This is Doctor Ferry,
and this is an extra work problem, on the
frequency response using Bode plots.
Suppose we're given a Bode plot that looks
like this right here.
This is the magnitude in decibels, and
this is the phase angle.
And we want to understand how.
A input signal, input voltage is processed
through a circuit that has this frequency
response.
So suppose this is my input signal here.
It's a sine wave, a sinusoid.
And we'll call that V in.
And
we want to know what, what is V out.
And in particular, we want to plot V out
on the same set of scales as we do V in.
So, the basic relationship that we're
going to
need here is that for a linear system.
If I've got an input that's a, a sinusoid,
with an input amplitude, A sub i.
The corresponding output is also a
sinusoid at the same frequency.
With a change in amplitude and a change in
phase.
Now, the amplitude change is A naught is
equal Ai times the magnitude of the
transfer
function at that frequency, and the phase
is
the angle of the transfer function at that
frequency.
Now this is all on a linear scale because
we've got the magnitude, here, on a linear
scale.
How to re, re-relate that to a log scale.
Well from the log scale, we have this, all
of this information.
The phase is directly.
This is in degrees.
So I can just read off the plot at any
given frequency and I read the phase.
But the magnitude, on here, what I have is
20 times a log of
H, in decibels.
And suppose I read that to be some
constant.
You know?
Some value, C.
You know, at any given particular
frequency, I have some value, C.
So I want to find out what H is.
H is equal to, solving for, this magnitude
here, I have 10 to the C over 20.
And that tells me what H is, if I've
measured C directly off of here.
C in decibels.
So that's the basis that this is, this
whole process is, is built upon.
Let's look at, for this particular signal.
This is my via input signal.
I've gotta identify the frequency and the
amplitude.
Well, the amplitude, A sub i, is equal
to 1, because it's going between one and
zero.
There's no DC component to it, because it
oscillates about zero volts.
So it's just a single sine wave.
And if I look at the period here.
From here to here is 0.01 seconds.
So the period of this sine wave is 0.01.
The frequency is 1 over T.
Which is 100 hertz.
So this is in hertz.
And my Bode plot's in radians per second.
So I have to convert this to radians per
second.
So I'll call it omega sub 1 is equal to 2
pi f.
Call this f sub 1.
This is the particular frequency of my
input signal.
That will be 628 radians per second.
Okay.
So, I have to figure out what H is
at that particular frequency.
So, 628.
Well, this is 100 right here.
200, 300, 400, 500, 600, 700.
So it's between this and this.
Going up here to between that and that.
So it's roughly going to be about there.
And that is minus three decibels
right there.
And the angle is approximately minus 45
degrees.
So, if I want to find the output, the
input
is equal to the cosine amplitude 1 cosine
of 628t.
The output is equal to, A out,
cosine of 628 t plus some phase angle.
The output
amplitude is equal to the input amplitude
which is 1, times H at that.
Frequency times A naught, which is, which
is one.
So we have to find what H is.
And going back to this, we measured C to
be minus 3 decibels, so H is equal to.
10 to the minus 3 over 20.
If I solve that, that is equal to 0.707.
So that's what the amplitude is and the
angle we
measure here might be negative 45 degrees.
So, our output
is equal to 0.707.
Cosine of 628t minus 45 degrees.
Okay.
So I need
to plot this on here.
Minus 45 degrees.
And an amplitude of 0.707.
Let me mark this.
We're going
to be oscillating between plus 0.707 and
minus 0.707.
Now another thing to, to notice is that.
When we do, go through a sine wave, we go
through 360 degrees for every cycle.
So this is 360 degrees, if this is zero
degrees, that's 360.
So this would be 180, 90 and 45 is right
there.
So when we have a faze line of 45 degrees.
It's about that much.
So, from here, this goes through a zero
crossing here.
The output will go through a zero crossing
right there.
And this will go through a corresponding
zero crossing with the lag right there,
there.
This distance right there.
And correspondingly, the peak and the
valleys
will be delayed by about that much.
So this would be, be delay.
This is, this would be 90 degrees, this
would be 45.
Should be about right there.
Looks like it's maybe just to the right of
these.
Signals.
So all I need to do is then take my
marker, and
fill
this in.
And that's a corresponding V out.
So summarizing what we've done here, is
everything is based on this input-output
relationships for linear circuits and
linear
systems, where the cosine output is the
sinusoidal output results in sinusoidal
output
in steady state at the same frequency
but a change in amplitude and phase and
that's given by this relationship.
This being a Bode plot we
have to convert the Bode, the magnitude in
decibels back
to a magnitude on a linear scale and we do
that with this conversion right here.
Where C is the value in decibels that our
plot is.
And looking at our input signal, we had to
first find what V in is.
We had to find the, the frequency.
And we had to find the amplitude in order
to
get this expression.
Once we have this expression, we have the
input amplitude, and the phase, and the
output amplitude.
We have to look at the magnitude at that
frequency, covert it to a linear value.
And then we're able to write the output
frequency, ou-, output Voltage.
And then just plot it.
Remembering of course that going
through one cycle is 360 degrees and
whatever this value
is here, we would do that as sort of a
proportion.
So, 45 degrees would be here if this is
360.
And then we just plotted it.
Alright, thank you.