Downloaded from www.pencilji.com LECTURE NOTES OF ENGINEERING MATHEMATICS–III (Sub Code: MAT31) COURSE CONTENT om 1) Numerical Analysis 2) Fourier Series 3) Fourier Transforms & Z-transforms 4) Partial Differential Equations 5) Linear Algebra 6) Calculus of Variations pa jin w .re Reference Book: Advanced Engineering Mathematics by E. Kreyszig (8th Edition – 2001) John Wiley & Sons, INC. New York ul .c Text Book: Higher Engineering Mathematics by Dr. B.S.Grewal (36th Edition – 2002) Khanna Publishers,New Delhi w FOURIER SERIES DEFINITIONS : fro m w Dr.A.T.Eswara Professor and Head Department of Mathematics P.E.S.College of Engineering Mandya -571 401 oa d A function y = f(x) is said to be even, if f(-x) = f(x). The graph of the even function is always symmetrical about the y-axis. do w nl A function y=f(x) is said to be odd, if f(-x) = - f(x). The graph of the odd function is always symmetrical about the origin. x = f(x) and the For example, the function f(x) = x in [-1,1] is even as f(-x) = x function f(x) = x in [-1,1] is odd as f(-x) = -x = -f(x). The graphs of these functions are shown below : Downloaded from www.pencilji.com ul .c om Downloaded from www.pencilji.com Graph of f(x) = x pa Graph of f(x) = x 1. If f(x) is even and g(x) is odd, then w w h(x) = f(x) x g(x) is odd h(x) = f(x) x f(x) is even h(x) = g(x) x g(x) is even w .re jin Note that the graph of f(x) = x is symmetrical about the y-axis and the graph of f(x) = x is symmetrical about the origin. fro m For example, 1. h(x) = x2 cosx is even, since both x2 and cosx are even functions 2. h(x) = xsinx is even, since x and sinx are odd functions 3. h(x) = x2 sinx is odd, since x2 is even and sinx is odd. oa d a 2. If f(x) is even, then a f ( x)dx 2 f ( x)dx a 0 nl a 3. If f(x) is odd, then f ( x)dx 0 do w a For example, a a cos xdx a 2 cos xdx, as cosx is even 0 a and sin xdx 0, as sinx is odd a Downloaded from www.pencilji.com com PERIODIC FUNCTIONS :A periodic function has a basic shape which is repeated over and over again.2.3. The fundamental range is the time (or sometimes distance) over which the basic shape is defined. etc. The Fourier series is named after the French Mathematician and Physicist Jacques Fourier (1768 – 1830)..…. It may be verified that a linear combination of periodic functions is also periodic. ul . Fourier series has its application in problems pertaining to Heat conduction.2.Downloaded from www. Sines and cosines are the most fundamental periodic functions.pencilji. The subject matter may be divided into the following sub topics.. do w Note that the graph of the function between 0 and 2 is the same as that between 2 and 4 and so on. FOURIER SERIES A Fourier series of a periodic function consists of a sum of sine and cosine terms.