Simulation Analysis Of Motor Shaft Position Control Using Ziegler Nichols And Alternative PID Tuning Methods In Matlab (Essay).

Apply Ziegler Nichols PID tuning method and a second alternative PID tuning method for this motor shaft position control problem using Matlab simulations. Explain

Perform an analysis of controller performance using performance metrics of your own choice for both of your PID tuning methods. Evaluate the robustness of this controller to external disturbances.

Provide your conclusions from this simulation analysis.

The DC motor system of Section 2 is now discretised. Use Matlab to determine the open loop and closed loop Z-domain transfer functions of the discretised system.

Perform a root locus analysis of the discretised system in the Z domain. Select three different values of the sampling period to evaluate the effect of discretisation on stability.

Apply PID control in discrete time using Matlab simulations with the same sampling time periods as in 3b.

Compare the performance of PID control of 3c with an equivalent continuous time PID implementation using performance metrics of your own choice.

Performance Metrics for PID Tuning Methods

1. The following table gives the absolute, relative and cumulative relative frequencies of number of passengers travelling at each train station in Melbourne during the peak time, 7am to 9:29 am. The data has been divided into 10 intervals.

Class Interval	Mid-point of Interval	Count	Relative Frequency	Cumulative Relative Frequency
248-547	365.72	18	37.50%	37.50%
548-847	662.67	6	12.50%	50.00%
848-1147	985.22	9	18.75%	68.75%
1148-1447	1271.43	7	14.58%	83.33%
1448-1747	1601.00	2	4.17%	87.50%
1748-2047	1848.00	2	4.17%	91.67%
2048-2347	2268.00	1	2.08%	93.75%
2648-2947	2830.00	1	2.08%	95.83%
2948-3247	2958.00	1	2.08%	97.92%
7448-7747	7729.00	1	2.08%	100.00%

Table 1: Frequency Table of Number of Passengers at Peak time

1. The figure shows the histogram obtained using the above table for the number of passengers.

The average of the number of passengers who are present in a train station in Melbourne during the peak times between 7am and 9:29 am is 1062.6875. The value of the median for the raw data was obtained after having sorted the data in ascending order. Then, as the total number of observations is 48, the median corresponds to 24^thobservation in the sorted data and is equal to 733. The mode is the observation with highest frequency. Considering the data in raw ungrouped form, the observation 401 had highest frequency of 2. Thus the mode is 401.

1. The data provided in this problem is a sample from the larger population of data. A population refers to the sum of all the cases or set of instances of the event under study. It accounts for each and every individuals, subjects, objects or data points. A sample is only a representative of the population. It is a subset of the larger set that is the population. The data here clearly does not account for all the students attending Holmes each day.
2. The standard deviation for the sample of the weekly attendance at Holmes was computed using the following:

Here, n = number of sampled observations,  = i^th observation in the sample, 1 ≤ i ≤n,

 = mean weekly attendance.

The calculations for the sample standard deviation are shown in the following table:

X	X-	(X-)2
472	-12.5714	158.0408
413	-71.5714	5122.469
503	18.42857	339.6122
612	127.4286	16238.04
399	-85.5714	7322.469
538	53.42857	2854.612
455	-29.5714	874.4694

Table 2: Calculation for Standard Deviation of Weekly attendance

The standard deviation, using the formula specified above, was then computed to be equal to 68.559.

The inter quartile range is the difference of the third quartile from the first quartile. The value of the first quartile was computed 6014. The value of the third quartile was computed as 7223. Thus the inter quartile range was computed as 1209.

The inter quartile range is unaffected by outliers that may be present in the data. Hence unlike standard deviation which depends on the mean and hence is influenced by outliers, the inter quartile range can be particularly useful in measuring variation of a dataset which has extreme values.

The correlation coefficient computed using observations on attendance and the sales of chocolate bar was computed as 0.967. Therefore a strong positive relation exists between the two variables.

