0:00:00.000,0:00:04.800
>> In this video, we're going to show[br]how phasors can be used to perform

0:00:04.800,0:00:11.715
a certain subset of[br]trigonometric calculations.

0:00:11.715,0:00:13.890
The phasors don't always lend themselves,

0:00:13.890,0:00:17.010
we can't always replace trig with phasors.

0:00:17.010,0:00:19.230
But under certain circumstances we can,

0:00:19.230,0:00:22.170
and it turns out that[br]those circumstances are very common

0:00:22.170,0:00:25.965
in dealing with sinusoidal[br]steady state circuits.

0:00:25.965,0:00:29.070
So I have defined two different voltages,

0:00:29.070,0:00:35.850
V1 is equal to 5 times the[br]cosine of 300t plus 50 degrees.

0:00:35.850,0:00:40.965
A second one V2, three cosine[br]of 300t plus 25 degrees.

0:00:40.965,0:00:45.230
The first thing we note is[br]that the frequencies of

0:00:45.230,0:00:47.210
the sinusoids have got to[br]be the same in order for

0:00:47.210,0:00:50.470
the phasor approach to work.

0:00:50.470,0:00:53.505
But given in, that in RLC circuits,

0:00:53.505,0:00:55.570
driven by a single frequency,

0:00:55.570,0:00:57.350
all the voltages and currents in

0:00:57.350,0:01:00.215
that circuit will be oscillating[br]at that same frequency.

0:01:00.215,0:01:03.875
So this limitation that[br]the frequencies have to be the same,

0:01:03.875,0:01:07.145
isn't a big deal for us in[br]what we're going to be doing.

0:01:07.145,0:01:10.550
Each of these is oscillating[br]at different frequencies,

0:01:10.550,0:01:13.590
at the same frequency but[br]with different phase offsets.

0:01:13.590,0:01:15.440
What I wanted to do here,[br]is show you how we can

0:01:15.440,0:01:17.239
use phasors to add and subtract,

0:01:17.239,0:01:20.575
and multiply and divide[br]these trigonometric functions.

0:01:20.575,0:01:25.340
To do so, we're going to be using[br]the phasor representations of these.

0:01:25.340,0:01:29.210
So I've gone ahead and define[br]those first, define those here.

0:01:29.210,0:01:34.735
The phasor representation[br]of V1 is in polar form,

0:01:34.735,0:01:37.665
is equal to 5e to the j50,

0:01:37.665,0:01:39.765
five is the amplitude,

0:01:39.765,0:01:42.495
50 is the angle.

0:01:42.495,0:01:44.340
To get it to rectangular coordinates,

0:01:44.340,0:01:49.080
it's just 5 times the cosine of[br]50 plus 5 times j sine of 50.

0:01:49.080,0:01:53.055
Or 3.21 plus j3.82.

0:01:53.055,0:01:59.180
Similarly, the phasor representation[br]of this time function.

0:01:59.180,0:02:01.345
The amplitude is three,

0:02:01.345,0:02:03.510
the phase angle is 25,

0:02:03.510,0:02:06.585
so we have 3e to the j25 in its polar form,

0:02:06.585,0:02:08.445
or in rectangular coordinates,

0:02:08.445,0:02:14.695
this phasor would be 2.72 plus j1.27.

0:02:14.695,0:02:19.615
So phasors are the complex representations[br]of the trigonometric functions.

0:02:19.615,0:02:22.189
Phasors can be written in[br]either their polar form,

0:02:22.189,0:02:23.615
or the rectangular form.

0:02:23.615,0:02:29.355
It turns out that when we're[br]adding two sinusoids together,

0:02:29.355,0:02:32.105
it's easier because we're going[br]to be adding their phasors.

0:02:32.105,0:02:35.270
It's easier to add with[br]rectangular coordinates.

0:02:35.270,0:02:39.290
Because when you're adding, you're[br]adding two complex numbers,

0:02:39.290,0:02:41.425
it's just the real part plus the real part,

0:02:41.425,0:02:43.755
and the imaginary part plus imaginary part.

0:02:43.755,0:02:53.190
So then we can say[br]phasor V1 plus phasor V2,

0:02:53.190,0:02:58.805
we'll add the phasors and then switch[br]the result back to the time domain.

0:02:58.805,0:03:09.160
Phasor V1 plus phasor V2 is[br]equal to 3.21 plus j3.83.

0:03:09.160,0:03:11.850
That's phasor V1 in[br]rectangular coordinates,

0:03:11.850,0:03:20.765
plus phasor V2 in rectangular[br]coordinates is 2.72 plus j1.27.

0:03:20.765,0:03:22.640
You add those up and you

0:03:22.640,0:03:33.960
get 5.93 plus j5.10.

0:03:33.960,0:03:35.680
Now, let's just convert those back to

0:03:35.680,0:03:38.529
polar coordinates because[br]in polar coordinates,

0:03:38.529,0:03:41.950
it is easiest to see what the magnitude and

0:03:41.950,0:03:46.495
the phasors that we use then to[br]rewrite it in its time domain form.

0:03:46.495,0:03:49.315
So in polar coordinates,

0:03:49.315,0:03:52.460
this is then equal to 7.82e

0:03:52.460,0:03:59.910
to the j40.67 degrees.

