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Welcome
to the lecture titled Steady State Gain from Asymmetrical Relay Test. In this lecture,
we shall derive some explicit and exclusive expressions meant for getting the steady state
gain from the relay test. The asymmetrical relay test will result in asymmetrical output,
and we have attempted earlier to measure four parameters on the asymmetrical output. Now,
we can make use of the asymmetrical output and input to the system to estimate one more
parameter associated with the transfer function model.
Now, the transfer function model is G s is equal to k plus minus T 0 s plus 1 times e
to the power minus theta s upon T 1 s plus minus 1 T 2 s plus 1. So, the transfer function
model has got five unknowns; and those are the steady state gain k, the 0 t 0, the time
delay theta, and the two time constants T 1 and T 2. So, in all we have got five unknowns;
but so far, we have been able to make four measurements on the asymmetrical output of
the relay system. Now, those measurements are A p, the peak
amplitude of the output signal; A v, the negative peak amplitude of the output signal and two
zero crossings those are namely t 2 and t 4; and we have also found that at time if
2, x t 2 is equal to 0 and at time t 4, x t 4 is equal to 0. Thus the four measurements
have enabled us to estimate four unknowns associated with the transfer function model
using four non-linear equations. In this lecture, we shall try to develop analytical
expressions, so that can be used to estimate the steady state gain associated with the
transfer function model. What is steady state gain? When s tends to 0 or the frequency of
a system output becomes 0 at that time G 0 becomes k and that is known as the steady
state gain. How can we find the steady state gain? For that, we have to concentrate on
the asymmetrical output and if I concentrate on one period of the asymmetrical output,
the area of the asymmetrical output can be obtained conveniently using the analytical
expressions we have derived earlier. So, what will be the area of the asymmetrical
output for one period, this will be the area for the asymmetrical output . Let us denote
the asymmetrical output by the symbol a y, thus a y can be given in the form of an integral
which is starting from time t 0 to t 4 with y t d t. So, that will give us the area of
the asymmetrical output signal whereas, the area of the input signal can be also obtained,
whereas the input signal area of the input signal can be shown as this one.
So, then we have to make use of this to find the two areas, where a u will be the areas
for the input signal area for the input signal and next the ratio of the areas will give
you the steady state gain. So, k will be equal to a y upon a u. So, all these things will
be developed in sequence.
Now, I will start with the asymmetrical output first. So, the output will assume this form
for one period of time. Now, y t is the output and I will denote different time instants
to find the area of this signal. Now, I will start from time t equal to 0 till different
span depending on the type of input signal we have for the system. So, this is our input
signal, which has got a number of piecewise constant inputs during different instant of
time. Now, what is this, this is our theta, so this
span is denoted by theta. And here, we get the period tau and for the full period we
have got tau p starting from time t equal to 0; to find the area of this signal from
time t equal to 0 to tau p, now I have to write the expression y t is equal to integral
from 0 to tau p y t dt sorry not this is not the output, this is the area of the output.
So, I write a y is equal to integral from 0 to tau p y t d t.
Now, how to find y t for different time segments, because we have got different type of inputs,
a number of piecewise constant input to the system; therefore, the output during that
particular piecewise constant input has to be found initially. So, this has to be expanded
and written in the form of integral from 0 to theta y t dt plus theta to I will start
from here, till we go to this point, so that will be our tau plus theta; so, we go from
theta to tau plus theta y t dt plus tau plus theta to tau p y t d t; this will give the
three terms combined together will give us the area of the output signal.
To find the area of the output signal, I need to find the state of the system at different
instant of time, because we know that the output can be found using the state variables
x t, where y t becomes c x t, c is the constant of the dynamics of the second order plus dead
time system.
Now, for that I will start deriving the state equations for different segment of time. So,
we know that the solution of a state equation is given by the general expression, x t is
equal to e to the power A t minus t 0 x t 0 plus integral from t 0 to t e to the power
A t minus tau B u tau minus theta d tau. This is what we get for the dynamics of a system
at any instant of time, where x t stands for the state of the system as you know.
Now, since in our case luckily we have got u tau minus theta is a constant either h 1
or minus h 2 using that it is very convenient to find the second part of the x t, which
becomes t 0 to t e to the power A t minus tau B h 1 d tau or minus h 2 d tau. So, let
us take u tau minus theta is equal to h 1, which results in the integral part integral
from t 0 to t e to the power A t minus tau B h 1 d tau.
