Triangle function Fourier transform

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Question:Triangle function Fourier transform

Posted: tsunamiBTP 292 Product: Maple

November 30 2017

I know that the Fourier transform of the triangle function is the sinc function squared, but I cannot seem to reproduce this in MAPLE. How do I get from R2 to the square of the sinc function or do I have a screw up in defining the integral? I need another set of eyes. I have been at this for awhile & cannot seem to breakthrough.

plot(piecewise(-1 < x and x < 0, 1+x, 0 < x and x < 1, 1-x), x = -2 .. 2, linestyle = solid, thickness = 5, scaling = constrained, title = "Symmetric Triangle Wave")

(int((1+t)*exp(-I*(2*Pi*n*t/T)), t = -(1/2)*T .. 0)+int((1-t)*exp(-I*(2*Pi*n*t/T)), t = 0 .. (1/2)*T))/T = ((-(2*I)*n*Pi-T+I*T*n*Pi)*exp(I*n*Pi)+(2*I)*n*Pi+T-(2*I)*n*Pi+T+((2*I)*n*Pi-T-I*T*n*Pi)*exp(-I*n*Pi))/(2^2*n^2*Pi^2)

(int((1+t)*exp(-I*(2*Pi*n*t/T)), t = -(1/2)*T .. 0)+int((1-t)*exp(-I*(2*Pi*n*t/T)), t = 0 .. (1/2)*T))/T = ((-T+I*T*n*Pi)*exp(I*n*Pi)+2*T-(T+I*T*n*Pi)*exp(-I*n*Pi))/(2^2*n^2*Pi^2)

R1 := ((-T+I*T*n*Pi)*exp(I*n*Pi)+2*T-(T+I*T*n*Pi)*exp(-I*n*Pi))/(2^2*n^2*Pi^2)

(1/4)*((-T+I*T*n*Pi)*exp(I*n*Pi)+2*T-(T+I*T*n*Pi)*exp(-I*n*Pi))/(n^2*Pi^2)

(1)

R1 = (1/4)*(2*T-T*(exp(-I*n*Pi)+exp(I*n*Pi))-I*(exp(-I*n*Pi)-exp(I*n*Pi))*Pi*T*n)/(n^2*Pi^2)

R2 := (1/4)*(-2*cos(Pi*n)*T-2*sin(Pi*n)*T*n*Pi+2*T)/(n^2*Pi^2)

(1/4)*(-2*cos(Pi*n)*T-2*sin(Pi*n)*T*n*Pi+2*T)/(n^2*Pi^2)

(2)

R2 = 2*T*(1-cos(Pi*n)-Pi*n*sin(Pi*n))/(2^2*n^2*Pi^2)

(1/2)*(-exp(I*n*Pi)+1+I*n*Pi)/(Pi^2*n^2)-(1/2)*(I*exp(I*n*Pi)*Pi*n-exp(I*n*Pi)+1)*exp(-I*n*Pi)/(Pi^2*n^2) = (1/2)*(2-exp(I*n*Pi)-exp(-I*n*Pi))/(Pi^2*n^2)