0:03:59.910,0:04:02.455
I'd suggest that you stop the video now.

0:04:02.455,0:04:03.970
Just go ahead make that conversion from

0:04:03.970,0:04:06.740
the rectangle coordinates back to this.

0:04:06.740,0:04:09.575
Now, once we've got it in this form,

0:04:09.575,0:04:13.060
it's very easy to write[br]the time-domain version.

0:04:13.060,0:04:19.825
V1 of t plus V2 of t is equal to,

0:04:19.825,0:04:25.570
the amplitude will be[br]7.82 times the cosine.

0:04:25.570,0:04:27.550
It's going to be oscillating[br]at the same frequency,

0:04:27.550,0:04:34.405
300t plus 40.67 degrees.

0:04:34.405,0:04:40.630
I'll leave it to you to attempt to add[br]these two sinusoids in the time domain.

0:04:40.630,0:04:43.630
It involves a number of[br]trigonometric identities.

0:04:43.630,0:04:47.330
It's not undoable, but it is[br]computationally intensive.

0:04:47.330,0:04:50.050
So when we're adding,

0:04:51.230,0:04:56.425
we add the phasors in[br]rectangular coordinates.

0:04:56.425,0:04:59.120
Now, let's look at multiplying.

0:04:59.740,0:05:03.820
We want to multiply V1 of t times V2 of

0:05:03.820,0:05:07.645
t. We're going to do[br]that using the phasors.

0:05:07.645,0:05:14.345
So we'll multiply phasor[br]V1 times phasor V2,

0:05:14.345,0:05:23.805
which is 5e to the j50 times 3e to the j25.

0:05:23.805,0:05:26.889
Well, when multiplying things like[br]this is just the coefficients,

0:05:26.889,0:05:29.490
5 times 3 is 15.

0:05:29.490,0:05:36.370
Then we add the exponents e to the J75,

0:05:36.550,0:05:39.500
that in it's already in its polar form,

0:05:39.500,0:05:42.230
we can then see that the product of

0:05:42.230,0:05:45.650
V1 times V2 is going to[br]have an amplitude of 15.

0:05:45.650,0:05:52.760
So touching this back to[br]V1 of t times V2 of t,

0:05:52.760,0:05:58.334
that then is equal to 15 Cosine of

0:05:58.334,0:06:07.320
300 t plus the angle of 75 degrees.

0:06:07.320,0:06:09.824
You'll notice that in both these examples,

0:06:09.824,0:06:12.905
both of the time domain functions[br]were in terms of cosines.

0:06:12.905,0:06:14.210
When we're using phasors,

0:06:14.210,0:06:16.250
because we're talking about the real part,

0:06:16.250,0:06:20.180
or using the real part of[br]the complex exponential,

0:06:20.180,0:06:23.525
the real part is related to the cosine.

0:06:23.525,0:06:26.540
So underlying all this[br]is the assumption that

0:06:26.540,0:06:30.245
the time domain functions that we're[br]going to be representing as phasors,

0:06:30.245,0:06:32.915
are going to be written[br]in their cosine form.

0:06:32.915,0:06:36.630
Here I've got V3, it's[br]written in its sine form,

0:06:36.630,0:06:39.440
so before we can use it[br]or convert it to phasors,

0:06:39.440,0:06:42.320
we need to translate this,

0:06:42.320,0:06:46.130
the sine term into a cosine term.

0:06:46.130,0:06:49.140
To do that, we're going to just look at,

0:06:49.140,0:06:53.060
and remind ourselves the relationship[br]between the sine and cosine.

0:06:53.060,0:06:56.855
The cosine waveform looks[br]something like this,

0:06:56.855,0:07:03.025
and the sine waveform[br]looks something like this.

0:07:03.025,0:07:05.510
If it was drawn better[br]you'd be able to see,

0:07:05.510,0:07:14.675
but hopefully this is good enough to[br]observe that the sine of theta here,

0:07:14.675,0:07:17.515
is simply the cosine wave form,

0:07:17.515,0:07:20.415
delayed by 90 degrees.

0:07:20.415,0:07:21.925
Or the sine of theta,

0:07:21.925,0:07:28.895
is equal to the cosine of[br]theta minus 90 degrees.

0:07:28.895,0:07:35.090
So to change this or to transform[br]this into a cosine term,

0:07:35.090,0:07:40.550
we can simply say then that V3 of[br]t is equal to 5 times the cosine

0:07:40.550,0:07:49.010
of 300t minus 10, minus 90 degrees.

0:07:49.010,0:07:51.350
We take the original sine term,

0:07:51.350,0:07:53.075
delay it by 90 degrees,

0:07:53.075,0:07:55.760
and that gives us the cosine term.

0:07:55.760,0:08:00.800
We can now convert it to phasor V3,

0:08:00.800,0:08:11.160
phasor V3 would equal 5e to[br]the j -10 -90 is -100.

0:08:11.990,0:08:16.350
Its polar form would then[br]be 5e to the minus j100.

0:08:16.350,0:08:20.765
In rectangular coordinates, that then[br]would be just to, for completeness.

0:08:20.765,0:08:29.460
That would be negative 0.868 plus j4.92.