In the last lecture we found that the
phase margin test, is a way to determine the stability of the closed-loop transfer functions, based on the loop gain T and its phase, at the crossover frequency. In this lecture, we're going to ask how
much phase margin is enough. Is 1 degree of phase margin, sufficient? Certainly, what we found in the last
lecture is that, if we had a phase margin of plus 1 degree, then our phase margin test
would say that the, the closed loop system is
stable. Now, of course, perhaps we want more
margin than that in, in a good robust design. And, so we need some amount of additional
phase margin, to make sure we have a, a sufficient cushion for reliability and
accounting, for variations in components. But, there's actually more to, it than
that. The phase margin also determines, the
closed loop response. At least, it's, it's a measure of the closed loop transient response, in
some sense. And so, in this lecture, we're going to
discuss the relationship between the phase margin
and the q factor, of the closed loop poles, of t
over 1 plus t, that occur at the, the crossover
frequency. And what we're going to find is that, the
larger the phase margin, then the better the response is, as far as having less
overshoot and ringing and having a lower q factor for these poles, that are near the
crossover frequency. We're going to actually make a dominant
pole approximation and characterize the closed-loop response of t over 1 plus t,
and 1 over 1 plus t, as having two poles near the crossover
frequency, that have some Q-factor. And we'll find a relationship between that
Q factor, and the phase margin of t. So, this is a nice design tool that really tells us, how much phase margin we need,
and then provides a, a design criterion, for how to
shape the loop gain to give us a good
closed-loop response. I'm going to explain this relationship,
using a simple second-order system in which, in which we assume, that the
loop gain T, has this form. So, there is a low frequency pole that
we're going to approximate for simplicity, as a pole at the origin.
So this 1 over s over Omega knot term, is a term that goes to infinity
magnitude at DC, and then, it rolls off with a minus 20
db per decade slope, and it has a magnitude of 1 or zero db, at
omega equals omega knot. So it has this characteristic, or this asymptote, and it
goes through 0 dB at f equals f naught, or omega equals
omega naught. And then, we'll have a second pole at high frequency, at frequency omega equals omega
2, or f equals f 2, that will cause the slope to change to minus 40 dB per decade, after
that. Now, the real loop gain may be more complex, but if we
can approximate the loop gain T in the vicinity of a
cross-over frequency like this, then this should be a good
approximation, of what happens in t over 1 plus t, in the vicinity of the
crossover frequency. So perhaps, the real t actually has
multiple poles and zeros, that are happening at very low
frequency, with all kinds of things going on. But, when we get near, say, within a
decade or so of the crossover frequency, we can
approximate t like this. So this in fact, is a good approximation
for many of the systems, that we want to
design. And in fact, the results will apply to
even more cases, simply, than this. How does this behave, as far as phase is
concerned? Well, this, this s over omega naught term in the denominator, gives us a
phase of minus 90 degrees. And then the second pole at omega 2, gives
us a phase asymptote, that decreases towards minus 180 degrees,
at high frequency. And so, the phase margin then, will lie
somewhere between 0 and 90 degrees, depending on exactly, where the crossover
frequency is, relative to f 2. Let's use this approximation for t, and
calculate the closed loop response. So, we can work out, what t over 1 plus t
is, when we plug this in for t. And in fact, the easy way to do that, is to first divide the numerator and
denominator by t, and what we get is, 1 over 1 plus 1 over t,
and 1 over t is really easy to calculate, we, we plug this in. So, what we find is that, t over 1 plus t becomes this expression, that has
two poles. Alright, we can write these two poles, or
this denominator in our standard form, to identify a corner
frequency, omega c, and a q factor. What we find, when we do that is that, omega 0 omega 2, this coefficient of s
squared, is equal to this co-efficient, omega c
squared. So we get that, omega c is root omega, not
omega 2. We'll find that, when we plot omega, our
frequency on a log scale, that this square root of the product, is this geometric
mean, which on the log scale, lies half way
between. So omega c is halfway between omega 0 and omega 2. So, if we go back to this plot, here is f
0 and f 2, and f c is halfway in between. Okay, we can also work out, what q is.
