So in this video, we're going to continue 
our discussion of circuits that have any 
kind of element we want, capacitors, 
inductors and resistors. 
In the previous video, we found the 
solution of the circuit when the source 
was a complex exponential. 
Well, that's a mathematical We don't 
really, we can't really produce forces 
that look like that in the lab in the 
real world. 
So, is that just a mathematical thing 
that's kind of fun to play with, but it 
turns out it's much more important than 
that. 
We're going to see how to apply that 
knowledge we developed last time, and the 
source is a sinusoid. 
Turns out it's an easy extension of what 
we already have done. 
So let's review very quickly the way the 
so called Impedance method works. 
We start with our circuit. 
We pretend the source is a complex 
exponential and that means we can now 
only worry about complex amplitudes. 
We replace every element by, it's 
appropriate impedance, whatever it may 
be, 
and then we can use, what we learned from 
resistor circuits. 
Voltage divider, current divider, series 
parallesl relations, to figure out how 
the output variables, complex amplituded, 
is related to the complex amplitude of 
the source. 
Okay. 
So that's what we learned last time, 
except like now we're going to handle a 
bit more real world sources than complex 
exponentials. 
So what we did is we applied it to our 
simple RC circuit. 
we replaced it I thought about it as an 
impedence kind of circuit and want to 
make sure we note that this is a 
lowercase v and over here's supposed to 
be v and a t. 
This is lowercase v out of t, but over 
here we're only talking about the complex 
amplitude. 
So I'm thinking about this as being a 
complex exponential source. 
And now everything in this circuit is a 
complex amplitude. 
And what we discovered when we used our 
voltage divider for this circuit was that 
the amplitude of the source amplitude of 
the output rather, was equal to this 
quantity times the amplitude of the 
input, and I'm uncancelling the complex 
exponential so we get the entire answer. 
OK, so that's what happens when the 
source is a complex exponential. 
Well what I want now do is talk about the 
case where the source is a sinusoid. 
This sinusoid has a frequency F nought, 
and it has an amplitude of A. 
Okay, well, that's not the same as a 
complex exponential. So how does this 
result apply to the case where that's 
what the source really is? Well, the 
mathematical trick is where it's coming. 
We know from the Lewis formula that a 
cosine is equal to one half times e to 
the j theta, plus e to the minus j theta. 
A complex amp, ampli-, a complex 
exponential rather at an angle of, theta 
and a complex exponential adding angle of 
minus theta. 
That's a cosine. 
So that applies to the case where we 
have, theta here is just cosine 2 pi of 
naught t. 
So I think of this as being a positive 
frequency, complex exponential, 
and this being a negative frequency 
complex exponential. 
This may sound a little strange. 
Negative frequency things. 
But it turns out this is really very 
important, and the right way to think 
about it. 
So, we can figure out what the output is 
for each piece, at least. 
We know that if the source was that, is 
this term, that that's the output, and we 
know that if it's the negative frequency 
term, all we have to do is replace f by 
minus f, and we get the answer. 
So we know about each piece of the 
original cosine, and what the output is 
for each of those, 
but how about putting them together? How 
do we do that? Well, now we get to the 
really interesting conclusion. 
Now, so this is where we are. 
This, we used the complex exponential 
source to hop over to the impedance 
version. 
We used all of the tricks we've learned 
for resistance circuits, and now when the 
source is a I want to point out, that all 
cir-, the circuit elements we've been 
talking about, resistors, capacitors, 
inductors and the, interconnection laws, 
KVL and KCL are linear, 
which means superposition applies. 
Now, what the definition of linear was, 
that if the input consists of a sum of 
terms, the output also, the output 
consists of the sum of the outputs to 
each term considered separately, you just 
add them up and multiply by a constant if 
there are any out there. 
So, we're done. 
Based on what we've already know, we know 
the output, 
voltage. 
And this, notice this is lowercase v, is 
the output to the first term plus the 
output to the second term. Add it up and 
since there's a weighing factor of half, 
it goes there to, and this is because of 
superposition because the circuit 
elements we've been talking about are 
linear. 
This is why the linear concept is so 
important, really makes our ohms easy. 
Okay, so let's continue this. 
I'm going to show you how to simplify 
that answer. 
It looks little daunting as it seems 
right now, 
but there's little, another mathematical 
thing we can do. 
Now, so, this is Oilers formula for the 
cosine, 
and these source was 2 pi f naught t. 
