btec hnc - control systems and automation - use laplace transforms to determine system parameters

Use Laplace Transforms to Determine System

ParametersControl Systems and Automation

By Brendan Burr

Brendan Burr BTEC Higher National Certificate in ElectronicsControl Systems and Automation

Table of Contents

TABLE OF CONTENTS - 2 -

TASK 1 - 5 -

1.1 Explain the need for the following operator methods when solving engineering problems:- - 5 -

a) Laplace Transforms - 5 -Solution:- - 5 -

b) Inverse Laplace Transforms - 5 -Solution:- - 5 -

c) Partial Fractions - 6 -Solution:- - 6 -

d) Completing the square method - 7 -Solution:- - 7 -

TASK 2 - 8 -

2.1 (a) Derive the equations in the time domain (t) for the following signals: - - 8 -Solution (i):- - 8 -Solution (ii):- - 8 -

(b) Derive the equations in the frequency domain (s) using Laplace Transforms:- - 9 -

Solution (i):- - 9 -Solution (ii):- - 9 -

TASK 3 - 10 -

3.1 Use tables of Laplace Transforms to find the signal in the frequency domain:-- 10 -

(a) - 10 -Solution:- - 10 -

(b) - 10 -Solution:- - 10 -

(c) - 11 -Solution:- - 11 -

(d) - 11 -Solution:- - 11 -

2


3.2 Use tables of Inverse Laplace Transforms to find the signal in the time domain :- - 12 -

(a) - 12 -

Solution:- - 12 -

(b) - 12 -

Solution:- - 12 -

(c) - 13 -

Solution:- - 13 -

(d) - 14 -

Solution:- - 14 -

TASK 4 - 15 -

4.1 A first order control system is characterised by the following differential equation with initial conditions that at t = 0, i = 0. - 15 -

- 15 -

a) Determine the equation for the current i(t) when the applied voltage v(t) is a step input of 5 volts using Laplace Transforms. - 15 -

Solution:- - 15 -

b) Draw the response in the time domain. - 16 -Solution:- - 16 -

4.2 Solve the following Differential Equation with initial conditions that y = 0 at t = 0, using Laplace Transforms. - 17 -

- 17 -

Solution:- - 17 -

3


4.3 A second order control system is characterised by the following differential equation with the following initial conditions that at t = 0, v = -1 and at t = 0,

. - 18 -

- 18 -

(a) Determine the equation for the voltage v (t) using Laplace Transforms. - 18 -Solution:- - 18 -

(b) Comment upon the response in terms of damping and frequency. - 19 -Solution:- - 19 -

4.4 The complete response of a series RLC circuit is described by a second order differential equation in the time domain as follows :- - 20 -

- 20 -

If R = 1 k Ω , L = 100 mH and C = 0.1 μF. - 20 -

a) Determine the value of - 20 -Solution:- - 20 -

(b) Determine the value of . - 21 -Solution:- - 21 -

(c) Derive an expression for the transfer function in terms of the Laplace variable S assuming initial conditions are zero. N.B The expression must contain the actual values found in (a) and (b) above. - 22 -

Solution:- - 22 -

EVALUATION - 23 -

CONCLUSION - 24 -

BIBLIOGRAPHY - 24 -

Books - 24 -

Catalogues - 24 -

Websites - 24 -

4


Task 1

1.1 Explain the need for the following operator methods when solving engineering problems:-

a) Laplace Transforms

Solution:-

We use Laplace Transforms as standard modelling tools used in Control Engineering.Laplace Transforms allow engineers to transform differential equations into more easily solvable algebraic equations. This allows differential equations, which describe circuit behaviour through time, to be converted to a form which allows algebraic manipulations to be carried out.The Laplace Transform, F(s), of a Time Domain signal f(t) is define by an integral:

The s in F(s) is called the Laplace Variable, which has units of frequency.

b) Inverse Laplace Transforms

Solution:-

We use Inverse Laplace Transforms to reverse the effect and take the algebraic equation back into the time domain. This allows engineers to determine the circuit behaviour, after an equation has been manipulated.Engineers will use a table of the Laplace Transforms, to prevent the requirement of performing differential calculations where a simple conversion is available.I will be using the Laplace Transforms table for the majority of this assignment to enable me to change a function in the time domain into a function in the s domain.

5


c) Partial Fractions

Solution:-

This method enables us to convert an algebraic expression into simple fractions, for example:

into

There are basically three partial fractions, which are as follows:1. Linear factors in the denominator

Expression

Partial fraction

2. Repeated linear factors in the denominator

Expression

Partial fraction

3. Quadratic factors in the denominator, when the quadratic does not factorise without imaginary terms

Expression

Partial fraction

Or if there is also a linear in the denominator

Expression

Partial fraction

6


d) Completing the square method

Solution:-

This is a method used to solve quadratic equations if they cannot be factorised. It can be used to determine the value of “x” within a quadratic equation, with five steps.

Step 1 – Divide all terms by a (the coefficient of .

Step 2 – Move the number term, to the right side of the equation.

Step 3 – Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.

Step 4 – Take the square root on both sides of the equation.

Step 5 – Add or subtract the number that remains on the left side of the equation to find x.