So, by equating q omega c to omega 0 here, we can find that
oq is omega 0, divided by omega c. Or alternatively, we can write this, as
square root of omega 0 over omega 2. So then, here is the t over 1 plus t
transfer function. What we have, is two poles, at this f c
that, is halfway that between f 0 and f two, and
it has a q, that was omega 0 over omega c or, or f 0
over f c. And if you work out what that is, that turns out to, to follow the low frequency
asymptote here. If you have, want the equation the low
frequency asymptote, is f 0 over f, and if you plug in f as fc, you
get this value q, which is right there. So, the actual value of the function, as
I've drawn here, lies below the 0 db asymptote, and in fact, we have
the low q case. You can approximate, what the two poles
are then, using the low-q approximation, that we've
already talked about, where we'll have one pole at q times omega c, which
turns out to be omega 0 or right here, at f 0. And the other pole is at omega c over q, which turns out to
be omega 2 or f 2, right there. And the low Q approximation, then gives us
what we would get, by easing the approximate construction of the asymptotes, as we've previously
discussed. So, this is the low q case. Let's consider the high Q case. Suppose, you move f 2 down, to a lower
frequency. And here, we're going to choose f 2, to be
below f c. So, f 0 is again, the frequency where the low frequency asymptote passes through
0 db. We have to extend the low frequency
asymptot in this case, to find f 0, and f 2 is the second pole frequency.
Fc is again, have way between. In this case, the previous argument still
apply. q is omega 0 omega c or f 0 over f c,
which now, is a big number than one. We have a q larger than one, and the exact
function goes up and touches this slow frequency
asymptote and then, comes down. So, this is the high q case and we have this q, that grows. It's larger and larger, as f2 goes to a
lower and lower frequency. Well, what is happening, as far as phase
margin goes? If we move f2 to a lower frequency, you
can see that, that will move this phase asymptote, to a lower frequency, and our
phase of t at the crossover, gets lower. And as f 2 goes to dc, our phase margin goes to zero,
and, and as we, as this happens, we find that the
closed loop q, grows. We can work out an expression for the
exact phase margin, of our original t, and we can plug that into our expression
for, the q of the closed loop poles. And with some algebra, we can find this expression, for the relationship between
the q of the closed loop poles and the phase margin
from t, there's what it turns out to be. Takes some algebra to do that. Or alternatively, you can solve for the
phase margin, in terms of q. And here's an expression for, what phase
margin it takes, in order to get a certain closed loop q, for
this example. Here's a plot of that result. So, this is phase margin of t, on the
horizontal axis, and q of the closed loops poles of t over 1 plus t, on the vertical
axis, with q expressed in decibels. So, that's the answer. You can see that, if you want a q of 1,
which is 0 db, right here. It takes this phase margin, which works out to be 52 degrees. Or, if you like Q of a half for your
closed loop poles, which means we have real poles, that takes
a phase margin of 76 degrees. And the smaller the phase margin, the
higher the q factor is. So, this phase margin of 1 degree, would
give us a very high Q, and a very large resonance in
our closed-loop transfer functions, and that's a very good
reason to require more phase margin. So, we need phase margin, not just for a
yes, no stability test, but we need phase margin, to keep the resonance
in our close loop transfer functions, at a controlled and low value, 'kay? The next question is, how much q can we
allow? That really depends on your application,
but let's see, what the impact of having a higher q is. Well, this is really, an exercise in
constructing inverse laplace transforms. Let's suppose that, we have this closed
loop transfer function, with these complex poles, and what we want to do is,
work out, say the step response. So, we'll have some input to our transfer
function, that will take a step, and it's laplace
transform. Then we can express, let's say, a unit
step function can be written into laplace transform,
domain as, 1 over s. You can multiply that by this second order response, and then, take the inverse
laplace transform, to work out the exact time domain response, in the output that we
get. So, this is also an exercise in algebra
and inverse laplace transforms, but, the re, the result, in the case where
q is greater than a half, so that we have
complex poles, is this. So, if our input is, is a unit step function, then our output is this time
domain response. In the case of real poles, we get decaying exponential time, type responses and
here's what, what that is. Okay, and that's a pretty big expression,
here is a plot though. So again, our input is a unit step, that's
doing this. We would like the output to follow that,
and it eventually does settle down and follow that, but there is some trangent that happens, following the, the
step. So, for low q, we have these decaying
exponential type responses, but very monotonically and
