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.