# 0.3 Modelling corruption  (Page 5/11)

 Page 5 / 11

## The delta “function”

One way to see how a system behaves is to kick it and see how it responds.Some systems react sluggishly, barely moving away from their resting state, while othersrespond quickly and vigorously. Defining exactly what is meant mathematicallyby a “kick” is trickier than it seems because the kick must occur over a very short amount of time,yet must be energetic in order to have any effect. This section defines the impulse (or delta) function $\delta \left(t\right)$ , which is a useful “kick” for the study of linear systems.

The criterion that the impulse be energetic is translated to the mathematical statement that itsintegral over all time must be nonzero, and it is typically scaled to unity, that is,

${\int }_{-\infty }^{\infty }\delta \left(t\right)dt=1.$

The criterion that it occur over a very short time span is translated to the statement that, for every positive $ϵ$ ,

$\delta \left(t\right)=\left\{\begin{array}{cc}0\hfill & t<-ϵ\hfill \\ 0\hfill & t>ϵ\hfill \end{array}\right).$

Thus, the impulse $\delta \left(t\right)$ is explicitly defined to be equal to zero for all $t\ne 0$ . On the other hand, $\delta \left(t\right)$ is implicitly defined when $t=0$ by the requirement that its integral be unity. Together, these guarantee that $\delta \left(t\right)$ is no ordinary function. The impulse is called a distribution and is the subject of considerable mathematical investigation.

The most important consequence of the definitions [link] and [link] is the sifting property

${\int }_{-\infty }^{\infty }w\left(t\right)\delta \left(t-{t}_{0}\right)dt=w\left(t\right){|}_{t={t}_{0}}=w\left({t}_{0}\right),$

which says that the delta function picks out the value of the function $w\left(t\right)$ from under the integral at exactly the time whenthe argument of the $\delta$ function is zero, that is, when $t={t}_{0}$ . To see this, observe that $\delta \left(t-{t}_{0}\right)$ is zero whenever $t\ne {t}_{0}$ , and hence $w\left(t\right)\delta \left(t-{t}_{0}\right)$ is zero whenever $t\ne {t}_{0}$ . Thus,

$\begin{array}{ccc}\hfill {\int }_{-\infty }^{\infty }w\left(t\right)\delta \left(t-{t}_{0}\right)dt& =& {\int }_{-\infty }^{\infty }w\left({t}_{0}\right)\delta \left(t-{t}_{0}\right)dt\hfill \\ & =& w\left({t}_{0}\right){\int }_{-\infty }^{\infty }\delta \left(t-{t}_{0}\right)dt\hfill \\ & =& w\left({t}_{0}\right)·1=w\left({t}_{0}\right).\hfill \end{array}$

Sometimes it is helpful to think of the impulse as a limit. For instance, define the rectangular pulseof width $1/n$ and height $n$ by

${\delta }_{n}\left(t\right)=\left\{\begin{array}{cc}0,\hfill & t<-1/2n\hfill \\ n,\hfill & -1/2n\le t\le 1/2n\hfill \\ 0,\hfill & t>1/2n\hfill \end{array}\right).$

Then $\delta \left(t\right)={lim}_{n\to \infty }{\delta }_{n}\left(t\right)$ fulfills both criteria [link] and [link] . Informally, it is not unreasonable to think of $\delta \left(t\right)$ as being zero everywhere except at $t=0$ , where it is infinite.While it is not really possible to “plot” the delta function $\delta \left(t-{t}_{0}\right)$ , it can be represented in graphical form as zero everywhereexcept for an up-pointing arrow at ${t}_{0}$ . When the $\delta$ function is scaled by a constant, the value of the constant is often placedin parenthesis near the arrowhead. Sometimes, when the constant is negative, the arrow is drawn pointing down.For instance, [link] shows a graphical representation of the function $w\left(t\right)=\delta \left(t+10\right)-2\delta \left(t+1\right)+3\delta \left(t-5\right)$ .

What is the spectrum (Fourier transform) of $\delta \left(t\right)$ ? This can be calculated directly from the definitionby replacing $w\left(t\right)$ in [link] with $\delta \left(t\right)$ :

$\mathcal{F}\left\{\delta \left(t\right)\right\}={\int }_{-\infty }^{\infty }\delta \left(t\right){e}^{-j2\pi ft}dt.$

Apply the sifting property [link] with $w\left(t\right)={e}^{-j2\pi ft}$ and ${t}_{0}=0$ . Thus $\mathcal{F}\left\{\delta \left(t\right)\right\}={e}^{-j2\pi ft}{|}_{t=0}=1$ .

Alternatively, suppose that $\delta$ is a function of frequency, that is, a spike at zero frequency.The corresponding time domain function can be calculated analogously using the definitionof the inverse Fourier transform, that is, by substituting $\delta \left(f\right)$ for $W\left(f\right)$ in [link] and integrating:

${\mathcal{F}}^{-1}\left\{\delta \left(f\right)\right\}={\int }_{-\infty }^{\infty }\delta \left(f\right){e}^{j2\pi ft}df={e}^{j2\pi ft}{|}_{f=0}=1.$

Thus a spike at frequency zero is a “DC signal” (a constant) in time.

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!