The following presentation is a part of the level 4 module -- Electrical and Electronic Principles. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1st year undergraduate programme. The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

1. Transient Responses (Laplace Transforms) Electrical and Electronic Principles © University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License .

3. Transient Responses So far all the calculations we have performed have led to a Steady State solution to a problem i.e. the final value after everything has settled down. There are in fact two parts to the total response of a system to an input, these are: The Steady State which lasts indefinitely and The Transient Response, which decays to zero, leaving only the steady state. The steady state values can be determined using circuit laws and complex number theory. The transient is more difficult as it involves differential equations.

4. Consider the simple charge up circuit below. When the switch is closed we can say that: The equation for a capacitor is Giving us: This is called a first order linear equation with constant coefficients. If we want to know Vc we cannot solve it using simple circuit theory, but it can be solved using some knowledge of Calculus. Transient Responses (Laplace Transforms) V i R C

5. The voltage V is known as the driving function or forcing function and is responsible for the final steady state value. To solve the function we set the forcing function to zero We ask the question what form of Vc will allow us to equate it with its derivative? The form used is This gives us

6. putting this in the equation gives us: this is called the Auxiliary Equation. From which giving us this is known as the Complementary Function . This is part of the total solution but we still need to evaluate the constant A. In order to obtain the complete solution we need to add the Particular Integral. This is the steady state value and for our example this is Transient Responses (Laplace Transforms)

7. So the complete solution is: We still need to determine A. This can be done by using initial conditions: when t = 0, Vc = 0 So giving Our solution is therefore If we start with Vc charged to V and then allow it to discharge we have the same complementary function but the steady state will be 0 and the initial conditions will be Vc = V at t = 0. A would equal V and we would have: Transient Responses (Laplace Transforms)

8. If C = 1  F and R = 1 M  what will happen to the capacitor voltage if a 10v supply is connected across the two components when t = 0?

10. 6.32v >9.8v >9.9v Area = 10   4  5  Transient Responses (Laplace Transforms)

11. Second order equations. Consider the diagram below. For an inductor: For the circuit: but giving us: second order. V i R C L

12. Once again we can use as a solution: Substituting in and removing the forcing function we get: This gives us This can be solved but begins to become complicated especially if the value in the square root is negative and we end up with complex values.

13. Solving for transient conditions is therefore not easy. To make the solving of these problems easier we use Laplace Transforms. Laplace Transformation. What we are able to do is to take a problem in the time domain (t) and to convert it into the Laplace domain (s). The conversion is carried out using a simple set of rules. Transient Responses (Laplace Transforms)

14. Rules 1. If a function of time is multiplied by a constant then the Laplace transform is multiplied by the same constant. e.g. a step of 6v to an electrical system is the same as 6 times a unit step and therefore has the value 6/s. 2. If an equation contains the sum of two separate quantities that are functions of time then the transform is the sum of the individual transforms. 3. The Laplace transform of a first derivative of a function is: Transform of where is the value of the function at t=0 [initial conditions are normally 0] Transient Responses (Laplace Transforms)

15. Rules 4. The Laplace transform of a second derivative of a function is: Transform of where is the value of the derivative of the function at t=0 5. The Laplace transform of an integral of a function is: Transform of Transient Responses (Laplace Transforms)

16. Consider the first order equation for the RC network. Translating gives us: which means that Vc is given by: This is now converted back to the time domain using the reverse Laplace transforms. Is there a transform? If not, we must split the function into simpler parts. Use partial fractions: Transient Responses (Laplace Transforms)

17. from which s = 0 giving equating s giving transform What about a ramp input into the network?

18. Consider the second order equation for the RLC network. Translating gives us: which means that Vc is given by: This is now converted back to the time domain using the reverse Laplace transforms. Is there a transform? If not, we must split the function into simpler parts. Use partial fractions. The method depends upon whether the bracket will factorise.

19. Example V = 10, R = 290 L = 0.1 C = 10 x 10 -6 a = 1, b = 2900, c = 1000000 10v i 290  10  F 0.1H Vc

20. Use Partial Fractions to simplify the denominator. Make s = 0:

21. Make s = -2500: Make s = -400: transform When t = 0 Vc = 10 + 1.905 – 11.905 = 0 Over the page shows the plot:

22. Transient Responses (Laplace Transforms)

23. Example R is reduced to 100  10v i 100  10  F 0.1H Vc V = 10, R = 100 L = 0.1 C = 10 x 10 -6 a = 1, b = 1000, c = 1000000

24. This will not factorise due to the negative square root. We need to take the quadratic equation and complete the square – ( this then matches one of the transforms ): Use Partial Fractions to simplify the denominator.

25. Make s = 0: Equate s values Equate s 2 values We now need to ensure that the fraction is in the correct form for inverse transformation. Transient Responses (Laplace Transforms)

26. transform When t = 0 Vc = 10 - 10– 0 = 0 Over the page shows the plot:

27. Transient Responses (Laplace Transforms)

28. If R was further reduced e.g. 25  this process could be repeated and the following result would be obtained: The plots for the three values of R are shown on the next slide. Transient Responses (Laplace Transforms)

29. R=290  R=100  R=25 

30. This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2009 University of Wales Newport Except where other wise noted, this work is licensed under a Creative Commons Attribution 2.0 License . The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The name and logo of University of Wales Newport is a trade mark and all rights in it are reserved. The name and logo should not be reproduced without the express authorisation of the University. Transient Responses (Laplace Transforms)