[PDF] The Fourier transform of the Heaviside function: a tragedy





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  • .x0a/dx0D H.x a/: That is, the Heaviside step function is the cumulative area under the delta function curve. 2.The Heaviside step function is that function of x that is zero to the left of a and one to the right of a , H.x a/  ( 0 if x < a , 1 if x  a . This is pictured in the ?rst of ?g. 0.2. Shear V for a concentrated load.

What is the relationship between Heaviside function and delta function?

  • The delta function can then be de?ned as ?(x)= ( ? if x =0, 0 if x 6= 0. (12) and the relationship between Heaviside function and delta function is given by dH(x) dx =?(x) (13) and H(x)=. Z x ??. ?(x)dx = ( 0 if x <0, 1 if x >0.

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  • 5.4 Heaviside’s Method This practical method was popularized by the English electrical engineer Oliver Heaviside (1850–1925). A typical application of the method is to solve 2s (s+1)(s2+1) = L(f(t)) for the t-expression f(t) = ?e?t+cost+sint. The details in Heaviside’s method involve a sequence of easy-to-learn college algebra steps.

What is the Fourier transform of the Heaviside function?

  • The Fourier transform of the Heaviside function: a tragedy Let (1)H(t) = ( 1; t >0; 0; t <0: This function is theunit steporHeaviside1function. A basic fact aboutH(t) is that it is an antiderivative of the Dirac delta function:2 (2)H0(t) =–(t): If we attempt to take the Fourier transform ofH(t) directly we get the following statement: H~(! ) = 1 p
Math 611 Mathematical Physics I (Bueler) September 28, 2005 The Fourier transform of the Heaviside function: a tragedy Let (1)H(t) =(1; t >0;

0; t <0:

This function is theunit steporHeaviside1function. A basic fact aboutH(t) is that it is an antiderivative of the Dirac delta function: 2 (2)H0(t) =±(t): If we attempt to take the Fourier transform ofH(t) directly we get the following statement:

H(!) =1

p

2¼Z

1 0 e¡i!tdt= limB!+11 p

2¼1¡e¡i!B

i! The limit on the right, and the integral itself, does not exist because lim

B!1e¡i!Bdoes not exist.

One might propose that theaveragevalue ofe¡i!B, for ¯xed!andB! 1, is zero. It might then make sense to conclude: (3) ~H(!)?=1 p

2¼1

i!

Certainly this formula is a good candidate for

~H, but it would be nice to get some kind of con¯r- mation because we are clearly on shaky ground here.

Another possibility for accessing

~H(!) is to de¯ne3 H

L(t) =(1;0< t < L;

0; t <0 ort > L

and consider the Fourier transform:

HL(!) =1

p

2¼Z

L 0 e¡i!tdt=1 p

2¼1¡e¡i!L

i! =r 2 e¡i!L=2sin(!L=2) This integral and transform make sense becauseLis ¯nite. But this calculation doesn't help us. We still have the same di±culty of limits asL! 1. In fact, we have merely replaced \B" with \L" in the argument which led to (3).

Let's consider less direct routes to get

~H(!). From equation (2) we know, by easy and uncontro- versial calculation, that ~±(!) = 1=p

2¼. But then, fromF[f0] =i!~f(!),

(4) ~H(!)?=1 i!

F[H0] =1

p

2¼1

i! This seems to con¯rm (3); we are getting somewhere. For further con¯rmation we form a di®erent approximation toH(t) and take the limit. Let H

®(t) =(e¡®t; t >0;

0; t <0:

1

FromWikipedia: Oliver Heaviside (1850{1925) was a self-taught English engineer, mathematician and physicist

who adapted complex numbers to the study of electrical circuits, developed techniques for applying Laplace transforms

to the solution of di®erential equations, ...[and] recast Maxwell's mathematical analysis from its original quaternion

form to its modern vector terminology, thereby reducing the original twenty equations in twenty unknowns down to

the four di®erential equations in four unknowns we now know as Maxwell's equations.

2This statement is precise ifH(t) and±(t) are regarded asgeneralized functions, also known asdistributions.

3This is what I was doing in class. I made a mess of it.

2

Note that

lim®!0+H®(t) =H(t) for everyt2R. Furthermore, if® >0 then the Fourier transform exists:

H®(!) =1

p

2¼Z

1 0 e¡(®+i!)tdt=1 p

2¼1

®+i!:

Thus we see the same formula for

~Happear in the®!0+limit:4 (5) ~H(!)?= lim®!0+~H®(!) =1 p

2¼1

i! Unfortunately, formula (3) is wrong. Note the formula is unde¯ned at zero. In some sense, as

I now show, we have the value at!= 0 wrong.

Formula (3) doesn't stand up to applying the inverse transform to get back toH(t). Indeed, F

¡1·1

p

2¼1

i! =1

2¼iZ

1

¡1e

i!t

2¼iµ

Z0 ¡1

¢¢¢+Z

1 0 =1

2¼iZ

1 0e i!t¡e¡i!t d!=1 Z 1

0sin!t

d!=1 2 There are several comments to make on the above calculation; it is correct with certain caveats.

¡1:::

andR1

0:::give real parts of1and¡1, respectively, but once we change variables and combine

revealed. This splitting of an integral at zero is calledcomputing the Cauchy principle valueof the an extra negative; the result is¡1=2. Finally, the integralZ1 0sin! d!=¼ 2 is by no means trivial. There is no antiderivative of sincx= sinx=xexpressible in elementary functions, but contour integral techniques (chapter 20) give the value here.

Thus we have

F

¡1·1

p

2¼1

i! (t) =(1=2; t >0;

¡1=2; t <0:

This is notH(t) but ratherH(t)¡1=2. Thus

(CORRECT!) ~H(!) =1 p

2¼1

i! +1 2 p

2¼±(!) =1

p

2¼1

i! +r 2 becauseF¡1[±(!)](t) = 1=p

2¼.

The only di®erence between this correct result and our previous calculations (3), (4), and (5) is the value at!= 0. In fact, it is clear, at least in retrospect, that those calculationsdid not apply to the value at!= 0. Exercise/Application (optional). Suppose we know the transform~f(!) off(t). Letg(t) = H(t)f(t), sog(t) is the \same signal asf(t) but turned o® until time zero." Show that ~g(!) =1

2¼iZ

1

¡1~

f(!¡x) x dx+1 2 ~f(!):

Hint:F[u(t)v(t)] = (1=p

4

In this argument we come close to theLaplacetransform ofH(t). The following transform is easy and uncontro-

versial:L[H(t¡t0)] =e¡st0=s.quotesdbs_dbs12.pdfusesText_18
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