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Welcome to this lecture, a wealth of techniques
is available for analysis of linear time invariant
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systems and design of controller for SISO
systems; what a SISO system, it is a system
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that has got single input and single output.
In this lecture, we shall try to design a
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powerful technique for such SISO process,
in spite of the wealth of techniques available
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still search is on to find suitable, and most
acceptable technique for PID controller design.
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We shall see how a simple but powerful control
design technique can be developed.
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Coming to the PI controller for SISO process,
here we have got a SISO process given by the
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transfer function 4 upon 2 s square plus s,
and we have got a PI controller given by the
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proportional gain 2, and the integral gain1,
the PI controller can also be written in the
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form of G c s equal to 2 plus 1 by s, which
in the standard form becomes 2 times 1 plus
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1 upon 2 s; so, the integral time constant
becomes 2 second.
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Now, why this process is known as a SISO or
single input single output process, the process
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gives us one output, and the process is subjected
to one input, where we get such type of processes;
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let us consider this room, the room can be
treated as one such process, if we try to
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control the room temperature, in that case
the input to the room could be the full air
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that can be injected and the output from the
room will be the room temperature. So, we
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can develop a controller for control of room
temperature, and in this case, we have got
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the process G s which dynamics is given by
4 upon 2 square plus s, as we have said earlier,
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but when this controller is employed, we get
an oscillatory output from the process, why
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do we get such oscillatory output to understand
the logic behind that we need to analyze the
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systems dynamics.
Let us consider the systems closed loop transfer
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function, closed loop transfer function which
can be given in the form of GGC upon 1 GGC,
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for analysis let me assume the dynamics of
this controller G c to be in the standard
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PI controller form K p times 1plus 1upon T
i S, then that will give me the closed loop
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transfer function as K p 1 plus one upon T
i S times T i S 4 by 2 square plus s divided
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by 1 K p 1 plus 1 upon T i S by T i S times
4 upon 2 square plus s, which can ultimately
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be available in the form of 4 K p times T
i S plus 1 in the numerator, and 2 T i S cube
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plus T i S square plus 4 K p T i S plus 4
K p in the denominator. So, the stability
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of the closed loop transfer function or the
closed loop system can be ascertained from
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the denominator polynomial of the closed loop
transfer function.
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If we form the Routh table for the closed
loop transfer function, in that case the Routh
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table assumes the form of S cube, S square,
S 1, S 0 with the coefficient S of the row
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having s cubed as 2 T i 4 K p T i for this
T i, and 4 K p are the coefficients, and next
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from this cross multiplication, and manipulation
we get the coefficient of s 1 row as 4 K p
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times T i minus 2 0, and then 4 K p in the
last row. What we get from this Routh table,
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when the coefficients of the first column
possess the same sign, in that case the system
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is stable the closed loop system must be stable.
When T i equal to 2, what happens at that
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time, both the coefficients of this row will
be 0, when T i equal to 2, I get the coefficients
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at 0 0, what that implies that implies that
the closed loop transfer function will have
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2 poles located on the imaginary axis; that
means, the system response will be oscillatory,
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there must be a pair of complex conjugate
poles in the s plane, and which implies that
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the system output, closed loop system, output
might be oscillatory, exactly that is what
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is happening when T i equal to 2 as we have
seen.
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Since the integral time constant is T i equal
to 2 giving us, again of one here as we see,
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because we can express the PI controller in
the form of 2 times 1 plus 1 upon 2, as we
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have we have seen. So, when T i equal to 2,
we get an oscillatory output from the system,
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but when T i will be greater than 2 or T i
is very large compared to compared to 2, in
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that case we must obtain a stable response
from the closed loop system, but when T i
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is less than 2, the closed loop system will
yield us an unstable response or the the response
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output response will explode, this is how
one can design a PI controller for the process
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given as 4 upon 2 square plus s, the T i Should
be greater than 2 to obtain some stable response
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not like this.
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Next, let us attempt to design a PID controller
for the same process; a PID controller can
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be designed with the same analysis, using
the same analysis let me start with the closed
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loop transfer function assuming the form of
the PID controller.
