In this screencast we will look at how a first order transfer function responds to a step
change in the input. We will do this in two parts. The first part we will look at a generic
transfer function which has a gain K, time constant tau. So we want to see how a transfer
function with this generic gain and time constant reacts to a change in magnitude M as a step
function of our input variable x. The reason why we are keeping things in generic terms
K, tau, and M, instead of numbers is because if we develop a generic relationship, this
relationship will hold for all first order transfer functions, not mattering both the
magnitude of the values or what type of process it is. Once we have done that we will use
the answer from part (a) to allow us to asses the transfer function part (b), relating the
height of fluid in a tank to the inlet flow rate. Part (a) is asking us to look at how
a step function of magnitude M for our input variable here x influences our output variable
here, y. So what we first have to do is since our transfer is in the Laplace domain, we
have to turn our variable, our step function of magnitude M, into the Laplace domain so you
can use a Laplace transform sheet and know the fact that the Laplace transform of a step
is just M over S. So what can now do is substitute in M over S for x, and we end up with y over
M over S equals the generic first order transfer function in standard form. If we move the
M over S over, we end up with y equals KM over S times tau s plus one. We cannot take
a Laplace transform of this function quite easily so we have to use either partial fractions
or the heavy side method. Here we'll use partial fractions, and this will just become A over
S plus B over tau s plus one. We then find a least common denominator for our process,
which will be S times tau s plus one, and we know the fact that the numerator has to
equal KM. If we look at our terms on the left hand side of the equation we see the fact
that the highest power of S here we have is S to the first. If we factor A out we end
up with tau times A, and the S to the first power, and then we also have A for S of the
zeroth power. For B we just have the S, there is no S to the zeroth term, just an S to
the first term. So now we have to match this to what our right hand side is. So there is
no S term, so for the S to the first term it will just be zero, and our S to the zeroth
term will have to equal KM. Well this makes our mathematical lives not terribly difficult
because we have already shown what A equals, A equals KM, and what we then have is tau
times KM plus B equals zero, and that allows us to solve for B which is equal to negative
KM times tau. Substituting these values in leaves. Now that we have separated out the
two terms, we can easily take the inverse Laplace transform of both of these. The inverse
Laplace transform of KM over S is just going to be KM, and the inverse Laplace transform
of KM tau over tau s plus one is going to be KM tau the constant multiplied by the inverse
Laplace transform of one over tau s plus one, which is one over tau times e to the negative
t over tau. The taus will cancel, and we will have a minus sign coming back down. The inverse
Laplace transform of the left hand side will just simply be the y of t. If we factor out
the KM we end up with an answer of KM times one minus e to the negative t over tau. So
if we want to see what this means graphically we can see this is an exponential of some
sort, so if we sketch this out where our x axis will be time and our y axis will be x
of t and y of t. X of t will just be a step function of magnitude m, and if we were to
sketch this KM times one minus e to the negative t over tau it would look something like what
is drawn here. And there are a couple important values here: The first is the fact that this
function will converge to KM. Why it will converge to KM is the fact that at long times
the e to the negative t over tau term will approach zero because t will get high, so
e to the negative infinity, which is zero. So therefore at long time this function to KM, so in
other words the steady state of the system. Another point to note here is the fact the
curve here was drawn such that the steady state value is larger than the step function.
That would indicate the fact that the gain here is greater than one. But note the fact
that it could be drawn such that the steady state ends at the same place of x of t, which
where the gain would be one. A value less than x of t which could mean one of two things:
the fact the gain is less than positive one, and also could mean the fact that the value
could go negative, meaning the fact that the steady state gain is negative. One more important
landmark is when y of t equals sixty three point two percent, or point six three two
of KM. What that corresponds to is a time that is equal to tau. So with this information
that we have obtained we can now begin to solve part (b), which gives us information
about a first order transfer function relating the height of fluid in the tank to inlet flow
rate, so therefore our input variable here would be the inlet flow rate, which we will
call here q i, and the out variable will be the height of fluid inside of the tank of
h. It says that it is a first order transfer function with a gain of ten, in units of
height per volumetric flow rate. Our time constant is given as fifty seconds. That gives
us that our transfer function is ten over fifty s plus one. It gives us information
about the steady state system, where the inlet flow rate is point four and the fluid height
in the tank is three meters. When the inlet flow rate is increased instantaneously to
point four one cubic meters per second. So the first question is what is the height of
the fluid in the tank a hundred seconds after the increase has occurred? So in order to
analyze this we will substitute in our variables here which are K, which is the steady state
gain of the process, that is ten, M, which is the magnitude of the step change. So here,
the problem statement states that it has increased instantaneously from point four to point four
one, therefore our change in magnitude is point zero one, and that our tau in this problem
is given as fifty seconds. We are interested in what is going to happen at a time of one
hundred seconds. So if we substitute everything in, we can find some information about y of
t, where in this case y is the output variable, which is H. When we substitute all of these
numbers in we get an answer of H equals point zero eight six meters, but at a first glance
this may look strange because our initial steady state height inside the tank as given
in the problem statement was three meters, yet all of a sudden our answer is point zero
eight six meters, this doesn't make any sense, but it makes sense when we understand how
we developed that y of t function. So to gain some better insight on this lets quickly re-sketch
our process here. For a first order transfer function we notice the fact that at the end
it converges to KM. At time equals zero, the system equals zero, not the initial height,
even though it is often not explicitly show, the y of t relationship represents a deviation
from the steady state. So what that means is that the final height is really h r plus
the h we found, which is going to equal three point zero eight six meters. So again, an important
point to note here is the relationship that is provided here is basically in deviation
variables. The second part asks what will the steady state value be for the height,
and if you recall from our analysis before is that as t goes to infinity, the exponential
term goes to zero. In other words y of t is going to equal KM as t approaches infinity.
So what that means is that our steady state height is going to equal our original starting
height, h bar, plus KM. So here, that will be three point one meters.