Welcome back to our continuing series on 
linear circuits. 
And today we will be talking about 
transfer functions. 
The aim of today's lesson is to 
understand how linear systems react it to 
inputs with different frequencies. 
In the previous lesson we looked at AC 
analysis. 
So we were able to apply the techniques 
that we learned the beginning of this 
course to phasors systems where we have 
sinusoidal inputs. 
And using phasors those complex numbers 
to do basically the same things that we 
did before. 
And then there was the sinusoidal respond 
lab. 
Today we're going to take those tools and 
apply them to transfer functions. 
In transfer functions are, well first of 
all we're going to describe, describe 
what they are and how they're used and be 
able to plot them. 
Whenever we have a linear system with a 
sinusoid input we're going to have a 
sinusoid output and output will be at 
that same frequency. 
So here if I take my input equal to this 
Ain times the cosine of Omega t plus 
Theta n. 
We put it in phasor format we get this 
and for the output we get this for our 
phaser for the output. 
But the thing about linear systems is 
that as the frequency changes so will 
these quantities. 
So if I look at this example I'm going to 
instead of using Vs for voltages I'm 
going to let them be arbitrary inputs of 
X and outputs Y. 
And just for simplicity we're going to 
let the input phase be 0 since what we're 
really interested in is the difference 
between the input phase and the output 
phase. 
So here I have set up an input function, 
what's particular frequency that has this 
amplitude and some phase, which will be 
0. 
And then here I have an output so you can 
see that it has a certain phase to it and 
then a certain amplitude associated with 
it as well. 
Now suppose that I change the frequency, 
let's see what changes. 
So now I've doubled the frequency and in 
this particular case we see that the 
output now has a smaller amplitude than 
the previous example. 
So even though my input has the same 
amplitude as the previous input the 
change in frequency lead to a different 
output amplitude. 
And we also we that where the previous 
example the output started down, this one 
starts up. 
So we see that it has a different phase 
as well. 
Transfer functions is the way that we 
describe this change in amplitude and 
phase. 
And so the way we're going to define it 
is. 
Lets say that the transfer function with 
respect to a func, a frequency f is going 
to be equal to the output phasor divided 
by the input phasor at a particular 
frequency. 
So that's what this little subscript f 
indicates. 
If you use complex numbers and calculate 
using these formats. 
You see this, it's essentially just going 
to be the the amplitude's going to be the 
relationship between the output amplitude 
and the input amplitude. 
And then the phase angle is going to be 
the phase of the output minus the phase 
of the input. 
This might be a little bit confusing so 
we're going to do a few examples to kind 
of illustrate the point. 
In our AC analysis section we looked at 
this circuit and we discovered that at 
the frequency f = 780 Hz that the output 
Vc was equal to 1 over the square root of 
2. 
[UNKNOWN] minus Pi fourths times our 
input in phasor format. 
And so if I wanted to know the the 
transfer function at that frequency if I 
take this Vc and divide it by V. 
We get 1 over the square root of 2 phasor 
angle minus pi fouths. 
And so if we go and look at the value of 
transfer function 780, we see that it's 
indeed the case. 
well actually this is approximate but 
it's pretty close. 
Now typically we're not going to want to 
do these calculations for each individual 
frequency and then compile them all 
together to get our final format. 
Normally what we're going to want to do 
is come up with a, a functional 
expression in terms of the frequency or 
possibly the angular frequency. 
To identify what the transfer function 
will be. 
So that's what we're going to do in this 
example. 
Here we have a simple RC circuit and it 
is basically going to use the exact same 
tools that we used in the RLC example. 
So if I let this be a phasor V, here the 
impedance for the resistor is just R. 
And the impedance for this capacitor is 
going to be 1 over j omega C. 
So to do our calculation here, we see 
that our phasor Vc is going to be equal 
to 1 over j omega C impedance this. 
Divided by R plus 1 over j omega C, which 
is the impedance of both of these 
together. 
And we're going to multiply all of that 
by our input voltage, the voltage here. 
But transfer function, which here since 
I'm using omegas I'll let them do the 
transfer function with respect to omega. 
This is going to be the ratio of this to 
V, which means these V's cancel out and 
we'll multiply the top and bottom by J 
omega C just to simplify the expression a 
little bit. 
So it becomes 1 over 1 plus J omega RC. 
Then you see that we have one expression 
in terms of an angular frequency here 
that explains the way that this transfer 
function will behave. 
Now I define the transfer function with 
respect to frequency so it's really to do 
the conversion there. 
We'll change our omega to 2 pi f so here, 
this would be j2pifRC. 
Let's all clear all of that and put up 
the expression again here. 
Typeset's what's nice and easy to read. 
