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>> In this video, we're going to show
how phasors can be used to perform

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a certain subset of
trigonometric calculations.

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The phasors don't always lend themselves,

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we can't always replace trig with phasors.

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But under certain circumstances we can,

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and it turns out that
those circumstances are very common

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in dealing with sinusoidal
steady state circuits.

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So I have defined two different voltages,

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V1 is equal to 5 times the
cosine of 300t plus 50 degrees.

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A second one V2, three cosine
of 300t plus 25 degrees.

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The first thing we note is
that the frequencies of

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the sinusoids have got to
be the same in order for

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the phasor approach to work.

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But given in, that in RLC circuits,

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driven by a single frequency,

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all the voltages and currents in

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that circuit will be oscillating
at that same frequency.

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So this limitation that
the frequencies have to be the same,

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isn't a big deal for us in
what we're going to be doing.

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Each of these is oscillating
at different frequencies,

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at the same frequency but
with different phase offsets.

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What I wanted to do here,
is show you how we can

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use phasors to add and subtract,

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and multiply and divide
these trigonometric functions.

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To do so, we're going to be using
the phasor representations of these.

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So I've gone ahead and define
those first, define those here.

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The phasor representation
of V1 is in polar form,

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is equal to 5e to the j50,

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five is the amplitude,

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50 is the angle.

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To get it to rectangular coordinates,

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it's just 5 times the cosine of
50 plus 5 times j sine of 50.

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Or 3.21 plus j3.82.

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Similarly, the phasor representation
of this time function.

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The amplitude is three,

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the phase angle is 25,

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so we have 3e to the j25 in its polar form,

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or in rectangular coordinates,

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this phasor would be 2.72 plus j1.27.

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So phasors are the complex representations
of the trigonometric functions.

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Phasors can be written in
either their polar form,

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or the rectangular form.

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It turns out that when we're
adding two sinusoids together,

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it's easier because we're going
to be adding their phasors.

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It's easier to add with
rectangular coordinates.

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Because when you're adding, you're
adding two complex numbers,

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it's just the real part plus the real part,

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and the imaginary part plus imaginary part.

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So then we can say
phasor V1 plus phasor V2,

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we'll add the phasors and then switch
the result back to the time domain.

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Phasor V1 plus phasor V2 is
equal to 3.21 plus j3.83.

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That's phasor V1 in
rectangular coordinates,

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plus phasor V2 in rectangular
coordinates is 2.72 plus j1.27.

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You add those up and you

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get 5.93 plus j5.10.

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Now, let's just convert those back to

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polar coordinates because
in polar coordinates,

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it is easiest to see what the magnitude and

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the phasors that we use then to
rewrite it in its time domain form.

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So in polar coordinates,

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this is then equal to 7.82e

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to the j40.67 degrees.

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I'd suggest that you stop the video now.

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Just go ahead make that conversion from

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the rectangle coordinates back to this.

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Now, once we've got it in this form,

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it's very easy to write
the time-domain version.

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V1 of t plus V2 of t is equal to,

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the amplitude will be
7.82 times the cosine.

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It's going to be oscillating
at the same frequency,

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300t plus 40.67 degrees.

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I'll leave it to you to attempt to add
these two sinusoids in the time domain.

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It involves a number of
trigonometric identities.

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It's not undoable, but it is
computationally intensive.

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So when we're adding,

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we add the phasors in
rectangular coordinates.

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Now, let's look at multiplying.

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We want to multiply V1 of t times V2 of

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t. We're going to do
that using the phasors.

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So we'll multiply phasor
V1 times phasor V2,

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which is 5e to the j50 times 3e to the j25.

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Well, when multiplying things like
this is just the coefficients,

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5 times 3 is 15.

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Then we add the exponents e to the J75,

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that in it's already in its polar form,

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we can then see that the product of

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V1 times V2 is going to
have an amplitude of 15.

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So touching this back to
V1 of t times V2 of t,

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that then is equal to 15 Cosine of

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300 t plus the angle of 75 degrees.

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You'll notice that in both these examples,

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both of the time domain functions
were in terms of cosines.

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When we're using phasors,

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because we're talking about the real part,

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or using the real part of
the complex exponential,

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the real part is related to the cosine.

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So underlying all this
is the assumption that

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the time domain functions that we're
going to be representing as phasors,

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are going to be written
in their cosine form.

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Here I've got V3, it's
written in its sine form,

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so before we can use it
or convert it to phasors,

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we need to translate this,

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the sine term into a cosine term.

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To do that, we're going to just look at,

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and remind ourselves the relationship
between the sine and cosine.

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The cosine waveform looks
something like this,

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and the sine waveform
looks something like this.

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If it was drawn better
you'd be able to see,

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but hopefully this is good enough to
observe that the sine of theta here,

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is simply the cosine wave form,

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delayed by 90 degrees.

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Or the sine of theta,

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is equal to the cosine of
theta minus 90 degrees.

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So to change this or to transform
this into a cosine term,

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we can simply say then that V3 of
t is equal to 5 times the cosine

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of 300t minus 10, minus 90 degrees.

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We take the original sine term,

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delay it by 90 degrees,

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and that gives us the cosine term.

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We can now convert it to phasor V3,

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phasor V3 would equal 5e to
the j -10 -90 is -100.

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Its polar form would then
be 5e to the minus j100.

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In rectangular coordinates, that then
would be just to, for completeness.

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That would be negative 0.868 plus j4.92.