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I'm currently studying Stability of Flight, and there's a topic in this course that treats aircraft responses to certain maneuvers of the pilot.

Consider for example, the impulsive deflection of the rudder of an airplane, so that the angle of the deflection in time is given by:

\begin{equation} \delta_r(t)=% \begin{cases} \delta_r &\text{if $t=0$}\\ 0 &\text{if $t\neq 0$}. \end{cases} \hspace{15pt}(1) \end{equation}

where $\delta_r$ is a constant. Normally this type of action is expressed as:

$$\delta_r(t)=\delta_r\delta(t)\hspace{15pt}(2)$$

where $\delta(t)$ is the Dirac delta function. My problem is that these two expressions aren't equivalent, because the Dirac delta function is a distribution so that:

\begin{equation} \delta(t)=% \begin{cases} \infty &\text{if $t=0$}\\ 0 &\text{if $t\neq 0$}. \end{cases} \hspace{15pt} (3) \end{equation}

So the expression (2) can only express the density of the action in time. To $(1)$ and $(2)$ be equivalent, the $\infty$ in (3) should be 1. I've seen this consideration in others areas of science, for example on Electronics, an applied impulsive voltage on a circuit is expressed in the same way as the deflection of the rudder of the airplane. Usually, the Laplace transform of (2) is computed and the response of the system is then obtained through the transfer function. So to compute the Laplace transform we need to use the signal density in time and not the actual signal?

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    Your equation (3) is not the definition of the Dirac delta. It's at best a rough approximation. On the other hand, as far as I can tell, (1) and (2) really are _not_ equivalent. I think (2) probably is useful, while (1) seems nonsensical and useless, even harmful. – 2017-02-06
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    @DavidK I agree with you relatively to the definition of the Dirac delta. The Dirac delta is a continuous function, smoother than the one that I wrote in my question, that tends to infinity in certain point. I think that Dirac delta models correctly the variation associated with the inpulse input, but not its amplitude. In my perspective the right model should tend to 1 instead of infinity. I believe there's an explanation for the use of Dirac delta, maybe I'm missing something. – 2017-02-17
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    The Dirac delta isn't strictly "a" function but rather a distribution. Its purpose is to be multiplied by some function and then integrated. If you want something of amplitude 1, try the Heaviside step function, which is what you get if you integrate the Dirac delta by itself. – 2017-02-17

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