Delta Function and Heaviside Function - IIST

1 Delta Function and Heaviside FunctionA. SalihDepartment of Aerospace EngineeringIndian Institute of Space Science and Technology, Thiruvananthapuram 12 February 2015 We discuss some of the basic properties of the generalized functions , viz., Dirac- Delta func-tion and Heaviside step step functionThe one-dimensional Heaviside step Function centered atais defined in the following wayH(x a) =(0ifx<a,1ifx>a.(1a)Fora=0the discontinuity is atx=0, thus we haveH(x) =(0ifx<0,1ifx>0.(1b)The Heaviside Function is displayed in Fig. (x a)1a0xH(x)1 Figure 1:The Heaviside functionsH(x a)andH(x).Dirac- Delta functionTo understand the behaviour of Dirac- Delta Function (or Delta Function , for short) (x), weconsider the rectangular pulse Function (x,a) = hifa 12h<x<a+12h,0otherwise.(2)10x (x,a)haa 12ha+12hFigure 2:The pulse figure 2, it can be seen that ash , the amplitude of pulse becomes very large andits width becomes very small so that for any value ofh, the integral of the rectangular pulseZ (x,a)dx=1if the the integral of definition(a 12h,a+12h)lies in the interval( , ), and zero if range ofintegration does not contain the pulse.))

2 Now, we can define theDirac- Delta Function (x a)located at the pointx=aas (x a) =limh (x,a) =limh (x a).(3)To understand the significance of (x a), let us consider the integralZ f(x) (x,a)dxwheref(x)is an arbitrary continuous Function defined over <x< . From mean valuetheorem, we haveZ f(x) (x a)dx=a+12hZa 12hf(x) (x,a)dx= a+12h a 12h f( )D( ) =1hh f( ) =f( )where is an unknown point within the interval(a 12h,a+12h). Ash , we have (x a) (x a), and the point in the interval(a 12h,a+12h)moves closer toa, and hencef( ) f(a). Thus we have the fundamental property of the Delta functionZ f(x) (x a)dx=(f(a)if <a< ,0otherwise.(4)For example,Z61(3x 1) (x 2)dx= shows the filtering property of the Delta Function when it occurs under the integral sign,because from all the values off(x)in the interval of integration, Delta Function (x a)hasselected the valuef(a)at the location where it is acting. Delta functions are not ordinaryfunctions in the sense that we can ask for the value of (x a)at sayx=7.)

3 They are examplesof what are called generalized functions , and they are characterized by their effect on otherfunctions through integral(4).Iff(x) =1, we obtain the following relationZ (x a)dx=(1if <x< ,0otherwise.(5a)where the limit of the integration can be extended from to . Thus, we haveZ (x a)dx=1.(5b)The fact that the Delta Function is not an ordinary Function and thus cannot be representedon a graph is clearly apparent from definition(2), because (x a) =( ifx=a,0ifx6=a.(6)such thatZ (x a)dx= we replace the upper limit of the integral , by a finite valuex, then we have the followingpropertyZx (x a)dx=(0ifx<a,1ifx>a.(7)Comparing equations(1a)and(7)we get the following relation between Heaviside Function anddelta functionH(x a) =Zx (x a)dx.(8)Differentiation of equation(8)with respect tox, yields the following relationdH(x a)dx= (x a).(9)If the Delta Function is acting at the origin, , ifa=0, we have the fundamental propertyof the Delta functionZ f(x) (x)dx=(f(0)if <0< ,0otherwise.))))

4 (10)and iff(x) =1in the above equation, we haveZ (x)dx=(1if <0< ,0otherwiseandZ (x)dx=1.(11)3 The Delta Function can then be defined as (x) =( ifx=0,0ifx6=0.(12)and the relationship between Heaviside Function and Delta Function is given bydH(x)dx= (x)(13)andH(x) =Zx (x)dx=(0ifx<0,1ifx>0.(14)Regularized Dirac- Delta functionInstead of using the limit of ever-narrowing rectangular pulse of unit area when defining deltafunction, any similar functions can be used, provided theirintegral is unity and their amplitudeincrease as their pulse-like property narrows. For example, a regularized (smeared-out) deltafunction in an interval(a ,a+ )is given by (x a) = 12 h1+cos (x a) iifa <x<a+ ,0otherwise(15)where is a parameter that determines the size of the width of smearing. The variation of (x)withxfor different values of is shown in figure. Note that the Function value of thepeak (which is at the pointx=a) is1/ .The property given by equation(5)is also valid for regularized Delta Function .)))

5 To show this,we integrate (x)over the interval[a ,a+ ];Za+ a (x)dx=Za+ a 12 1+cos (x a) dx=Z 12 h1+cos y idy(puttingy=x a)=12 "y+sin y / # =12 [( +0) ( +0)]= useful property of the regularized Delta Function is givenbylim 0Z f(x) (x a)dx=f(a).(16)4If the Delta Function is acting at the origin, , ifa=0, the regularized Delta functiondefined by(15)becomes (x) = 12 1+cos x if <x< ,0otherwise.(17)Another example of regularized Delta Function is a sequenceof bell-shaped pulses defined as k(x a) =1k 2 e 12(x ak)2(18)wherekis a parameter. This regularized Delta Function approachesto Delta Function (x a)ask 0. That is, (x a) =limk 01k 2 e 12(x ak)2.(19)Note that the integral of k(x a), ,Z 1k 2 e 12(x ak)2=1for all values ofk>0, and the bell-shaped pulses defined in this way becomes narrower ask 0as displayed in Fig. 3. If the Delta Function is acting at the origin, , ifa=0, the regularizeddelta Function defined by(18)becomes k(x) =1k 2 e 12(xk)2.

6 (20)k= k(x)Figure 3:The regularized Delta Function as defined in(20).5