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Welcome to the lecture titled Effects of Measurement
Noise and Load Disturbances. Unwanted responses
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in the output of the system are often known
as noise responses and load disturbance responses.
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Noise responses are generally of high frequencies,
whereas load disturbance responses are of
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low frequencies.
It is often very difficult to design a controller,
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which can reduce the effects of ill effects
of noise and load disturbances simultaneously.
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We shall see in this lecture, how a noise
canceller can be designed using standard form.
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Consider a process, given as G s equal to
K upon s square plus alpha 1 s plus alpha
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0. So, this all-pole second-order process
has got poles located at, suppose at far off
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from the left half of the s plane. Then in
that case, it becomes very easy to design
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controllers whereas, we shall consider a typical
case, where alpha 1 will be less than alpha
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0, which signifies that the process is of
under damped type.
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So, let us consider the process dynamics as
shown over here, for this process when the
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PI controller dynamics is defined as G c s
equal to K p plus K i upon s, then the closed
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loop transfer function T s which is nothing
but, T s is equal to G G c s upon 1 plus G
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G c s can be retained and simplified and obtained
in a form shown over here, where the closed
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loop transfer function numerator can have
the terms K p upon K i times s plus 1 divided
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by s cubed upon K Ki plus alpha 1 s square
upon K Ki plus alpha 0 plus K Kp s upon K
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Ki plus 1.
So, when the transfer function has been obtained
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in this typical form, then assuming K Ki is
equal to beta cubed and s equal to beta s
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n, where s n stands for the normalized s normalized
s a Laplace on domain. Then, the closed loop
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transfer function can be written as, T s n
is equal to c 1 s n plus 1 upon s n cubed
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plus d 2 s n square plus d 1 s n plus 1.
Thus, it has been possible to obtain the closed
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loop transfer function in the standard third-order
transfer function form. This standard transfer
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function, third-order transfer function has
got three coefficients namely c 1, d 2 and
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d 1 and we know that, for any given c 1, it
is not difficult to find d 2 and d 1 by optimize
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optimization or optimization of some standard
criterion often known as integral square time
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error criterion.
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Now, we have got the d 2 expressed as alpha
1 upon beta, d 1 expressed as alpha 0 plus
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K Kp upon beta square and c 1 is equal to
K Kp beta upon K Ki. These are the three variables
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we have in the standard transfer function,
now manipulating this three variables, using
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the three variables, one can write d 1 is
equal to alpha 0 plus K Kp upon beta square
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is equal to alpha 0 upon beta square plus
K Kp upon beta square.
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Again, this can be written as alpha 0 upon
beta cube times beta in the numerator plus
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K Kp K Kp beta upon beta cubed. Now, beta
cube we know is beta cubed is equal to K Ki.
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So, substitution of K Ki will give you, K
Ki plus K Kp beta upon K Ki. So, again K p
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beta is equal to c 1 Ki. So, alpha 0 beta
upon K Ki plus here, one can write as K c1
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K i upon K Ki. So, K Ki cancellation will
be there, this can be written as c 1 plus
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alpha 0 beta upon K Ki. So, manipulating this,
it is not difficult to write the expression
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for K Kp as K Kp is equal to alpha 0 c 1 upon
d 1 minus c 1.
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Similarly, using the three expressions, the
three expressions for d 2, d 1 and c 1 it
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is not difficult to write the expressions
for K Ki as alpha 1 upon d 2 to the power
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3. So, basically using the variables d 2,
d 1 and c 1, we have been able to write the
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final expression, explicit expressions for
the 2 unknowns K p and K i in the form of
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K Kp is equal to alpha 0 c 1 upon d 1 minus
c 1 and K Ki is equal to alpha 1 d 2 upon
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to the power 3.
So, what is the beauty of a obtaining the
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explicit expressions for the two unknowns
K p and K i in this explicit form? Since alpha
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0 is known, alpha 1 is known then, for any
given c 1, using the standard form the standard
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coefficients d 2 and d 1 can be obtained.
Thus c 1, d 1, d 2 all those quantities will
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be known to us, thus enabling us to estimate
the unknowns K p and K i. So, with the powerful
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explicit expressions given over here, it is
possible to find unique values for the unknowns
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K p and K i using the optimize coefficients
d 2 and d 1 for any given c 1.
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Now, using the plot which gives us optimum
values for d 2 and d 1 for various c 1 by
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minimizing the ISTE criterion, ISTE criterion
it is not difficult to obtain the values for
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d 2 and d 1 when c 1 is equal to 0.5. When
c 1 is equal to 0.5, then your d 2 will be
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of 1.595 and d 1 the upper one, d 1 is of
2.12. So, for various c 1 it is possible to
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find various combinations of d 2 and d 1 using
this plot, which is about the plot of d 2
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d 1 versus c 1 optimizing the ISTE criterion.
