Effects of Gain Constants Proportional Integral (PI) Control Experiment
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8th Feb 2020 Mechanics Reference this
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Abstract:
The usage and effects of gain constants proportional integral (PI) control were explored throughout this experiment. The experiment adjusted the proportional and integral constants to perform five different tests. The group tested using only proportional feedback, only integral feedback, both proportional and integral feedback with no set point parameter, traditional PI control with a set point of one, and lastly PI control with an adjusted optimal set point parameter. The group tested all of these settings for the speed controller and concluded that in order to have the best response, both proportional and integral control should be used along with an adjusted optimal value for the set point parameter that provided a critically damped response. When done this provided the best response with the smallest percent overshoot, lowest settling time, along with no steady state error.
Nomenclature:
${b}_{\mathit{sp}}$
Set point parameter
k Gain
${k}_{p}$
Proportional constant
${k}_{i}$
Integral constant
${T}_{s}$
Settling time $$
$\tau $
Time constant
${\omega}_{n}$
Natural frequency
$\zeta $
Damping ratio
Introduction:
Control systems are used in just about everything these days. They are in vehicles, they are in manufacturing robots, and they are used in many different applications all over the world. This report explores the factors that affect and determine the response for a specific type of control system. This control system examined is PI control or proportional integral control. This system is a loop control, where once the signal is sent; it goes through a loop back to the PI controller where they verify that the value is the correct value. If not the controller corrects the value and the process repeats. The process is then repeated until the correct value is reached.
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Find out morePI control is extremely useful for speed control. Applying PI to motors or other objects that require a desired speed works well as it is extremely hard to get the speed correct on the first attempt for most processes and therefore it is good to have a controller that will correct the speed value until the desired speed is reached.
Analysis:
The PI control is applied to the DC motor that was modeled in the previous lab. In the previous lab, the output function of a DC motor was determined to be
$G\left(s\right)=\frac{k}{\mathit{\tau s}+1}$ 
(1) 
However, for this lab a low pass filter was added to reduce the noise as the noise can greatly influence the accuracy and effectiveness of the PI controller. The low pass filter used was
$F\left(s\right)=\frac{1}{{t}_{c}s+1}$ 
(2) 
This is represented by the block diagram below,
Figure 1.
Where ${k}_{p}$
is the proportional control constant, $\frac{{k}_{i}}{s}$
is the integral control component, and ${b}_{\mathit{sp}}$
is the set point parameter. When ${b}_{\mathit{sp}}$
is equal to one the controller is just the standard PI controller. In algebraic form the controller is
$\widehat{u\left(s\right)={k}_{p}\left({b}_{\mathit{sp}}\widehat{r}\left(s\right)\u2013\widehat{y}\left(s\right)\right)+\frac{{k}_{i}}{s}(\widehat{r}\left(s\right)\u2013\widehat{y}\left(s\right))}$ 
(3) 
Then the transfer function is
$\widehat{y}\left(s\right)=H\left(s\right)\widehat{r}\left(s\right)$ 
(4) 
while
$\widehat{y}\left(s\right)=G\left(s\right)F\left(s\right)\widehat{u}\left(s\right)$ 
(5) 
Where
$G\left(s\right)=\frac{k}{\mathit{\tau s}+1}$ 
(6) 
As found in the previous lab. Then when equation 3 is substituted into equation 5 we get
$\widehat{y}\left(s\right)=G\left(s\right)F\left(s\right)\left({k}_{p}{b}_{\mathit{sp}}+\frac{{k}_{i}}{s}\right)\widehat{r}\left(s\right)\u2013G\left(s\right)F\left(s\right)\left({k}_{p}+\frac{{k}_{i}}{s}\right)\widehat{y}\left(s\right)$ 
(7) 
Then when solved for y(s) we find
$\widehat{y}\left(s\right)=\frac{G\left(s\right)F\left(s\right)\left({k}_{p}{b}_{\mathit{sp}}+\frac{{k}_{i}}{s}\right)}{1+G\left(s\right)F\left(s\right)\left({k}_{p}+\frac{{k}_{i}}{s}\right)}\widehat{r}\left(s\right)$ 
(8) 
Then when plugging in equations 1 and 2
$\widehat{y}\left(s\right)=\frac{\frac{k}{\left(\mathit{\tau s}+1\right)\left({t}_{c}s+1\right)}\left({k}_{p}{b}_{\mathit{sp}}+\frac{{k}_{i}}{s}\right)}{1+\frac{k}{\left(\mathit{\tau s}+1\right)\left({t}_{c}s+1\right){(k}_{p}+\frac{{k}_{i}}{s})}}\widehat{r}\left(s\right)$ 
(9) 
Then getting rid of the complex fraction and finding H(s) from equation 5
$H\left(s\right)=\frac{k{k}_{p}{b}_{\mathit{sp}}S+k{k}_{i}}{\tau {\tau}_{c}{S}^{3}+\left({t}_{c}+\tau \right){S}^{2}+\left(k{k}_{p}+1\right)S+k{k}_{i}}$ 
(10) 
The goal of this lab is to determine the values of parameters ${k}_{p}$
and ${k}_{i}$
that provide no steady state error, no overshoot, and a settling time of roughly 2%. This is done using five different strategies. For most of these methods, the third order polynomial in the denominator of H(s) is reduced to a second order polynomial. This is done to be able to compare it to the wave equation
${S}^{2}+2\zeta {\omega}_{n}+{\omega}_{n}^{2}$ 
(11) 
Where the fastest settling time with no overshoot occurs when $\zeta =1$
and the wave is critically damped. Two percent settling time will occur when ${\tau}_{s}=\frac{4}{\zeta {\omega}_{n}}$
. It is also recalled that using the final value theorem when H(0)=1 then there is no steady state error.