So, let me find the integral which comes out in the form of A inverse then, we will have
integral e to the power A t minus tau with the limits t 0 to t B h 1; when the limit
is put then we get it in the form of A inverse I minus e A t minus t 0 B h 1. So, thus we
get the integral in this form. Now, the integral can also be obtained using one simpler expression
which can be given in the form of 0 to t minus t 0 e to the power A s B u s d s.
So, either one can use the upper one or the bottom one, because the bottom one gives you
very simpler expression compared to the earlier one. Now, the bottom one also can be solved
and found as, A inverse e to the power A s with the limits 0 to t minus t 0 and you have
got u s equal to h 1, so B h 1; and that will give you A inverse e to the power A t minus
t 0 minus I B h 1. So, there is difference in sign only. So, the you needs not worry
about that, because here the integral will be found with respect to e to the power minus
A tau so obviously, there will be a minus here and ultimately you get here minus and
which gives you finally, in the form of A inverse e to the power A t minus t 0 minus
I B h 1. So, I get the same expression using the upper
or the lower integral. So, I shall use henceforth, the lower integral in place of the upper one
for finding the state of the system at any instant of time.
Thus I can write x t h e to the power A t minus t 0 x t 0 plus integral from 0 to t
minus t 0 integral from 0 to t minus t 0; please keep in mind the limits of this integral,
which always starts from 0 to the time span for that segment of the input, t minus t 0
e to the power A s B u s d s. Thus I shall make use of the lower one, the lower expression
to find the output of the symmetrical sorry output of the relay control system for different
time spans. Now, I I shall start finding the output for different time spans.
Let me redraw the output signal once more. So, that it will be possible for you to follow
the different time spans conveniently. So, the output is given in this form, I will draw
the input to the system y t is the output, t is the time axis. And the input to the system
becomes like this for one period of the output signal, where the input has got heights h
1 and minus h 2. Now, as you know this span is the theta time delay associated with the
dynamics of a system whereas, this span we take as tau from here; whereas, this span
again becomes theta and finally, period of the output is given by tau p.
So, the output y t for different segments can be used using the expression y t equal
to c x t. Now, when the y output is considered part of the output is considered for the time
range 0 is less equal to t is less equal to theta; then, the output for this time span
can be found using the expression c x t, where your time span is from t ranging from 0 to
theta. So, x t again can be obtained using the general expression which is nothing but,
e to the power A t minus t 0 x t 0; then I have the integral for the second part which
is obtained in the form of A inverse, so I will have c again c plus c A inverse e to
the power A t minus t 0 minus I times B h 1.
So, let me write once more. So, the output signal will be given by the expressions c
e A t minus t 0 x t 0 plus c A inverse e to the power A t minus t 0 minus I B h 1; because
the input to the system at that time is this piecewise constant input h 1 for which we
have got the output starting from time t equal to 0 till time t equal to theta. So, this
part of the output of this part of the signal is given by this expression y t. Now, to find
the area under these let me designate that area by symbol p 1. So, p 1 will be from time
0 to theta y t dt then, the area can be found now using the expression yes I will I will
find the area later later on. So, let me find the state where the output for different segments
initially then, we will start finding the area later on. Then for this time range we
have got the state expression as this one, which is multiplied with c to give the output
y t.
Similarly, for another time range which is starting from y t. So, for this time range
where I start from time t equal to theta and I will go up to time t equal to tau plus theta
now; in that case, the state variable for that time range for the time range, theta
is less equal to t is less equal to tau plus theta. The state variable x t is given as
e to the power A t minus theta x theta plus A inverse e to the power A t minus theta minus
I now B; what will be the input at this time, the input at this time is minus h 2 therefore,
we will have minus h 2 over here. So, upon simplification which again gives
us the expression e A to the power t minus theta x theta plus or I can write here minus
A inverse e to the power A t minus theta minus I B h 2. So, this is what you get for the
state variable for the second time range for the time range t between theta to tau plus
theta. Then the area of that time span can be obtained using the output and in which
case the area can be obtained as shown here in the see that part of the output signal.