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Let the PID controller be given as G c s equal
to K p 1 plus 1 upon T i S plus Td s, and
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we know our process to be 4 upon s square
plus 2 square plus s, thus that will give
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us the closed loop transfer function as GGC
upon 1 plus GGC, which can be obtained after
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simplification in the form of 4 K p times
T i Td s square plus T i S plus 1 in the numerator,
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and 2 T i S cubed plus T i plus 4 K p T i
Td times s square in the denominator along
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with 4 K p T i S plus 4 K p, now when a PID
controller of this form is used in the series
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compensation scheme, we get the denominator
polynomial of this form. Now, again, let me
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form the Routh table, so forming the Routh
table as S cube S square S 1 S 0 with coefficients
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as 2 T i 4 K p T i, then here we have got
T i plus 4 K p T i Td and here 4 K p, now
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the next coefficient can be expressed as 4
K p T i minus 8 K p upon 1 4 K p Td, and this
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is 0, then the cross multiplication like this,
and this divided by this will give us 4 K
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p as the coefficient of the last row. So,
the closed loop system will give us a stable
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response, time response when all the coefficients
of the first column have same sign. So, let
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me assume, let me assume T i is greater than
0, then this becomes positive, when K plus
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greater than 0, than this becomes positive,
and when K p T i Td is greater than 0, Td
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is also greater than 0, we get this coefficient
as positive; what about this coefficient,
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this will be positive provided T i is greater
than 2 upon 1 4 K p Td; so, these can be expressed
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in this form, when T i is greater than 2 upon
1 4 K p Td, in that case the coefficients
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of the column will be positive, thus implying
that the closed loop system will have a stable
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time response, it will not blow out.
Now, with the choice of K p greater than 0,
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Td greater than 0; obviously, one can get
some constant on T i, T i is greater than
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some value, if that is made, if this condition
is made certainly, we will get a stable response
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from the closed loop system, for the time
being let us assume K p equal to 4, let Td
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1, when K p equal to 4 and Td 1, then the
condition gives us T i has to be greater than
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2 upon 17, that is same as T i is greater
than 0.118 approximately. So, if we choose
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a T i value which is greater than 0. 118 with
the choice of K p equal to 4, and Td equal
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to 1, certainly we will get a stable time
response from the closed loop system.
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Let us assume T i to be 4, we are far off
from this value. So, certainly it will ensure
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stability of the closed loop system. So, when
T i equal to 4 along with K p equal to 4 and
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Td 1 what sort of response.
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We expect from the closed loop system, we
expect such a time response from the closed
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loop system, the controller K p is 4, now
this is same as K p Td which is equal to 4,
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in this case Td is 1 K p equal to 4, and Td
1, therefore, K p Td is 4, in this case, it
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is actually K p by T i equal to 1, since K
p equal to 4 and T i equal to 4, thus we are
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getting K p upon t of i as 1, because the
form of the controller is K p 1 plus 1 upon
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T i S plus Td s which gives us upon multiplication
K p plus K p by T i S plus K p Td s; therefore,
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we have written this K p value K p by T i
divided by s and K p Td derivative term giving
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us S.
So, this PID controller is now giving us a
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satisfactory response with certain overshoot
response time and so on; let us try to see
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how much we get for the choice of K p equal
to 4, T i equal to 4 and Td 1, when the controller
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is implemented, then the time response, we
get is having some overshoot of 18 percent,
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overshoot of approximately 18 percent, and
settling time of approximately 10 second,
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this is what we get, and similarly, when the
disturbance load, disturbance is applied to
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the system, after time t equal to 20 second,
we get satisfactory disturbance rejection
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as we see, because the disturbance is dying
down after some time, and it is not having
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very much peaking. So, what we have got from
these with the choice of the soon K p T i
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Td gains or values the PID controller is giving
us a satisfactory time response.
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Now, question may arise, why we choose K p
to be 4 T i to be 4 or Td to be 1, can we
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choose better than those values to give us
still better time response of the closed loop,
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yes, it is possible, it all depends on how
we are designing a controller for a given
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SISO process; now, how to choose appropriate
K p T i Td, such that one can meet the design
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specifications, that can be answered by designing
the controllers by some appropriate techniques,
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one such technique powerful technique will
be discussed in this lecture.