So, using this plot we have found for c 1
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equal to 0.5, d 2 is equal to 1.595 and d
1 is equal to 2.12.
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So, using K equal to 1, let us assume the
process model is having the dynamics given
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by s square plus s plus 4. So, we consider
an all-pole transfer function of the form
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G s is equal to 1 upon s square plus s plus
4 and the damping ratio for this one is equal
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to 1 upon 2 root of 2 root of 4 is equal to
1 upon 4 is 0.25. So, obviously, the damping
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ratio is 0.25 and we have got an under damped
process for designing a PI controller for
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the system, for the second-order under damped
system.
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So, the second-order under damped process
has got parameters K equal to 1, alpha 1 is
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equal to 1 and alpha 0 equal to 4. And for
c 1 1 equal to 0.5, we have d 2 is equal to
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1.595 and d 1 is equal to 2.12 then using
the explicit expressions for K p, which is
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K Kp is equal to alpha 0 c 1 upon d 1 minus
c 1, which gives us alpha 0 is 4. So, 4 times
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0.5 upon 2.12 minus 0.5, which gives us K
Kp to be of value 1.2346 thus, giving us K
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p is equal to 1.2346 since K is equal to 1.
Similarly, using the next expression K i is
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equal to alpha 1 upon d 2 to the power 3 upon
K, which gives us K i as 0.2464. Thus the
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PI controller is designed as G c s is equal
to K p plus K i upon s is equal to 1.2346
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plus 0.2464 upon s. So, this gives us optimum
PI controller for the all-pole under damped
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process, under damped process. Let us try
to simulate and see the responses we get from
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this PI controller.
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So, the simulation diagram can be met in this
form, where we have a reference input which
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is nothing but, the set point input with value
r is equal to 1. A unit step input we have
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some disturbance input d, which is of step
input occurs at time t equal to 20 seconds
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and this is of magnitude minus 0.5 and we
have got also to the system band-limited white
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noise to see the effects of disturbances on
the performance of the PI controller. Let
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us look at the controller dynamics of the
process and so on.
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Now, the process dynamics is shown over here,
which is given as 1 upon s square plus s plus
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4, this is obtained when we assume this specified
values for K is equal to 1 and alpha 1 is
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equal to 1 and alpha 0 is equal to 4 because
the standard form of the all-pole second-order
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transfer function is K upon s square plus
alpha 1 s plus alpha 0.
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Now, the process dynamics is shown over here,
now the PI controller is employed with K p
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as 1.2346 and K i as 0.2464. So, this gives
us the PI controller dynamics. So, we have
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got the PI controller over here, in the feed
forward path for the process. What else we
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have to see the impact of load and measurement
noise disturbances.
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Let us set r equal to 0 that is no need for
any set point input because we have interest
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in the disturbance responses. Therefore, let
us set the disturbance d of magnitude 0 minus
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0.5 which occurs at time t equal to 20 seconds.
Similarly, the band-limited white noise, let
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this occur from the beginning of the simulation;
that means, at time t equal to 0, we have
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band-limited white noise for the system.
Now to clearly see the effects of these disturbances,
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initially what we shall do? We shall not apply
any sort of disturbances or measurement noise
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to the system and see the impacts of these
disturbances on the system. So, for that when
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the system is not subjected to any disturbances,
what sort of output response we get, that
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we shall see first.
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So, when the system is not subjected to band-limited
white noise, but it is subjected to a static
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load disturbance of magnitude minus 0.5, which
occurs at time t equal to minus t equal to
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20 seconds is shown over here. So, when a
static load disturbance of magnitude minus
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0.5 occurs at time t equal to 20 second, the
response is shown over here. As it is evident
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from the response, we have got a very sluggish
response because the settling time the disturbance
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occurs here and it takes almost more than
100 seconds to go to the steady state.
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Therefore, the response takes almost 80 seconds
to go to the steady state, although the response
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is fast as is apparent from here and the response
magnitude load disturbance rejection is not
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instantaneous, but still it has a faster load
disturbance excursion means response whereas,
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the response is very sluggish, it takes almost
80 seconds to go to the steady state. This
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is the type of load disturbance response we
get from the system, closed loop system in
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the absence of measurement noise.