……. A general periodic function f(x) of period T satisfies the condition om f(x+T) = f(x) Here f(x) is a real-valued function and T is a positive real number. n=1. nl oa d fro m w w The graph of the function is shown below : w .pencilji..re jin The function f(x) = sinx is periodic of period 2 since Sin(x+2n ) = sinx. Downloaded from www. = f(x+nT) f(x) = f(x+nT). pa Thus. it follows that f(x) = f(x+T) = f(x+2T) = f(x+3T) = …. The length of the fundamental range is called the period.c As a consequence.3. acoustics. n=1.com . f(x) = 0 (5) 2 n1 l l Suppose f(x) is discontinuous at x. It can be proved that the sum of the series (4) is f(x) if f(x) is continuous at x.a+2l).2. the real numbers a0.bn are called the Fourier coefficients of f(x)..pencilji. and b1. f(x) is defined in an interval (a.com FOURIER SERIES Series with arbitrary period Complex series Harmonic Analysis om Half-range series FORMULA FOR FOURIER SERIES ul .. 3. f(x) is continuous or has only a finite number of discontinuities in the interval (a.2. 2.a+2l). f(x) has no or only a finite number of maxima or minima in the interval (a..com . let a 2l a 1 l a 2l 1 l a 2l w f ( x)dx n l xdx.pencilji.. …. a2. (3) a oa d fro bn (1) f ( x) cos a m an 1 l w a0 Then.re jin pa 1. the infinite series do w nl a0 n n an cos x bn sin x (4) 2 n1 l l is called the Fourier series of f(x) in the interval (a.an. Thus we have a n n an cos x bn sin x …….3..3. Also. Also. b2 . n 1.. (2) and (3) are called Euler’s formulae..a+2l). a1.. then the sum of the series (4) would be 1 f (x ) f (x ) 2 where f(x+) and f(x-) are the values of f(x) immediately to the right and to the left of f(x) respectively.a+2l).. (2) f ( x) sin n l xdx. and f(x+2l) = f(x) so that f(x) is a periodic function of period 2l. n 1. Downloaded from www.Downloaded from www. …. The formulae (1)...c Consider a real-valued function f(x) which obeys the following conditions called Dirichlet’s conditions : w . . If we set l = . (7) n=1. …… Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (-l .2. ). Formulae (6) reduce to an cos nx bn sin nx (8) n 1 do w nl Suppose a=-l. Then f(x) is defined over the interval (-l . Formulae (1). xdx.2. n=1.. then f(x) is defined over the interval (. l). 1 2 1 2 f ( x)dx 0 f ( x) cos nxdx 0 f ( x) sin nxdx a0 2 m Also..2l). in this case. l). then f(x) is defined over the interval (0.2.. Formulae (1). w 0 w bn 2 1 .re an jin a0 = pa If we set l= .2l).pencilji. ul ..…. (2) (3) reduce to n =1.com Particular Cases Case (i) Suppose a=0.…… l a0 1 f ( x)dx l l an 1 n f ( x) cos l l l (9) l bn l xdx 1 n f ( x) sin l l l xdx.. (3) reduce to 2l 1 a0 f ( x)dx l0 2l an n 1 f ( x) cos l0 l bn n 1 f ( x) sin l0 l (6) n 1. (5) becomes fro f(x) = oa d Case (ii) n=1.2.Downloaded from www. Then f(x) is defined over the interval (0...2 ). w .2. 2l om xdx.c Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (0.pencilji.com . Formulae (9) reduce to a0 = 1 f ( x)dx Downloaded from www.. (2).…. 