- #1
Ali 2
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The Dirac detla or unit impulse function is defined as :
[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 ..
But IF we 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 ?
[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 ..
But IF we 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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