What type of control signal we will get, we
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get a control signal of this form, what information
we get from here? It shows that much control
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effort much or energy spent to give closed
loop performance of the system. So, much control
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effort is here because the time taken is very
large, you see to go to the steady state,
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we take almost more than 100 seconds. So,
then this is the amount of energy we spent,
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this is the control effort we spent for the
closed loop dynamics. So, next we shall see
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the impact of the measurement noise on the
system.
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Now, when both the static load disturbance
of magnitude minus 0.5 and band-limited white
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noise is present in the system, then the output
of the system is obtained as shown over here.
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So, we get the load disturbance and noise
responses, keep in mind that we have set the
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reference input or set point input to be 0,
because we have interest in the noise responses
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or load disturbance responses. So, when the
noise power is of 10 to the power minus 6
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and the sample time is 0.01 second, then the
output of the disturbance output of the system
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can be obtained of in this form. Now, it is
very much noisy, not only the response is
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sluggish, the output is very much noisy and
it is very difficult to find out the exact
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output from the system unless some filtering
is used.
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Similarly, the control effort also can be,
control signal can also be plotted and seen
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to be of this form, where we have got the
effect of noise and the control effort is
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very high. So, the measurement noise is quite
evident from the control signal as well as
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the response, noise responses of the system.
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Next, effort will be made to design a disturbance
rejection controller. Where is that controller
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in this loop? We have got a controller placed
in the feedback path. So, the controller now
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is known as a noise filter, this noise filter
the primary job of this noise filter is to
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filter out the measurement noise. This process
the measurement noise can be shown in the
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form when some sensor is put over here; the
noise introduced by the sensor is shown by
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the band-limited white noise injected over
this point. Now if some measurement noise
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filter is put in the loop in this form, then
it can give some improved performance than
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the earlier PI controller. So, the effort
will be made now to design a noise filter
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for the closed loop system; apart from the
PI controller, the PI will be there we shall
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have noise filter in the closed loop.
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Then all the process we have all the steps
we have made used earlier can be repeated
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with the changes that, a noise filter dynamics
of the form G f s is equal to 1 upon T f s
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plus 1 is injected, this is new. So, when
the noise filter dynamics is injected is inserted
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then, the analysis gives a closed loop transfer
function of the form T s is equal to T i s
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plus 1 upon s cubed T i upon K Kp plus alpha
1 s square T i upon K Kp plus alpha 0 s T
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i upon K Ki plus 1. How do we get this form
of closed loop transfer function? This we
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get with the help of a new type of PI controller.
Please keep in mind, the type of PI controller
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we had considered earlier was your GG s equal
to K p plus K i by s whereas, we have consider
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a parallel PI controller of the form G s is
equal to K p times 1 plus one upon T i s.
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So, we have got different parameter here TT
i in place of K i. So, T i has been introduced
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in place of K i, this has been done intentionally
for is in analysis of the closed loop transfer
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function, one can make use of the earlier
form of course with some difficulties.
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Now, again consider the all-pole second-order
transfer function of the process given by
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G s is equal to K upon s square plus alpha
1 s plus alpha 0 and the PI controller of
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the form G c s equal to K p 1 plus 1 upon
T i s. Let the noise filter dynamics be given
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by G f s is equal to 1 upon T f s plus 1.
Why this has been, why we have we are using
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we have interest in this particular type of
PI controller form? The reason is that, one
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pole 0 cancellation can be initiated and we
can get a simpler closed loop transfer function.
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So, this closed loop transfer function one
obtains provided T f is equal to T i. Please
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keep in mind, unless you make this assumption
you do not get a closed loop transfer function
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of this form. Why I say so, because if you
carefully observe look at the closed loop
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transfer function, we do not get any term
with T f. So, term with T f is missing here;
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that means, certainly some cancellation has
been made, some approximation has been made
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otherwise, T f must appear in the closed loop
transfer function.
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Now, the closed loop transfer function for
the system for the closed loop system, for
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this system has to be given in the form of
T s is equal to G G c s upon one plus G G
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c H s. Then here in the denominator, do not
mind about the numerator; in the denominator,
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it can be written as G is your K upon s square
plus alpha 1 s plus alpha 0, then G c is K
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p T i s plus 1 upon T i s, then H s is this
noise filter dynamics we give it by H s.
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Here we have got one upon T f s plus 1. So,
this when we approximate T f is equal to T
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i then this approximation enables us to cancel
one pole with one 0. Thus giving us in the
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denominator terms like 1 plus K Kp upon s
square plus alpha 1 s plus alpha 0 times T
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i s only. So, that is the that is why, we
have got a closed loop transfer function which
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is divide of the T f term.