2 are x x S1 ( x) a0 a1 cos b1 sin l l x x 2 x 2 x S2 ( x) a0 a1 cos b1 sin a2 cos b2 sin l l l l If we draw the graphs of partial sums and compare these with the graph of the original function f(x). However. are the successive derivatives of u and v1 vdx. x3 e2 x 2 3x 2 e2 x 4 6x e2 x 8 2 6 cos nx n3 e2 x 16 The following integrals are also useful : Downloaded from www. (10) n=1. through the following examples : cos nx sin nx x 2 sin nxdx x 2 2x n n2 x 3e 2 x dx 2. oa d Some useful results : nl 1. we can find the sum through the partial sum S N defined as follows : fro m w w integral values.... uvdx uv1 u 'v2 u ''v3 .com .2.. The following rule called Bernoulli’s generalized rule of integration by parts is useful in evaluating the Fourier coefficients. the partial sums for N=1.…. v2 v1dx. it may be verified that SN(x) approximates f(x) for some large N. We illustrate the rule. we get f(x) = a0 2 an cos nx bn sin nx om Putting l = f ( x) cos nxdx n 1 ul . It uses an infinite number of terms which is impossible to calculate.....….... do w Here u ..pencilji.. f ( x) sin nxdx in (5).re n 1 n l jin pa The Fourier series gives the exact value of the function.com 1 an 1 bn n=1.Downloaded from www. In particular.….c PARTIAL SUMS n N S N ( x) a0 an cos x bn sin n l x where N takes positive w . .2.pencilji.. u .. If ‘n’ is integer. Downloaded from www. then sin n = 0 .< x < pa f(x) = ul .Downloaded from www.c 1.pencilji. an and bn in the Fourier expansion f ( x) a0 2 an cos nx n 1 bn sin nx n 1 we get. so that 1 0 2 0 oa d 1 do w nl bn cos nx n2 1 ( 2 1 2 x) sin nxdx cos nx ( 1) n x sin nx n2 ( 1) n n Using the values of a0 .pencilji. 1 2 x sin nx n ( 1) fro an m Here we use integration by parts.re 1 a0 jin We have. cos2n =1 om Examples 1 2 x in . cosn = (-1)n .com .com e ax e ax cosbxdx a cosbx b sin bx a 2 b2 e ax a sin bx b cosbx a 2 b2 e ax sin bxdx 3. Obtain the Fourier expansion of = 1 2 f ( x)dx x2 2 x 1 f ( x) cos nxdx 1 ( 2 x)dx 1 1 ( 2 x) cos nxdx w w an 1 w . sin2n = 0. pencilji.c a 2 sinh a a ax e ax jin a 1 an a e 1 e dx w . f(x) = sinh a a 2a sinh a n For x=0. Deduce that n om ( 1) n 2 1 2n 2 cos ech Here.com n( 1) n sin nx 2 n2 1a . a=1.com n ( 1) sin nx 2 n1 n This is the required Fourier expansion of the given function. f ( x) 2. the series reduces to sinh 2 sinh f(0)=1 = n ( 1) n cos nx 2 n2 1 a 2 sinh a n ( 1) n 2 1 1 n or 1= sinh 2 sinh 1 2 n ( 1) n 2 1 2n Downloaded from www.re ea ax e w 1 a0 2n ( 1) n sinh a = a 2 n2 do w nl Thus. ).pencilji. Obtain the Fourier expansion of f(x)=e -ax in the interval (.. 1 an e ax a cos nx n sin nx n2 ax sin nxdx 1 e a2 ax n2 a sin nx n cos nx oa d = fro m e w 2a ( 1) n sinh a a 2 n2 1 pa cos nxdx a2 bn = ul .Downloaded from www. ). Hence 1 2 a0 = f ( x)dx = f ( x)dx = 2 2 x3 3 x 2 dx 0 = f ( x) cos nxdx w 2 f ( x) cos nxdx. ul .pencilji.. Obtain the Fourier expansion of f(x) = x2 over the interval (.com or 2 sinh 1= n ( 1) n 2 1 2n Thus..re 1 an w 3 m or 0 2 2 a0 jin 0 x 2 cos nxdx oa d 0 Integrating by parts... Thus Downloaded from www.com . since f(x)cosnx is even 0 2 fro = w .. 