For example:

Solving

Step 1 -

Step 2 -

Step 3 -

Step 4 - (to 3 d.p)

Step 5 -

7


Task 2

2.1 (a) Derive the equations in the time domain (t) for the following signals: -

(i) (ii)

Solution (i):-

Solution (ii):-

8


(b) Derive the equations in the frequency domain (s) using Laplace Transforms:-

Solution (i):-

Solution (ii):-

1

9


Task 3

3.1 Use tables of Laplace Transforms to find the signal in the frequency domain:-

(a)

Solution:-

(b)

Solution:-

1

1

10


(c)

Solution:-

(d)

Solution:-

11


3.2 Use tables of Inverse Laplace Transforms to find the signal in the time domain :-

(a)

Solution:-

(b)

Solution:-

1

12


(c)

Solution:-

Denominator:

So:

So:

13


(d)

Solution:-

Denominator:

So:

Numerator:

So:

14


Task 4

4.1 A first order control system is characterised by the following differential equation with initial conditions that at t = 0, i = 0.

a) Determine the equation for the current i(t) when the applied voltage v(t) is a step input of 5 volts using Laplace Transforms.

Solution:-

i(0) is the value of the current when t=0, therefore i(0) = 0

So:

[Time Domain]

15


b) Draw the response in the time domain.

Solution:-

Seconds

16


4.2 Solve the following Differential Equation with initial conditions that y = 0 at t = 0, using Laplace Transforms.

Solution:-

Y(0) is the value of y when t=0, therefore y(0)=0

17


4.3 A second order control system is characterised by the following differential equation with the following initial conditions that at t = 0, v = -1 and

at t = 0, .

(a) Determine the equation for the voltage v (t) using Laplace Transforms.

Solution:-

Resistor = 0 Ohms, i.e. No Damping.

So;

Rads/sec

Using Laplace Transforms we get:

Using tables of Laplace Transforms we get:

18


(b) Comment upon the response in terms of damping and frequency.

Solution:-

Rads/sec

Hz

Secs

Secs (proven by graph 1)

There is no damping as R=0, you can see this through the second graph above as the waveform has a constant peak value. Zeta also equals zero.

19


4.4 The complete response of a series RLC circuit is described by a second order differential equation in the time domain as follows :-

If R = 1 k Ω , L = 100 mH and C = 0.1 μF.

a) Determine the value of

Solution:-

So:

Rads/sec

20


(b) Determine the value of .

Solution:-

(Zeta has no units)

21


(c) Derive an expression for the transfer function in terms of the Laplace variable S assuming initial conditions are zero. N.B The expression must contain the actual values found in (a) and (b) above.

Solution:-

1 st Term

2 nd Term

So:

Extracting the “Vo” as a common factor on the LHS, giving:-

So the transfer function G(s) of a second order control system is as follows:

Therefore:

22


EvaluationThis assignment tested my understanding of the information given to me in class. For Task 1, I had to explain a number of methods used in control engineering calculations. I used the Control Engineering book written by W. Bolton, as well as notes taken from class, to help me explain the Laplace Transforms as well as the Inverse Laplace Transforms. I then used the book to also help me explain the Partial Fractions, even though we had gone through it in class.I couldn’t remember doing much work on the completing the square method, with regards to the Laplace Transforms, however it did fit into the explanation given for the Partial Fractions. I completed a bit of research in the Higher Engineering Mathematics book, written by J. Bird, to help refresh my memory and soon remembered that we have used this method in analytical methods in previous years.Task 2 involved me performing some basic transpositions between the time and the s domains. Once you get the hang of using the Laplace Transforms Table this task was very straightforward. I was able to work out the answers, transforming from the time domain to the s domain and then inverting it for the second part of the task.For Task 3, it took me a little bit longer to come up with a solution, however the working out was very similar to Task 2. Using the Laplace Transforms Table I was able to come up with the answer. I couldn’t find a method of checking the answer, other than ensuring that all my workings were correct.Using the examples given to me in class along with the existing knowledge gained from the rest of the unit, I was able to complete Task 4. For Task 4.1, I used a couple of the Laplace Transforms taken from the table. This enabled me to convert the equation into the s domain and then manipulate it, before turning it back into the time domain. The manipulation involved some basic transposition of formula to put the values of omega and alpha into a similar place to match up with the LT Table. I used a similar approach to Task 4.2, manipulation on the numbers gave me values for alpha, which I could then convert from one domain to the other with relative ease.I had a bit of difficulty with the “+” and “-“ symbols in Task 4.3. At first I wasn’t sure whether they got dropped in the conversion to the time domain at the end of the workings, but after thinking about it for a bit I made the decision to keep them in. I used Graphmatica to see the waveform and check it matched up with my workings, which it did, however I found that this wasn’t a suitable check for the question.Tasks 4.4a and b were very straightforward. They simply involved inputting some values into equations. Task 4.4c involved some careful manipulation but in all this was also relatively easy.

23


ConclusionI found this assignment quite time consuming in the write up. I have noticed in previous years of Analytical Methods assignments that typing up formula can be extremely time consuming because of the level of detail required. It is for this reason that it is also easy to make a typographical mistake on the assignment, even though the rough workings are accurate.Because of the level of tutoring we received for this assignment, I found it generally easy. At times during lessons I did find myself having to concentrate quite hard to make sense of the new material, however it did go in eventually.I am confident that the answers I have provided are accurate even though I was unable to provide a checking method for the majority of them.

Bibliography

Through guidance from my lecturer, the following text books, catalogues and websites I was able to complete this assignment:

Books

Higher Engineering Mathematics (John Bird) ISBN: 0-7506-8152-7

Control Engineering (W. Bolton) ISBN: 0-582-32773-3

Catalogues

N/A

Websites

N/A

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btec hnc - control systems and automation - use laplace transforms to determine system parameters

Documents

solution i

time domain t

electronicscontrol systems

frequency domain s

following signals

initial conditions

following operator methods

order control system