Let X denote the weekly attendance at Holmes and let Y be the number of chocolate bars that are sold. Taking Y as dependent and X as independent variable, the regression model of the following form  is to be determined:

Y= a+ b. X +

The estimate of the parameters after solving the normality equations for solving the linear regression model are:

a =

b=, where,   = mean of Y, = mean of X

Computing the value of a and b, table 3 shows the computations:

X	Y	X-	Y-	2	2
472	6916	-12.5714	113.4286	158.0408	12866.04082	-1425.96
413	5884	-71.5714	-918.571	5122.469	843773.4694	65743.47
503	7223	18.42857	420.4286	339.6122	176760.1837	7747.898
612	8158	127.4286	1355.429	16238.04	1837186.612	172720.3
399	6014	-85.5714	-788.571	7322.469	621844.898	67479.18
538	7209	53.42857	406.4286	2854.612	165184.1837	21714.9
455	6214	-29.5714	-588.571	874.4694	346416.3265	17404.9

Table 3: Computations for Regression

The mean of weekly attendance was found to be 484.571, the mean number of chocolate bars sold in a week was found to be 6802.57. Then

Discretisation and Root Locus Analysis of the Discretised System in the Z Domain

b= = 10.977 and a = 6802.57 – 10.977× 484.571 = 1628.6889

The regression model was therefore determined to be:

Y= 1628.6889 + 10.9772X

The sales are expected to increase by 10.9972 units with unit increase in weekly attendance and with zero attendance the expected sales is 1628.6 or 1629 units approximately.

The coefficient of determination or 1-R² is the proportion of variation explained by the model. It is given by the formula: 1- . The value of the denominator was obtained as 4004031.714. The following table gives the computations done for the um of squared error:

X	Y	Predicted Y	Error( =Y-Predicted Y)	square of Error
472	6916	6668.343	-247.657	61333.82
413	5884	6038.387	154.3866	23835.21
503	7223	6999.338	-223.662	50024.87
612	8158	8163.156	5.156081	26.58517
399	6014	5888.905	-125.095	15648.69
538	7209	7373.041	164.0408	26909.38
455	6214	6486.83	272.8304	74436.41

Table 4: Computations for Coefficient of Determination

The sum of squared error was found to be 252214.9601 Then the coefficient of determination was found to be 1- = 0.937.

Therefore it is seen that the model here explains 93.7% of the total variation in the dependent variable Y,  that is the number if chocolate bars sold in a week. This is quite a good fit.

Let A be the event that a student is enrolled in Holmes, let B denote the event that a student is given grass root training, let C be the event that student is an external student and let D be the event that student is given scientific training. The following table gives the observed frequencies of the four events, A, B, C, D.

	Scientific training	Grassroots training	TOTAL
Recruited from Holmes students	35	92	127
External recruitment	54	12	66
TOTAL	89	104	193

Table 5: Table of Observed frequencies

• Then the probability that A occurs or B occurs is given by the formula:

P (A  B) = P (A)+P(B)–P(

Here P(A) and P(B) are the marginal frequencies of “Recruited from Holmes students” and “Grassroots training” divided by total frequency.

P (A) = = 0.658

P (B) = = 0.538

P ( = = 0.476

Then P (A  B) =0.658+0.538-0.476 =0.720

• The probability that a randomly selected player is both an external studentAND is in scientific training is given by P( C D) = = 0.279
• Given that a player is from Holmes, the probability that he is in scientific training is given by the probability of P( D| A) , where

P (D|A) =

Here P (D, A) = P( A  D) = = 0.181 and P( A) = 0.658 (from part a)

Then P (D|A) =   = 0.2775

• The test for independence between training and recruitment is done using Chi-squared statistic for independence. The chi squared statistic is computed using the formula:

Here  = Expected frequency of i^th case, = Observed frequency i^th case, 1 ≤ i ≤ 4 (2 cases for each of 2 variable)

The expected cell frequencies computed using the ratio of the product of the row and column totals with that of the total frequency is:

	Scientific training	Grassroots training
Recruited from Holmes students	58.56476684	68.4352332
External recruitment	30.43523316	35.5647668

Table 6: Expected Cell Frequencies for Chi-squared Test

Then the following table breaks down the computation for the Chi-squared statistic:

Observed(Oi)	Expected(Ei)	Oi- Ei	(Oi- Ei)2 / Ei
35	58.56476684	-23.564767	9.48178
54	30.43523316	23.5647668	18.24524
92	68.43523316	23.5647668	8.114216
12	35.56476684	-23.564767	15.61372

Table 7: Computations for Chi-square Statistics

The chi square statistic was then obtained as per the specified formula as 51.454. The degrees of freedom of the distribution is then (2-1) × (2-1) =1. The critical value of the chi-square statistic is 3.841. Thus the test concludes that the variables are not independent.

