Welcome back to linear circuits.
This lesson is a continuation on frequency response.
This time were looking at Bode plots. .
So again it, it still fits in the category of frequency response.
In our previous lesson we introduced frequency response as a way of showing
how a sig, how a system
or circuit processes signals of different frequencies.
In that particular case, we just used the linear scale.
The only difference we're doing in this lesson
is that we're now going to use a Bode
plot, which is a different scale, logarithmic scale,
as a way of showing that the frequency response.
So, Bode plot.
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We use these scales.
Again, it's logarithmic in both the, this axis as well as this axis.
In this axis, we usually plot it on semi log paper.
So, this scale is logarithmic here. And let me define something here.
Let me define this distance from here to here as a decade.
It's a distance to go from sum frequency f to 10 times the frequency.
Call it f sub one. A particular frequency to 10 times that.
So this is also a decade.
Because I'm going from this frequency to this frequency.
10 times it.
the same thing here.
If I went from this frequency to this frequency, same thing is also decade.
Now, frequency is often shown in terms of radiance per second, or hertz.
Recall that, two pi f is equal to omega.
So that's it, remember the conversion to go between those two.
Now on this scale, it is actually a linear scale here.
And we often mark these in terms of 20 degree or 20 decibel differences.
So that might be 20, 40, 60, and so on. The db stands for decibels.
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And we get that value, we get the
Bode plot, by simply taking the magnitude function,
the frequency response function and taking 20 times
the log that's logged base ten of it.
The angle plot is the same way we take the angle in
degrees typically and we plot it using a logarithmic scale along this axis.
Let's compare the linear plot and the Bode plot side-by-side.
On the left is the linear plot, so this is a function of a magnitude of h of omega.
And this is the angle of H of omega.
Now every point along here, at every
frequency, I compute the, the magnitude in decibels.
So over here, 92 in decibels, I get by taking
20 times the log of it. So for example, at Omega equals
zero, I have a value of one. Well, 20 times a log of one
is zero, and that's where i get zero decibels right here.
Right over here I have a value of 0.1. So, 20 times a log
of 0.1 Is minus 20, and that's
where I get this point here.
So that's true of every point along the way.
And the angle I get, really by simply moving it over.
But I'm stretching out the scale, the frequency scale, like this.
So, I mean, counting it, in terms of the
logarithmic scale here, this is the value of 10, this
would be 20, 30, 40, 50, 60, 70, 80, 90 and 100.
Then this would be 200, 300, 400 and so on.
So that's the difference between scales, but it's the same information.
I'm plotting this, but this time I'm taking 20 times the log of the magnitude.
I'm plotting that and that's the only difference
between the Bode plot and the linear plot.
Let's look at a first order system characteristics on a Bode plot.
This is an RC circuit.
This is a first order system we had looked before what the magnitude looks
like and the magnitude function is like this, the angle function is like this.
So if we took the magnitude and take 20 times the
log of it and plot that and then plot the angle,
all in the logarithmic scale, we get these two plots.
And a couple of things that I wanted to point out here.
One is at high frequency, if I look at
this slope here, it looks like a straight line slope.
And it is, it turns out it's minus 20 decibels per decade.
because a decade is between a frequency and
10 times that frequency, so that's one decade.
And then we went down, if I follow this line up here, kind of
stretch this out a bit
It's going from here down to here.
In fact if this kept going it would keep following that line.
So our curve is actually asymptotic to a minus 20 decibels
per decade line, and that's true of all first order systems.
Now the angle, it starts out at zero And then it goes down to minus 90.
It's asymptotic to minus 90.
So, it goes from 0 degrees to minus 90 degrees.
And that's, those are the characteristics of it, a first
order system like a, RC network in this sort of configuration.
Now lets look at a Bode Plot of an RLC Circuit.
An RLC Circuit is what I'll call a second order system.
Second order system because, if I were to find, look at
the differential equation it would have a second derivative in there.
So I call it second order.
So, this particular RLC circuit has this transfer function.
If I find the magnitude, and then take 20 times
the log of it and plot it, I get this.
And, if I take the angle, I get this.
What are the characteristics here?
Well, at high frequency, if I look at this slope.
This slope is minus 40 decibels per decade,
and that's because this is a second order,
a second order circuit. I'll get minus 40 db per decade.
And also with this RLC, or second order circuit, I look at the angle.
And the angle also is different.
It goes from zero degrees to minus 180 degrees.
Now the curve here, this comes to about, right around one over the square root of
LC, is what I'll call the cross-over frequency, one over the square root of LC.
That's often called the resonate frequency.
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Now a second order systems in RLC circuits don't always look like this.
You notice the resonate frequency didn't involve R.
And the R is clearly is in this function, so it does affect what it will look like.
And the effect of R is really shown better by looking at
a case where this is, this is the value of very, large value
of R. So I'll say plot for large R.
