Minimum Sampling Frequency for a 20-term expression [closed]

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]]

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Bob Hanlon Nov 1 '17 at 13:23
• 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). – 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

FullSimplify[ComplexExpand[ReIm[e]]]

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