The first method is using only the proportional feedback only to adjust the signal. For this method, we set ${k}_{i}=0$
as we are not using any integral control. This reduces $H\left(s\right)$
to become
$H\left(s\right)=\frac{k{k}_{p}{b}_{\mathit{sp}}}{{t}_{c}\tau {S}^{2}+\left({t}_{c}+\tau \right)S+(k{k}_{p}+1)}$ 
(12) 
Where
$H\left(0\right)=\frac{k{k}_{p}{b}_{\mathit{sp}}}{k{k}_{p}+1}\ne 1$ 
(13) 
Therefore, this method will have steady state error. To determine what ${k}_{p}$
provides critical damping. Poles:
${S}^{2}+\frac{\left({t}_{c}+\tau \right)}{{t}_{c}\tau}S+\frac{(k{k}_{p}+1)}{{t}_{c}\tau}$ 
(14) 
Where the coefficient to S is $\zeta $
and the last term is ${\mathit{\omega n}}^{2}$
. Plugging into
$2\zeta {\omega}_{n}=2{\omega}_{n}$ 
(15) 
Which then becomes
$2\sqrt{\frac{(k{k}_{\mathit{pc}}+1)}{{t}_{c}\tau}}=\frac{({t}_{c}+\tau )}{{t}_{c}\tau}$ 
(16) 
And then solving for ${k}_{\mathit{pc}}$
${k}_{\mathit{pc}}=\left(\frac{1}{4}\frac{{\left({t}_{c}+\tau \right)}^{2}}{\tau {t}_{c}}\u20131\right)*\frac{1}{k}$ 
(17) 
The second method uses only integral control by setting ${k}_{p}$
equal to zero, and therefore H(s) becomes
$H\left(s\right)=\frac{k{k}_{i}}{{t}_{c}\tau {S}^{3}+\left({t}_{c}+\tau \right){S}^{2}+S+k{k}_{i}}$ 
(18) 
Then when applying the FVT (final value theorem)
$H\left(0\right)=\frac{k{k}_{i}}{k{k}_{i}}=1$ 
(19) 
Therefore, there is no steady state error for this method. The denominator for this method is a third order polynomial not a second order. However, when comparing the coefficient of the cubed term to the other coefficients it tends to be significantly smaller and is ignored, reducing this to a second order polynomial.
${S}^{2}+\frac{1}{({t}_{c}+\tau )}S+\frac{k{k}_{i}}{{t}_{c}+\tau}$ 
(20) 
Then we recall from equation 11 we know that
$\frac{1}{({t}_{}+\tau )}=2\sqrt{\frac{k{k}_{i}}{{t}_{c}+\tau}}$ 
(21) 
Then solving for ${k}_{i}$
we determine
${k}_{i}=\frac{1}{4k(\tau +{t}_{c})}$ 
(22) 
Which corresponds with
${T}_{s}=8({t}_{c}+\tau )$ 
(23) 
The next method used involves both proportional and integral control, however the ${b}_{\mathit{sp}}$
is set to zero. When the ${b}_{\mathit{sp}}$
is zero
$H\left(s\right)=\frac{k{k}_{i}}{\tau {\tau}_{c}{S}^{3}+\left({t}_{c}+\tau \right){S}^{2}+\left(k{k}_{p}+1\right)S+k{k}_{i}}$ 
(24) 
And when the FVT is applied it is found that H(0) does equal one so there will be no steady state error. Now if we once again assume that the coefficient to ${S}^{3}$
is much smaller than the rest then it can once again be reduced to a second order polynomial.
