3.1 Introduction.
. . The author made an attempt at obtaining the impulsive response of transfer function by use of the numerical solution of differential equation, because he thought that the method might be better and more simple than inverse Laplace transformation in case of higher-order system. However the attempt had to solve the two problems which are as for the accuracy of numerical calculus and the value at t=0 for Laplace transformed function. The work came to introduction of the operational calculus and the rigorous definition of Laplace transform.
. . The Laplace transform of the derivative of f(t) is expressed in Eq.(3.1) so it should be the same with the derivative of impulse function. However Laplace transform makes an exception of the relation in Eq.(3.2) because δ(0) is infinite. Some theories define the value at t=0 for unit step function u(t) as one or one half. These cause some troubles when we will solve the impulsive response of transfer function by differential equation.

L f ' (t)=sL f(t)−f(0)

(3.1)

L δ' (t)=s L δ(t)−δ(0)

(3.2)

. . The author draws the conclusion that δ(0) and u(0) are equal to zero and other Laplace transformable functions are f(t)u(t), where f(t) is not constantly zero over the range t≤0. Then he shows the method obtaining the impulsive response of transfer function by differential equation.

3.2 Impulse function.
. . The Dirac delta function does not satisfy Eq.(3.2) because δ(0) is infinite. The impulse function should have zero for δ(0), δ' (0), …… and δ⁽ⁿ⁻¹⁾(0) in order that δ⁽ⁿ⁾(t) and δ⁽ⁿ⁻¹⁾(t) satisfy the same relation as Eq.(3.2). The author defines the impulse function with these properties as follows.

. . [Definition 1]. .

. . The impulse function starts up increasing at t=0 and reaches to the peak at t_p=nε then decreases as an exponential function. Its integrated value is always unity for any value ε.

tp→+0,

,

δ(tp+0)→0,

δ(0)=0

(3.5)

. . [Theorem 3.1]

f(t)δ(t)=f(0)δ(t),

if f(+0) is equal to f(0).

(3.7)

[Proof]. . f(t)δ(t) is zero everywhere except at t_p=+0 and f(t_p) is equal to f(0).

• • •

for k=0, 1, 2, ……

∴ f(t)δ(t)=f(0)δ(t)

[Q.E.D.]

. . In the case of δ(t−d), the impulse function has the peak at t_p=d+0 and is zero everywhere except at t=t_p. Its integrated value is unity and it is evident that

3.3 Unit step function.
. . The unit step function must be the integral of the impulse function over the range zero to t with zero for u(0). So it is defined as follows.

. . [Definition 2]

. . By the definition, the Laplace transform of unit step function is

∴ 2u(tp)δ(t)=δ(t)

∴ 2u(tp)=1

. . [Theorem 3.2]

[Proof]. . Letting F' (t)=f(t), the left side of Eq.(3.14) is

__________={F(t)−F(d)}u(t−d)

[Q.E.D.]

. . By substituting e^−st for f(t) and limiting t→∞, the Laplace transform of u(t−d) is

Hence u(t−d) is the convolution integral of u(t) and δ(t−d).

3.4 Laplace transform of any function.
. . If f(t) is a known function for t>0 and zero for t≤0, the Laplace transform is defined by the equation

. . [Definition 3]. . The Laplace transform is the Fourier transform of the function f(t) multiplied by u(t)e^−at, where f(t) is not always zero for t≤0, that is,

sL f(t)u(t)=L f ' (t)u(t)+f(0)

(3.20)

. .[Theorem 3.3]

L {f(t)u(t)}(n)=snL f(t)u(t)

(3.22)

[Proof]. .f(0)u(0)=0 is evident. {f(0)u(0)}' =0 is also evident by Eq.(3.18). By differentiating Eq.(3.18),

3.5 Laplace transform and Fourier transform.
. . The function u(t)e^−at in Eq.(3.17) picks up the portion of the function f(t) from zero to about 1/a for t so it is required that the value a is positive and greater than the value b if the function f(t) is e^bt, where it is the conditions for the existence of the integral from zero to infinite. Therefore the Laplace transform represents the frequency response of the portion.
. . If the function f(t) with zero for t≤0 is shifted from t to t+d, the Laplace transform is expressed in Eq.(3.23). It represents that f(t−d) is the convolution integral of f(t) and δ(t−d). In case of a=0, It represents that the Fourier transform does not vary with the value d except the phase response. On the other hand, the magnitude response of the Laplace transform is e^−ad times that of Eq.(3.16).

f(t)u(t)=t2u(t)−(t−1)2u(t−1)

because of

(3.26)

3.6 Impulsive response and its differential equation.
. . When y(t) is the impulsive response of a transfer function expressed in Eq.(3.28), the values for t≤0 are all zero because the values of input function for t≤0 are all zero. Therefore y(t) is expressed in Eq.(3.29) because the transfer function has the coefficient b_n, if it is not zero. δ(0) to δ⁽ⁿ⁻¹⁾(0) are all zero by Definition 1 for δ(t) then y(0) to y⁽ⁿ⁻¹⁾(0) are all zero from the proof of Theorem 3.3. Hence Eq.(3.28) is transformed into the differential equation expressed in Eq.(3.30) by Theorem 3.3.

