Welcome back to Linear Circuits.
This is Doctor Ferri.
This lesson is on the Frequency Response.
We're actually breaking the idea of
frequency response into 2 lessons, one
that contains linear plots and the other
that will be on boaty plots.
So, this topic is the Frequency Response.
Topic.
In the previous lesson, we introduce the
transfer function.
And we also introduce the frequency re,
the frequency spectrum.
This particular lesson is aimed at
combining
those two concepts and using them
together.
To introduce the frequency response, it's
a way
of showing how a system processes signals
of
different frequencies, and understanding
the
frequency spectrum helps in understanding
transfer
function helps in order to be able to do
this lesson.
So, let's go back to the transfer
functions.
A, an RLC circuit in this particular
configuration has this transfer function.
Remember we got this transfer function,
for
example, by doing a voltage divider law.
Finding the voltage v sub c as a function
of v sub s.
So it turns out that v sub c is equal h
times v sub s.
Where this is the phaser for v sub s.
This is the phaser for v sub c, and this
is the relationship.
That's the transfer function.
So the transfer function, we've calculated
already to be this.
If I want to find the magnitude of it.
[SOUND]
It's going to be the magnitude of the
numerator,
which is 1, divided by the magnitude of
the denominator.
Well, the magnitude of the denominator,
this is a complex number.
It's the square root of the sum of the
squares of the
real parts squared, which is 1 squared,
plus the imaginary part squared.
So now I have a function which is the
magnitude of h as a function of omega.
The angle of h, can also be written as a
function of, of omega.
In this case, the angle of the numerator
is one,
the angle of the denominator is the angle
of this.
Since it's in the denominator, we're going
to have a negative sign there.
And it's
negative arctan, of the imaginary part
over the real part.
So it's mega RC over one.
And the
negative is because this part is in the
denominator.
So just make it a little bit clearer
there.
Since
I now have these functions of a mega.
I can plot this.
This is a real function of omega.
I can plot it versus omega.
And for this particular case, with
particular values of r and
c, I would get a plot that looks sort of
like this.
A magnitude here versus omega.
The angle and degrees versus omega.
And this is a frequency response, so
we're using terms a little bit
interchangeably here.
when people use the word transfer
function, sometimes they'll, they'll talk
about transfer function as a little bit
more general term then this.
And in our case in as far as this course
is
concerned, we're using the term transfer
function to mean this function here.
This H of Omega, and then the, when we go
to the plots I'm going to call the plots
the frequency response.
So an example of how we use the frequency
response, is shown here when we look at
an RC circuit.
The input voltage to this RC
circuit, is this signal right here.
And this signal is clearly a sine wave.
Super imposed with a high frequency sign
wave, so low
frequency sine wave added together with
the high frequency sine wave.
If that's my input voltage,
then my output voltage to this circuit
looks like this, if I plot it.
Now clearly you can see something's
happened.
My low frequency signal has not changed
very much.
But my high frequency signal is now much
smaller in amplitude.
And it helps for us to understand this,
better in the frequency domain.
So in the frequency domain, I can plot the
spectrum of this input signal.
So I've got a low frequency signal, that's
at
50 radians per second, and an amplitude of
1.
If you look at this, you can see.
That the average value here is a
low-frequency signal with an amplitude of
1.
And then the high-frequency signal so on
top of
that, is another sign wave again, with an
amplitude
of 1.
So I plot those to the frequency spectrum.
This is the frequency response of this
system.
Of this circuit.
For a particular value in R and C.
To figure out what this output voltage is
going to look like, I take the input
voltage.
Spectrum at 50 radians
per second.
Now go to this plot and look at 50 radians
per second, which is right here,
and I go up, follow up this, and I find
out what this plot, this value is.
That's about point, say 0.95.
I take 0.95 and multiply it
by this amplitude of 1.
And then I
get the spectrum of the output.
If that's 1, this
is 0.95.
Now at 800 hertz, or
800 radians per second I have an amplitude
of 1.
And, in this case, I get a value of about,
I'd say, 0.1 0.13.
So I take 0.13, multiply it by this input
amplitude,
to give give me the output amplitude,
which is
about 0.13.
And what we found in the output,
I've, the input amplitude is almost the
same for the low frequency signal.
But the high frequency signal has a much
smaller amplitude.
So what's happened is, I've used the
frequency response
to tell me how this circuit.
Processes sinusoidal at different
frequencies.
At low frequencies, it passes them through
without much change.
At high frequencies, it, it attenuates
them, or makes the amplitude much smaller.
And that's a big benefit for the frequency
response.
Now let's look at an example of how to
use a frequency response in solving an
actual problem.
Suppose the circuit has the frequency
response plot shown here.
Where I'm showing both the magnitude and
the angle plot.
And the question is, what is the steady
state response v0 to input of this.
Okay to, in order to be able to analyze
this,
it really helps us to go back to the
transfer function.
I'm doing
a little bit more detail here than I did
in the previous slide.
Because, I'm also including the effect of
the angle.
So, going back to the, what we learned in
transfer functions, we said if an in, if a
there's a input Ai cosine of Omega 1 T
into my circuit, the output of my circuit
will have the form in steady state
of A out cosine at that same frequency.
But shifted in phase and a change in
amplitude.
So cosine same frequency, different
amplitude, different phase.
And we calculate the amplitude and the
phase through this transfer function.
Out is equal the magnitude of h.
The transfer function,
evaluated at that frequency times A n /g.
And the angle,
beta is simply the angle of h at that
frequency.
So if I look at this particular example.
Now we've got numbers here.
We, we've got an input of two which is
constant, and we've got this input, 200.
Now at two, that corresponds to a DC
input.
Which corresponds
to a frequency, let's look at here, a
frequency of 0.
So at omega equals 0, at omega
equals 0, and I'll mark this is DC input.
At omega equals 0, I'll look at this plot
at omega equals
0 I have a magnitude of 1 and an angle of
0.
So h at 0 is 1 angle of h at 0 is 0
degrees.
So the corresponding output to an input of
2, so an input of 2 going
into my system.
Gives me output of 2, because it scales by
1.
At omega equals 200, and this is in
radians per second,
I look over at 200, and I follow this up.
And calculate that value.
That's at
approximately
0.45.
And that is an angle
of about minus 65.
So the corresponding
value, if I put, the corresponding output
the, A out, is equal to 0.45
times the input amplitude, which is one,
and the
angle Is this angle minus 65°.
So let me put it
altogether.
V out is
equal to the output due to this part by
itself to the
DC part.
And that output is 2 plus the output
corresponding to this part by itself.
And that output would be A out 0.45 times
the cosine of that frequency is 200 t.
The angle is minus 65 degrees.
So, what we're able to do is use the
frequency response to be able to determine
what the output is to a particular
input involving a DC component.
Or sine wave and it could be we could have
multiple sine waves and we do the same
procedure for every sine wave and then we
just add them altogether at the end.
So, in summary a frequency response is
a plot of the transfer function versus
frequency.
A frequency response can be used to
determine the steady-state sinusoidal
response of a circuit at different
frequencies.
So for our next lesson, we will do the
same sort of thing in analyzing
the frequency response, but we will be
using Bode plots rather than linear plots.