Chapter 17 : Fourier Series Page 1 of 12

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1 Chapter 7 : Fourier Series Page of SECTION C Further Fourier Series By the end of this section you will be able to obtain the Fourier series for more complicated functions visualize graphs of Fourier series In the last section we found Fourier series of rectangular waveforms. In this section we will obtain Fourier series of other waveforms such as sawtooth. The reason for placing non-rectangular waveforms in this section is because it is more difficult to find the Fourier series for such functions. The integration is more complicated, it generally involves the technique of integration by parts. You need to be very familiar with this method otherwise you will find the C section tough going. However we do return to rectangular waveforms in section C3. C Fourier Series of Sawtooth Waveforms The integration by parts formula is given by d uv x uv u v dx The verbal form of integration by parts is stated as differentiate one (u) and integrate the other (v ). As stated in the last section, obtaining a Fourier series means that you will need to have a thorough understanding of algebra, trigonometry, integration and setching graphs. Application of integration by parts formula should be at your finger tips. For Example 5 we will use the following well nown trigonometric result: (*) sin This means sineven multiple of where is a whole number. which we can visualize from the graph: Figure 7

2 Chapter 7 : Fourier Series Page of Also (**) cos This means coseven multiple of where represents a whole number. which can be visualized from the graph: Figure 8 Example 5 Determine the Fourier series of the sawtooth waveform f t f t shown in Fig. 9: Figure 9 Solution How do we find the Fourier series of Since f t f t? has period so we obtain the values of the Fourier coefficients A, and A B given by the formulae from the last section: f t ] A f t cos t [These are the Cosine coefficients] (7.3) A f t d t [Average value of (7.4) B f t t [These are the Sine coefficients] (7.5) sin What are we going to use for f t in these formulae? We examine the graph between and. Why? Because the period of the given function is and we only need to cover one complete period of the function. 4 6 t

3 Chapter 7 : Fourier Series Page 3 of Since the graph shown in Fig. 9 is a straight line between and, we now it is of the form y mt c where m is the gradient and is given by What does c represent in y mt c? m gradient Cancelling It is the value where the line crosses the vertical axis. Hence from the graph we have c. Substituting these values, m and c, into y mt c gives f t y t Therefore the equation of the line between to is given by Also by observing the graph we can say that f t f t t. has a period of because it repeats itself every interval. Let s find the average value A first. The formula is A d f t t (7.3) By examining the graph below, the integral the line the line f t and the t axis from f t t f t from to as shown shaded in Fig.. represents the area between to. This area is the same as the area under f t t Figure How can we write this area from to in terms of integration? Substituting this into (7.3) we have t d t f t t between to

4 Chapter 7 : Fourier Series Page 4 of (7.3) A f t Replacing f t t and t changing limits t Integrating Substituting the limits 4 and taing out 4 Cancelling 4 Hence the average value of f t is A. The value of A is the average value of the function over one period. We have a straight line going for to, so we would expect the average value to be half way between this, so. This can be a helpful chec when evaluating the constant term A of the Fourier series of a function. How do we find the cosine coefficients We use formula (7.4) and for This gives A? f t we substitute f t A f t cost t cos t d t How do we evaluate the right hand integral cos t and the limits to. t t? We need to use integration by parts formula. In our integrand we let u t v cost : u t v cos t and sin t u Differentiating v cost uv uv uv Substituting these into integration by parts formula,, gives t sint sint t cos t Using uv uv u v Integrating by sin cost cost sint Substituting limits and using cos cos By (*) * sin * * cos by (**)

5 Chapter 7 : Fourier Series Page 5 of Substituting this What does A A t cos t into gives A mean with respect to the Fourier series we are trying to find? means that there are no cosine terms in the Fourier series. What else do we need to find? The values of B. (The sine coefficients of the Fourier series). By integrating (7.5) between to rather than to respectively gives us B f t sint t sint Remember f t t between to B t sint d t How do we evaluate this integral sin t t? We need to use integration by parts again to find sin t t : u t v sin t cos t u Differentiating v sint uv dx uv uv dx Substituting this into integration by parts formula,, gives t cos t cos t t sint Substituting limits and cos cost taing out common factor sint sint cos By cos t by (**) sin sin Substituting limits by (*) Simplifying t sin t into gives Substituting this (*) sin (**) cos B t sin t