6 2 32 pa The function f(x) is even.pencilji. Deduce that 2 1 1 1 2 . we get 2 do w nl an Also..c om 3. bn 4( 1) n2 1 x2 sin nx n 2x cos nx n2 2 sin nx n3 0 n f ( x) sin nxdx 0 since f(x)sinnx is odd. cos ech 2 n ( 1) n 2 1 2n This is the desired deduction.Downloaded from www. . . f ( x) ul ...0 x 2 x..pencilji. Hence the function f(x) is an even function.. 1 22 1 6 1 32 ... AB represents the line f(x)=(2 -x) and AC represents the line x= . 1 an f ( x) cos nxdx 2 f ( x) cos nxdx 0 2 = x cos nxdx 0 Downloaded from www.com . om f ( x) x 2 pa x.. w .Downloaded from www.com 2 4 3 n 2 2 4 3 1 1 n2 n ( 1) n cos nx n2 1 1 2 1 n 2 6 2 Hence.c 4. which in turn is parallel to y-axis.. do w nl a0 = 2 f ( x)dx = 2 f ( x)dx 0 xdx 0 since f(x)cosnx is even. Note that the graph is symmetrical about the line AC.re 1 jin 2 fro m w w Here OA represents the line f(x)=x. 1 oa d Here. Obtain the Fourier expansion of Deduce that 1 1 8 32 52 The graph of f(x) is shown below.pencilji. .. we get nl oa d fro 5..re 2 f( ) pa For x= .com 2 sin nx n x = cos nx n2 1 0 2 ( 1) n 1 2 n Also. m 1 w n 1 w 8 jin or 2 2 1 ( 1) n 1 cos n 2 n 1n 2 cos(2n 1) (2n 1) 2 w .. x 0 f(x) = x.. 2 1 (2n 1)2 2 1 8 32 This is the series as required.. since f(x)sinnx is odd 2 1 ( 1) n 1 cos nx 2 1n n ul .. a0 1 0 dx xdx 2 0 an 1 0 cos nxdx x cos nxdx 0 1 ( 1) n 1 2 n Downloaded from www.. 8 3 52 do w Here..0 x Deduce that 2 1 1 1 2 .c 2 f ( x) om Thus the Fourier series of f(x) is 2 n 1 Thus. or 1 52 . Obtain the Fourier expansion of .Downloaded from www..pencilji.pencilji. 1 bn f ( x) sin nxdx 0 .com . Here f(x+) =0. f(x-)=.c om f(x) = 1 8 1 32 1 52 . we get an 2 1 x 2 sin n x n ( 2 x) cos n x (n ) 2 ( 2) sin n x ( n )3 Downloaded from www.at x=0.. Obtain the Fourier series of f(x) = 1-x2 over the interval (-1.re 2 jin Simplifying we get.. w . w 6. Also period of f(x) is 1-(-1)=2 1 2 oa d = 2 (1 x )dx x3 3 2 x 0 1 0 do w nl 4 3 1 1 f ( x) cos(n x)dx 1 1 an 1 as f(x) cos(n x) is even 2 f ( x) cos(n x) dx 0 1 = 2 (1 x 2 ) cos(n x)dx 0 Integrating by parts..pencilji.1).com 0 1 bn sin nxdx x sin nxdx 0 1 1 2( 1) n n Fourier series is 1 1 2( 1) n n nx ( 1 ) 1 cos sin nx 2 n 4 n 1n n 1 Note that the point x=0 is a point of discontinuity of f(x)...Downloaded from www. 1 1 Hence [ f ( x ) f ( x )] 0 2 2 2 The Fourier expansion of f(x) at x=0 becomes 1 1 [( 1) n 1] 2 2 4 n 1n 1 or 4 n 1 [( 1) n 1] 2 1n pa 2 ul .pencilji. Here 1 1 1 f ( x)dx = 2 f ( x)dx 1 1 0 fro a0 = m w The given function is even.com 1 0 . as f(-x) = f(x). re Deduce that 1 pa 7.. Downloaded from www. The Fourier series of f(x) is 2 n ( 1) n 1 cos(n x) n2 1 a0 w 3 3 2 Hence f(x) is even 3/ 2 1 3/ 2 f ( x)dx 3/ 2 3/ 2 oa d 4 3 4x dx 3 1 0 1 3/ 2 nl an do w m f(-x) = f(x).com 4( 1) n 1 n2 2 1 1 bn f ( x) sin(n x)dx 1 1 = =0.c 2 3 f(x) = 2 3/ 2 f ( x) cos 3/ 2 f ( x) cos 0 3/ 2 f ( x)dx 0 0 3/ 2 3/ 2 2 3/ 2 n x dx 3/ 2 2n x dx 3 3/ 2 4 4x 1 3 3 = 4 n 2 2 2n x sin 3 2n 3 4 3 2n x cos 3 2n 3 2 0 1 ( 1)n Also. Obtain the Fourier expansion of 4x 3 1 in x 0 3 2 f(x) = 4x 3 1 in0 x 3 2 2 om 4 ul .Downloaded from www. since f(x)sin(n x) is odd.com . fro Also 3 2 1 52 w 8 The period of f(x) is 1 32 jin w .pencilji...pencilji. ... 