I have read : "An impulse is a generalized function having infinite length and finite area. It is also known as the Dirac Delta function. Magnitude of an impulse doesn’t make sense , only its area does.

A pulse is defined over a finite time interval. Both the area under the pulse and magnitude of a pulse at an instant of time is well defined."

What is the difference between area and magnitude of a pulse? what do they physically mean? and is there a relation between them?

  • 2
    \$\begingroup\$ Think of a pulse as a rectangular function. The magnitude is the amplitude of the rect. The area is just the integral of the rect. Also Dirac’s delta is not a function. \$\endgroup\$
    – user110971
    Aug 21, 2018 at 9:15
  • \$\begingroup\$ @user110971 The Dirac delta is an extended function. It depends how a function is defined. There are extended function definitions in higher mathematics (mostly useless for real word engineers) where the Dirac delta is a function. \$\endgroup\$ Aug 21, 2018 at 9:23
  • \$\begingroup\$ @HorrorVacui most people think of a function as a map from two domains, e.g. a map from the reals to the reals. If this is your definition, Dirac’s delta is not a function. In addition, it has applications in actual engineering. I encountered such an application in signal processing recently. \$\endgroup\$
    – user110971
    Aug 21, 2018 at 9:30
  • \$\begingroup\$ Generally 'strength' is used for the area of an impulse, rather than 'magnitude'. \$\endgroup\$
    – Chu
    Aug 21, 2018 at 12:45
  • \$\begingroup\$ @user110971 You have written that Dirac delta is not a function. I am surprised that you now claim that it is a function. Sure the Dirac delta is used heavily in Fourier transformations, but all you need to remember is that its integral is unity at t=0. Anybody who studied the basics of (electrical) engineering know it, since it simplifies so much many differential equation calculations. I haven't stated that Dirac delta is useless, only that its not needed to be familiar with the realm of extended functions to use the Dirac delta. \$\endgroup\$ Aug 22, 2018 at 20:16

1 Answer 1


You have it backwards.

Some mathematician somewhere said 'let's define something over an infinite time, that is only defined by its area. Let most of it be zero, let the peak amplitude tend to infinity, let the duration of the non-zero part tend to zero, such that the area under the curve is constant. And we'll call that a Dirac Delta, aka an impulse'. And a lot of other people agreed, because it's useful mathematically, and it's a good asymptotic model for a number of physical things.

An engineer somewhere said 'let's define something over a finite time, with a finite amplitude and duration, and we'll call that a pulse'. And a lot of people agreed, because it's useful.

Whether those are good names for the things so defined, or sufficiently different from each other, is a matter for history. That's where we are.

The amplitude, or magnitude, is the instantaneous value.

The area under the curve is the integral of the value over time.

Sometimes we only need to know the area. For instance when a cricket ball hits a bat, the momentum transferred is equal to the integral of the force with time. With leather on willow, the force will be much lower, and last much longer, than with a steel bat and a steel ball. However, the change in momentum for the two will be broadly similar. What matters is not the amplitude or duration of the contact, but the area under the force curve.

Similarly in electrical engineering, anything that depends on an integral, like the current in an inductor after a voltage pulse, or the charge on a capacitor after a current pulse, depends on the area under the curve.

  • \$\begingroup\$ I think you are actually wrong about the history of the Dirac delta. As far as I know, it first appears in the Fourier integral theorem. Later Cauchy expressed the theorem in exponential form, which yielded the modern definition of Dirac’s delta. \$\endgroup\$
    – user110971
    Aug 22, 2018 at 3:25
  • \$\begingroup\$ @user110971 I look forward to your better-researched answer. But would it add anything to the OP's understanding? \$\endgroup\$
    – Neil_UK
    Aug 22, 2018 at 4:54
  • \$\begingroup\$ No, it wouldn’t. That’s why I posted a comment rather than an answer. I mean, the history of Dirac’s delta is not related to the question. \$\endgroup\$
    – user110971
    Aug 22, 2018 at 5:01

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