And like I said I thought about this as a 
positive frequency term and a negative 
frequency term. 
There's another way of thinking about 
this. 
Suppose this is a complex number z, this 
is z conjugate. 
I'm adding them up, and dividing by 2. 
Well, that's the formula for the real 
part, and so we're actually writing the 
source as the real part of a complex. 
Pretty much. 
And on the output. 
Looks very similar, right? There's the 
positive frequency term. 
There's the negative frequency term. 
And that still is, something plus it's 
complex conjugate, divided by 2, that's 
the real part. 
So, this, if you will, is the trick. 
We, can think about, 
whne we have a source that is a sinusoid, 
what you think about is the real part of 
the complex exponential. 
The output is going to be the real part 
of the same complex exponential with the 
amplitude modified by what the circuit 
does. 
That we found using, voltage divider. 
Well, how do you find this real part? 
Well, mathematically, we have a situation 
where we have something in cartesan form, 
and something in polar form. 
And you need to, figure out how that real 
part, the easiest possible way. 
And your answer is, is to convert the 
Cartesian, form to Polar form. 
So, that's what I've done. 
So, the, so this is the ratio of 
something in Cartesian coordinates. 
1, and the denominator. 
J2 pi F nought RC plus 1, and as I showed 
you, the magnitude of a ratio is the 
ratio of the magnitudes. So one obviously 
has magnitude of one, and the magnitude 
at the bottom is the imaginary part 
squared plus the real part squared, and 
you take the square root of result. 
Okay, that's where that comes from. 
The angle of a ratio is the angle of the 
numerator minus the angle of the 
denominator, and the angle of 1 is 0, and 
so what we're left with is, that's the 
angle. 
Okay, so, 
mpw we need to simplify further. 
All I'm going to do is merge the two 
complex exponential terms, and the term, 
because the exponents add, and now taking 
the real part is easy. 
Because, this is just some real number, 
so far as the mathematics is concerned, 
the only thing complex is, is this, and 
that's e to the j theta. 
And what the real part of the e to the j 
theta? That's just a cosine, so our 
answer is, we're in the source. 
It goes a cosine 2 pi f of t Now put is 
also a sine wave with a same frequency 
but with a different phase and a 
different amplitude. 
Very nice, 
and it turns out that result is quite 
general. So let us preview again how we 
found it. 
I really want to make sure we get this 
down. 
So, the source was a sinusoid. 
I thought about it as the real part. 
I know by converting to impedances, that 
this is the way the amplitude of the 
output is related to the amplitude of the 
source, and over here, that is VN. 
Right, because that's the amplitude of 
that complex exponent. 
We now know that the alpha is the real 
part, where you put in how the circuit 
transformed the voltage amplitudes, and 
just finding real part is just like using 
the procedure you did. 
converting the polar form, gives us our 
final answer. 
Looks a little complicated, but we'll, 
understand what this means in just a 
second. 
We'll have a clearer interpretation of 
it. 
Now, let's just handle in a little bit 
more general case. 
Suppose there was a phase, 
a source out of phase. 
Well, the way that works out is now V in 
is Ae to the j phi. 
So it really is a complex number now. 
Using impedances, we're always going to 
find that the complex amplitude of the 
output is equal to something what that is 
the circuit does, 
times complex amplitude of the source. 
This something it does is called a 
transfer function. 
And, if you look back over what we've 
done, it always depends on frequency. 
So that's why I had to put the frequency 
of the source, f naught. 
And I'm just going to define that In 
general, it's the ratio of the complex 
amplitude and the output divided by the 
complex amplitude of the source. 
This captures, the transfer function 
captures what the circuit does to the 
amplitude and what it does to the phase. 
So, here's what's going on. 
We've found using impedances that that's 
the transfer function in Cartesian form, 
and as we've seen it's, much easier to 
deal with mathematically if we convert 
that to Polar form. 
This is routinely what we do for any 
circuit. 
Using impedances, we're going to wind up 
with a result in Cartesian form. 
We then have to do some Calculations to 
figure out what it is in polar form. 
And then, it gets very easy to figure out 
what the output is. 
So, when the input is that, the output is 
always this. 
For any circuit containing resistors, 
capacitors, and inductors. 
That's always the answer. 
This is very very general. 
So we now can see that assuming this 
source was a complex exponential, it 
actually had a big payoff because the 
circuit was linear. 
