,0:04:09.58,Default,,0000,0000,0000,,Now, once we've got it in this form,
Dialogue: 0,0:04:09.58,0:04:13.06,Default,,0000,0000,0000,,it's very easy to write\Nthe time-domain version.
Dialogue: 0,0:04:13.06,0:04:19.82,Default,,0000,0000,0000,,V1 of t plus V2 of t is equal to,
Dialogue: 0,0:04:19.82,0:04:25.57,Default,,0000,0000,0000,,the amplitude will be\N7.82 times the cosine.
Dialogue: 0,0:04:25.57,0:04:27.55,Default,,0000,0000,0000,,It's going to be oscillating\Nat the same frequency,
Dialogue: 0,0:04:27.55,0:04:34.40,Default,,0000,0000,0000,,300t plus 40.67 degrees.
Dialogue: 0,0:04:34.40,0:04:40.63,Default,,0000,0000,0000,,I'll leave it to you to attempt to add\Nthese two sinusoids in the time domain.
Dialogue: 0,0:04:40.63,0:04:43.63,Default,,0000,0000,0000,,It involves a number of\Ntrigonometric identities.
Dialogue: 0,0:04:43.63,0:04:47.33,Default,,0000,0000,0000,,It's not undoable, but it is\Ncomputationally intensive.
Dialogue: 0,0:04:47.33,0:04:50.05,Default,,0000,0000,0000,,So when we're adding,
Dialogue: 0,0:04:51.23,0:04:56.42,Default,,0000,0000,0000,,we add the phasors in\Nrectangular coordinates.
Dialogue: 0,0:04:56.42,0:04:59.12,Default,,0000,0000,0000,,Now, let's look at multiplying.
Dialogue: 0,0:04:59.74,0:05:03.82,Default,,0000,0000,0000,,We want to multiply V1 of t times V2 of
Dialogue: 0,0:05:03.82,0:05:07.64,Default,,0000,0000,0000,,t. We're going to do\Nthat using the phasors.
Dialogue: 0,0:05:07.64,0:05:14.34,Default,,0000,0000,0000,,So we'll multiply phasor\NV1 times phasor V2,
Dialogue: 0,0:05:14.34,0:05:23.80,Default,,0000,0000,0000,,which is 5e to the j50 times 3e to the j25.
Dialogue: 0,0:05:23.80,0:05:26.89,Default,,0000,0000,0000,,Well, when multiplying things like\Nthis is just the coefficients,
Dialogue: 0,0:05:26.89,0:05:29.49,Default,,0000,0000,0000,,5 times 3 is 15.
Dialogue: 0,0:05:29.49,0:05:36.37,Default,,0000,0000,0000,,Then we add the exponents e to the J75,
Dialogue: 0,0:05:36.55,0:05:39.50,Default,,0000,0000,0000,,that in it's already in its polar form,
Dialogue: 0,0:05:39.50,0:05:42.23,Default,,0000,0000,0000,,we can then see that the product of
Dialogue: 0,0:05:42.23,0:05:45.65,Default,,0000,0000,0000,,V1 times V2 is going to\Nhave an amplitude of 15.
Dialogue: 0,0:05:45.65,0:05:52.76,Default,,0000,0000,0000,,So touching this back to\NV1 of t times V2 of t,
Dialogue: 0,0:05:52.76,0:05:58.33,Default,,0000,0000,0000,,that then is equal to 15 Cosine of
Dialogue: 0,0:05:58.33,0:06:07.32,Default,,0000,0000,0000,,300 t plus the angle of 75 degrees.
Dialogue: 0,0:06:07.32,0:06:09.82,Default,,0000,0000,0000,,You'll notice that in both these examples,
Dialogue: 0,0:06:09.82,0:06:12.90,Default,,0000,0000,0000,,both of the time domain functions\Nwere in terms of cosines.
Dialogue: 0,0:06:12.90,0:06:14.21,Default,,0000,0000,0000,,When we're using phasors,
Dialogue: 0,0:06:14.21,0:06:16.25,Default,,0000,0000,0000,,because we're talking about the real part,
Dialogue: 0,0:06:16.25,0:06:20.18,Default,,0000,0000,0000,,or using the real part of\Nthe complex exponential,
Dialogue: 0,0:06:20.18,0:06:23.52,Default,,0000,0000,0000,,the real part is related to the cosine.