8 3 52 f(0) = jin NOTE w . we get ul . ) is 4 3 ( 1) n cos nx 3 n2 n 1 The partial sums corresponding to N = 1...2. 2 pa or 4 om 1 1 ( 1) n 2 n 1n 8 1 1 ..c Thus.Downloaded from www.com ... 1= 2 1 2 3 52 2 1 1 1 2 .6 2 3 4 cos x oa d S1 ( x) n N 4 fro S N ( x) m 2 w n Let us define ( 1) n cos nx n2 1 w 2 f ( x) 2 are S 2 ( x) 4 cos x cos 2 x nl 3 2 1 4 1 4 cos5 x cos5 x cos 4 x cos3 x 9 25 4 9 3 The graphs of S1 .... We recall that the Fourier expansion of f(x) = x2 over (. S2 .com 3 1 3 bn 2 n x dx 3 2 f ( x) sin 3 2 0 Thus f(x) = 4 2 n 1 2n x 1 ( 1) n cos 2 3 1n putting x=0..….pencilji. …S6 against the graph of f(x) = x2 are plotted individually and shown below : do w S6 ( x) 4 cos x cos 2 x Downloaded from www..re Here verify the validity of Fourier expansion through partial sums by considering an example...pencilji. ) given by Downloaded from www.l) which is regarded as half interval.pencilji. The definition can be extended to the other half in such a manner that the function becomes even or odd.c om Downloaded from www.com w w On comparison. Then we define f(x) = .l).l). This will result in cosine series or sine series only. Putting l= in (11).pencilji. do w nl oa d HALF-RANGE FOURIER SERIES The Fourier expansion of the periodic function f(x) of period 2l may contain both sine and cosine terms. This verifies the validity of Fourier expansion for the function considered. we obtain the half-range sine series of f(x) over (0.0). fro m Exercise Check for the validity of Fourier expansion through partial sums along with relevant graphs for other examples also. Hence f(x) becomes an odd function in (-l .re jin pa ul .com . l).w .(-x) in (-l. Sine series : Suppose f(x) = (x) is given in the interval (0. The Fourier series then is f ( x) bn sin n 1 n x l (11) l 2 n x where bn f ( x) sin dx l 0 l The series (11) is called half-range sine series over (0. we find that the graph of f(x) = x2 coincides with that of S6 (x). Many a time it is required to obtain the Fourier expansion of f(x) in the interval (0. re Then the Fourier series of f(x) is given by a0 n x f ( x) an cos 2 n1 l where.. ).pencilji..3..Downloaded from www. . ( x) ( x) f ( x) l 2 n x f ( x) cos dx l 0 l The series (12) is called half-range cosine series over (0. l 2 a0 f ( x)dx l 0 ul . 