Now I'm going to point out some 
mathematical details that are very 
important, and that is, suppose we had a 
sine wave, a real sine? 
Okay, well the easiest way to think about 
sin is as the imaginary part of 
something. 
When everything goes through the 
superposition, still goes through because 
of the [INAUDIBLE] formula, and you also 
use imaginary part, and this is very 
important. 
So, however you define, the, source is 
being related to a complex exponential, 
either by an imaginary part or a real 
part. 
You use the same representation for the 
ouput. 
Sum when it influence that, now put as 
this, and and that's a very, again a 
general result. 
You can use either the real part or 
imaginary part, whichever one is easier, 
on the point out that you can actually 
work this the hard way if you will, you 
can assume a cosine is the imaginary part 
of something. 
Notice the presence of this because of 
the phase shift by power of 2, 
represent the sign as the real part of 
something if you want to. 
Should actually go back, over our little 
example, RC circuit example, once you try 
and using the imaginary part and see, see 
if you get the same answer. 
We've got using the real part except we 
have to change the amplitude of the 
source by e to the j pi over 2. 
You better get the same answer once you 
simplify things, and trust me, you will. 
Okay. 
So, here's the big picture. 
for any circuit we have, pretend the 
circuit is a complex exponential 
[INAUDIBLE] although it really isn't. 
Use impedances to find the transfer 
function. 
And now we express the source as the real 
part or imaginary part of a complex 
exponential. 
This works for sinusoidal sources. 
And the output is going to be the real 
part or the imaginary part, whichever one 
we assumed. 
Don't try to switch them, 
of the transfer function times the 
complex exponential of the source. 
That's the procedure using impedances to 
solve any circuit that has a sinusoidal 
source. 
So, let's go through this for an example, 
and we ought to try something bit more 
interesting than, RC circuit. 
And, what I am showing you here, is 
what's known as a RLC circuit. 
So, our circuit now has a capacitor, 
a resistor, and an inductor. 
The source is this, and I'm going to ask 
you two questions. 
I'm going to find the transfer function, 
and I'm going to find the output for this 
input. 
Okay. 
So we now know, we know what to do. 
So here we go. 
So the first thing we do is we take our 
circuit and replace it by one having 
impedances, 
for again these are conflict amplitudes. 
Now I've done a little notational thing 
for you here. 
The impedance of this one ferrite 
capacity is 1 over j 2 pi f. 
Well, to simplify the mathematics and 
manipulation of these formulas. 
As you'll see, going to see in a second, 
it gets pretty complicated. 
I like to think of j2 pi f as a single 
variable s, 
and then write the impedance that way as 
just being, of the capacitors being 1 
over S. 
What I'm going to do is, use current 
divider, voltage divider. 
those tools. 
And then, at the end, replace S as J2 pi 
F. 
So, the impedance of a 1 farad capacitor 
is 1 over s. 
The impedance of R2 of 2 ohm resistors is 
2, and we, impedance of our 4 Henry 
conductor is 4s. 
Okay, so we now what to find V out. 
Okay, so The complex amplitude V-out , is 
pretty clear that this is a parallel 
combination series with that impedance. 
So, I use voltage divider, and I find 
that V-out over V-in is equal to 2, and 
parallel with 4s divided 2 and parallel 
with 4s plus 1 over s. 
Okay? So, now we multiply everything all 
out. 
We get 8s over 2 plus 4s divided 8s over 
2 plus 4s plus 1 divided s. 
Bet you didn't know that electrical 
engineers have to very fluent with 
fractions. 
This turns out to be one of the things 
that happens all the time. 
You can see now why it did not want to 
write j2 pi f over it, 
because it's complicated enough as it is. 
This happens all the time obviously. 
Alright. 
So, we can, do some things to simplify 
this. 
The first thing is, notice that this 
ratio has some 2's in it, we can cancel. 
We got 4 we got 1, make that 2, we got 4, 
we got 2. 
Now we can simplify this. 
So, I'm going to multiply top and bottom 
by s. Okay. you get rid of that. 
That puts an S square there. 
Oops, squared. 
And that's a squared there, 
and we multiply top and bottom by 1 plus 
2 S, 
which gets rid of it down and there and 
up there and this becomes 1 plus 2s. 
Okay, so, what we get, I know you can't 
read that. 
It's very difficult to read, so let me 
write it for you. 