${S}^{2}+\frac{\left(1+k{k}_{p}\right)}{({t}_{c}+\tau )}S+\frac{k{k}_{i}}{({t}_{c}+\tau )}$ 
(25) 
We then say that we want the case where the settling time is no more than a quarter of a second and that we want the critically damped case. To apply this we use
${T}_{s}=\frac{4}{\zeta {\omega}_{n}}$ 
(26) 
Where zeta is equal to 1 for the critically damped case and the settling time is equal to 0.25 seconds and solve for the natural frequency. Then once we know the natural frequency, we use that in equations 15 & 16 in order to solve for ${k}_{i}$
and then use that to solve for ${k}_{p}$
For the fourth method, we set the ${b}_{\mathit{sp}}$
value equal to one, which gives us traditional PI control. With the ${b}_{\mathit{sp}}$
equal to one the FVT still holds and we still get H(0) equal to ,. For this method, we instead use the ZieglerNichols approach as opposed to analyzing the transfer function in the manner we did previously. We use this approach because it can be used even when the transfer function is not known. The first step is to find the P becomes unstable by varying the ${k}_{p}$
value. The value of ${k}_{p}$
at which the time response becomes unstable is referred to as ${k}_{\mathit{pu}}$
. Another value we found was ${T}_{u}$
which is the time from peak to peak during the instability. For the ZieglerNichols method we know that
${k}_{p}\approx 0.4{K}_{\mathit{pu}}$ ${k}_{i}\approx \frac{{k}_{p}}{0.8{T}_{u}}\approx 0.5\frac{{k}_{\mathit{pu}}}{{T}_{u}}$ 
(27) (28) 
For the last approach of using PI control with an optimal ${b}_{\mathit{sp}}$
value, we did not analyze the transfer function to find the optimal ${\mathit{b}}_{\mathit{sp}}$
value. The optimal ${b}_{\mathit{sp}}$
value can be determined by varying it from zero to one until the best response is found.
Experimental Equipment and Procedure:
For the experiment a DCMCT motor unit was used. The lab teaching assistant powered up the unit. Once the unit was turned on the program LabVIEW was downloaded with its contents being placed on a folder on the desktop. Next Labview.exe was double clicked and opened. To verify the program was working the group input a value for ${k}_{p}$
and noticed the motor respond with motion.
For the first method of proportional control without integral control the group set the signal to a square wave. It is very important that the control signal does not saturate. There is a green box in the center of the screen that will turn red if this occurs. The group then adjusted the offset of the slider so that the sign convention, or direction, of the velocity did not change to avoid coulomb friction affecting the results. The group then used the values of 25 $\raisebox{1ex}{$\mathit{rads}$}\!\left/ \!\raisebox{1ex}{$s$}\right.$
for the amplitude, 0.6 Hz for the frequency, and an offset of 50 $\raisebox{1ex}{$\mathit{rads}$}\!\left/ \!\raisebox{1ex}{$s$}\right.$
. The group then varied the frequency from 0.1 to 1 Hz in increments of 0.1 to determine the best value to observe a steady state behavior. The group ended up staying with 0.6 Hz. In the previous lab the group obtained values for K and $\tau $
for the system. However in this experiment the group used K=16 and $\tau =0.13$
for consistency along with ${t}_{c}=0.03$
for the filter to eliminate unwanted noise. Next the group set ${k}_{i}=0$
and ${b}_{\mathit{sp}}=1$
to verify that the proportional control was the only thing affecting the signal. At this point the group started with a value of 0.01 for ${k}_{p}$
and increased it with a step of 0.01 until a second order response was noticed. After this the group adjusted the ${k}_{p}$
value until it produced a critically damped solution. The value was then further increased until the system went unstable to find the value for ${k}_{\mathit{pu}}$
along with the period of instability ${T}_{u}$
.
For the next method of Integral control, a similar procedure was used. The proportional gain was set to zero, and the integral gain was varied from 0 $\raisebox{1ex}{$\mathit{Vs}$}\!\left/ \!\raisebox{1ex}{$\mathit{rad}$}\right.$
to 5 $\raisebox{1ex}{$\mathit{Vs}$}\!\left/ \!\raisebox{1ex}{$\mathit{rad}$}\right.$
. The group then found the value for integral gain that provided a critically damped solution.
