How can I determined what the minimum sampling frequency should be for the 20-term expression for the function

f[t_] = (4/Pi) Sum[(1/n) Sin[2 Pi n t], {n, 1, Infinity, 2}]

where n= 1,3,5,10,20[![enter image description here][1]][1]

How can I sample all plots at ten times that frequency


closed as unclear what you're asking by Daniel Lichtblau, Coolwater, MarcoB, m_goldberg, C. E. Feb 1 '18 at 17:03

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  • $\begingroup$ EDIT: If you truncate the infinite series at m terms then the highest frequency component has frequency of 2 Pi (2m-1). That is, you have band limited the signal to a frequency of 2 Pi (2m-1). The minimum sampling frequency for the band limited signal is the Nyquist rate (twice the highest frequency component) which is then 4 Pi (2m-1). Ten times this would be 40 Pi (2m-1). In your case you specified m = 20. $\endgroup$ – Bob Hanlon Nov 1 '17 at 13:23
  • $\begingroup$ Following what @BobHanlon says, just take samples of all the functions with 1,3,5,10,and 20 terms, at spacings $1/(1560\pi)$ in (0,1). $\endgroup$ – José Antonio Díaz Navas Nov 1 '17 at 16:14

I am not clear what you are after. This is an investigation rather than an answer. If I do

e = (4/Pi) Sum[(1/n) Sin[2 Pi n t], {n, 1, Infinity, 2}]

I get a closed form for the expression

(2 I (ArcTanh[E^(-2 I π t)] - ArcTanh[E^(2 I π t)]))/π

Which appears to be a complex expression. So I do


and get

{(Arg[1 - E^(-2 I π t)] - Arg[1 + E^(-2 I π t)] - 
  Arg[1 - E^(2 I π t)] + Arg[1 + E^(2 I π t)])/π, 0}

Which shows it is actually real. I can plot this and get

Plot[Re[e], {t, 0, 10}]

Mathematica graphics

Can you explain where your 20 terms are? This just seems to be a square wave with period 1.


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