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Let us go to some analysis, now before going
to the powerful technique, when the process,
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when the process is assumed to have the form
G s equal to k upon s square plus alpha 1
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s plus alpha 0, and when the controller G
c s is assumed to have the transfer function
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K p 1 plus 1 upon T i S a PI controller f
or a all pole single input single output process,
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we get the closed loop transfer function T
s expressed as GGC s upon 1 plus GGC s as
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K k p T i S plus 1 divided by T i S times
s square plus alpha 1 s plus alpha 0 plus
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K k p T i S plus 1; now, this can be simplified
further, and expressed in the form of T i
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S plus 1 by T i upon K k p s cubed plus alpha
1 T i by K k p s square plus alpha 0 plus
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K k p T i by K k p s plus 1, why are you doing
like this, so, that with the assumption of
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s equal to beta s n and beta cube equal to
K k p by T i, it is easy to get the closed
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loop transfer function in the normalized form
given as T s n as GGC s n upon 1 plus GGC
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s n in some convenient form, which can be
written as T i beta, sorry, this will be T
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i beta s n 1 upon s n cubed plus alpha 1 by
beta s n square plus alpha 0 plus K k p by
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beta square s n 1, which can be written in
the generalized form C 1 s n 1 divided by
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s n cubed plus d 2 n square plus d 1 s n 1;
so, why are you we doing all this analysis.
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When we have some all pole process transfer
function and a PI controller of this form,
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the closed loop transfer function can always
be expressed in this simple convenient generalized
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form, had there been a 4th order, all pole
process dynamics, in that case what we will
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get the closed loop transfer function, in
that case T s n will be simply of some C 1
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s n plus 1 in the numerator along with s n
5 plus d 4 s n 4 plus d 3 s n 3 plus d 2 n
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2 plus d 1 s n 1; so, this is why I am calling
we are getting some generalized transfer function,
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because irrespective of any order of the process
dynamics, if the process dynamics is available
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in the all pole form, in that case it is always
possible to get the standard transfer function
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which is expressed in the form of standard
closed loop transfer function expressed in
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the form of having one numerator in the transfer
function, and having denominator in this convenient
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form, and where we will have the denominator
expressed as some s n to the power n plus
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d n minus 1 s n to the power n minus 1 and
so on till 1.
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So, this closed loop transfer function is
assuming some typical form, if it is possible
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to find a convenient a satisfactory closed
loop transfer function, which is of the form
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C 1 s 1 upon s n cubed plus d 2 n square plus
d 1 s n plus 1, if it is possible to get said
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of d 2 and d 1 values for some chosen C 1,
by optimizing some performance index, in that
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case what will happen, we may get a transfer
function which will give us desired time response
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of a closed loop system that is the objective.
Then from back calculation, since d 2, d 1
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are made of nothing but the coefficients or
the parameters of the controller, one can
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easily estimate the controller parameters.
So, what we have been doing, we are trying
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to find some standard transfer function which
are known to us, and which are tested also
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that they are time response will be highly
satisfactory, then the closed loop transfer
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function can be used by back calculation to
calculate the unknown parameters of a controller,
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which will definitely yield very satisfactory
closed loop performance of a closed loop system.
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So, we will see how to calculate some standard
transfer function, how to find some suitable
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standard transfer functions of different order.
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Consider a third order transfer function,
which can be given in the form of T s equal
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to y upon r s equal to C 1 s plus 1 divided
by s cube plus d 2 square plus d 1 s plus
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1, this is called a standard transfer function,
because we have seen that with appropriate
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d 2 and d 1 with a choice of C 1, this transfer
function closed loop transfer function gives
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us quite satisfactory time responses, what
we wish to have from a closed loop control
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system, then the error function can be obtained
as R minus Y upon R which is same as E upon
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R s equal to s cubed plus d 2 square plus
d 1 minus C 1 s divided by s cube plus d 2
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square plus d 1 s 1, than the error function
we obtain in that form.
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Now, this error can be minimized using the
performance indices, in this case how using
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some ISTE performance index, one can minimize
and find the optimum values of C 1, d 2 and
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d 1 is explained the ISTE criterion is expressed
as one upon 2 phi J integration from minus
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J infinity to J infinity F s times F minus
S d s, where F s equal to partial differentiation
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of E s, how to compute the coefficients C
1, d 2, d 1 using some minimization routine,
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one can easily use lustrums refer ship algorithm
writing some simple met lab core, it is easy
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to minimize these performance index minimization
of this performance index yields a said of
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C 1, d s and C 1. So, a said of d s means
all the d coefficients d 2, d 1 and so on,
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and C 1 can be obtained, and that ensures
us the minimum value of the ISTE criterion,
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one can use the ISE criterion also, but often
it is found that the IST criterion, minimization
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of IST criterion often leads to very satisfactory
closed loop performances of a closed loop
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system, how such values are obtained, we can
see. So, using the ISTE optimization or minimization,
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one can obtain d 2 and d 1 coefficient for
various C 1; this gives the X axis goes from
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0 to 8, and the Y axis goes from 0 to 10;
so, this is the beginning point for the Y
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axis.