It's 2s squared, I'm sorry, 4s squared 
divided by 4s squared plus 1 plus 2s. 
Okay? So, this is the name of the game. 
Use the parallel and the series rule. 
Here we use voltage divider, parallel 
rule, series rules and then we simplify 
it, and really makes your life a hole lot 
simpler to use s for j 2pi f. 
So, this is what you get, 
and now I can write it in terms of f, 
which is what you need to do in order to 
solve for our sinusoidal source. 
We need this h of f, and here it is in 
all its glory. 
So s squared is going to be 4 pi minus 4 
pi f squared times the 4 gives us the 16 
etc. 
So this is what it is. 
So, at this point, what I would like to 
do is find the frequency of the source. 
So, the, the frequency, this has to be 
written as 2 pi. 
F, t. 
So, 2pi f, has to be a half for this 
example, 
which means the frequency, is 1 over 4pi. 
Kind of a strange frequency, but it 
really simplifies the calculation for 
this example. 
I now want to find the value of the 
transfer function at that frequency. 
And, like I said, this is really, makes 
this assumption of that frequency value, 
really makes, the calculations very 
simple, 
because the numerator just turns out to 
be minus 1. 
Put that same term here. 
This becomes j, and the 1 is 1. 
So we get minus 1 over J, and reciprocal 
of J is minus j. 
So we get J for an answer. 
So our transfer function at the frequency 
of the source, that's all we need to 
know, is what the transfer function is at 
the frequency of the source, is J, 
through the phase shift. 
So, now we get to figure out what the 
output is. 
So, I'm going to think about our sinusoid 
as being the imaginary part Since it's a 
sine wave, probably makes it easier. 
And it's 10, e to the jt over 2. 
Well, we know that the output is going to 
be the imaginary part of 10j, e to the j, 
t over 2. 
Where j is that transfer function 
evaluated at the frequency of the source. 
And, j, again, it's in Cartesian form. 
Convert to polar form, and that's our 
answer. 
And, I know from the trig identities, 
that that's just cosine. 
So, we stick in 10, it turns out it's a 
very special frequency. 
because what this really does,at that 
frequency is convert the sine to a 
cosine, and it as the same amplitude. 
That's really pretty surprising, but 
again I have chosen a frequency that 
makes the calculations easy, and that's 
basically why that happened. 
In general, as we know, it's going to 
produce a sine wave at the same frequency 
with a different phase, and a different 
amplitude. 
We choose a different, frequency, this 
amplitude, will, definitely be different 
than 10. 
Okay. 
Now. 
I've taught this course for many years, 
and I've learned what students do that 
isn't correct. 
So I want to talk little bit about what 
not to do. 
How to use impedances so here's the name 
of the game. You start with a circuit. 
We know the input, 
provided it is the imaginary part of 
something, How do we find the transfer 
function, and we evaluate it at the 
frequency of the source, and we now know 
since I used imaginary part here, that 
that is the imaginary part here and 
that's the correct approach. 
Okay. 
Transfer function evaluated at the 
frequency of the source. 
We get that from that formula times the, 
complex amplitude representing, the, 
source, with, along with it's complex 
exponential frequency, we take the 
imaginary part. 
Well, I have seen students do the 
following. 
They just say well, let's assign this 
little source, you told me it's just the 
transfer function times that. 
Well, that's clearly wrong. At least in 
this example know that the transfer 
function of that frequency is j. So why 
in the world would you put in a real 
value input and come out with something 
that's complex valued? 
This is wrong, you have to think of the 
source as a complex exponential, and then 
things go through. 
So this is not correct. 
Another thing that isn't very good is to 
mix the two things together. 
So, you just stick in the source, 
multiply by the transfer function of the 
right frequency, and take the measuring 
part. 
Now that's not what you have to do. 
There's this intermediate step, remember, 
pretending the source is a complex 
exponential. 
That's the right formula. 
This is wrong. 
Okay. 
So, I hope this helps. 
We now know a very general result. 
The output for any sinusoidal input of a 
circuit is going to be the, 
a sinusoid at the same frequency, and the 
way we find it is by converting to 
complex exponential form and then taking, 
multiplying by the transfer function, and 
taking the real or imaginary part, 
depending on how we chose to do. 
Learn more about, how to use impedances 
to figure out what these kind of transfer 
functions do, and to get a much more 
general, broader picture of circuits.