This is Dr. Ferri and this is an
extra problem worked on frequency response using linear plots.
Let's take a look at this system, this particular circuits got a transfer
function that can be plotted like this to give you this frequency response.
Now, this is on a linear scale.
And, what's plotted is versus frequency and hertz.
And, we're asked to find what is the steady
state response v not to an input like this.
Now, doing a problem like
this, we're basing it all in the basic concept of the input output relationship.
From a transfer function.
So suppose my circuit has this transfer function and
my input is the signusoid like this. Then my output.
Will be a sine wave at that frequency.
With a phase shift.
The output amplitude is related to the input amplitude, and this is in steady
state by this transfer function, in particular
the transfer function magnitude at that frequency.
The angle here, phase angle, is a phase angle of h at that frequency.
Now this particular thing is all done in terms of radians per second.
Frequency or radians per second.
When we deal with hertz, it's, the relationship is omega is equal to 2 pi f.
So f is typically in hertz, and when
we talk about omega, it's in radians per second.
So in this particular case, we just have to realize that
we have to do this conversion between Radians per second, and hertz.
Because this plot is given in hertz.
So we, we're given now, a system with three
different frequency components.
One a DC value, and then this frequency here, and then this frequency here.
So we can apply this whole analysis at each frequency.
And analyze the response individually of these components.
And then at the end, sum them together.
So let's look at them at f
equals zero, which is zero radians per second, this is the DC value.
I look at my plot at that frequency. So that's right here, I
look at the plot at that frequency and look at the
magnitude which is 0.8. So h at zero has a
magnitude of 0.8. And that
means that if I have a DC input of 1, so DC
input of 1, it gets multiplied by that
magnitude, gives me the output. So
that's my DC value. Corresponding to this input DC value.
Now let's look at the next component right here.
That's at f equal to 10,000, so
this is f right there, it gets multiplied by two pi to give me omega.
Let's look at h at that corresponding frequency.
Well this is ten to the fourth right there, so this right here is 10,000.
So H at 10,000 and I'm going to write this in terms of hertz.
The magnitude right there is zero.
Okay. So this 10,000 hertz.
If I were to plot this in radiance per second,
would be 10,000 times 2 pi, which is this omega.
So this particular frequency. We've got a magnitude of zero.
That means my output magni-, my output amplitude is equal to zero.
And since the output amplitudes equal zero, I don't care what
the phase is because this whole thing is going to be zero.
Now let's look at this last term right here, which is a frequency
of 15,000 hertz, or 15,000 times 2 pi in terms of radiance per second.
But we're looking at 15,000 because Hertz is what this plot is.
So at 15,000 is right here, I go up here and
look at this, and that is a value of 0.5. So H
at 15,000 Hertz.
Is equal 0.5. That means my output amplitude
is equal to the input amplitude, which is 1, times this
magnitude of H, which is 0.5, and that will be 0.5.
And my angle.
Is from this angle plot right here, and this is in degrees.
I look at 15,000, and that's right
there. That's minus
130. So, my output phase will be theta
is equal to minus 130 degrees. So let's put them all together, v out
is equal to the response due to this DC, which found to be 0.8.
Plus the response due to this frequency component, which we found to be zero.
Plus response due to this frequency component which
we found to have an amplitude of 0.5.
At that frequency
and then with this angle.
So this is the output there that's the output amplitude and that's a phase.
And this phase was minus 130.
And that's the result.
Now, just to summarize what we did
here, is everything was based on this relationship
between an input sinusoid to a linear system
and how it's related by the transfer function.
It just effects the amplitude and the phase through
this relationship right here. And we use that component by component.
So we had three frequency components here.
And at each frequency component, we looked at the
magnitude and the phase, and we computed them individually.
And then we added up the corresponding responses,
to find the total response due to all three.
Thank you.