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So, this is what we have got d 2 and d 1.
So, for any C 1, suppose C 1 equal to 4, I
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can find the Optimum values for d 2 and d
1, from here Optimum values for d 2 and d
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1, when C 1 equal to 0, the Optimum values
of d 2 is this much, and the Optimum values
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for d 1 is this much. So, this set of Optimum
C 1, d 2, d 1 gives us a standard third order
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transfer function, and when the closed loop
transfer function with the use of a controller
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becomes a standard third order transfer function,
then in that case we get definitely a quite
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satisfactory time response from the system.
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What sort of time responses is expected from
the standard transfer functions? Let us see,
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when C 1 equal to 0, the closed loop time
Response will be of this form, when C 1 equal
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to 0.5, we get, you see when C 1 is equal
to 0, we have got finite d 2 and d 1 value;
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similarly, for each C 1 values, we have got
the said of these values are d 2 is something
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and d 1 equal to something, all those values
can be obtained from this plot. So, for any
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given C 1, always we get some d 2 and d 1,
these Optimum values of d 2 and d 1 for any
186
00:28:51,110 --> 00:28:58,110
given C 1 always yields a quite satisfactory
time response of the closed loop system, why
187
00:28:59,940 --> 00:29:06,940
do I call this quite satisfactory time response,
if we look at the responses, we see that the
188
00:29:07,730 --> 00:29:14,270
responses are having at most 5 percent of
over shoot.
189
00:29:14,270 --> 00:29:21,270
Similarly, the responses have got settling
time, settling time from 5 to 7 second, and
190
00:29:25,620 --> 00:29:32,620
the responses has got 0 steady state error,
these are quite desirable while designing
191
00:29:32,620 --> 00:29:39,620
a closed loop control system or while designing
a controller, so the standard transfer functions
192
00:29:42,210 --> 00:29:49,210
are giving us quite satisfactory time responses
of the closed loop system, then from the back
193
00:29:51,510 --> 00:29:58,510
calculation, we can find out the parameters
of a controller using the expressions d 2,
194
00:29:59,000 --> 00:30:03,620
d 1, C 1 and expressions for beta.
195
00:30:03,620 --> 00:30:09,910
We shall see in the form of some example,
in some, in the simulation example, how convenient
196
00:30:09,910 --> 00:30:16,460
controller can be designed, before going to
that, we have seen that the standard transfer
197
00:30:16,460 --> 00:30:22,020
function can be expressed in this form, where
T s is the closed loop transfer function having
198
00:30:22,020 --> 00:30:29,020
the numerator C 1 s 1 upon s cube plus d 2
square plus d 1 s 1, it is loop gain can be
199
00:30:32,820 --> 00:30:39,820
obtained, how do we get loop gain, because
we know that T s is nothing but GGC s upon
200
00:30:40,820 --> 00:30:47,820
1 GGC s. So, which can be used to get GGC
s as T s minus 1 upon T s with little manipulation,
201
00:30:54,590 --> 00:31:01,510
it is easy to see that, one can obtain GGC
s as T s minus one upon T s.
202
00:31:01,510 --> 00:31:08,510
So, when T s is given in this form, then GGC
s will be available, it can be obtained in
203
00:31:08,610 --> 00:31:15,610
the form of C 1 s 1 divided by s time s square
plus d 2 plus d 1 minus C 1, where you getting
204
00:31:17,650 --> 00:31:24,650
these, loop gain from the loop gain, it is
easy to get the gain margin phase margin of
205
00:31:25,799 --> 00:31:31,330
the closed loop system. So, this gain and
phase margins will give us information about
206
00:31:31,330 --> 00:31:38,330
frequency response of the system. So, when
the staid up C 1, d 2 and d 1 which are Optimum
207
00:31:40,350 --> 00:31:46,630
as far as the standard transfer function is
concerned are good in this expression, and
208
00:31:46,630 --> 00:31:53,380
the gain and 5 phase margins are found, definitely
one will get very satisfactory phase and gain
209
00:31:53,380 --> 00:31:54,700
margins.
210
00:31:54,700 --> 00:32:01,700
Let us go to the simulation study, where we
will make use of the standard form based controller
211
00:32:03,700 --> 00:32:09,730
design, a third order standard transfer function
will be used to design a PI controller for
212
00:32:09,730 --> 00:32:16,730
a all pole, for an all pole SISO process.