So, after getting the closed loop transfer
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function, again making use of the approximation
that, K Kp upon T i is equal to beta cubed
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and s is equal to beta times s n. The same
closed loop transfer function can be written
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in the form of c 1 s n plus 1 in the numerator,
having a denominator of s n cubed plus d 2
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s n square plus d 1 s n plus 1, thus we get
a standard third-order transfer function with
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the coefficient c 1, d 2 and d 1.
Now, like the earlier case with the help of
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simplification, T i is not difficult to find
explicit expressions for the two unknowns
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of the controller K p and T i. And since we
have T f is equal to T i therefore, there
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is need for estimating two unknowns for the
controller, K p and T i are the two unknowns
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then all other things are known and the closed
loop system will have a controller in place.
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Now, we have d 2 is now alpha 1 upon beta,
if you look at carefully if you look at carefully
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then, this d 2 can be written as alpha 1 upon
beta and d 1 is equal to alpha 0 upon beta
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square and c 1 is equal to beta T i.
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Now, since d 1 is equal to alpha 0 upon beta
square T i is same as alpha 0 beta upon beta
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00:27:38,670 --> 00:27:45,670
cubed and beta cubed is nothing but, beta
cubed is K Kp upon T i. So, you can write
188
00:27:46,390 --> 00:27:53,390
alpha 0 beta T i upon K Kp again beta T i
is c 1. So, we get in the numerator alpha
189
00:28:00,960 --> 00:28:07,960
0, in the numerator alpha 0 beta T i is c
1 alpha 0 c 1 upon K K p. So, d 1 is equal
190
00:28:11,960 --> 00:28:18,960
to K K alpha 0 c 1 upon K Kp implies that
K Kp is equal to alpha 0 c 1 upon d 1. So,
191
00:28:21,730 --> 00:28:26,980
this is how we obtain explicit expressions
for the unknown K p.
192
00:28:26,980 --> 00:28:33,980
So, K Kp is equal to alpha 0 c 1 upon d 1;
similarly, making use of the second expression
193
00:28:35,400 --> 00:28:42,400
d 2 expression for d 2, which is given as
d 2 is equal to alpha 1 upon beta. Again you
194
00:28:47,830 --> 00:28:54,830
can write down this as alpha 1 beta square
upon K Kp upon t times T i in the numerator,
195
00:29:01,690 --> 00:29:08,690
then you will get here alpha 1. So, if I substitute
now beta T i, so I have got alpha 1 beta time’s
196
00:29:16,000 --> 00:29:23,000
c 1 upon K K p. So, K Kp is nothing but, now
alpha 0, c 1 and d 1, so this type of manipulation
197
00:29:27,150 --> 00:29:34,150
will give you c 1 c 1 cancellation.
So, therefore, we will get this is equal to
198
00:29:35,110 --> 00:29:42,110
d 2 then yes, now, d 2 is equal to alpha 1
beta d 1 upon alpha 0, with further manipulation
199
00:29:54,000 --> 00:30:01,000
will get alpha 1 beta d 1 upon alpha 0. So,
with little manipulation it will not be difficult
200
00:30:04,880 --> 00:30:11,880
to obtain T i in the form of d 2 c 1 upon
d 1. So, ultimately with substitution of d
201
00:30:13,410 --> 00:30:20,410
1 now here and c 1 will further d 1, substitution
of d 1 rather will enable because there will
202
00:30:20,620 --> 00:30:24,860
be cancellation.
So, that will give us T i ultimately in the
203
00:30:24,860 --> 00:30:31,860
form of T i is equal to d 2, c 1 upon d 1.
So, the final expressions we have got from
204
00:30:33,050 --> 00:30:40,050
the analysis of the coefficients of the third-order
standard transfer function is that, K Kp can
205
00:30:45,370 --> 00:30:52,250
be obtained in the form of alpha 0 c 1 upon
d 1 and T i in the form of d 2 c 1 upon d
206
00:30:52,250 --> 00:30:59,250
1. The main reason for obtaining in this convenient
form is that, since for any c 1, it will not
207
00:31:00,240 --> 00:31:06,870
be difficult to get d 2 and d 1 therefore,
assume any c 1 and obtain optimum d 2 and
208
00:31:06,870 --> 00:31:12,540
d 1.
