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somebody please explain unit step function ...how come it is defined at zero...though there is a jump ...but still were are able to differentiate and integrate it from zero ....

Mathematics
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there are not really called functions ... rather generalized function and derivatives are called generalized derivatives.
it would be great if you could make that generalisation more lucid...actually i am stuck and thinking of finding it other way..
|dw:1346491565254:dw| this is your step function f(t) = u(t-t_0)

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the derivative at t=t_0, would be delta(t-t_0)
|dw:1346491624874:dw|
yeah ... that works as a switch.
how you define derivative at 0?
exactly that works as a switch with constant supply
at 0, the derivative will be equal to ... value of jump times delta(t)
well don't we count for jump..i mean definition of derivative
at this particular case ... you jump is just 1
http://en.wikipedia.org/wiki/Dirac_delta_function
your slope is infinite ... so this acts as inpulse. there a very nice course from mit ocw. particularly exercise.
slope is zero i guess
it's a horizontal line
at zero ... you have infinite slope
impulse function has infinite slope at zero....actually point of confusion arises when i apply this concept in electrical circuit how come they consider derivative at zero or may be i am not able to correlate that properly
working with Green's functions?
no..working with impulse ramp and unit step function....i feel contradictory
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iii-fourier-series-and-laplace-transform/step-and-delta-functions-integrals-and-generalized-derivatives/
okay ...i'll have to check this out
don't miss exercise.
sure !! thank you ...

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