Inverse Fourier Transform of | cos[(2 pi f)/100)] |

Question

I would like a help calculating the Inverse Fourier Transform of

Absolute cos[(2 pi f)/100]

Please provide the code that you have tried and found not to work well. — bbgodfrey, Dec 12 '15 at 06:47
InverseFourierTransform [Abs[cos([Omega]/100)], [Omega], t, FourierParameters -> {0, -2pi}] — Nick, Dec 12 '15 at 06:53
The integral do not converge. There is no Inverse. The function $f(t)$ must satisfy $\int_{-\infty}^{\infty} |f(t)| < \infty$ to have an inverse Fourier — Nasser, Dec 12 '15 at 06:56
That means that X(f)= | cos[(2 pi/100)*f} | can't be transformed to x(t) ? — Nick, Dec 12 '15 at 07:12
@Nick The result will be a distribution. The only representation I can think of is a series of delta functions with some coefficients. — Andrew, Dec 12 '15 at 07:21
http://i.stack.imgur.com/IIZ5d.jpg I also made that calculations.Are they correct? — Nick, Dec 12 '15 at 08:27
Based on the images you posted, it appears that your question belongs on [math.se], which deals with mathematics. The present site deals with Mathematica software. — bbgodfrey, Dec 12 '15 at 13:05
@bbgodfrey I guess you are 100% right ..but i was puzzled why i couldn't run it on Mathematica. Since I am here and compared to the (zero) responses i get from Mathematics section any help (off the record) would be appreciated . — Nick, Dec 12 '15 at 16:56
This problem definitely is solvable, both analytically and with Mathematica, but InverseFourierTransform[Abs[Cos[Omega/100]], Omega, t, FourierParameters -> {0, -2 Pi}] is not sufficient. I shall investigate a few other options when I have time. — bbgodfrey, Dec 12 '15 at 18:21
Thank you very much...and it seams so easy... $X(f)= | cos [(2 π/100)*f] |$ — Nick, Dec 12 '15 at 18:26

score 1 · Answer 1 · answered Dec 13 '15 at 01:09

As I noted in a comment above, this problem can be solved with Mathematica (or, of course, analytically). The function is periodic, so a single period of the function can be transformed, and then the transforms of all periods summed. It is convenient to begin by subtracting the average value of the function over the period, which is 2/Pi.

Integrate[(Cos[ω/100] - 2/Pi), {ω, -50 Pi, 50 Pi}]
(* 0 *)

Mathematica quickly can integrate the central period of the function.

InverseFourierTransform[
    Piecewise[{{Cos[ω/100] - 2/π, -50 Pi < ω < 50 Pi}}, 0], ω, t]
(* (2 Sqrt[2] (-50 π t Cos[50 π t] + (1 - 10000 t^2) Sin[50 π t]))/
   (π^(3/2) t (-1 + 10000 t^2)) *)

and one would hope that it could solve for any period by

InverseFourierTransform[Piecewise[{{Cos[ω/100] - 2/π, 
    100 (n - 1/2) Pi < ω < 100 (n + 1/2) Pi},
    {Cos[ω/100] - 2/π, -100 (n + 1/2) Pi < ω < -100 (n - 1/2) Pi}}, 0], ω, t, 
     Assumptions -> n > 0]] // FullSimplify

but it seems unable to do so. However, for specific values of n, for instance, n = 1, it readily determines

(* (4 Sqrt[2] Cos[100 I π t] (50 π t Cos[50 π t] + (1 - 10000 t^2) 
    Sin[50 π t]))/(π^(3/2) t (-1 + 10000 t^2))

from which one can generalize that the term for n > 0 is

(* (4 n Sqrt[2] Cos[100 I π t] (50 π t Cos[50 π t] + (1 - 10000 t^2) 
    Sin[50 π t]))/(π^(3/2) t (-1 + 10000 t^2))

Thus, the sum of this term over n > 0 plus the n = 0 term obtained earlier, plus the delta function corresponding to the constant value 2/Pi is the desired transform.

I found it 50. Here is how: cos(ω/100)=cos[(2πf)/100)=cos[(π/50)*f] The period should be T=(2π/1) / (π/50)=100 and since we have absolute value we don't have negatives so the period becomes half --> T=100/2=50 — Nick, Dec 13 '15 at 03:10

Inverse Fourier Transform of | cos[(2 pi f)/100)] |

1 Answers1