2 bn f ( x) sin nxdx 0 2 ( x x 2 ) sin nxdx 0 Integrating by parts. do w 0 Examples : 1. We have.. given in (-l.com f ( x) bn sin nx n 1 2 bn f ( x) sin nxdx 0 Cosine series : om Let us define pa jin (12) w ..2.l) .0) ….in order to make the function even. we get an cos nx fro Putting l = w w an oa d 2 n 1 a0 f ( x)dx 0 2 nl an f ( x) cos nxdx n 1.c in (0.pencilji.. we get Downloaded from www.l) a0 2 f ( x) where m in (12). Expand f(x) = x( -x) as half-range sine series over the interval (0.com . . n 2.0 2 xdx ( 0 2 2 2 2 x cos nxdx ( 0 x) cos nxdx w an x)dx 2 w 2 w .com 2 bn cos nx n x x2 2x sin nx n2 ( 2) cos nx n3 0 om 4 1 ( 1) n 3 n The sine series of f(x) is 4 1 f ( x) 1 ( 1)n sin nx 3 n n 1 x 2 f ( x) 2 x jin x. Obtain the half-range cosine series of f(x) = c-x in 0<x<c Here c a0 2 (c x)dx c0 c c 2 n x (c x) cos dx c0 c Integrating by parts and simplifying we get...6.. 52 do w nl oa d 8 .re Here a0 ul ... Obtain the cosine series of fro m Performing integration by parts and simplifying.Downloaded from www..pencilji..c 2. over(0.com . we get 2 n an 1 ( 1) n 2 cos 2 n 2 cos10x .10.. ) pa x..pencilji. an Downloaded from www. the Fourier cosine series is 2 cos 2 x cos6 x f(x) = 4 12 32 3. n2 Thus. . 1. Deduce that 1 4 2 5.. f(x) = over (-2 . f(x) = x over (.re x ) jin x2 over (.. 2 x 0 9. f(x) = ) 2 over (0. f(x) = (2 x). 12 32 w .Downloaded from www.... ) x.com 2c 1 ( 1) n 2 2 n The cosine series is given by an f(x) = c 2 2c 2 n 1 n x 1 ( 1) n cos 2 c 1n Exercices: 2 ul .... Deduce that 8 x.. 2 ) 4... 2) a. ) .3) 11.... x 0 7..pencilji. Deduce that 1 3 2 1 4 n 2n 0.. f(x) = x sinx over 0 x 2 . ) 1... 4 3 5 8.1 x 2 1 do w nl Deduce that Downloaded from www.0 x oa d fro m w w 6...com .. f(x) = 1 1 . 4 2.... x 0 over (. f(x) = x(2-x) over (0.1) x. 8 12 32 1..0 x 2 10. f(x) = 0. 3 5 1 1 . f(x) = x2 over (-1.0 x 1 12. f(x) = 2x + 3x2 over (.. f(x) = x over (... f(x) = x om Obtain the Fourier series of the following functions over the specified intervals : 1 1 ..0 x Deduce that 2 1 1 . x 0 over (.pencilji.c pa 3. ) . f(x) = cosx over (0.com . ) 15. (2).pencilji. l) pa ul . (3) may be redefined as f ( x) cos n x dx l 2 f ( x) cos f ( x) sin n x dx l 2 f ( x) sin a a n x l n x l Using these in (5). f(x) = xsinx over (0. The amplitude of the first harmonic is harmonic is a22 a12 b12 and that of second b22 and so on. f(x) = lx-x2 over (0 .Downloaded from www. f(x) = sin3x over (0. The method described cannot be employed when f(x) is not known explicitly. the term a 2cos2x+b2sin2x is called the second harmonic and so on. Harmonic analysis is the process of finding the Fourier coefficients numerically.re The Fourier series of a known function f(x) in a given interval may be found by finding the Fourier coefficients.com Obtain the half-range sine series of the following functions over the specified intervals : 13. the integrals in Euler’s formulae cannot be evaluated. b aa 1 2l a 2l f ( x)dx 2 bn 2 2[ f ( x)] a 1 2l a 2l 1 2l a 2l do w an 2 nl a0 oa d The Fourier coefficients defined through Euler’s formulae.b) denoted by [f(x)] is b 1 defined as f ( x) f ( x)dx . ) 17. x l 2 jin HARMONIC ANALYSIS w w . The term a 1cosx+b1sinx is called the first harmonic or fundamental harmonic. ) 14. (1).pencilji. we obtain the Fourier