In this screencast we will take a look at lead-lag systems, with a particular focus
on first-order lead-lag systems. So what is presented on the screen here is the standard
form of the transfer function for a first order lead-lag system. Lets first take a look at
why this transfer function gets its name. Lets first start with the denominator, the
denominator represents , as the subscript indicates, this represents the lag portion,
and to understand this think about when we develop regular first-order transfer functions
of the form, lets say K over tau s plus one. So when we draw the response to a step function
we see something that looks like this, this being time and this being the variable y.
So what we notice is a few things. One is the fact that the system does not immediately
respond to a step change of some magnitude, it takes some time. Also remember the fact
that after one tau the system reaches sixty two point three percent of its maximum, or
at five tau the system reaches about ninety nine percent of its maximum, so therefore
what its saying is the fact that the higher tau is, the slower the response. So in other
words, the dynamics of the denominator, that tau value, act as a means to slow down the
response, which his indicated by the fact of its presence in the denominator. So therefore
if you were to see a term with a respect to s in the numerator, you would expect the opposite,
the fact that its presence would be there to speed up the dynamics, which is the case,
hence y is referred to as a unit. From a sense of a chemical process, seeing a first-order
lead-lag system isn't terribly common, however lead-lag systems are crucially important when
developing control systems. Most notably lead- lag compensators, and developing appropriate
transfer function for feed forward controllers. So therefore there is an importance from a
ChemE standpoint, even though it might not be exactly seen in the mathematical model. Speaking
of the mathematical model, looking at this equation, what type of differential equation
would we need in order to get a first order lead-lag response. If we think about it for
a second the denominator term always comes from the dynamics with respect to the y value,
so therefore this would be something on the order of tau lag dy dt plus y, but since
we have this tau lead s plus one in the denominator, what that is indicative of is actually the
fact that the input variable, in our case here x, also has some derivative relationship
with respect to time. In that tau s plus one form we have a similar functional form here
on the left hand side and the right hand side. With K, the steady state gain, the
same steady state gain that for any first order transfer function or most other transfer
function that we've seen before, providing that ratio between the change in output and
the change in input. So this is our differential equation, which allows us to get the above
transfer function. So what we'll now do is take a look at how a lead-lag process responds
to a step input change. So in order to do that we would have y, the step increase in
the Laplace domain, or decrease, be represented by M over s, and then we would just bring
back our transfer function. We move the M over s over and we end up with. From here
we can break down into its specific roots, so s and tau lag s plus one, and we can use
partial fractions to solve for A and B. So we now have our equation, so if we take the inverse
Laplace transforms, the inverse Laplace transforms of KM over s will just be KM, and the inverse
Laplace transform of the second term will be the coefficient multiplied by one over
tau lag, multiplied by e to the negative t over tau lag. Also, if we look at our output
variable y, the inverse Laplace transform is just the variable of the time domain, so
this will be our final answer. After doing some simplification rearrangement, we can
restate this as y of t equals KM multiplied by one minus one minus tau lead over tau
lag multiplied by the exponential. What does this actually look like? So we see the
fact that we have some decaying exponential behavior here to give us the idea of what
our curve will have some look like, in fact it will something that will decay to a steady
state, in either a positive or negative manner. So in order to analyze this, we will take
a look at our initial and final conditions. So y of zero would be KM multiplied by one
minus one minus tau lead over tau lag multiplied by this exponential. Well the exponential
of zero is going to be one, so therefore the inside of the brackets will become one minus
one, minus a minus becomes a plus, tau lead over tau lag. The ones cancel, and that shows
the fact that our initial value will be KM multiplied by the ratio of tau lead over tau
lag. So an interesting point to note about this type of function is the fact that its
initial value is in fact not zero. Now we will evaluate the system as y approaches infinity
to figure out where we'll end up, so we will do the same thing except we substitute in infinity
for time, and what that will mean is the fact that the exponential goes to zero, meaning
that this whole second term will go to zero. That means the fact that this system will
converge in the end to KM. What we'll now do is begin to graph this function, but what
we will do for our y axis is we'll graph the normalized value of y, which is divided by
KM. Our x axis will still remain time. So an important point to note here is the fact
that since we've normalized by KM, the system will start at the ratio, if we go back looking
up here, it will start at the ratio of tau lead over tau lag, whatever that may be, and
they will all converge to KM. So since we have normalized by KM that means that here
this system, the systems will all converge to one. There are three possibilities
that exist. The first possibilities are for cases where tau lead is greater than tau lag,
and what that means is the fact that our initial condition will be at some value greater than
one, but will then will decay is some exponential like fashion towards the steady state value
of one. So these are the cases when tau lead is greater than tau lag. When tau lead equals
tau lag what you'll notice in the equation is the fact that this term here will go to
zero, so in other words the process will equal one, and what that means is the fact that
the lead and the lag both, in some ways, cancel each other out. Remember the fact that the
lead is quickening the process up, while the lag is slowing it down. So when they are equal
both the quickening actions and the slowing actions are cancelling out. Finally, for cases
where tau lead is less than tau lag, which can also include tau lead being a negative
number, you see the opposite of what we saw where it was greater than tau lag. The system
will start below one and works itself towards the steady state value of one. And these two
here show what values are negative but it also can very well be the fact that the lag values
can also be greater than the lead and both being positive numbers. This screencast shows
the nuts and bolts associated with understanding lead lag processes, why they exist, and why
they are important in chemical engineering processes.