Given a periodic function f with period 27: if -n
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![Given a periodic function f with period 27:
if -n <I<0
f(x) =
(x – n)², if 0 < r <a
(a) Show that the Fourier series expansion of the function f is
sin [(2n – 1)a]
(2n – 1)3
f(a) =D +Σ[ cos(nur) +
(-1)"_
sin(næ) -
cos(n.r) +
n=1
n=1
(Hint: Fourier series:
f(x) :
() + b, sin (")]
a, cos
n=1
where
| f(r)dr
do
f(x) cos
dr n = 1,2, ...
an =
and
b
| f(x) sin () d.x n = 1,2, ...)
(b) Use the result of (a), to show that
(c) Using (a) to show the following:
(-1)n+1
and
n2
12
n=1
||
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6364ca9-15f9-4b1e-884f-665511fe53c6%2F1325be72-c8ba-4199-a127-98ae58940358%2F4m7q1p_processed.png&w=3840&q=75)
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- if the Fourier series coefficients for the signal X[n] with a period N are "ax", calculate the Fourier series coefficients of the following expressions in terms of "ax". (a) x[n-2]+ x[n-(N-4)/2] (N even) (b) -1"x[-n] (N even) (c) yin]={ x[n], it n even 0, otherwiseConsider the 27-periodic function f(x) defined on [-, π] by the formula f(x) = cos(). (a) Find the Fourier series expansion of f(x). (b) Use part (a) to find the sum of the following series: A = n=1 (-1) ¹+1 4n² - 1 and B=24²-1A periodic function, f(x) with period 4x is defined as - 2n sx<-1 - nSX<0 2n, %3D f(x) Osx<* 2n, Sketch the graph of f(x) on the interval [-5x, 57). Determine if f(x) is an even, odd or neither even nor odd function. a) b) Find the Fourier series of f(x).
- = 1) The function f(x) periodic on the interval [0, 2л] has complex Fourier series f(x): Σ(1/n²) einx where the sum over n goes from - infinity to infinity. Convert this to cosine and sine Fourier Series by finding the values of A's and B's in the expression Ao + ΣAn cos(nx) + Σ Bn sin(nx) where each sum goes from 1 to infinity. Hint: consider the n and -n term together in the complex Fourier Series or use Euler's identity.Find the trigonometric Fourier series for the function f(x): [-T/2, π/2] → R given by the expression: ƒ(2) - {0 = O о O O cos 2x if x = [-π/2, 0] 0 if x = (0, π/2] O ∞ FS(x) = -2 cos(2x) + 1 n=1 FS(x) = −cos(2x) + Σ2 FS(x) = −sin(2x) + Σn=2 FS(x) = cos(2x) + Σn=0 FS(x) = cos(2x) + Ex-1 - n² cos² (n²-1)π 2n cos² n cos² (=) 2 n cos² (n²-1)T (+) 2 2(n²-1)π NE (+) (n.²-1) 2n cos² * ( =) 2 (n²+1)π -sin(2nx). -sin(2nx). -sin(nx). sin(2nx). -sin(2nx).Consider f(t)=9+2t+6t², for -Determine the Fourier Series of f(x) = x², over the interval - < x < and has period 2 ! f(x) = 3r [c cos x + cos 3x + cos 5x + .] sina+sin 2x + sin 3x +... 2 1 1 f(x) sin x + sin 3x + sin 5x + 1 π 3 5 1 1 -{sin z + sin 2x + sin 3x + x ...} 2 2 3 $ 1 f(2)=-4 [cos 2-2008 22 +00832 - 008 42 + cos cos cos 4x 4] 3 4² = # f(x) 12 = TIf f(x) is an even function on [-pi, pi], then the Fourier series of f is of of the form a0 + E-1 a, cos(jx) 2 A True В FalseFind the Fourier series for the following periodic time function 0Recommended textbooks for youAdvanced Engineering MathematicsAdvanced MathISBN:9780470458365Author:Erwin KreyszigPublisher:Wiley, John & Sons, Incorporated
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