So, known quantities will be alpha 0, alpha
209
00:31:12,540 --> 00:31:19,540
1 is not there, now for any c 1, we get d
1, c 1 we get d 1 and d 2; therefore, all
210
00:31:20,780 --> 00:31:27,240
the quantities in the right half of the two
expressions are known to us. Thus it will
211
00:31:27,240 --> 00:31:34,240
be possible to estimate K p and T i values
making use of the value c 1, d 2 and d 1 for
212
00:31:35,770 --> 00:31:42,770
a given process transfer function model. So,
if the transfer function model of a process
213
00:31:43,300 --> 00:31:50,300
is known and if T i is available in the all-pole
form, then it becomes very easy to design
214
00:31:50,860 --> 00:31:56,980
PI controller with a noise filter in the loop
for the closed loop system.
215
00:31:56,980 --> 00:32:03,980
Now, the closed loop system is having K equal
to 1, we consider the same process model as
216
00:32:07,360 --> 00:32:14,360
we have considered in the earlier case, where
K is equal to 1, alpha 1 is equal to 1 and
217
00:32:15,020 --> 00:32:22,020
alpha 0 equal to 4. That means the process
G s is now given in the form of 1 upon s square
218
00:32:22,610 --> 00:32:29,610
plus s plus 4. So, we consider the same second-order
under damped process with the steady state
219
00:32:31,500 --> 00:32:38,500
gain K is equal to 1.
Now, c 1 equal to 0.5 is d 21.595 and d 1
220
00:32:43,100 --> 00:32:50,100
is of 2.12, this we get from the minimization
of ISTE criterion. So, please do not forget
221
00:32:52,660 --> 00:32:59,660
from the minimization of ISTE criterion, we
get the standard values d 2 and d 1, for any
222
00:33:03,640 --> 00:33:10,640
given c 1. Now, putting those values the explicit
expressions for K Kp which is nothing but,
223
00:33:12,990 --> 00:33:19,990
K Kp is equal to alpha 0 c 1 upon d 1 is equal
to 4 times 0.5 c 1 is 0.5 alpha 0 is 4; therefore,
224
00:33:25,580 --> 00:33:32,580
4 times 0.5 upon 2.12 gives us K Kp as 0.9434,
where K p becomes 0,934. Now, T f is equal
225
00:33:41,050 --> 00:33:48,050
to T i is given as d 2 c 1 upon d 1, now which
is nothing but, d 2 is 1.595 times 0.5 upon
226
00:33:51,840 --> 00:33:58,800
2.12, which gives us T f is equal to T i is
equal to 0.3762.
227
00:33:58,800 --> 00:34:05,800
So, thus we design the PI controller as well
as the first-order noise filter for the closed
228
00:34:10,510 --> 00:34:17,510
loop system. For the noise controller, dynamics
is given by G f s is equal to 1 upon T f s
229
00:34:20,230 --> 00:34:27,230
plus 1 is equal to 1 upon 0.3762 s plus 1
and G c s obtained in the earlier form for
230
00:34:32,820 --> 00:34:39,820
the sake of comparison is G c s is equal to
K p plus K i upon s is equal to 0.9434 plus
231
00:34:43,360 --> 00:34:50,360
2.5078 upon s. How do you get that one? It
is not difficult to obtain that value for
232
00:34:51,350 --> 00:34:58,350
the K I, making use of the comparison that,
when the G c s is written in different form,
233
00:35:02,040 --> 00:35:09,040
K p 1 plus 1 upon T i s this gives us K p
plus K p upon T i s.
234
00:35:12,520 --> 00:35:19,520
So, this is again expressed in the form of
K p plus K i upon s therefore, this value
235
00:35:20,690 --> 00:35:27,690
2.5018 has been obtained from the ratio of
K p and T i. So, if you take the ratio of
236
00:35:29,010 --> 00:35:36,010
K p and T i; that means, 0.9434 upon 0.3762
will be equal to 2.5078. So, thus we get a
237
00:35:42,810 --> 00:35:48,940
PI controller of the earlier form for the
sake of comparison or I can say that, to make
238
00:35:48,940 --> 00:35:55,940
use of the sense simulation model, I have
obtained the K i in that earlier form, then
239
00:35:56,380 --> 00:35:59,250
let us go to the simulation diagram.
240
00:35:59,250 --> 00:36:06,250
So, this simulation diagram has got a noise
filter and rest of the things remains as it
241
00:36:07,210 --> 00:36:13,650
is. Earlier, we have made use of this simulation
diagram, now this simulation diagram has got
242
00:36:13,650 --> 00:36:20,650
an additional transfer function in the loop
which is nothing but, the noise filter. So,
243
00:36:23,950 --> 00:36:30,950
this is our PI controller, PI controller process
band-limited white noise and the step load
244
00:36:35,730 --> 00:36:42,730
disturbance, which occurs at time t equal
to 20 second and the magnitude of the static
245
00:36:43,550 --> 00:36:50,550
load disturbance or static load disturbance
magnitude is equal to minus 0.5 like the earlier
246
00:36:53,740 --> 00:37:00,740
case, for the sake of comparison. Again we
set r is equal to 0, because we have interest
247
00:37:00,920 --> 00:37:05,190
in the disturbance responses of the closed
loop system.