Y(s)=

bnsn+bn−1sn−1+bn−2sn−2+………+b0 sn+an−1sn−1+an−2sn−2+………+a0

•1

(3.28)

y=z(t)u(t)+bnδ(t)

(3.29)

y(n)+an−1y(n−1)+an−2y(n−2)+………+a0y=bnδ(n)+bn−1δ(n−1)+bn−2δ(n−2)+………+b0δ

(3.30)

Y(s)=

s+1999 s2+1001s+1000

•1

(3.31)

y" +1001y' +1000y=δ' (t)+1999δ(t)

(3.32)

δ' (t)+1999δ(t)=	1 ε2		1−	t ε	e	−t ε	+	1999t ε2	e	−t ε	(3.33)

y=Ae−t+Be−1000t+C1e

−t ε

+C2te

−t ε

A=

2 (ε−1)2

B=

−1 (1000ε−1)2

(t≥0)

(3.34)

,

,

,

C1=

−1999999ε2+3998ε−1 (1000ε−1)2(ε−1)2

,

C2=

1999ε−1 ε(1000ε−1)(ε−1)

(3.35)

z" +1001z' +1000z=0

,

z(0)=1

,

z' (0)=998

(3.36)

. . [Theorem 3.4]. . When y(t)=z(t)u(t)+kδ(t) is the solution of the differential equation in Eq.(3.30), z(t) is expressed in the homogeneous differential equation Eq.(3.37) with the initial values in Eq.(3.38).

z(n)+an−1z(n−1)+an−2z(n−2)+………+a0z=0

(3.37)

a0

a1

a2

an−1

1

k

=

b0

(3.38)

a1

a2

an−1

1

0

z0

b1

a2

an−1

1

0

0

z0'

b2

an−1

1

0

0

z0(n−2)

bn−1

1

0

0

z0(n−1)

bn

y(t)=	z(t)u(t)+kδ(t)
y' (t)=	z' (t)u(t)+z0δ+kδ'
y" (t)=	z" (t)u(t)+z0' δ+z0δ' +kδ"
y(3)=	z(3)•u(t)+z0" δ+z0' δ' +z0δ" +kδ(3)
y(n)=	z(n)u(t)+z0(n−1)δ+ +z0δ(n−1)+kδ(n)

	1000	1001	1			k		=		1999
	1001	1	0			z0				1
	1	0	0			z0'				0
∴		k=0,		z0=1,				z0' =998

Y(s)=

Ts Ts+1

•1

y' +

1 T

y=δ' (t)

(3.39)

1 T

. .1

k

=

0

∴

k=1,

z0=

−1 T

∴

y=

−1 T

e

−t T

u(t)+δ(t)

(3.40)

1

. .0

z0

1

y' +	1 T	y=δ' (t)+	1 T		1−e	−t T		δ(t)	(3.41)

a0

a1

a2

an−1

1

0

=

b0

−k

a0

(3.42)

a1

a2

an−1

1

0

z0

b1

a1

a2

an−1

1

0

0

z0'

b2

a2

an−1

1

0

0

z0(n−2)

bn−1

an−1

1

0

0

z0(n−1)

bn

1

3.7 The numerical simulator of impulsive response.
. . Any impulsive response is obtained by the numerical solution of differential equation, because it is able to transform the impulsive response of any transfer function into the homogeneous differential equation with the initial values by Theorem 3.4. So the numerical solution of differential equation is the simulator of any impulsive response. This is very simple and easy method because it does not use the poles of Laplace transform Y(s).
. . The inverse Laplace transformation requires the numerical method obtaining the poles, the partial fraction expansion of Y(s) and the numerical procedure of complex number. The numerical calculus of poles causes troubles in the accuracy of repeated pole, nearly equal poles and very small pole among very large others, which are well known as the trouble of a quadratic equation.
. . When Y(s) is expressed in Eq.(3.43), Eq.(3.44) is the numerical expression in single precision. The discriminant D² has the negative value depending on the rounded errors of two constants at least significant bit by binary system. The shown value is the result of BASIC program given below. Therefore the numerical poles are not double pole but a pair of complex poles and the inverse Laplace transform comes to Eq.(3.45), if the difference of nearly equal complex numbers is obtained correctly without cancellation. It is very different from the analytical result in Eq.(3.46).