So, Let us assume that the process dynamics
213
00:32:19,120 --> 00:32:26,120
is having all poles, means, having poles only
with a static gain of k, and its denominator
214
00:32:27,860 --> 00:32:34,860
polynomial expressed as s square plus alpha
1 s plus alpha 0, then we will go back to
215
00:32:35,600 --> 00:32:42,600
some specific form of, form of transfer function,
and design the PI controller. Let the PI controller
216
00:32:43,470 --> 00:32:50,470
be expressed in the transfer function as K
p plus K i upon s, which can again be expressed
217
00:32:51,039 --> 00:32:58,039
in the standard form as K p 1 K i by K p s.
So, for each in analysis we have expressed
218
00:33:01,690 --> 00:33:08,690
the transfer function of the controller in
the form of K p plus K i s. So, with this
219
00:33:08,820 --> 00:33:15,820
process dynamics, and PI controller dynamics,
the closed loop transfer function becomes
220
00:33:16,159 --> 00:33:23,159
T s equal to K p upon K i s plus 1 divided
by s cubed by K K i plus alpha 1 s square
221
00:33:32,150 --> 00:33:39,150
by K K i plus alpha 0 plus K K p times s by
K K i 1; so, the closed loop transfer function
222
00:33:45,799 --> 00:33:52,799
can be obtained in this form.
Now, assuming K K i as beta cubed and setting
223
00:33:56,799 --> 00:34:03,799
s as beta s n, the above transfer function
can be expressed as T s n equal to C 1 s n
224
00:34:08,669 --> 00:34:15,669
1 divided by s n cube plus d 2 n square plus
d 1 s n 1, where d 2 equal to alpha 1 by beta,
225
00:34:20,690 --> 00:34:27,690
if you see d 2 equal to this alpha 1 by beta,
then d 1 equal to alpha 0 plus K k p by beta
226
00:34:31,719 --> 00:34:38,719
square and C 1 equal to K p beta upon K i.
So, all these comparing this transfer function,
227
00:34:40,029 --> 00:34:47,029
with this one definitely one can write the
expressions for d 2 d 1 and C 1, in this fashion;
228
00:34:48,509 --> 00:34:54,409
why we are writing in this fashion, now you
see the beauty of this method, now the d 2
229
00:34:54,409 --> 00:35:01,409
parameter is having the plant parameter or
process parameter alpha 1 and beta d 1 is
230
00:35:03,019 --> 00:35:10,019
continuing the unknown controller parameter
K p and C 1 is controlling the unknown controller
231
00:35:11,170 --> 00:35:18,170
parameter K p and K i both.
So, using the Optimum values of d 2, d 1 and
232
00:35:20,940 --> 00:35:27,940
C 1, it is possible to use back calculation,
and get the design values for K p and T i,
233
00:35:29,670 --> 00:35:36,670
this is how one designs a controller using
standard form. Now, let us go to the particular
234
00:35:39,839 --> 00:35:46,839
case, for our case, as we have seen when the
process dynamics is given as 0.5 upon s square
235
00:35:51,309 --> 00:35:58,309
1.595 s plus 1.62 at that time k becomes 0.5
alpha 1 becomes 1.595 and alpha 0 becomes
236
00:36:06,170 --> 00:36:06,710
1.62.
237
00:36:06,710 --> 00:36:13,710
That is what we have got, then substituting
these values and using the plot for the optimized
238
00:36:18,489 --> 00:36:25,489
coefficients, when C 1 equal to point 5 d
2 can be obtained as 1.595, and d 1 equal
239
00:36:26,869 --> 00:36:33,869
to 2.12, where from we get when C 1 equal
to 0. 5, what are the Optimum values for d
240
00:36:35,430 --> 00:36:42,430
1 and d 2, when C 1 equal to 0.5, I can find
the d 2 to be of this 1.595, when C 1 equal
241
00:36:49,200 --> 00:36:56,200
to 0.5, d 2, 1.595 and d 1 equal to 2.12;
so, this what we get from the plot; next,
242
00:37:03,009 --> 00:37:10,009
but d 2 can also be expressed as alpha 1 by
beta, alpha 1 is known to us, thus giving
243
00:37:13,950 --> 00:37:20,950
us beta as 1 alpha 1 is 1.595, whereas d 2
is 1.595, therefore, beta 1. Next expression
244
00:37:26,400 --> 00:37:33,400
we have got that, we have got one more expression
K K i equal to beta cubed, that we have assumed
245
00:37:38,719 --> 00:37:45,660
K K i equal to beta cube, then since beta
1; therefore, K i can be obtained as beta
246
00:37:45,660 --> 00:37:52,660
cubed by K K is given, that is one of the
process parameter 1 upon 0.5, so k become
247
00:37:53,029 --> 00:37:57,440
2.