Dialogue: 0,0:06:23.52,0:06:26.54,Default,,0000,0000,0000,,So underlying all this\Nis the assumption that
Dialogue: 0,0:06:26.54,0:06:30.24,Default,,0000,0000,0000,,the time domain functions that we're\Ngoing to be representing as phasors,
Dialogue: 0,0:06:30.24,0:06:32.92,Default,,0000,0000,0000,,are going to be written\Nin their cosine form.
Dialogue: 0,0:06:32.92,0:06:36.63,Default,,0000,0000,0000,,Here I've got V3, it's\Nwritten in its sine form,
Dialogue: 0,0:06:36.63,0:06:39.44,Default,,0000,0000,0000,,so before we can use it\Nor convert it to phasors,
Dialogue: 0,0:06:39.44,0:06:42.32,Default,,0000,0000,0000,,we need to translate this,
Dialogue: 0,0:06:42.32,0:06:46.13,Default,,0000,0000,0000,,the sine term into a cosine term.
Dialogue: 0,0:06:46.13,0:06:49.14,Default,,0000,0000,0000,,To do that, we're going to just look at,
Dialogue: 0,0:06:49.14,0:06:53.06,Default,,0000,0000,0000,,and remind ourselves the relationship\Nbetween the sine and cosine.
Dialogue: 0,0:06:53.06,0:06:56.86,Default,,0000,0000,0000,,The cosine waveform looks\Nsomething like this,
Dialogue: 0,0:06:56.86,0:07:03.02,Default,,0000,0000,0000,,and the sine waveform\Nlooks something like this.
Dialogue: 0,0:07:03.02,0:07:05.51,Default,,0000,0000,0000,,If it was drawn better\Nyou'd be able to see,
Dialogue: 0,0:07:05.51,0:07:14.68,Default,,0000,0000,0000,,but hopefully this is good enough to\Nobserve that the sine of theta here,
Dialogue: 0,0:07:14.68,0:07:17.52,Default,,0000,0000,0000,,is simply the cosine wave form,
Dialogue: 0,0:07:17.52,0:07:20.42,Default,,0000,0000,0000,,delayed by 90 degrees.
Dialogue: 0,0:07:20.42,0:07:21.92,Default,,0000,0000,0000,,Or the sine of theta,
Dialogue: 0,0:07:21.92,0:07:28.90,Default,,0000,0000,0000,,is equal to the cosine of\Ntheta minus 90 degrees.
Dialogue: 0,0:07:28.90,0:07:35.09,Default,,0000,0000,0000,,So to change this or to transform\Nthis into a cosine term,
Dialogue: 0,0:07:35.09,0:07:40.55,Default,,0000,0000,0000,,we can simply say then that V3 of\Nt is equal to 5 times the cosine
Dialogue: 0,0:07:40.55,0:07:49.01,Default,,0000,0000,0000,,of 300t minus 10, minus 90 degrees.
Dialogue: 0,0:07:49.01,0:07:51.35,Default,,0000,0000,0000,,We take the original sine term,
Dialogue: 0,0:07:51.35,0:07:53.08,Default,,0000,0000,0000,,delay it by 90 degrees,
Dialogue: 0,0:07:53.08,0:07:55.76,Default,,0000,0000,0000,,and that gives us the cosine term.
Dialogue: 0,0:07:55.76,0:08:00.80,Default,,0000,0000,0000,,We can now convert it to phasor V3,
Dialogue: 0,0:08:00.80,0:08:11.16,Default,,0000,0000,0000,,phasor V3 would equal 5e to\Nthe j -10 -90 is -100.
Dialogue: 0,0:08:11.99,0:08:16.35,Default,,0000,0000,0000,,Its polar form would then\Nbe 5e to the minus j100.
Dialogue: 0,0:08:16.35,0:08:20.76,Default,,0000,0000,0000,,In rectangular coordinates, that then\Nwould be just to, for completeness.
Dialogue: 0,0:08:20.76,0:08:29.46,Default,,0000,0000,0000,,That would be negative 0.868 plus j4.92.