248
00:37:05,190 --> 00:37:11,870
Then we get the load disturbance responses
of this form, please keep in mind the static
249
00:37:11,870 --> 00:37:18,870
load disturbance is of magnitude l is equal
to minus 0.5 occurring at time t equal to
250
00:37:22,620 --> 00:37:29,620
20 seconds. Now, this is what we get from
the current scheme. Now, with the inclusion
251
00:37:33,400 --> 00:37:40,400
of a noise filter in the loop, the static
load response has improved significantly,
252
00:37:42,870 --> 00:37:49,870
it is easily observable from this plot that,
the response is not only fast, it settles
253
00:37:51,100 --> 00:37:58,100
down within some 40 no, 20 means 30 within
some 30 second . So, it take almost 30 seconds,
254
00:38:03,860 --> 00:38:10,540
in place of 80 seconds for the earlier case
to settle down to the steady state and the
255
00:38:10,540 --> 00:38:16,510
magnitude of the load disturbance response
is also not higher compared to the earlier
256
00:38:16,510 --> 00:38:22,060
case. You see the magnitude is not changing
there is little bit of over shoot or under
257
00:38:22,060 --> 00:38:29,060
shoot, but those are insignificant. So, overall
the static load disturbance response for the
258
00:38:29,310 --> 00:38:35,430
second scheme is quite satisfactory. Now,
I can say this response is quite satisfactory
259
00:38:35,430 --> 00:38:40,820
from the point of speed of response as well
as settling time.
260
00:38:40,820 --> 00:38:47,730
Let us see the control signal, the control
effort one has to provide with the inclusion
261
00:38:47,730 --> 00:38:53,210
of the static field with the noise filter
in the loop is very less, if you see the excursion
262
00:38:53,210 --> 00:38:58,070
the amount of energy you spent is very less
compared to the energy you spent, you take
263
00:38:58,070 --> 00:39:05,070
the area of the lower curve and area of the
upper curve, then below the line, zero line
264
00:39:06,990 --> 00:39:11,260
then that gives a significant improvement
in the second case.
265
00:39:11,260 --> 00:39:18,260
So, with the inclusion of a noise filter in
the loop, one the unit provides, the unit
266
00:39:18,900 --> 00:39:25,900
requires much control effort and there will
not be actuator saturation also for the second
267
00:39:27,130 --> 00:39:32,250
case. So, actuator saturation is a very big
problem when the control effort is very higher
268
00:39:32,250 --> 00:39:38,870
and the control signal is of high magnitude,
then control saturation occurs and actuator
269
00:39:38,870 --> 00:39:45,220
or valves may get saturated. In that case,
there are problems to overcome, those problems
270
00:39:45,220 --> 00:39:52,220
often it is desirable to provide suitable
noise disturbance rejecters in a closed loop
271
00:39:52,510 --> 00:39:59,510
system. Now, let us investigate the effects
of measurement noise on the system.
272
00:40:00,920 --> 00:40:07,920
So, when a noise with power 10 to the power
minus 6 and a sampling time of 0.01 second
273
00:40:08,860 --> 00:40:15,860
is used, then the load disturbance response
and the noise responses can be obtained of
274
00:40:18,170 --> 00:40:25,170
this form. You see we have got quite satisfactory
load disturbance and noise responses compared
275
00:40:26,580 --> 00:40:33,580
to the earlier case. Let us go back to the
first case where, the response of the system
276
00:40:36,140 --> 00:40:43,140
is shown in this form. This is what we get,
when we do not have a noise filter in the
277
00:40:44,650 --> 00:40:51,440
loop, this is the output response we get.
So, please observe it, the magnitude is from
278
00:40:51,440 --> 00:40:58,440
minus 0.2 to 0.05 with high value of excursion
of measurement noise. Now, compared to that,
279
00:40:59,880 --> 00:41:06,880
we have got a response of this form. So, excursion
is not only less, earlier it was minus 0.2,
280
00:41:09,900 --> 00:41:16,900
we have come up to minus 0.13 and here also,
it was very high. So, both ways not only the
281
00:41:20,810 --> 00:41:26,650
improvement in the magnitude of the load responses
has taken place, there is significant reduction
282
00:41:26,650 --> 00:41:30,640
in the noise level as well.