Y(s)=

1 s2+(44/7)s+(484/49)

(3.43)

Y(s)=

1 s2+6.285714s+9.877551

(3.44)

D2=6.2857142−4×9.877551=−3.814697×10−6	jω=D/2=9.765625×10−4j
B=44/7:C=484/49:D2=B*B−4*C:D=SQR(−D2)

y=	1 ω	e−3.142857tsinωt	(3.45)	yt=te	−22t 7	(3.46)

Y(s)=

1 s2+(72/13)s+(1296/169)

(3.47)

Y(s)=

1 s2+5.538462s+7.668639

(3.48)

D2=5.5384622−4×7.668639=1.907349×10−6	D=1.381068×10−3
B=72/13:C=1296/169:D2=B*B−4*C:D=SQR(D2)

y=	1 D	e−2.768540t−e−2.769921t	(3.49)	yt=te	−36t 13	(3.50)

Table 3.2 The case D2>0 by rounded error.
IB	EE	t	Laplace	Simulator	Accurate	exp
1 2 3 4 5 6 7 8 9 10 15 20 25 30 35 40 45 50 60 70 80 90 100 150 200 250 260 270 280 290 300	4 8 8 16 8 8 8 4 4 4 4 2 2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4	0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 15.0 20.0 25.0 26.0 27.0 28.0 29.0 30.0	7.5829260 1.1493075 1.3070513 1.3215093 1.2518086 1.1388416 1.0074244 8.7298602 7.4442796 6.2703721 2.3553682 7.8642648 2.4618818 7.3976151 2.1610798 6.1853134 1.7424429 4.8490278 3.6486452 2.6692597 1.9131514 1.3496389 9.4054035 1.3681575 1.7693020 2.1450040 1.3987099 9.1085960 5.9234641 3.8477601 2.4961132	7.5811274 1.1494700 1.3071419 1.3212813 1.2521005 1.1390801 1.0074764 8.7289229 7.4446961 6.2710218 2.3555843 7.8651421 2.4619882 7.3983805 2.1614885 6.1860657 1.7427557 4.8491265 3.6490809 2.6697400 1.9133717 1.3498644 9.4055813 1.3682550 1.7692994 2.1449140 1.3988892 9.1099043 5.9244493 3.8479376 2.4962690	7.58112819 1.14947008 1.30714199 1.32128144 1.25210048 1.13908007 1.00747644 8.72892306 7.44469664 6.27102248 2.35558508 7.86514460 2.46198784 7.39837479 2.16148535 6.18604995 1.74275013 4.84910729 3.64906330 2.66972510 1.91336070 1.34985690 9.40553661 1.36825368 1.76928238 2.14486002 1.39884840 9.10960246 5.92423189 3.84778125 2.49615753	-02 -01 -01 -01 -01 -01 -01 -02 -02 -02 -02 -03 -03 -04 -04 -05 -05 -06 -07 -08 -09 -10 -12 -17 -23 -29 -30 -32 -33 -34 -35

Table 3.1 The case D2<0 by rounded error.
IB	EE	t	Laplace	Simulator	Accurate	exp
1 2 3 4 5 6 7 8 9 10 15 20 25 30 35 40 45 50 60 70 80 90 100 150 200 250 260 270 272	8 8 8 8 8 8 8 4 4 4 4 2 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4	0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 15.0 20.0 25.0 26.0 27.0 27.2	7.3031031 1.0667063 1.1685400 1.1378624 1.0387408 9.1032371 7.7562213 6.4736538 5.3187482 4.3159299 1.3449397 3.7254493 9.6744415 2.4118152 5.8455844 1.3878954 3.2437426 7.4875703 3.8778914 1.9526134 9.6312208 4.6763544 2.2425307 5.0372520 1.0057567 1.8826088 8.4501330 3.7872693 0.0000000	7.3031038 1.0667065 1.1685402 1.1378627 1.0387410 9.1032393 7.7562220 6.4736553 5.3187493 4.3159310 1.3449405 3.7254544 9.6744596 2.4118229 5.8456051 1.3879015 3.2437570 7.4876061 3.8779149 1.9526267 9.6313131 4.6764090 2.2425613 5.0374123 1.0058035 1.8827438 8.4508017 3.7875821 2.0350776	7.30310346 1.06670639 1.16854005 1.13786250 1.03874094 9.10323861 7.75622124 6.47365522 5.31874923 4.31593093 1.34494024 3.72545195 9.67444862 2.41181899 5.84559528 1.38789922 3.24375241 7.48759648 3.87791391 1.95262767 9.63132129 4.67641321 2.24256404 5.03742439 1.00581870 1.88279114 8.45103635 3.78769387 2.03513807	-02 -01 -01 -01 -01 -02 -02 -02 -02 -02 -02 -03 -04 -04 -05 -05 -06 -07 -08 -09 -11 -12 -13 -20 -26 -33 -35 -36 -36

Continued to. .3.8 The problem having to be cautioned.