The Dirac detla or unit impulse function is defined as :(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\delta (t) = \left \{ \begin {matrix} \infty \quad \ t = 0 \\ 0 \quad : \ t \neq 0 \end{matrix} [/itex]

and the unit step function :

[itex] u(t) = \left \{ \begin {matrix} 1 \quad \ t \geqslant 0 \\ 0 \quad : \ t < 0 \end{matrix} [/itex]

It is said that the

[tex] \frac d {dt} u(t) = \delta (t) [/itex] ..

but .. if we came to the definition of the derivative , we should find the limit from left and right .. :

[tex] { \displaysytle u'_+ (0) = \lim { h \to 0^+ } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^+ } \frac {1-1} h = 0 }[/tex]

[itex]{ \displaystyle u'_- (0) = \lim { h \to 0^- } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^- } \frac {0-1 }h = \infty} [/itex]

We see that the derivative is infinity from left only .. and it is not equal to the right limit , so the derivative doesn't exist .. and it is not an imuplse ..

ButIFwe define u (0) to be between 0 and 1 , For instance 0.5 , then :

[itex]u'_+ (0) = \lim { h \to 0^+ } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^+ } \frac {1-0.5} h = \infty[/itex]

Which makes the derivative from both sides infinity , which gives us the impulse ..

As a result , we should chane the definition of the unit step function ..

Do you agree with me ?

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# Dirac's Delta as a derivative of the unit step

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