unj fmipa-fisika

1. Kalkulus II Teguh Budi P, M.Si Sesion#11-12 JurusanFisika FakultasMatematikadanIlmuPengetahuanAlam

2. Outline Fourier Series Trigonometric form of Fourier Series Solving a problem using Fourier series 09/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 2

3. Infinite Series (part 3) © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 3 09/01/2011

4. FOURIER SERIES Jean-Baptiste Fourier (France, 1768 - 1830) proved that almost any period function can be represented as the sum of sinusoids with integrally related frequencies. The Fourier series is one example of an orthogonal set of basis functions, as a very important example for engineers. Trigonometric form of Fourier Series Let us map the functions 1, and by the following : The purpose of this nothing deeper than to map the conventional Fourier series onto the notation we have derived for orthogonal functions 09/01/2011 4 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

5. The Fourier series is a special case of the more general theory of orthogonal functions. Now calculate the value of lm from ie The value of lm for is simply the power in a sinewave (cosine wave) The value of l0 is the power in a DC signal of unit amplitude. Now we can derive immediately the Euler formula from equation by substituting in the values of and lm from the above equations then 09/01/2011 5 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

6. In the Fourier series, instead of using the term a-m as the coefficient of the cosine terms in the Fourier expansion we usually use the term am with bm reserved for the sine terms. The important point to realize here is that the Fourier series expansion is only a special case of an expansion in terms of orthogonal functions There are many other function (e.g. Walsh function), so using the Fourier series as an example, try and understand the more general orthogonal function approach When we write a periodic function using a Fourier series expansion in terms of a DC term and sine and cosine terms the problem which remains is to determine the coefficients a0 , am and bm 09/01/2011 6 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

7. Remove all u(t) y(t) frequencies > 5.5 Hz 2V t t=0 1s Solving a problem using Fourier series Consider a sawtooth wave which rises from -2V to 2V in a second. It passes through a linear time invariant communication channel which does not pass frequencies greater than 5.5 Hz. What is the power lost in the channel ? Assume the output and input impedance are the same. (Use sine or cosine Fourier series). 09/01/2011 7 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

8. The first step is set up the problem mathematically. The time origin has not been specified in the problem. Since the system is time invariant it doesn't matter when t = 0 is located since it is will not change the form of the output. Choose time t = 0 at the center of the rise of the sawtooth because it makes the function which we now call u(t), into an odd function. Since the system is specified in terms of it frequency response, i.e. what it will do if a sinwave of a given frequency is input, it makes a lot of sense to express as a sum of either sines and cosines or complex exponentials since as we know what happens to these functions. If it's a sinewave or cosine wave and has a frequency less than 5.5 Hz it is transmitted, otherwise it is eliminated. The situation with complex exponential is a little trickier, if its in the range [-5.5,5.5] Hz then will be transmitted otherwise will be eliminated. It would do no good to find the response to each sinewave individually, because we could not then add up these individual response to form the total output, because that would require superposition to hold. 09/01/2011 8 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

9. Let us calculate the Fourier series for a sawtooth wave or arbitrary period and amplitude. Now with the choice of t = 0, we can write the input as mathematically within the period as We don't need to worry that outside the range [-T/2 < t < T/2] the above formula is incorrect since all the calculation are done within the range [-T/2 < t < T/2]. As we strict to the range given the mathematical description is identical to the sawtooth and all will be well. We want the input to written in the form 09/01/2011 9 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

10. Go back to Euler formula for Fourier series which have derived earlier from the general orthogonality conditions Now for the DC value of the sawtooth For an the coefficients of the cosine terms 09/01/2011 10 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

11. The next step is use integration by parts, i.e. Therefore 09/01/2011 11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

12. This is a lot of work for 0, when it is fairly obvious that the integral of the product of an odd function (sawtooth) and an even function (the cosine term) is always zero when we integrate from [-v,v] whatever value of v, as shown below The procedure for calculating for is almost identical, the final answer is Now we know a0 , an and bn , we can write down the Fourier series representation for u(t) after substituting T = 1 and A = 2 The series has all the sine terms present, i.e.bm is never zero and there are no cosine terms. 09/01/2011 12 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

13. How can be check that our calculations for u(t) is correct ? We can calculate the power in the signal by and also by Parseval's theorem when applied to sine and cosine functions Where the 0.5 came from ? Remember that is lm which has the basis function was equal to sin(t) and the power in the sinewave, lm = 0.5 . Having calculated the Fourier series and having checked it using the Parseval's theorem it only remains to calculate the power in the first 5 harmonics, i.e. those with a frequency less than 5.5 Hz 09/01/2011 13 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

14. Thus the power transmitted by the channel is The power loss is therefore 1.3333 - 1.1863 = 0.1470, and the power gain in dB is thus -0.51 dB. The final answer is that the channel attenuates the signal by 0.51 dB. 09/01/2011 14 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |

15. Thank You 09/01/2011 © 2010 Universitas Negeri Jakarta | www.unj.ac.id | 15