Similarly, using the expression C 1 equal
248
00:37:57,440 --> 00:38:04,440
to K p beta upon K i, K p unknown, since we
known C 1 beta, and K i are known the only
249
00:38:06,119 --> 00:38:13,119
unknown K p can be estimated as 1, thus we
have been able to design a PI controller,
250
00:38:13,579 --> 00:38:20,579
which is given as G c s equal to K p plus
K i upon s as 1 2 by s; what is expected from
251
00:38:23,719 --> 00:38:30,719
this controller? This controller must give
us a time response which is as we have seen
252
00:38:31,019 --> 00:38:34,549
for the standard transfer function.
253
00:38:34,549 --> 00:38:41,549
Now, going to the simulation diagram in the
simulation diagram, what we have, we will
254
00:38:43,979 --> 00:38:50,979
apply unit step input. So, r t the reference
equal to some unit step input 1, then let
255
00:38:54,880 --> 00:39:01,880
the load be also given by some magnitude l
equal to 0.1 with the time at which it occurs
256
00:39:07,589 --> 00:39:14,589
at time t equal to 20 second; now, the PI
controller, we have designed for the given
257
00:39:14,599 --> 00:39:21,599
process is s plus 2 upon s which is nothing
but s plus 2 upon s can be expressed as one
258
00:39:27,809 --> 00:39:34,809
time 1, one upon 0.5 s, no, in this case,
it will be different, let us go back to the
259
00:39:46,630 --> 00:39:53,630
exact form we have got, I have got 1 2, so,
in this case since I am making use of this
260
00:39:53,759 --> 00:40:00,759
one. So, I will have one 0.5 over here, than
in that case, it is giving us 0.5 s 1, and
261
00:40:03,769 --> 00:40:10,769
we are not getting that way how can we express,
this in the standard form the form, I have
262
00:40:11,349 --> 00:40:18,140
written is not correct actually. So, to get
it in the standard PI transfer function form
263
00:40:18,140 --> 00:40:25,140
s plus 2 by s can be written as 1 2 by s 2
by s which is nothing but 1 one upon 0.5 s.
264
00:40:30,579 --> 00:40:37,579
So, it is in the form of G c s as K p 1 plus
1 upon 0.5 s, where we have got 0.5, in the
265
00:40:47,569 --> 00:40:54,569
form of again K p 1 one upon T i S. So, I
can say that I have designed a PI controller,
266
00:40:54,839 --> 00:41:01,839
where the proportional gain is one and the
integral time constant equal to 0 .5 giving
267
00:41:03,630 --> 00:41:08,029
us a PI controller in the form of s plus 2
by s.
268
00:41:08,029 --> 00:41:15,029
So, this is how we have realized the PI controller
G c s for the process having parameters k
269
00:41:17,400 --> 00:41:24,400
equal to 0.5 and alpha 1 1.595 and alpha 0
1.62. So, what sort of output we will expect
270
00:41:30,029 --> 00:41:37,029
from this one, when this is stimulated, definitely
we will expect a quite satisfactory time response
271
00:41:38,349 --> 00:41:45,349
from the closed loop system. As we expect
the output here is having almost settling
272
00:41:49,029 --> 00:41:56,029
time of 5 percent and of overshoot of, sorry
overshoot of overshoot of 5 percent settling
273
00:42:02,809 --> 00:42:09,809
time is of 8 second as expected. So, what
we get basically, we get a closed loop transfer
274
00:42:12,729 --> 00:42:19,150
function response, that is much similar to
that we get from a standard closed loop transfer
275
00:42:19,150 --> 00:42:20,219
function.
