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.