Infinite sum of heviside functions

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SUMMARY

The discussion centers on calculating the infinite sum of Heaviside functions, specifically the function defined as f(t) = ∑[n=-∞ to ∞] (-1)^n [u(t) - u(t-T)]. The ambiguity in notation is clarified, distinguishing between two interpretations of the summation. The function g(t) = u(t) - u(t-T) is identified as being equal to 1 for the interval 0 ≤ t < T and 0 otherwise, providing a basis for further analysis of the infinite series.

PREREQUISITES
  • Understanding of Heaviside step function (u(t))
  • Familiarity with infinite series and summation notation
  • Basic knowledge of periodic functions
  • Concept of piecewise functions
NEXT STEPS
  • Explore the properties of the Heaviside step function in detail
  • Learn about convergence criteria for infinite series
  • Investigate Fourier series and their relation to periodic functions
  • Study piecewise function analysis and applications
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Students and educators in mathematics, particularly those studying calculus and signal processing, as well as anyone interested in the applications of Heaviside functions in engineering and physics.

erezb84
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Homework Statement


how can i find the infinite sum of heviside functions?...

i have this func:
f(t) = sum[n=-infinity to infinity] (-1)^n*[u(t)-(u(t-T))]
where T is a period time..
 
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erezb84 said:

Homework Statement


how can i find the infinite sum of heviside functions?...

i have this func:
f(t) = sum[n=-infinity to infinity] (-1)^n*[u(t)-(u(t-T))]
where T is a period time..

The notation is ambiguous. Do you mean [tex]f(t) = \sum_{n=-\infty}^{\infty} (-1)^n [u(t) - u(t-T)],[/tex] or do you mean [tex]f(t) = \sum_{n=-\infty}^{\infty} (-1)^{n [u(t) - u(t-T)]} ?[/tex] In any case, the function g(t) = u(t) - u(t-T) is 1 for 0 <= t < T and is zero for other values of t. That should allow you to easily examine your infinite series.

RGV
 

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