276
00:42:20,219 --> 00:42:27,219
Coming to the other analysis, frequency analysis
or frequency response of the closed loop system,
277
00:42:27,729 --> 00:42:33,559
the Nyquist diagram of the closed loop system
can be drawn in this case, for which we need
278
00:42:33,559 --> 00:42:40,559
to get the loop gain G c G s the loop gain
for this will be s plus 2 upon s time, we
279
00:42:45,819 --> 00:42:52,819
have got 0.5 by s square 1.595 s plus 2.1162,
we have got 1.62, 1.62. So, using this loop
280
00:43:05,959 --> 00:43:12,959
gain or loop transfer function, one can obtain
the polar plotter Nyquist diagram what information,
281
00:43:13,109 --> 00:43:20,109
we get from Nyquist diagram, the faith crossover
point is somewhere around here, and the gain
282
00:43:21,609 --> 00:43:28,609
cross over point will be somewhere around
here, as we see measure this span, and inverse
283
00:43:30,920 --> 00:43:37,269
of this will give us the gain margin, this
span is very small, because this span total
284
00:43:37,269 --> 00:43:43,779
span from here to here one.
So, we will have a very good gain margin for
285
00:43:43,779 --> 00:43:50,779
the system; similarly, if I draw a line from
here to here, and measure this angle, I will
286
00:43:50,779 --> 00:43:57,779
get a very good phase margin for the system.
What are those values, let us see using the
287
00:43:59,209 --> 00:44:06,209
analysis, one can obtain those values as some
gain margin of 12 in absolute value term,
288
00:44:13,599 --> 00:44:20,599
and the phase margin of some 65 degree, we
know that a gain margin of greater than 2,
289
00:44:24,190 --> 00:44:31,190
and a phase margin of greater than 30 degree
ensures a robust control system or a quite
290
00:44:32,789 --> 00:44:39,329
stable closed loop system. So, as far as stability
of closed loop transfer function is concerned,
291
00:44:39,329 --> 00:44:46,329
we are far off from those values, and therefore,
we have got a quite robust closed loop control
292
00:44:46,650 --> 00:44:47,269
system.
293
00:44:47,269 --> 00:44:54,269
Now, same information can be derived using
the bode diagram. So, using the bode diagram,
294
00:44:54,779 --> 00:45:01,170
we can track the phase and gain cross over
frequencies, and then the phase and gain margin.
295
00:45:01,170 --> 00:45:08,170
So, phase cross over frequency, if I draw
a line over here at180 degree, I will get
296
00:45:10,549 --> 00:45:17,259
somewhere like this, and that will give me
from the phase cross over frequency, we can
297
00:45:17,259 --> 00:45:23,849
draw a line vertically, and find how much
you are away from 0 degree, this much will
298
00:45:23,849 --> 00:45:29,380
give you the gain margin. So, minus of this
value will give us the gain margin; so, we
299
00:45:29,380 --> 00:45:36,380
get a quite high gain margin for the systems;
similarly, let us track the gain cross over
300
00:45:36,729 --> 00:45:41,989
frequency. So, this is the gain cross over
frequency, the frequency at which the gain
301
00:45:41,989 --> 00:45:48,039
of the system is 1 or in d b the gain of the
system is 0 d b.
302
00:45:48,039 --> 00:45:53,640
So, we will get the gain cross over frequency
of this magnitude, for which let us find what
303
00:45:53,640 --> 00:45:58,999
is the phase angle at that time, how far you
are from the minus 180 degree, we are far
304
00:45:58,999 --> 00:46:05,999
of which gives us s phase margin of approximately
some 6o to 65 degree, thus what we see, that
305
00:46:08,539 --> 00:46:15,539
the closed loop system is not only giving
us a quite satisfactory time response, also
306
00:46:17,319 --> 00:46:22,219
we are getting a quite satisfactory frequency
response from the system.
307
00:46:22,219 --> 00:46:29,219
Now, there is one problem associated with
this method that one has to make use of series
308
00:46:31,059 --> 00:46:37,959
form of PID controller in place of the parallel
form of PID controller. What a parallel form
309
00:46:37,959 --> 00:46:44,670
of PID controller is we write a controller
given by the transfer function G c s as K
310
00:46:44,670 --> 00:46:51,670
p times 1 plus one upon T i S plus Td s as
the transfer function; for a parallel form
311
00:46:53,299 --> 00:47:00,269
of PID controller, when a parallel form of
PID controller, then the controller itself
312
00:47:00,269 --> 00:47:07,269
is introducing 2 0 E s, if you simplify this
one, it is giving us an expression in the
313
00:47:09,089 --> 00:47:16,089
form of K p T i Td s square plus T i S 1 upon
T i S. So, from this expression we say that
314
00:47:22,239 --> 00:47:27,609
the transfer function of the PID controller,
which is in parallel PID controller form gives
315
00:47:27,609 --> 00:47:34,609
us 2 0 E s introduces 2 0 Es in the closed
loop system and one pole at the origin.