eventually, settle down to follow the, the input. And for high Q, we have these ringing or
sinusoidal type of responses, that have overshoot and ringing, but the
ringing decays in amplitude and eventually, it
settles down. So, one has to decide, whether overshoot
is allowed. One also has to decide, how fast we want
the response to be. Q of a half is, the case where we have two
real and repeated roots, this is actually, [COUGH] the fastest response that has no
over-shoot in ringing, and it's this response, right
here. This is called the critically damped case. Cube greater than a half, gives us complex poles, this is called the under damped
case. These responses are faster, but they have
over-shoot and ringing, and take some time to settle down, to the
final value. Whereas q less than a half gives us,
what's called the over-damped case, where we have two real roots that
are at different values. And we get decaying exponential type responses that
are slower, but don't overshoot. So often, we think that the critically
damped case for q of a half is the best, because it's the
fastest response with no overshoot. On the other hand, we might allow a little
bit of overshoot, so a q of 1, say, has maybe, 15 or 20% overshoot, which we have
to decide, if that's acceptable or not, but it is a faster
response, and requires less phase margin. Generally, in power applications, we can't allow overshoots heading towards twice the
final value, like we get in the lightly damped
case, with very little phase margin. This type of overshoot and ringing,
usually will cause higher stresses on, on our
components, where the voltage will go to twice the, the final value, and usually, we can't allow
that. Here's a nice little function, that is the
peak. We can solve this function to find the
peak value, and this is what it becomes, as a function
of q. So, this peak for the under-damped case,
is the value of the first peak, right up here.
Let's say right here, as a function of q. So, if we decide, how high we can allow
the peak to be, then we can solve this to see, how
much q that requires, and then we can go back
to the previous plot here to see, how much bayes
margin is required. One other thing about this plot is that,
the horizontal axis here, is labeled in units of radions, of
omega c, times time. So, if you have a requirement that you
want the settling time or the rise time to be so many, say microseconds or
milliseconds, then we can plug that time in here, for t, and
actually work out, what the cross over frequency,
omega C needs to be. For example, if we have q of a half, and
we want a settling time of some number of seconds, we may define settling time, as
the time that it takes the wave form, to get within some
amount. Like say, within 10% of the final value,
and stay there. So, we can draw some boundary, that's 1 plus or minus 0.1, or
10%. And then see that okay, if we have q of a
half, then right there, is where we get to, within 10% of the final value, and that's maybe, four
radiants. So, you can plug in omega c equals four,
and then solve for, what omega c is required, to get the
given t, or, or time, that is the specification. You can also see that q of 1 would, will over, no, it will rise faster to this
value, but then, it will overshoot and go out of the
box, and it doesn't get back in the box, until
right there. So, it actually has a longer settling
time, but a faster rise time. So, one can decide exactly, what one needs
from the application and then use a plot like this, to choose, how big
the cross over frequency needs to be, and ultimately, what is the
bandwidth of the feedback loop. So, from this kind of plot then, we can
get a requirement on both, what omega c should be, and what
q should be. And then that in turn tells us, what the
phase margin should be, and what the feedback loop
crossover frequency should be. One last thing I need to point out is
that, I explained this with respect to a, an,
simply an example. I, I chose this somewhat arbitrary loop
gain, although I did argue that this loop gain,
is one that can approximate a great many
actual loop gains, in the vicinity of the
crossover frequency. It's possible to choose other loop gains, if their closed loop response
in well approximated, by a second order
system in the vicinity of the crossover frequency,
then in fact, you get the same answer, which is
this. And there's a, actually a good fundamental
reason for that, but as long as you correctly work out the magnitude and phase
of t, and relate it to q, you will get the same answer. If, if the system can be approximated by
this dominant pole or two poles, second order system, in the
vicinity of the crossover frequency. Now, of course, you can design more
complex feedback loops, whose closed loops response has more than two
poles, near the crossover frequency. And in that case, you can't characterize
the response, simply by this parameter q, and, and omega c, it takes more
parameters. So, in that case, you have to do more work and use more advanced, control techniques,
to work out what happens. But in fact, the vast majority of our
systems or feedback with systems, that we design in para electronics, do end
up being second order systems, and we can use these simple arguments
then, to design them.