Let A denote a person belonging to segment A, similarly let B, C denote a person belonging to segments B,C and D respectively. Again let X denote the phenomenon that a person prefers the product X over Y and Z.

It is given that 20% of the customer belonging to segment A prefers product X over Y and Z, 35% of those in segment B prefers X, 60% of those in segment C prefers X and 90% of those in D prefers X over Y and Z. The conditional probabilities that a person from the segment A, B and C prefer X are:

P(X|A) = 0.2, P(X|B)=0.35, P(X|C)= 0.6, P(X|D) = 0.9

The probabilities of a person belonging to the segments are:

P(A) = 0.55, P(B)=0.33, P(C)=0.1,P(D)=0.05

Then    P(A,X) = P(A)P(X|A) = 0.2×0.55 = 0.11

P(B,X) = P(B)P(X|B) = 0.33× 0.35 = 0.105

P(C,X) = P(C )P(X|C) = 0.1× 0.6 = 0.06

P(D,X) = P(D)P(X|D) =0.05× 0.9 =0.045

(b) Then the probability that a person favors X is,

P(X) = P(A,X)+P(B,X)+P(C,X)+P(D,X) = 0.32

• The probability that a person who favors X is from segment A is given by,

P(A|X) = == 0.3437.

The probability that the customer will buy is given to be 1/10. It is also given that 8 customers entered the store in a 1 minute time interval. The number of customers who make a purchase among them then follows a binomial distribution with p being 1/10 and n being 8. Then the probability that 2 out of the 8 make a purchase is given by the probability mass function of Binomial(8, 1/10) distribution, given as:

Then the probability that two of the customers make a purchase is given as,

 P(X=2) = =  = 0.148.

• It is given that 4 people enter a store per minute within the time slot of 10 am to 11 am. Then the number of people entering the store in 2 minutes within the same time slot is 8. Hence it can be said that the number of people entering a store in 2 minutes between 10 am and 11 am follows a Poisson distribution with or mean 8, that is, an average of 8 people enter the store in 2 minutes. Then the probability that 9 people enter the store in the next 2 minutes is given by the probability mass function of Poisson(8) for x= 9. The probability is then calculated using the formula:

P(X=x) =, x ≥ 0

Then putting x=9, the probability was found to be = = 0.12

The average price of a Surfers Paradise apartment was determined to be $1.1 million and the standard deviation was found to be $385 000. It is then assumed that the price of a Surfers Paradise apartment follows a normal distribution and that the distribution parameters are specified by mean 1.1 and standard deviation 385000/1000000, that is, 0.358.

• Then the probability that an apartment would sell for over $2 million is given by the cumulative density function of normal distribution. If X denotes the price of an apartment then X follows normal with mean 1.1 and standard deviation 0.358. Then, Z= follows standard normal distribution.

 Then using the CDF of standard normal, the probability that price is at least $2 million is given by, ,  being the CDF of standard normal distribution. Using the standard normal table,  = = 0.99403. Then the probability that price is greater than $2 million is given by, 1- = 1-0.99403 =0.00597.

• The  probability that the apartment will sell for over $1 million but less than $1.1 million is given by, P (Z<) – P ( Z<)

• Even though the population of apartments are a mix of old and new apartments and hence not a normal distribution, the sample size drawn is 50 and that makes it a large sample, which then as per central limit theorem implies that the standardized data follows a Z or standard normal distribution.
• 11 out of 45 investors in the sample said that they would invest $1 million or more. Then the estimated probability that an investor would invest $1 million or more is 11/45 that is 0.244. Then the estimated population mean is 0.244. Then the chance that 30% of the investors are willing to commit can be computed by committing is obtained  the large sample approximation to normal with mean 0.244 and standard deviation , which is 0.064 and then by computing the probability  =  = 0.807. So the probability that 30% would agree to invest $1 million is 0.807.