316
00:47:36,819 --> 00:47:43,819
So, when we will have 2 0 Es given by the
controller, and the plant is all pole system
317
00:47:45,509 --> 00:47:52,509
in that case, what happens the closed loop
transfer function will be having 2 pod 2 0
318
00:47:53,440 --> 00:48:00,440
Es in the numerator; that means, the standard
transfer functions will have 2 0 Es, for which
319
00:48:00,609 --> 00:48:07,609
it is very difficult to find the coefficients
done d 2 and so on, the point is that we have
320
00:48:07,650 --> 00:48:14,170
got standard transfer functions expressed
in the form of C 1 s plus 1 upon, then here
321
00:48:14,170 --> 00:48:21,170
we can have any order s to the power n plus
d n minus 1 s to the power n minus 1 and so
322
00:48:21,839 --> 00:48:28,839
on, but in the numerator we have got 1 0 only,
so, how to handle those situations, to handle
323
00:48:29,329 --> 00:48:35,819
those situations, what we have to have, we
can always have a parallel form a a series
324
00:48:35,819 --> 00:48:42,819
form of the PID controller, in which case
one of the zero of the PID controller is to
325
00:48:43,269 --> 00:48:50,269
be cancelled with one of the pole of the all
pole transfer function. How to select, how
326
00:48:52,089 --> 00:48:57,979
to make cancellation, we should choose the
Td of the controller as one upon d 1, and
327
00:48:57,979 --> 00:49:04,979
this d 1 should be as small as possible d
1 should assume some small value, in that
328
00:49:05,920 --> 00:49:12,219
case, we are assuming that a pole is cancelled
a poled, which a pole which is located far
329
00:49:12,219 --> 00:49:18,359
off from the left half from the origin of
s plane is cancelled, it is getting cancelled
330
00:49:18,359 --> 00:49:25,359
with the derivative term of the series PID
controller, then we can make use of a series
331
00:49:25,880 --> 00:49:32,880
form of PID controller, and a all pole process
model to design controllers using standard
332
00:49:36,940 --> 00:49:43,940
form.
Now, what have we learnt from the lesson,
333
00:49:44,219 --> 00:49:51,219
we have seen how a standard transfer function
results in satisfactory time and frequency
334
00:49:52,160 --> 00:49:59,160
responses, and how a standard transfer function
can be used to design the parameters of a
335
00:49:59,819 --> 00:50:06,819
p PI or PID controller using back calculation,
but there is one limitation with the method
336
00:50:07,390 --> 00:50:13,670
the process dynamics has to be available in
the form of all pole transfer function for,
337
00:50:13,670 --> 00:50:19,809
and the form of controller, one can use, when
we have a PID controller is that a series
338
00:50:19,809 --> 00:50:21,390
PID controller.
339
00:50:21,390 --> 00:50:28,390
Now, we will go to the points to ponder, what
are some limitations of the method, what happens
340
00:50:32,160 --> 00:50:38,140
when a process dynamics is available in the
first order plus date time transfer function
341
00:50:38,140 --> 00:50:45,140
form, when the process dynamics is available
in this form, one can always get the process
342
00:50:45,180 --> 00:50:52,180
dynamics expressed in the form of k upon 1
minus 1 theta s t 1 s 1, of course, when theta
343
00:50:56,660 --> 00:51:03,660
is a small number small value, when that is
not so, make use of plant order reduction
344
00:51:06,630 --> 00:51:13,630
technique or process order reduction technique,
order reduction technique to get the plant
345
00:51:15,699 --> 00:51:22,699
dynamics available in the form of all pole
transfer function, second point to ponder
346
00:51:22,809 --> 00:51:29,809
is that, why do we design series PID controller,
as we have discussed that a series PID controller
347
00:51:31,180 --> 00:51:38,180
enables us to cancel one of the zero of the
controller with one pole of the all pole transfer
348
00:51:38,839 --> 00:51:45,839
function of the process, then thus giving
us 1, 0 in the standard transfer function,
349
00:51:46,109 --> 00:51:53,109
if that is not the case, in that case, when
we use one parallel PID controller 2, 0 Es
350
00:51:53,380 --> 00:51:59,430
appears in the standard transfer function,
and it is very difficult to find the Optimum
351
00:51:59,430 --> 00:52:06,430
values of d s and c 2 C 1, in the case of
parallel PID controller, that is why we need
352
00:52:07,709 --> 00:52:14,709
to design series PID controller, that is all
in the lecture .