Now, whereas why we have been able to get this expression, simply I have substituted
the u s by its appropriate input in this case it is minus h 2, rest of the things will remain
as it is. And what changes have been made to this state equation? We start the output
signal from time t equal to theta till time t equal to tau plus theta therefore, the t
0 becomes theta. Please keep in mind t 0 is theta therefore,
you have got x theta over here, but x theta again we know that x theta is equal to e to
the power A theta x 0 plus A inverse e to the power A theta minus I B h 1. How do you
get this one, this can easily be obtained if you substitute t equal to theta here. So,
when t equal to theta is substituted over here and t 0 equal to 0, one obtain x theta
conveniently. So, when this x theta is used over here finally,
I get an expression for the x t as x t is equal to e to the power A t minus theta, x
theta will be substituted here e to the power A theta x 0 plus A inverse e to the power
A theta minus I B h 1 B h 1 then minus A inverse e to the power A t minus theta minus I B h
2. So, this is the expression for x t for the time spanning from time t equal to theta
to time t equal to tau plus theta. So, let me repeat and rewrite this expression after
simplification.
So, that x t for the time range for the time range theta is less than equal to t is less
than equal to tau plus theta, x t will have the final form given by e to the power A tau
plus theta x 0 now plus A inverse e to the power A tau plus theta minus e to the power
A tau B h 1 minus A inverse e to the power A tau minus 1 B h 2. Similarly, we are left
with one more segment of the output signal; therefore, we have to make use of this concept
to find the state variable for the last segment of the output signal.
So, the last segment of the output signal can be again shown as time t y t, where the
asymmetrical output again is shown in this form and the input to the system is like this;
therefore, the last segment starts from this instant time instant therefore, I have to
concentrate the output from here till the zero crossing and for that, the area again
all we obtained by this shaded part. Now, to find the state variable for the system
for this time range time t, this is your tau plus theta. So, this span is now given as
tau plus theta whereas, this is our tau p.
So, for the time range for the time range for the time range tau plus theta is less
than equal to t is less than equal to tau p; for this time, the state variable will
be given by x t is equal to e to the power A t minus tau plus theta x tau plus theta
then plus A inverse e to the power A t minus tau plus theta minus I B h 1 now. Again what
is x tau plus theta? x tau plus theta is nothing but, e to the power A tau plus theta x theta
plus A inverse e to the power A tau plus theta minus e to the power A tau B h 1 minus A inverse
e to the power A tau minus 1 minus I B h 2; what is I? I is the identity matrix of the
order of A. So, x t upon simplification can be found in
the form of e to the power A t minus tau minus theta x tau plus theta will be substituted
over here. So, finally giving us in the form of e to the power A tau plus theta x 0 sorry
this is not theta. So, this is equal to x 0 then this is your x 0 plus A inverse e to
the power A tau plus theta minus e to the power A tau B h 1 minus A inverse e to the
power A tau minus identity matrix time B h 2 plus, the remaining part as it is e to the
power A tau plus theta minus e to the power A tau B h 1 minus A inverse e to the power
A tau minus I B h 2.
So, we have got a horrible expression which can further be simplified and written finally
in the form of x t for the time range 0 for the time range tau plus theta to tau p. The
state variable will assume the form e to the power A t x 0 plus A inverse e to the power
A t minus e to the power A t minus theta B h 1 minus A inverse e to the power A t minus
theta minus theta t minus theta minus e to the power A t minus tau minus theta B h 2
then plus A inverse e to the power A t minus tau minus theta minus the identity matrix
time B h 1. So, as expected the state for the final segment
of the output is involving two inputs, two linear piecewise constant inputs h 1 and h
2. As you see here, the input to the system is assuming this form; whereas, during this
input we have got the output which is going from negative to positive. Therefore, we have
got the inputs h 1 and h 2 present in this expression. Now, after finding these expressions,
what to do with these all these expressions? So, we have been able to find the expression
for x t the state variable for different time ranges then it is possible to find the area
of the output signal using the expression.
The area of the output signal asymmetrical output signal is equal to integral from 0
to tau p y t d t, which is nothing but for our case 0 to tau p c x t d t. Now, I will
divide this into different time ranges as you have seen, because you have got different
piecewise constant input. Therefore, we cannot have the expressions expressed by single expression.