Now typically we, typically one thing 
that we can do to simplify this 
expression further is to substitute it. 
So let Fb equal to 1 over let fB be equal 
1 over pi RC and when you do that 
substituion you get H of f equal to 1 
over 1 plus J, f over fB. 
But using some analysis with doing 
complex numbers, we can actually break 
this up into two pieces because it's 
really difficult to for example, graph 
this function of complex numbers. 
Because for each different f, each 
frequency f there's a complex number 
associated with the transfer function. 
So typically what we will do is we'll 
break it up into a magnitude and a phase. 
Since that gives us two real numbers and 
it will lead to us having two different 
plots. 
And we can actually calculate that 
magnitude in face fairly simply. 
That turns out the magnitude of h of f is 
going to be equal to 
Here the top is 1 and we think of this 
this bottom as maybe being some A arg 
theta. 
If I put 1 over that then the result is 
going to be a function with an output. 
Or a complex number with an amplitude of 
1 over A. 
And a phase of 0 minus theta. 
Or 1 over A with an amplitude of minus 
theta. 
And so using that we can discover that 
the magnitude of H of f is going to be 
one over the square root of one plus f 
over fV quantity squared. 
And the phase angle will be equal to the 
minus arc tangent of f over fB. 
Now this, you might be lost on some of 
the details of exactly how to get this. 
And one of the reasons for that is people 
do complex numbers in high school and 
then they don't really touch it again for 
a very long time. 
So we will provide a, an appendix lesson. 
It's not part of the normal course flow 
that covers complex numbers and doing 
these types of calculations in case you 
were getting lost. 
So I'll put all that and I'll put up our 
two solutions here. 
And you can go through and actually plot 
these two functions with respect to 
frequencies. 
And so you notice that, that the 
magnitude of our input really didn't 
matter because it basically gets factored 
out in the end. 
So if I plot this out, we see that I get 
plots that kind of look like this. 
Now the first one for small values of fB, 
this first term dominates so it's going 
to be about 1. 
But as our frequency here increases, this 
one becomes less and less important. 
And this is going to start dropping down. 
This is going to start basically looking 
like 1 over, well fB over f kind of thing 
and in this example this is just our 
minus arc tangent behavior here. 
This bottom line corresponds to the brake 
frequency since if I put in fB here for 
the f, we discover that the magnitude is 
going to be 1 over the square root of 2. 
And the phase is going to be minus pi 
fourths for our transfer function. 
If instead we wanted to know what the 
transfer function was for a functioin 
that were operting at a 4 kilohertz 
frequency we could come to our plot here. 
And find 4 kilohertz per frequency range, 
find out what this point happens to be in 
so maybe about 0.4 and this is maybe 
around minus 3 pi eighths approximately. 
So we can quickly see what kind of 
magnitude and phase response that we will 
see for a particular input using these 
methods. 
In a final example we will again look at 
this RLC circuit and to do an analysis to 
see what the transfer function of this 
would be like. 
If instead of calculating it for a single 
frequency we do it as an expression of 
all frequencies. 
And in this case I will leave my transfer 
function in terms of omega's angular 
frequency just because it's going to be a 
little bit easier to write. 
Again we're going to be using the voltage 
divider so our Vc is going to be equal to 
1 over j omega C. 
Divided by R plus j omega L plus 1 over j 
omega C multiply top and oh, almost 
forgot. 
Times Vs. 
If I take this Vc and divide it by Vs and 
multiply the top and bottom by j omega c. 
It's discovered that H of omega is going 
to be equal to 1 divided by things around 
a little bit here. 
1 minus omega squared LC plus j omega RC. 
I'll go ahead and put this up here. 
And again we can use the same basic 
technique as we used before to find what 
the magnitude and phase of this will be. 
And so here the magnitude is going to be 
this expression, 1 over all of this in 
the square root and the phase is going to 
the minus arctangent of this ratio. 
You can actually put in different values 
for R, L and C and then plot out what 
this response will look like, what the 
magnitude and phase response will be 
like. 
Are using any kind of plotting software. 
And you'll discover that it kind of has 
some interesting behavior. 
Doctor Perry will in a later lecture go 
over what these transfer functions will 
look like and how to use them. 
In summary we introduced the concept of a 
transfer function. 
And we showed how to calculate the 
transfer function for a few different 
systems. 
And then we demonstrated the graph of the 
magnitude and the base response. 
And a little bit about how to read it. 
In the next lesson we will cover time and 
frequency domain. 
And then there will be a demo with a 
guitar string to show the frequency 
spectrum. 
When you pluck a guitar string it has a 
few different frequency components that 
contribute to give it the, that 
particular sound. 
And we'll start to see how we can then 
take these ideas of transfer functions 
and apply them to real world situations. 
Until next time.