series of f(x).0 x 2 19. f(x) = l k (l x). f(x) = (x-1)2 over (0.c om Obtain the half-range cosine series of the following functions over the specified intervals : 16. In such a case. but defined through the values of the function at some equidistant points. ) 18. we employ the following result : fro The mean value of a continuous function f(x) over the interval (a.1) l kx. m w To derive the relevant formulae for Fourier coefficients in Harmonic analysis. Downloaded from www. f(x) = x2 over (0. 0392 -1.2 1.0392 -1.299 -0.5 -0.0 1 1 0 0 1 60 1. y = f(x) cosx cos2x sinx sin2x ycosx 0 1.com or l= .7 120 1.4 0.1) 0.367 6 6 2 y cos 2 x 2( 0.7 1.3 3.0 Note that the values of y = f(x) are spread over the interval 0 x 2 and f(0) = f(2 ) = 1.5 -0.1 -0.0 1.1732 w w ysinx 0 m do w nl oa d fro Total We have -0. We prepare the following table to compute the first two harmonics.75 1.6454 -0.7 jin w .com Examples 1.re -0.4 1.5 -0.2124 1.5 0.866 ycos2 x pa x0 om 0 ul .1176 -0.7 -1 1 0 240 1.1 6 6 Downloaded from www.5 -0. Find the first two harmonics of the Fourier series of f(x) given the following table : 2 4 5 2 3 3 3 3 f(x) 1.0. we get a1 2[ y cos x] a2 2[ y cos 2 x] 2 y cos x 2(1. Hence the function is periodic and so we omit the last value f(2 ) = 0.pencilji.5 0.5 1.866 0.866 0.866 180 1.9 1.95 ysin2x -1.866 -0.5 -0.2 0.7 0 0 0.95 1.866 -0.Downloaded from www.5 1 0 0 1.3) 0.2.299 1.6 -0.866 0. n=1.pencilji.75 -0.7 1.2124 -0.866 300 1.c x n x l an 2 f ( x) cos bn n x 2 f ( x) sin l 2[ y cos nx] as the length of interval= 2l = 2 2[ y sin nx] Putting.5 -0.6454 -1.9 -0.6 -1. 0392 y sin 2 x 0.4.5 15 12.598 0 oa d y do w 2 nl 1 Total Downloaded from www. then 3 a0 y a1 cos b1 sin a2 cos 2 b2 sin 2 2 We prepare the following table using the given values : w Put 0 b3 sin 3 (1) m x 3 a3 cos a3 cos3 w = fro x 2 x 3 ycos ycos2 ycos3 ysin ysin2 ysin3 0 04 4 4 4 0 0 0 600 08 4 -4 -8 6.5 8 12. Express y as a Fourier series upto the third harmonic given the following values : 0 4 1 8 2 15 3 7 4 6 5 2 om x y x < 6. so that 2l = 6 or l = 3.196 0 5 3000 02 1 -1 -2 -1.5 -7.c The values of y at x=0.99 -12.com y sin x 2 b1 [ y sin x] b2 [ y sin 2 x] 6 2 1.99 -2. The Fourier series upto the third harmonic is w . That is (-0.2.pencilji.5 are given and hence the interval of x should be 0 length of the interval = 6-0 = 6.3.928 0 1200 15 -7.732 -1.1cos2x – 0.re or 2 x 3 b2 sin x .5 -4.pencilji.196 5.928 6.com .732 0 42 -8.0392 sinx) and (-0.367cosx + 1. The a0 2 a1 cos x l b1 sin x l a2 cos y a0 2 a1 cos x 3 b1 sin x 3 a2 cos 2 x l b2 sin 2 x l a3 cos 3 x l b3 sin 3 x l 3 x 3 b3 sin 3 x 3 jin y pa ul .0577 6 The first two harmonics are a1cosx+b1sinx and a2cos2x+b2 sin2x.0577sin2x) 2.99 0 3 1800 07 -7 7 -7 0 0 0 4 2400 06 -3 -3 6 -5.1.Downloaded from www. 5 6 2 ( 2.05 T/2 2T/3 5T/6 T 1. We have period T.98 oa d Show that there is a constant part of 0. Let us denote T a0 2[ A] a1 2 A cos 2 t T 2[ A cos ] b1 2 A sin 2 t T 2[ A sin ] We prepare the following table: Downloaded from www.99) 4.866sin 3 3 3 3 This is the required Fourier series upto the third harmonic.5 cos 0.598) 0. do w nl Note that the values of A at t=0 and t=T are the same. Hence A(t) is a periodic function of 2 t .33 6 2 ( 4. in the current A and obtain the amplitude of the first harmonic.5) 1.25 1. 