283
00:41:30,640 --> 00:41:37,640
So, the noise filter has given us satisfactory
load disturbance and noise measurement noise
284
00:41:38,970 --> 00:41:45,970
responses, the control signal is also quite
satisfactory. So, with much less control effort,
285
00:41:47,600 --> 00:41:54,600
one is possible to obtain satisfactory closed
loop performances from the closed loop system,
286
00:41:55,060 --> 00:42:02,060
of course with a noise filter in the loop.
So, before going to summary, let us do little
287
00:42:03,900 --> 00:42:09,600
bit of analysis over here, why that happens
so. Suppose the closed loop system is given
288
00:42:09,600 --> 00:42:16,600
like this, we have got a controller G c here
and we have got G, now this is the measurement
289
00:42:25,110 --> 00:42:30,920
noise, when we have a sensor over here as
we have been doing earlier H s. So, definitely
290
00:42:30,920 --> 00:42:37,920
there will be measurement noise. Now, without
putting a noise filter, what is the transfer
291
00:42:38,080 --> 00:42:41,880
function we get for different type of disturbance
inputs?
292
00:42:41,880 --> 00:42:47,710
Suppose the load disturbance d is occurring
is here, then y upon d, y upon d is given
293
00:42:47,710 --> 00:42:54,710
as G in the absence of H s. Now, actually
the output will be not here, here this is
294
00:43:00,380 --> 00:43:07,380
the sensor because the output is the output
of the system Y s. Then Y upon d is given
295
00:43:09,010 --> 00:43:16,010
as G upon 1 plus G G c H s whereas, as far
as the noise input is concerned. Suppose,
296
00:43:17,330 --> 00:43:24,330
here the noise input of noise power n s is
occurring, then y upon n is given as G G c
297
00:43:27,140 --> 00:43:34,140
h s upon 1 plus G G c h s.
So, if I look at the two transfer functions,
298
00:43:36,910 --> 00:43:43,910
what happens? To minimize the effects of load
disturbance responses, effects of load, effects
299
00:43:46,190 --> 00:43:53,190
of static load inputs, what has to be done
if G G c h s is very high a large number?
300
00:43:54,970 --> 00:44:00,130
In that case, this will be 0 that is our aim
because we do not have to have any effect
301
00:44:00,130 --> 00:44:07,000
of static load disturbances d, but when GG
c H s is very high, look at the second transfer
302
00:44:07,000 --> 00:44:14,000
function, then in that case it will be equal
to 1, it will be approximately 1. That means,
303
00:44:14,060 --> 00:44:20,880
the output y will be equal to n therefore,
the effects of measurement noise will be very
304
00:44:20,880 --> 00:44:27,230
much present in the output of the closed loop
system. Thus, it is not possible to design
305
00:44:27,230 --> 00:44:34,230
a controller G c, which will give us satisfactory
noise rejection as well as static load disturbance
306
00:44:36,270 --> 00:44:42,740
rejection. So, one has to have some compromise
while designing a controller for rejecting
307
00:44:42,740 --> 00:44:49,740
disturbance in the system.
Now, is there any other way we can handle
308
00:44:54,690 --> 00:45:01,690
this situation? Yes of course, if one designs
other type of controller, let us say a 2 degree
309
00:45:05,140 --> 00:45:12,140
of freedom controller, if I put a some filter
over here, F s yes, it is possible then the
310
00:45:14,350 --> 00:45:21,350
G c, G c can be purely designed for rejection
of the disturbances and F s later on can be
311
00:45:21,510 --> 00:45:28,510
designed for satisfactory closed loop performances.
In place of that, what we have done now? In
312
00:45:28,840 --> 00:45:35,840
place of injecting or putting a reference
filter in the reference path, one can put
313
00:45:39,650 --> 00:45:44,200
a filter, noise filter in the feedback path
that is what we have done.
314
00:45:44,200 --> 00:45:50,410
So, putting a noise filter of the form T f
s plus 1 in the feedback path, still we are
315
00:45:50,410 --> 00:45:55,960
able to provide a 2 degree of freedom controller
to the structure somehow. Now, the job of
316
00:45:55,960 --> 00:46:02,960
this is purely to reject disturbances whereas,
the job G c will be to provide overall satisfactory
317
00:46:04,610 --> 00:46:07,780
closed loop responses.