So, that way we have to make use of all the three time spans to find the final area of
the symmetric asymmetrical output signal. Now, this goes from the limits 0 to theta
c x t dt plus theta to tau p tau plus theta c x t dt plus tau plus theta to tau p c x
t d t. Now, I shall substitute the expressions x t, x t, x t we have found for different
time ranges. Now, the integral is for the limits 0 to theta, so the t is between 0 to
theta in this case, now for the second part it is your theta is less than equal to t is
less than equal to tau plus theta and for the third part it is tau plus theta is less
than equal to t is less than equal to tau p.
So, keep in mind one cannot make use of the single expression to find the area of the
output signal, because we have got three different piecewise constant input to the systems which
changes their magnitude at different instants of time. Therefore, the piecewise constants
input are to be considered definitely to find correct expression for the output of the asymmetrical
output of the system.
Now, I shall go on substituting the expressions for x t for different time range. So, let
us write this a a y by three parts, p 1 plus p 2 plus p 3. And try to find the parts individually
initially before combining and finding the final area of the output signal.
Then p 1 is given by the expression integral from 0 to theta c x t d t, which is again
written as now integral from 0 to theta c e to the power A t x t 0; and in this case,
t 0 equal to 0 therefore, directly I can write this as x 0 plus c A inverse e to the power
A t minus I B h 1 times d t. So, what is this, the c has been multiplied with x t the state
variables of the system and we have to integrate it for the time range t spanning from 0 to
theta. Now, when I start integrating this one I will
find the expression C A inverse e A t with limits 0 to theta with the multiplier x 0
again for the second part it will be C A to the power minus 2 e A t with the limits 0
to theta; then, the remaining part will have the expression minus c A inverse integral
from 0 to theta for t dt therefore, your left with t with limits 0 to theta times B h 1.
So, which upon simplification gives us, C A inverse e to the power A theta minus I times
x 0 plus C A minus 2 e to the power A theta minus I minus this will be simply C A inverse
B h 1 theta. So, the part p 1 has been found in this form p 1, let me repeat once more.
The p 1, the area part of the area of the output signal given by p 1 is given by the
expression p 1 is equal to C A inverse e to the power A theta minus I times x 0 plus C
A to the power minus 2 to the power times e to the power A theta minus I minus C A inverse
B h 1 theta. Then, similarly I can get an expression for
p 2, p 2 is from the integral from theta to tau plus theta c x t d t. Now, the correct
expression for this time range when put in this expression gives us p 2 as, p 2 is equal
to integral from theta to tau plus theta c e A t minus theta x theta minus C A inverse
e to the power A t minus theta minus I B h 2 d t. So, this integral can also be found
like the previous case.
And simplified to give us finally p 2 as, C A inverse e to the power A tau minus I times
e to the power A theta plus sorry e to the power A theta times x 0 plus A inverse e to
the power A theta minus I B h 1 minus C A to the power minus 2 e to the power A tau
minus I B h 2 plus C A inverse B h 2 tau. So, keep in mind the last term interestingly
the last term is found in the form of C A inverse h 2 tau; if you go back, see the last
term of p 1 again it is obtained in the form of minus C A inverse B h 1 theta. So, the
last term of the integrals of different parts are found to have the expression, C A inverse
B; so, C A inverse B that is one important observations we have made so far.
Now, I will try to find the third part, p 3 for which we have the integral starting
from time tau plus theta to tau p c x t d t. So, when this x t for this one will be
which one, the x t for this one is this one. So, the when this x t is substituted in p
3 then we obtain that expression in the form of integral from tau plus theta to tau p c
e to the power A t x 0 plus A inverse e to the power A t minus e to the power A t minus
theta B h 1, B h 1 will come here then minus A inverse e to the power A t minus theta minus e to
the power A t minus tau minus theta B h 2 plus A inverse c e to the power A c e to the
power A tau t minus tau minus theta minus I B h 1 d t.
So, C will not come here, because C is there already here. So, when again this is expanded
and simplified will get an expression for this p 3 is equal to I will write the final
expression of p 3, which will not be so simple.
But, let me try to write the whole expression for p 3 which comes out to be in the form
of c A inverse e to the power A tau p minus e to the power A tau plus theta times x 0
plus C A minus 2 e to the power A tau p minus e to the power A tau p no this will be tau,
A tau plus theta minus e to the power A tau p minus theta plus e to the power A tau B
h 1 plus sorry this is not plus minus C A to the power minus 2 with terms e to the power
A tau p minus theta minus e to the power A tau minus e to the power A tau p minus tau
minus theta plus I times B h 2, again plus C A to the power minus 2 e to the power A
tau p minus tau minus theta minus I B h 1 with the last term minus C A inverse B h 1
times tau p minus tau minus theta. So, again see the last term, interestingly when all
the parts are added together what do we get?
