TMA4213/4215 Matematikk 4M/N Vår 2013

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1 Norges teknisk naturvitenskapelige universitet Institutt for matematiske fag TMA43/45 Matematikk 4M/N Vår 3 Løsningsforslag Øving a) The Fourier series of the signal is f(x) =.4 cos ( 4 L x) +cos ( 5 L x). cos ( 6 L x) cos is -periodic, so the period of the three components are, respectively T 4 = 4/L = L, T 5 = 5/L = L 5, T 6 = 6/L = L 3. b) The frequency of a periodic signal is the inverse of the period, therefore the frequencies are f 4 = T 4 = L = Hz, f 5 = T 5 = 5 L = 75Hz, c) The keys are A3, C 4 (or D 4) and E4. f 6 = T 6 = 3 L = 33Hz. d) This is an A major chord. For musically interested: Notice that the major third corresponds to a frequency ratio of 5:4, while the perfect fifth corresponds to a frequency ratio of 3:. a) f(x) = x, f( x) = x = f(x). This is an odd function. b) f(x) =, f( x) = = f(x). This is an even function. c) f(x) = e x, f( x) = e x. This is neither equal to e x nor e x, so this function is neither even nor odd. d) f(x) = g(x) h(x), where both g and h are even functions. f( x) = g( x) h( x) = g(x) h(x) = f(x). This is an even function. e) f(x) = g(x) h(x), where both g and h are odd functions. f( x) = g( x) h( x) = ( g(x)) ( h(x)) = g(x) h(x) = f(x). This is an even function. f) f(x) = g(h(x)), where g is an even function and h is an odd function. f( x) = g(h( x)) = g( h(x)) = g(h(x)) = f(x). This is an even function. g) If a function f is both odd and even, we have to have f( x) = f(x) and f( x) = f(x) for all x. But then f( x) = f( x) + f( x) = f(x) + f(x) = Therefore f( x) = for all x and f(x) = for all x as well. In other words, the only way a function can be both odd and even is if it is always equal to zero. 7. januar 4 Side av 5

2 Løsningsforslag Øving 3 f(x) is -periodic and x, for < x <, f(x) = x, for < x <, x for < x <. a) Since f is odd, its Fourier series reduces to a Fourier sine series (Kreyszig Sec..). In other words, the Fourier coefficients of the even terms, that is a, a, a,... are all equal to zero. b) To simplify the calculations, we use equation (6**) from Kreyszig Sec.3 (Here L = ) b n = = = f(x) sin(nx) dx x sin(nx) dx + x sin(nx) dx + ( x) sin(nx) dx sin(nx) dx [ = n sin(nx) ] [ n x cos(nx) + [ n sin(nx) ] n x cos(nx) = 4 ( n ) n sin We insert n =,,..., 5 and get x sin(nx) dx n cos(nx) ] b = 4, b =, b 3 = 4 9, b 4 =, b 5 = januar 4 Side av 5

3 Løsningsforslag Øving 4 a) f(x) 3 3 b) According to Kreyszig Sec.4, the trigonometric polynomial of degree N which minimizes quadratic error, is exactly the N th partial sum of the Fourier series so F = sin x, F = sin x sin x, F 3 = sin x sin x + sin 3x, 3 F 4 = sin x sin x + 3 sin 3x sin 4x, F 5 = sin x sin x + 3 sin 3x sin 4x + sin 5x. 5 c) The simplest way to calculate the square error, is to use equation (6) from Kreyszig Sec.4. Here, f dx = x dx = 3 3, a =, a n = and b n = 4 n, we can therefore calculate for instance E f, = 3 3 (4 + 4 ) The numerical values for E f,n, N =,..., 5 are N E f,n januar 4 Side 3 av 5

4 Løsningsforslag Øving d) g(x) 3 3 e) Same as in c), the trigonometric polynomials which minimize the square error are the partial sums of the Fourier series. We note that when n is even, the term in the sum is simply zero, so G = G, G 4 = G 3 etc. G = G = 4 cos x, G 3 = G 4 = 4 cos x 4 cos 3x, 9 G 5 = G 6 = 4 cos x 4 4 cos 3x cos 5x. 9 5 f) We use equation (6) from Kreyszig Sec.4. Here g dx = x dx = 3 3, a =, a = 4, a 3 = 4 9, a 5 = 4 5, and all of the other Fourier coefficients for N 5 are equal to zero. The numerical values for E f,n, N =,..., 5 are N E g,n g) It is clear that E g,n is much smaller than E f,n for N =,..., 5. (In fact, the difference will become even larger as N grows.) The direct consequence of this is that G N is a much better approximation of the function x on x. The plot below, showing x, F 3 and G 3 confirms this. x F 3 (x) G 3 (x) 7. januar 4 Side 4 av 5

5 Løsningsforslag Øving The explanation is that the odd extension f(x) has a discontinuity at x =. To gain speed for the jump, the partial sum of the Fourier series has to oscillate around the discontinuity. The even extension g(x) has no such discontinuity, and the oscillations required to resolve the kinks at x = and x = are much smaller. We can also explain the difference from the coefficients using Parseval s identity. The short explanation is that the Fourier coefficients of the odd expansion go to zero as N, while the Fourier coefficents of the even expansion go to zero as. Therefore, N the Fourier series of the even expansion converge more rapidly. For E f,n, we can write E f,n = N N f dx b n == b n b n = n= n= n= n=n+ b n. Here b n = 4. The sum n n=n+ b n = 4 the bounds (recall the integral test from Calculus ) n=n+ n, we can approximate with N+ t dt < n=n+ n < N t dt, that is Therefore N + < n=n+ n < N. 4 (N + ) < E f,n < 4 N. For E g,n, we similarly get E g,n = n=n+ where a n = 6 when n is odd and zero if n is even. n 4 Again using integrals to bound this, we can obtain for N odd 6 N+ t 4 dt n=n+ a n < a n N 6 t 4, dt that is Therefore 8 3 (N + ) 3 < 8 3 n=n+ a n < 8 3 N 3. (N + ) 3 < E g,n < 8 3 N 3. Comparing the bonds we have obtained for E f,n and E g,n, we can see that E g,n goes to zero faster than E f,n. 7. januar 4 Side 5 av 5