I'm
going to redraw this, frequency response but for a very small value of R.
We'll show the difference.
So this is for a, small R.
[SOUND].
And what we see is that it peaks up here.
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And
that's really a resonant peak.
And what happens is that as R gets smaller, this peak gets higher.
That peak goes higher as R gets smaller.
Now the, the value of the high frequency is still minus 40 db per decade
[SOUND].
The angle still goes from zero to minus 180 degrees.
But it does so a little bit sharper.
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And what we see is this, this has to do with damping.
What
we call the damping ratio goes down as R goes down.
And so the peak goes up, causing a resonance.
What does it mean, this peak?
What it means is that at low frequency the output amplitude is
going to be about be about the same as the input amplitude.
But there becomes a frequency range in here where the output
amplitude is actually larger than the input.
Remember, zero db corresponds to an amplitude of one.
So that means the output amplitude is equal to the input amplitude.
When we're greater than one, when we're greater than zero, the
output amplitude is greater than the input amplitude in the resonance range.
And then it goes down even more so.
So the output just looks like it's, it's a bigger amplitude.
Bigger bigger sinusoid than the input amplitude.
Now, let's look at an example.
Trying to figure out what the output is to a particular input using the Bode plot.
So this is my input to my circuit.
Ant this is my Bode plot which is the, shows the transfer function in there.
I've got three different components to my input.
I want to analyse them each individually.
So we use the same thing while, if I look at the
output amplitude is equal to the input amplitude times the, transfer function
at that frequency the magnitude of it and the angle of the output is
equal to the angle of the transfer function at that frequency.
That's how we figure out what the steady state response is to a sine wave.
Well I've got three different components here.
The first being dc which corresponds to omega equals zero.
So I have to figure out what the amplitude
is, at or the magnitude is at omega equals zero.
Well I don't actually plot omega equals
zero here because it's a logarithmic scale.
So this is 10, and if I went out equal distance here, this would be one.
And then this would be 0.1, 0.001, and you can
see, I don't actually plot omega equals zero on this plot.
But what I see is that this, this is
approximately zero dB, and it looks like it's going to stay zero dB.
So I will say 20 times the log of h of zero is zero db.
And I have to
back out what h of zero is in order to use this formula here.
So h of zero, magnitude is equal to 10 raise to the zero over 20.
So I take 0 divided by twenty and then raise both sides by a
power of 10, and that equals what? So that particular thing
says that if I put a DC component of one I get out a DC component of one.
Now, let's look at the next frequency component
that's 100 and this is radiance per second.
At 100, again this also showing radiance per second, I should mark that.
At 100 I'm right here. So I look right there and right there.
My magnitude is,
and this is the magnitude I'll say in db which is
20 times the log of that is equal to zero db.
And what we already found above, that means that h.
That's going to equal one, and the angle was zero.
Okay now let me go to my last component,
3000, 20 times the log of h at
3000 is equal to. Let me go to 3000.
So this is 1000, 2000, this is 3000. So
following up to here, I get 3000. And that looks that is 10 decibels.
And over here, at 3000, that is right
here, that looks like that's about minus 70 degrees.
So we get 10 decibels.
If I want to find what h is,
that's equal to 10 to the, and then it's this number divide that number.
So it's 10 divided by 20, and that is
3.16. So putting
it all together, my output, I'll call it the out of T is the output
corresponding to the one by itself, which is one, plus the output
corresponding to this cosign by itself, and the magnitude.
the output amplitude is, the input amplitude times this
which is one so the amplitude stays the same.
Cosine 100 T, the original, the original phase
lag, or phase is minus 45. And I haven't changed it here.
So in other words this, these two components are passed through
my circuit without really being changed. And then the last component of my input,
the amplitude of the input is one.
I multiply it by the magnitude of the transfer function.
Magnitude of the frequency response at that point which is 3.16.
Cosine same frequency 3000, and this time my phase is minus 70.
So, what's interesting is at resonance if I put a sine wave at
resonance, my output amplitude is actually is larger than my, my ampli, amplitude.
So in summary, a frequency response is
a plot of the transfer function versus frequency.
In our last lesson we were looking at plotting
it first in the linear, as a linear plot.
This case it's a Bode plot.
Bode plot is just fr-, frequency response on a log scale.
The units are in decibels or dB.
An RC circuit, we found that the magnitude goes down by 20 dB per decade.
And the phase
goes from zero to minus 90.
An RLC circuit, we found the magnitude goes down by 40 decibels per decade.
And the phase goes from zero to minus 180.
And what we found is that an RLC circuit with low damping has a resonant peak and
that affects the output. The output would have a higher amplitude
at resonance than the input. In our next lesson we'll do a lab demo of
an RLC circuit frequency response. I'll see you online, Thank you.