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.