When p the area is found in the form of a y is equal to p 1 plus p 2 plus p 3. Then
one obtains, C A inverse e A theta minus I plus e to the power A tau plus theta minus
e to the power A theta plus e to the power A tau p minus e to the power A tau plus theta
times x 0 plus C A to the power minus 2 e to the power A theta minus I plus e to the
power A tau plus theta minus e to the power A theta minus e to the power A tau plus I
plus e to the power A tau p minus e to the power A tau plus theta minus e to the power
A tau p minus theta plus e to the power A tau plus e to the power A tau p minus tau
minus theta minus I times B h 1 with another term as C A to the power minus 2 I minus e
to the power A tau minus e to the power A tau p minus theta plus e to the power A tau
plus e to the power A tau p minus tau minus theta minus I B h 2 minus C A inverse B h
1 theta plus C A inverse B h 2 tau minus C A inverse B h 1 tau p minus tau minus theta.
So, this is the expression for area of the asymmetrical output signal, where we will
find that all other parts are terms then the last three terms becomes 0; when x 0 is substituted
over here, we have an expression for x 0 which had been found in the previous lecture. So,
when x 0 is substituted over here, then the first term first few terms get cancelled,
leaving us a y as a y is equal to minus C A inverse B h 1 theta plus C A inverse B h
2 tau minus C A inverse B h 1 tau p minus tau minus theta.
So, interestingly we get only simpler three expressions remaining in the expression for
the area of the output signal. This is an important observation. So, how the other parts
are getting cancelled, if you look carefully so many terms are cancelling out like this
one, this one cancelling out, e to the power A theta minus e to the power A theta, leaving
us only two terms in the first part. Similarly, e to the power A theta A theta
cancelling out, cancelling out minus I plus I cancelling out then, minus e A tau plus
e A tau cancelling out, and leaving three terms there; and here also e to the power
plus minus A tau. And you have got plus minus I then, whatever remaining terms you have
upon substitution of x 0 in that term.
What is x 0? x 0 is found to be in our earlier lecture we have seen that, x 0 is given by
the expression I e to the power minus A tau p inverse e to the power A tau p minus theta
gamma 1 h 1 minus e to the power A tau p minus tau gamma 2 h 2 minus e to the power A tau
p minus tau minus theta gamma 3 h 2 minus gamma 4 h 1; where again gammas are there,
when the gammas are substituted finally one obtains x 0 in the form of I minus e to the
power A tau p inverse with A inverse e to the power A tau p minus tau minus theta plus
e to the power A tau p minus e to the power A tau p minus theta minus I B h 1 with another
term minus A inverse e to the power A tau p minus theta minus e to the power A tau p
minus tau minus theta times B h 2. So, when this x 0 is substituted here, then
the first few terms apart from the last three terms cancels out and we are left with only
three terms.
Therefore, the output of the relay test has the final form given as, a y is equal to minus
C A inverse B h 1 theta plus C A inverse B h 2 tau minus C A inverse B h 1 tau p minus
tau minus theta. Then, what is C A inverse B? We know that, from the state space model
of the second order plus dead time transfer function model we have obtained C. C is equal
to k lambda 1 lambda 2 lambda 1 plus lambda 3 upon lambda 3 lambda 1 minus lambda 2; and
the other element is minus k lambda 1 lambda 2 lambda 1 lambda 2 lambda 2 plus lambda 3
upon lambda 3 lambda 1 minus lambda 2. So, this is the C vector with B vector we
know has got the elements 1 and 1; whereas, A is given by lambda 1 0 0 lambda 2 matrix.
Now, C A inverse B becomes C A inverse B becomes k lambda 1 lambda 2 lambda 1 plus lambda 3
times lambda 3 lambda 1 minus lambda 2 minus k lambda 1 lambda 2 lambda 2 plus lambda 3
upon lambda 3 times lambda 1 minus lambda 2 multiplied by A inverse B; so, A inverse
B will be 1 upon lambda 1 and 1 upon lambda 2.
