Then r (t) can be expanded into a linear combination of the complex exponential signals ( e j2π(kf 0)t ) k= as. c k e j2π(kf0)t + c k e j2π(kf 0)t

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1 .3 ourier Series Definition.37. Exponential ourier series: Let the real or complex signal r t be a periodic signal with period. Suppose the following Dirichlet conditions are satisfied: a r t is absolutely integrable over its period; i.e., r t dt <. b The number of maxima and minima of r t in each period is finite. c The number of discontinuities of r t in each period is finite. Then r t can be expanded into a linear combination of the complex exponential signals e j2πkf t as where r t = c k e j2πkf t = c + k= f = c k e j2πkft + c k e j2πkf t and 37 c k = α+ r t e j2πkf t dt, 38 α for some arbitrary α. We give some remarks here. { r t, if r t is continuous at t r t = rt + +rt 2, if r t is not continuous at t Although r t may not be exactly the same as rt, for the purpose of our class, it is sufficient to simply treat them as being the same to avoid having two different notations. Of course, we need to keep in mind that unexpected results may arise at the discontinuity points. The parameter α in the limits of the integration 38 is arbitrary. It can be chosen to simplify computation of the integral. Some references simply write c k = r t e jkω t dt to emphasize that we only need to integrate over one period of the signal; the starting point is not important. 5

2 The coefficients c k are called the k th ourier series coefficients of the signal r t. These are, in general, complex numbers. c = r t dt = average or DC value of rt The quantity f = is called the fundamental frequency of the signal r t. The kth multiple of the fundamental frequency for positive k s is called the kth harmonic. c k e j2πkf t + c k e j2πkf t = the k th harmonic component of r t. k = fundamental component of r t..38. Being able to write r t = c ne j2πkf t means we can easily find the ourier transform of any periodic function: r t = c k e j2πkf t Rf = The ourier transform of any periodic function is simply a bunch of weighted delta functions occuring at multiples of the fundamental frequency f..39. ormula 38 for finding the ourier series coefficients c k = α+ r t e j2πkf t dt 39 α is strikingly similar to formula 5 for finding the ourier transform: Rf = rte j2πft dt. There are three main differences. We have spent quite some effort learning about the ourier transform of a signal and its properties. It would be nice to have a way to reuse those concepts with ourier series. Identifying the three differences above lets us see their connection. 5

3 .. Getting the ourier coefficients from the ourier transform: Step Consider a restricted version r T t of rt where we only consider rt for one period. Step 2 ind the ourier transform R T f of r T t Step 3 The ourier coefficients are simply scaled samples of the ourier transform: c k = R T kf. Example.. Train of Impulses: ind the ourier series expansion for the train of impulses δ T t = δ t n drawn in igure 8. This infinite train of equally-spaced -functions is usually denoted by the Cyrillic letter shah igure 8: Train of impulses 52

4 .2. The ourier series derived in Example. gives an interesting ourier transform pair: δ t n = e j2πkf t A special case when = is quite easy to remember: δ t n δ f k Once we remember 2, we can easily use the scaling properties of the ourier transform 2 and the delta function 8 to generalize the special case 2 back to : δ at n = x at a δ δ t n a t n a f a X = a a a a a At the end, we plug-in a = f = /. 53 δ f ka δ f ka δ f a k

5 Example.3. ind the ourier coefficients of the square pulse periodic signal [5, p 57]. period : the scaled ourier transform of the restricted one period version of. c k = R T kf = = 2 sin k π 2 k π 2 = sin k π 2 kπ k k π 2 sin k π sinc 2π ck = sin k π 2 kπ T f f=kf = 2 sinc k π 2 5

6 Remarks: a Multiplication by this signal is equivalent to a switching ON-O operation. Same as periodically turning the switch on letting another signal pass through for half a period. O ON O ON O ON O ON O ON O b This signal can be expressed via a cosine function with the same period: {, cos 2πf t, r t = [cos 2πf t ] =, otherwise. c A duty cycle is the percentage of one period in which a signal is active. Here, duty cycle = proportion of the ON time = width period. In this example, the duty cycle is /2 = 5%. When the duty cycle is n, the nth harmonic c n along with its nonzero multiples are suppressed. 55

7 .. Parseval s Identity: P r = r t 2 = T r t 2 dt = c k ourier series expansion for real valued function: Suppose r t in the previous section is real-valued; that is r = r. Then, we have c k = c k and we provide here three alternative ways to represent the ourier series expansion: r t = = c + = c + 2 c n e j2πkf t = c + ck e j2πkft + c k e j2πkf t 3a k= a k cos 2πkf t + k= b k sin 2πkf t 3b k= c k cos 2πkf t + c k 3c k= where the corresponding coefficients are obtained from c k = α+ α a k = 2Re {c k } = 2 r t e j2πkf t dt = 2 a k jb k r t cos 2πkf t dt 5 b k = 2Im {c k } = 2 r t sin 2πkf t dt 6 T 2 c k = a 2 k + b2 k 7 bk c k = arctan 8 c = a 2 The Parseval s identity can then be expressed as P r = r t 2 = r t 2 dt = c k 2 = c T a k c k 2 k= 9 56

8 .6. To go from 3a to 3b and 3c, note that when we replace c k by c k, we have c k e j2πkf t + c k e j2πkf t = c k e j2πkf t + c ke j2πkf t = c k e j2πkf t + c k e j2πkf t = 2 Re { c k e j2πkf t }. Expression 3c then follows directly from the phasor concept: Re { c k e j2πkf t } = c k cos 2πkf t + c k. To get 3b, substitute c k by Re {c k } + j Im {c k } and e j2πkf t by cos 2πkf t + j sin 2πkf t. Example.7. or the train of impulses in Example., δ t n = e j2πkf t = + 2 cos kω t 5 Example.8. or the rectangular pulse train in Example.3, k= ourier series expansion: Trigonometric ourier series expansion: 2cos 2 2 cos cos cos

9 [cos ω t ] = π cos ω t 3 cos 3ω t + 5 cos 5ω t 7 cos 7ω t +... Example.9. Bipolar square pulse periodic signal [5, p 59]: sgncos ω t = cos ω t π 3 cos 3ω t + 5 cos 5ω t 7 cos 7ω t T t igure 9: Bipolar square pulse periodic signal - T t 58