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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.

[Note: This is an approximation of the real input. In reality the input is a continuous function, with a big variation (until $\delta_r$) on the instant of the action.]

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}

[Note: I know that this is also an approximation. Dirac delta function is a continuos function that tends to infinity in the instant of the action.]

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.

I think that a good model of an impulse input should have the same behavior of the Dirac delta, but with a normalization to 1. Am I missing something?

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  • $\begingroup$ en.wikipedia.org/wiki/Dirac_measure $\endgroup$ – lemon Feb 17 '17 at 12:39
  • $\begingroup$ Note that integral of $(1)$ along any interval is always $0$, while the integral of $(2)$ is $\delta_r$ if the integration interval includes $t=0$ and $0$ otherwise. Thus definition $(2)$ looks more useful than $(1)$. $\endgroup$ – Ruslan Feb 17 '17 at 13:03
  • $\begingroup$ Related question in QM: physics.stackexchange.com/q/89958/2451 $\endgroup$ – Qmechanic Feb 17 '17 at 13:13
  • $\begingroup$ uhm, i'm just an electrical engineer, but there is something wrong with your equation (1). the integral of that "delta" function is zero. no matter what $\delta_r$ is. $\endgroup$ – robert bristow-johnson Oct 4 '17 at 7:31
  • $\begingroup$ (Note: Dirac delta function is not a continuous function; it's not even a function) $\endgroup$ – user5174 Nov 3 '17 at 22:01
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First review the definition of the Dirac Delta distribution ('Delta function').

references:

a) "Mathematics for the Physical Sciences", Laurent Schwartz;

b) "Generalized Functions vol 1", I.M. Gelfand, G. E. Shilov.

These are classics and primary sources in the areas of generalized functions, b), and distributions, a). Generalized functions and distributions are the same thing.

So $δ(x-a)$ is defined by

I) $$f(a)=\int f(x)δ(x-a)dx ,$$

Where f, a 'test function', is adequately well behaved. So the Delta Distribution (or Delta Function) is defined by its action under the integral sign.

Sometimes, in addition, its value is given as zero everywhere except at the single point, a, where its value is infinity. But there are problems with this definition since a value at a single point should not change the value of a conventional Riemann integral. So instead sometimes a 'inner product' (or 'scalar product') is used for the definition (the argument of $\delta$ can be $x$ instead of $x-a$)

II) $f(0) = (f(x) , δ(x)) $, Gelfand p4

or

III) $f(0) = <f(x) , δ(x)> $, Schwartz p77.

From these $f(a) = (f(x) , δ(x-a)) $ follows.

So my answer is: your equation (1) is not good--it should be written similar to equation (2) with the understanding that ultimately it will be used under the integral sign. If we ignore the use of the inner product and assume $\delta_r$ is a constant which is the value of $\delta_r(t)$ at t=0, then:

IV) $\int \delta_r(t)\delta(t)dt = \delta_r(0)= \delta_r$

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Quite simply, we have using the first definition $\delta_r(t)$,

$$\langle f(t),\delta_r(t)\rangle = 0$$

while for the definition with $\delta_r \delta(t)$, one has that,

$$\langle f(t), \delta_r \delta(t)\rangle = \delta_r \langle f(t),\delta(t)\rangle = \delta_r f(0).$$

The reason is that $\delta_r(t)$ is a finite spike at $t=0$, and we are integrating over a null set of measure zero since it is a singleton, $\{0\}$, which yields zero.

On the other hand, the delta function can in a way be regarded as a measure which gives mass one to $\{0\}$ and zero otherwise, thus allowing us to obtain a finite, non-zero result.

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It's hard to say for sure without further context if that is the source of the confusion here, but equations (2) and (3) are used in analytical models with continuous time, while equation (1) is used in the context of a simulation with discrete time. You recover the same delta function properties in the discrete time models by replacing the integral with a sum. The reference to electronics in the question suggests that this may be the cause of confusion as similar notations will be used in the context for example of analog and digital filters/systems.

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