So, when this is simplified we get, k lambda 2 lambda 1 plus lambda 3 by lambda 3 times
lambda 1 minus lambda 2 minus k lambda 1 lambda 2 plus lambda 3 by lambda 3 lambda 1 minus
lambda 2, which is nothing but equal to minus k. So, interestingly we have found C A inverse
B to be minus k then, a y can be written as you see in all the three terms you have got
C A inverse B. So, why do not you take common then giving us a y as C A inverse B times
minus h 1 theta plus h 2 tau minus h 1 tau p minus tau minus theta; but, C A inverse
B is equal to minus k.
Therefore, allow me to write here minus k and writing like that I finally get the expression
for a y is equal to k times h 1 tau p minus h 1 plus h 2 tau. So, this is the expression
final expression for area of the output signal. So, whatever might be the area of the output
signal, the real output area of the output signal can be easily obtained using this simpler
expression; which involves the parameters relay heights and few measurements like first
sampling first zero crossing instant, second zero crossing instant from the beginning which
is considered to be 0 with the multiplication of k.
Now, let us try to find the area for the input signal now. Then what will be the area for
the input signal? As you have seen, the span for different time instants for the input
signal can be obtained looking at the output signal only. And for our case, the input signal
appears to be of this form it does not overshoot. So, we will have a zero crossing at this instant
of time. Now, I will write the difference spans now, what is this input signal? The
input signal is going from sorry y t and u t versus t. Now, this is the span x axis is not correct now, because I have
to start from 0. Let me redraw again, because it is getting confused.
To find the input signal let me plot the output signal first t, y t. So, the y t is like this
and we have got the input signal given by this zero crossing and this one. Thus, what
are the spans here, this is theta; what is this, this span is equal to tau; what is this,
this is equal to theta. And similarly, finally one period is given by tau p. Thus, if I try
to find the area of the input signal I have to consider the shaded area shaded part only.
So, to find the area of the shaded part, what I have to do? I have to consider, the heights
h 1 and minus h 2. Thus area of the input signal a u can be written as h 1 theta this
rectangle; then coming to the second rectangle the bottom one, what its width and height?
Height is equal to minus h 2 and width is equal to this will be tau plus theta minus
theta. So, this span will be equal to tau yes, and
the area of the upper rectangle now can be found as h 1 time tau p minus tau minus theta;
so, tau p minus tau minus theta. So, when this is simplified again what we get the expression
for a u as a u as h 1 tau p minus h 1 plus h 2 tau. So, if you look at the expression
for a y and a u, let me give you a y and a u once more.
So, a y is found to be k times h 1 tau p minus h 1 plus h 2 tau; and a u is found to be h
1 tau p minus h 1 plus h 2 tau. Thus the steady state gain k, the steady state gain state
gain can be obtained from the ratio of the areas of the output to the area of the input.
So, this is how the steady state gain of the transfer function model is found. So, when
you get the asymmetrical output signal measure the or find the area of the output signal
and area of the input signal take the ratio of the two; that will give you the steady
state gain of the transfer function model.
So, let me summarize my lecture now. Analytical expressions for the areas of input and output
signals, a y and a u have been found. Now, taking the ratio of these two areas, it is
possible to estimate the steady state gain of the system; thus k is equal to a y upon
a u. Why we have been doing so because we know the transfer function model the in our
transfer function model we have got five unknowns, earlier we had developed four non-linear equations
to estimate four unknowns; now, the fifth unknown the steady state gain of the system
can be estimated using the areas of the output and input signals.
Any points to ponder, what are the limitations any limitation of the proposed technique?
Yes, this technique is applicable only when the output the output is asymmetrical or I
mean the average output a y does not becomes 0 or a u does not become 0 especially. So,
this is one of the limitations of this method that, it cannot be used for sustained oscillatory
signal of symmetrical measure. That means when the output is symmetrical in that case,
we cannot make use of this technique to find the steady state gain of the system.
Is it possible to extend the technique for simple systems also? Yes, it is possible to
extend this technique for simple systems in which in which case, luckily for first order
plus dead time systems especially what happens, one can make the one can find the area of
half period signal, unlike the full period signal one can find the area of half period
signal. So, a y dash and similarly area of half period
of the input signal, a u dash and the ratio of the two will give you the steady state
gain that is particularly for first order plus dead time systems; that is whereas, this
technique may not be applicable for large order systems or systems with higher dynamics,
thank you.