Now the group applied both proportional and integral control while setting the ${b}_{\mathit{sp}}=0$
. For this case the calculations previously computed will provide the values of the integral and proportional gain. The group found and used ${k}_{i}=2.56$
and ${k}_{p}=0.2575$
. The Labview simulation was then run with these values. The settling time was then observed and compared. The group then adjusted both gains to attempt to get a better response.
Next the group will test proportional and integral control with ${b}_{\mathit{sp}}=1$
using the ZieglerNichols (ZN) method. The group used the results from the pure proportional section to calculate the proportional and integral gain values using the ZN method. LabVIEW was then run with the values obtained. From the response graph the overshoot and settling time were determined.
Lastly the group set the values for both gains to the values used for the method when the ${b}_{\mathit{sp}}$
was 0. The group then slowly increased the ${b}_{\mathit{sp}}$
from 0 to 1 and observed. The group then adjusted the value until they found the optimal value that provided the fastest settling time while allowing no overshoot.
Results:
The group found many things from performing the six different methods to control the DC motor. The group found that for only proportional control the response was of the same shape however not of the same magnitude with an extremely large steady state error shown in figure 6.2.3a.
Figure 6.2.3a
As the group increased the proportional gain constant the group noticed the response get closer to the correct values despite still a significant steady state error however becoming extremely unstable with an extremely large overshoot and no settling time as the response never reached steady state shown in figure 6.2.3b.
Figure 6.2.3b
For the second method where the group used solely integral control the group noticed a response that had little to no overshoot however a long settling time and no steady state error. This is seen in figure 6.3.3a.
Figure 6.3.3a
For the next method where the ${b}_{\mathit{sp}}$
value was set to zero the group found that despite using the estimated values for both of the gain constants the response looked extremely similar to solely integral control with a very long settling time and no overshoot, however with slightly more noise and not really any steady state error. (shown in figure 6.4.2)
Figure 6.4.2
The group then found that by using the theoretical gain values along with a ${b}_{\mathit{sp}}$
of 1 (traditional PI control) the group got a good result with a fairly short settling time, low noise, and no steady state error. This is shown in figure 6.5.2.
6.5.2
Discussion:
Overall, the results aligned with what was expected from the theory behind the experiment. The group noticed that for each of the methods there were issues. The first method used with solely proportional control at the theoretically optimal value provided a very bad response as the steady state error was extremely large. However, this was expected to occur due to the final value theorem providing a value other than 1. As the proportional gain constant was increased the response’s steady state error slowly reduced however the response became much more unstable at a ${k}_{p}$
value of 0.35 where it was hard to distinguish the correct value due to the noise.
For the second method using only integral control at the theoretical optimal values the group received a slightly better result with a time response that had no steady state error (shown in equation 19) and very little to no overshoot. This time response did however have a very large settling time which made this controller not as useful as it could have been. When the integral gain constant was increased the response shifted from an overdamped response (seen in figure 6.3.3a) to an underdamped response with a large overshoot and steady state error.
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View our servicesThe next method used was a combination of the previous two methods however altered the set point parameter to be a value of zero instead of 1. This provided a result that was extremely similar to the second method that used only integral control (no overshoot, long settling time, no steady state error). The group expected this to occur as when zero is used for the ${b}_{\mathit{sp}}$
value in equation ten it reduces to a very similar equation to that used in pure integral control with only one additional term in the denominator. Therefore, it makes sense that the response is very similar to the pure integral control response.
Now the group adjusted the value of the set point parameter to 1. This is known as traditional PI control. This combination of values proved to be extremely effective. The response had no steady state error and an extremely low settling time. The only downside to this method was that the percent overshoot was slightly large. This aligned with what the group expected as this set of parameters was the theoretical optimal values. With the set point parameter equal to one it essentially had no affect and allowed the optimal values to perform well.
The last method used uses the optimal gain constants for both integral and proportional control however uses a manually determined set point parameter value instead of 0 or 1. The group found that as the set point parameter increased from zero to one the settling time decreased, however that the amount of overshoot increased. This was accurate with what was previously noted for the methods where the ${b}_{\mathit{sp}}$
was set to zero and one. By increasing the ${b}_{\mathit{sp}}$
value from zero to one, the group found the response transition from overdamped at a set point parameter of zero to underdamped at a set point parameter of one.
Conclusion:
The following conclusions are supported by the results of this experiment.
 While both effective, proportional and integral control are most effective and accurate when coupled

The optimal
${b}_{\mathit{sp}}$
for a system is the value that produces a critically damped solution that happened at 0.1 for our system.
 It is better to have a set point parameter of one than zero.
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