L7 2 2 Example Trig using Phasors

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

a certain subset of
trigonometric calculations.

The phasors don't always lend themselves,

we can't always replace trig with phasors.

But under certain circumstances we can,

and it turns out that
those circumstances are very common

in dealing with sinusoidal
steady state circuits.

So I have defined two different voltages,

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

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

The first thing we note is
that the frequencies of

the sinusoids have got to
be the same in order for

the phasor approach to work.

But given in, that in RLC circuits,

driven by a single frequency,

all the voltages and currents in

that circuit will be oscillating
at that same frequency.

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

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

Each of these is oscillating
at different frequencies,

at the same frequency but
with different phase offsets.

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

use phasors to add and subtract,

and multiply and divide
these trigonometric functions.

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

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

The phasor representation
of V1 is in polar form,

is equal to 5e to the j50,

five is the amplitude,

50 is the angle.

To get it to rectangular coordinates,

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

Or 3.21 plus j3.82.

Similarly, the phasor representation
of this time function.

The amplitude is three,

the phase angle is 25,

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

or in rectangular coordinates,

this phasor would be 2.72 plus j1.27.

So phasors are the complex representations
of the trigonometric functions.

Phasors can be written in
either their polar form,

or the rectangular form.

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

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

It's easier to add with
rectangular coordinates.

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

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

and the imaginary part plus imaginary part.

So then we can say
phasor V1 plus phasor V2,

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

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

That's phasor V1 in
rectangular coordinates,

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

You add those up and you

get 5.93 plus j5.10.

Now, let's just convert those back to

polar coordinates because
in polar coordinates,

it is easiest to see what the magnitude and

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

So in polar coordinates,

this is then equal to 7.82e

to the j40.67 degrees.

I'd suggest that you stop the video now.

Just go ahead make that conversion from

the rectangle coordinates back to this.

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

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

V1 of t plus V2 of t is equal to,

the amplitude will be
7.82 times the cosine.

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

300t plus 40.67 degrees.

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

It involves a number of
trigonometric identities.

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

So when we're adding,

we add the phasors in
rectangular coordinates.

Now, let's look at multiplying.

We want to multiply V1 of t times V2 of

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

So we'll multiply phasor
V1 times phasor V2,

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

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

5 times 3 is 15.

Then we add the exponents e to the J75,

that in it's already in its polar form,

we can then see that the product of

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

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

that then is equal to 15 Cosine of

300 t plus the angle of 75 degrees.

You'll notice that in both these examples,

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

When we're using phasors,

because we're talking about the real part,

or using the real part of
the complex exponential,

the real part is related to the cosine.

So underlying all this
is the assumption that

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

are going to be written
in their cosine form.

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

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

we need to translate this,

the sine term into a cosine term.

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

and remind ourselves the relationship
between the sine and cosine.

The cosine waveform looks
something like this,

and the sine waveform
looks something like this.

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

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

is simply the cosine wave form,

delayed by 90 degrees.

Or the sine of theta,

is equal to the cosine of
theta minus 90 degrees.

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

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

of 300t minus 10, minus 90 degrees.

We take the original sine term,

delay it by 90 degrees,

and that gives us the cosine term.

We can now convert it to phasor V3,

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

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

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

That would be negative 0.868 plus j4.92.