7 2.98 1.88 -0.re y 2.5) 2.30 -0. we get 14 om 2[ y ] ul .667 cos x 0 T/6 A (amp) 1.pencilji.667 6 0 jin Using these in (1).com (1) .com a1 2[ y cos ] b1 2[ y sin ] a2 2[ y cos 2 ] b2 2[ y sin 2 ] a3 2[ y cos3 ] b3 2[ y sin 3 ] y 2 1 (42) 3 6 2 ( 8.75amp.833 6 2 (12.833cos w . The following table gives the variations of a periodic current A over a period T : 1.30 T/3 fro m t(secs) w w 3.c 2[ f ( x)] pa a0 x x 2 x 2 x (4.pencilji.33) sin 1.866 6 2 (8) 2.Downloaded from www. com .30 -1 0 -1.0717.88 -0.866 -0.5 -0.5 0.Downloaded from www.98 0 T/6 600 1.75 in it.866 5T/6 3000 -0. Downloaded from www.5 0. Also the amplitude of the first harmonic is a12 0.pencilji.12 a1 0.75 0.7621 pa ul .9093 T/2 1800 1.30 2T/3 2400 -0.866 Using the values of the table in (1).5 6 3 2 A cos 1.12 3. we get 2 0.c 0.866 0.44 -0.25 0.5 1.0137 b1 1.1258 T/3 1200 1.65 1.2165 1.0046sin T T The expression shows that A has a constant part 0.05 -0.com 2 t T t A cos sin Acos Asin 0 1.98 1 0 1.3733cos b12 = 1.0046 6 3 The Fourier expansion upto the first harmonic is a0 A a0 2 a1 cos 2 t T b1 sin 2 t T do w nl 2 t 2 t 1.3733 6 3 2 A sin 3.30 0.pencilji.5 -0.5 A 0 w .525 0.0137 jin 4.125 0.re Total om 0 oa d fro m w w 4. 16 1. 0 30 60 90 120 150 180 210 240 270 300 330 y 1.com . The displacement y of a part of a mechanism is tabulated with corresponding angular movement x0 of the crank. Obtain the constant term and the coefficients of the first sine and cosine terms in the Fourier expansion of y as given in the following table : 2 9.4 27.pencilji.10 0.8 27.0 3.com ASSIGNMENT : 1. Downloaded from www. where y is given in the following table : x 0 1 2 3 4 5 y 4 8 15 7 6 2 6. Find the Fourier series of y upto the second harmonic from the following table : do w 5. The turning moment T is given for a series of values of the crank angle 0 = 750 .2 4 6 8 10 12 24.25 1.4 2.50 1.0 nl Y 0 oa d x fro 4.8 2.5 22.c 2.re 3.16 1.0 18.0 -1. Express y as a Fourier series upto the third harmonic.30 1.4 pa 45 0 jin x0 ul .30 0.30 2.80 1.00 om x0 90 135 180 225 y 4.52 1.8 3. Obtain the Fourier series of y upto the second harmonic using the following table : y 9 18 2 w 1 24 w 0 3 4 5 28 26 20 m x w . Obtain the first three coefficients in the Fourier cosine series for y.5 270 315 360 2.76 2.pencilji.Downloaded from www.0 9. pencilji.com 0 0 30 60 90 120 150 180 T 0 5224 8097 7850 5499 2626 0 = 750 .pencilji.c om Obtain the first four terms in a series of sines to represent T and calculate T at Downloaded from www.com . do w nl oa d fro m w w w .Downloaded from www.re jin pa ul .