6 Chapter 7 : Fourier Series Page 6 of B Cancelling Putting these values, A, A No cosine terms and B into the generic Fourier series (7.) f t A A t A t B t B t gives cos cos... sin sin... f t sint sint sin3t sin4 t No cosine terms sint sin3t sin4t Taing out sin t This is the Fourier series of the sawtooth waveform f t as shown in Fig. 9. Note the long process of finding the Fourier series of a sawtooth waveform. Again it is very easy to mae a slip of signs in obtaining the Fourier coefficients. A slip of signs is the most common mistae made when evaluating Fourier series. C Setching Partial Sums of Fourier Series By using computer algebra system we can setch the Fourier series of f t Here is the MAPLE output for 5 terms: > f t and various partial sums of given in the previous example as shown below. > The graph of the first 5 non-zero terms of the Fourier series of the given sawtooth waveform f t. Fig Here is the MAPLE output for terms:

7 Chapter 7 : Fourier Series Page 7 of > > The graph of the first non-zero terms of the Fourier series of the given sawtooth waveform f t. Figure Notice how the shape of the graph is getting closer and closer to the given sawtooth function. C3 Applications of Fourier Series The next example is more complicated than the previous Example 5 because it uses multiples of in the argument of sine and cosine. However we do not need to use integration by parts formula to find the Fourier coefficients. For Example 6 we will need to use the following trigonometric result: if, 5, 9,... ($) sin if even if 3, 7,,... You can see this result from the following graph:

8 Chapter 7 : Fourier Series Page 8 of sint Figure 3 We also have if, 6,,... ($$) cos if odd if 4, 8,,... Again we can see this result from the following graph: cost Figure 4 These results are not well nown but you should be able to derive them by examining the graphs of sine and cosine functions shown above. We will assume these results and use them in Example 6.

9 Chapter 7 : Fourier Series Page 9 of Example 6 (Mechanical) A pulse force is applied to a mechanical system which has the following graph: f t f t (Newtons) Figure 5 Determine the Fourier series of Solution f t. We need to find all the sine, cosine and constant coefficients in the Fourier series. How do we find these? By using the given formulae. For the constant term (average value) A we use A f t d t (7.3) What is f t equal to between and? From the graph in Fig. 5 we can see that f t between to and / but f t between to /. Hence we only need to use f t and / because the other values are zero. A t Integrating Substituting limits 4 The average value of this function over a complete period is A / 4. What else do we need to find? The cosine coefficients 3 A by formula (7.4): 5 t to with limits secs

10 Chapter 7 : Fourier Series Page of A f t cost Using f t between cost the limits to sint Integrating cost Taing out a common factor sin o f and substituting limits What is sin equal to? We use the above stated result: if, 5, 9,... ($) sin if even if 3, 7,,... How do we find the values for A sin? By multiplying this ($) by / : What does this mean? if, 5, 9,... A sin if even if 3, 7,,... There are no even cosine terms in the Fourier series. If, 5, 9,... then the cosine coefficient A and we have (*) A, A, A, If 3, 7,,... then the cosine coefficient is negative, A, and we have (**) A, A, A, What is our next step? We have to find the sine coefficients, (7.5): B, of the Fourier series. This is given by

11 Chapter 7 : Fourier Series Page of B f t sint Using f t between sint the limits to cost Integrating sint Taing out a common factor cos cos of and substituting limits cos We use the above stated result: if, 6,,... ($$) cos if odd if 4, 8,,... And substitute these values of cos into the last line of the above derivation: B cos What is the value of B if, 6,,...? What is the value of B B if is odd? B What is the value of B if 4, 8,,...? B Hence the non-zero values of the sine coefficients,, 6,,... gives, 6,, is odd, 3, 5, 4, 8,, B, in the Fourier series is when and when is odd. Substituting these values of into and B, B, B, B, B, B, B, How do we obtain the Fourier series? Substitute the coefficients A and B into the generic Fourier series.

12 Chapter 7 : Fourier Series Page of We substitute the constant term A. What do we substitute for the cosine 4 coefficients A? Using (*) and (**) we have A, A, A, A, Remember there are no even cosine coefficients because A when even. Putting these and the above B values into (7.) f t A A t A t B t B t gives cos cos... sin sin... cost cos3t cos5t f t Cosine terms sint sint sin3t sin5t sin6t Sine terms cos3t cos5t cost 3 5 Taing out 4 sin3t sin5t sint sin t This is the Fourier series of the pulse force f t shown in Fig. 5. Even though we don t apply the integration by parts formula for the rectangular waveform of Example 6 it is still a long process to obtain the Fourier series. There are so many chances of maing a mistae in such a long calculation. SUMMARY The integration used to find the Fourier coefficients generally entails the method of integration by parts. If we have a waveform which involves fractions of then we need to be careful in covering all possible values of in evaluating the Fourier coefficients. (**) A, A, A (*) A, A, A 5 9

Trigonometric integrals by basic methods

Roberto s Notes on Integral Calculus Chapter : Integration methods Section 7 Trigonometric integrals by basic methods What you need to know already: Integrals of basic trigonometric functions. Basic trigonometric