318
00:46:07,780 --> 00:46:14,780
Now, so, the impact of the disturbance rejection
filter has been investigated here. But, let
319
00:46:27,020 --> 00:46:33,720
us try to analyze the system when you have
got a 2 degree of freedom controller. So,
320
00:46:33,720 --> 00:46:40,720
when we have got F s in the loop, in that
case G c G H the closed loop transfer function,
321
00:46:51,710 --> 00:46:58,710
where we have got Y upon r will be equal to
G c G f s upon 1 plus G c G h s. So, this
322
00:47:04,920 --> 00:47:11,920
F s can be designed in a suitable way to provide
satisfactory set point responses and G c can
323
00:47:12,310 --> 00:47:19,310
be designed in a way to provide satisfactory
disturbance rejections in the system.
324
00:47:21,250 --> 00:47:28,250
Now, in summary we can say, load disturbances
are typically of low frequencies. Therefore,
325
00:47:29,510 --> 00:47:35,790
focus on the behavior of the closed loop system
at low frequencies should be made to design
326
00:47:35,790 --> 00:47:42,790
suitable compensator for rejecting load disturbances
whereas, measurement noise disturbances are
327
00:47:44,550 --> 00:47:50,780
typically of very high frequencies. Therefore,
focus should be on to design controllers that
328
00:47:50,780 --> 00:47:56,750
would reject high frequency inputs or excursions
in the closed loop system.
329
00:47:56,750 --> 00:48:03,750
Now, disturbance rejection filters can overcome
both load and noise disturbances if design
330
00:48:07,160 --> 00:48:14,160
suitably, but often it is desirable to design
2 degree of freedom controllers for many closed
331
00:48:14,760 --> 00:48:21,760
loop system to overcome the ill effects of
disturbances, both sort of disturbances not
332
00:48:22,670 --> 00:48:29,000
only measurement noise, input to the system,
rather the static load input to the systems
333
00:48:29,000 --> 00:48:32,690
can be overcome by designing suitable filters.
334
00:48:32,690 --> 00:48:39,690
Now, coming to the points to ponder, we have
two important points to discuss about related
335
00:48:42,010 --> 00:48:49,010
to this topic. First is does the noise filter
provide load disturbance rejection? Yes, the
336
00:48:52,760 --> 00:48:59,760
noise filter which primary job is to reject
noise inputs to the system. The ill effects
337
00:49:01,540 --> 00:49:08,540
of noise inputs to the system can also provide
satisfactory load disturbance rejection, but
338
00:49:09,460 --> 00:49:15,609
that is not true that may not be true always.
Therefore, always it is necessary to design
339
00:49:15,609 --> 00:49:20,830
some load estimator and controller in a closed
loop system.
340
00:49:20,830 --> 00:49:26,770
So, noise filter has got its own limitation,
it can be of any order when we employ higher
341
00:49:26,770 --> 00:49:33,770
order noise filter, in that case the design
method will change. And we may not be able
342
00:49:34,060 --> 00:49:40,130
to straightforward we may not be able to find
explicit expression for the unknowns of the
343
00:49:40,130 --> 00:49:46,080
filter in a straightforward manner. So, those
are the limitations of designing higher order
344
00:49:46,080 --> 00:49:52,490
filter for a closed loop system for disturbance
rejection. But of course, with a higher order
345
00:49:52,490 --> 00:49:59,490
filter, often it is possible to design filters
which can provide not only control action
346
00:50:02,850 --> 00:50:03,220
for measurement noise inputs, rather for static
load disturbances as well.
347
00:50:03,220 --> 00:50:03,470
The second point is, are there other techniques
to deal with disturbances? Obviously, as we
348
00:50:03,420 --> 00:50:03,670
have seen towards the end of this lecture,
one can go for a 2 degree of freedom control
349
00:50:03,650 --> 00:50:03,900
structure for dealing with the disturbances,
disturbance inputs to a system. The beauty
350
00:50:03,820 --> 00:50:04,070
of the 2 degree of freedom control structures
are that, one controller can purely we designed
351
00:50:04,030 --> 00:50:04,280
for rejection of disturbances, disturbance
inputs to the systems whereas, the other controller
352
00:50:04,190 --> 00:50:04,440
can be designed for overall satisfactory performances,
time and frequency domain performances of
353
00:50:04,359 --> 00:50:04,609
a system.
So, we have got many more design techniques
354
00:50:04,490 --> 00:50:04,740
for dealing with disturbances in a system.
Now, one can design high gain, low gain, one
355
00:50:04,700 --> 00:50:04,950
can concentrate on high gain, low gain zones
of the bode magnitude and phase plots and
356
00:50:04,910 --> 00:50:05,160
Nyquist plots to design suitable controllers
for disturbance rejections, that is all in
357
00:50:05,070 --> 00:50:05,320
this lecture.