How to find Laplace of the second order nonlinear time delayed term?

[Sam Chak]

The Laplace transform can only be used on linear ordinary differential equations.

[David Goodmanson]

Hi Ponnada,

Is that

(x^3)'' or (x'')^3  ?  

Either way it's not really the delay that gets you so much as the ^3 power. In general, knowledge of L(x) does not give you any direct knowledge of L(x^3). But for the time delay, if what is meant by

L(x(t-tau))  is  L( x(t-tau))*h(t-tau) )  

with h being the heaviside function, then

L(x(t)) = f(s)   --->   L( x(t-tau)*h(t-tau) ) = f(s)*e^(-s*tau)

[Paul]

Hi David,

Knowledge of L(x(t)) does, in theory, provide direct knowledge of L(x(t)^3) if direct knowledge means the ability to find L(x(t)^3). Here's a simple example to illustrate, though I'm not sure how useful this would be in practice with a more complicated function.

Define a simple function (multiplication by heaviside(t) is implied)

syms f(t)
f(t) = exp(-t)

f(t) = 

and its Laplace transform

syms s
F(s) = laplace(f(t),t,s)

F(s) = 

The region of convergence of F(s) is

assume(real(s) > -1)

Now, the L(x(t)^2) can be found by the following integral where c is constant and taking the limit as T -> inf

syms sigma 
syms c T real
assumeAlso(T,'positive')
F2(s) = int(F(sigma)*F(s-sigma),sigma,c-1j*T,c+1j*T,'Hold',true)/2/sym(pi)/1j

F2(s) = 

The integral can be taken along any vertical line (sigma = c + 1j*w) that lies in the ROC of F(s), so choose c = 0 for simplicity

F2(s) = subs(F2(s),c,0)

F2(s) = 

Change of variables to integrate along the imaginary axis

syms w real
F2(s) = changeIntegrationVariable(F2(s),sigma,1j*w)

F2(s) = 

The SMT doesn't compute the integral

F2(s) = release(F2(s))

F2(s) = 

Nor does it compute the limit

limit(F2(s),T,inf)

ans = 

But, if we just use +-inf as the limits of integration, we (miraculously?) get the expected result

F2(s) = subs(F2(s),T,inf)

F2(s) = 

Repeating for another multiplication

F3(s) = int(F2(1j*w)*F(s-1j*w),w,-inf,inf)/2/sym(pi)

F3(s) = 

gives the expected result

laplace(f(t)^3)

ans = 

though in an altogether difficult way.

[David Goodmanson]

Hi Paul,

Interesting that convolutions in the s plane can get one there, so as you say there is a direct process. Of course how direct it is in practice, being able to actually calculate the integrals, is a whole different question. I wonder how it would go for, say,

f(t) = a^n Jn(at)          f(s) = (sqrt(s^2+a^2) -s)^n / sqrt(s^2+a^2)

[Paul]

I tried it; and it didn't go very well. I thought the approach was interesting from an academic standpoint, but "I'm not sure how useful this would be in practice." at least wrt to getting closed-form, symbolic expressions. Maybe it would be more useful if one just wants to evaluate the integral numerically on a grid of points of interest in the s-plane, whatever those might be. For example, if the ROC includes the imaginary axis (which is not true for f(s) above), then maybe it could be useful to evaluate the integral up the jw axis to get a frequency response. Pure speculation on my part.

[David Goodmanson]

Hi Paul, there is the usual Laplace transform convolution result,

h(t) = Int{0,t} f(t')g(t-t') dt'
Laplace(h(t)) = f(s)g(s)

but your comment above seem like one case of what might be a general convolution result going in the other direction,

Laplace(f(t)g(t)) = h(s)                         reverse
h(s) = 1/(2*pi*i) Integral f(s')*g(s-s') ds'

where the path of integration is -i*inf+c to i*inf+c as you have mentioned.

Do you have any references for such a result? My only Laplace transform book, the Schaum's Outline one, of course has the usual result but no mention of the other direction. Maybe the result is not general enough. With the inverse transform of f(s), c just has to be large enough for the path of integration to stay to the right of all the poles and branch points, etc. of f(s). For the reverse case I tried the example f = e^(-at), g = e^(-bt), and for the convolution integral in s', the path has to be to the right of 'a' but to the left of s+b. So I guess things could get complicated in general.

[Paul]

Hi David,

I don't have any bookshelf references handy; IIRC I just used the "Multiplication" entry in the properties table on the Wikipedia page for the unilateral Laplace transform (which has a reference contained therein). The bilateral Laplace transform has the same property, where again one has to integrate along a vertical line inside the ROC of F(s) (bilateral transform properties). (I'm kind of suprised the vertical line of integration is required to only be in the ROC of F(s), as opposed to the union of the ROC's of F(s) & G(s). I confess to being not all that familiar with this property of the Laplace transform.)

syms t real
f(t) = exp(-t);    % ROC: Re(s) > -1
g(t) = exp(-2*t);  % ROC: Re(s) > -2
h(t) = f(t)*g(t);

Standard approach for H(s)

syms s
H(s) = laplace(h(t))

H(s) = 

Complicated approach for H(s)

syms w real
F(s) = laplace(f(t));
G(s) = laplace(g(t));

We require c > -1, choose c = 0

H(s) = int(F(1j*w)*G(s-1j*w),w,-inf,inf)/2/sym(pi)

H(s) = 

assume(real(s)>-3)  % ROC of the Laplace of the product
H(s) = simplify(H(s))

H(s) = 

I can't wrangle H(s) into the simpler form 1/(s+3), but we can test at a few points

spoints = [-2, -1+1i, 0, 1, 5-3i];
[double(vpa(H(spoints)));1./(spoints+3)]

ans = 
   1.0000 + 0.0000i   0.4000 - 0.2000i   0.3333 + 0.0000i   0.2500 + 0.0000i   0.1096 + 0.0411i
   1.0000 + 0.0000i   0.4000 - 0.2000i   0.3333 + 0.0000i   0.2500 + 0.0000i   0.1096 + 0.0411i

[David Goodmanson]

That's a very good way of putting it, being in the union of ROCs of both f and g. I am not sure what you mean by being surprised that only the ROC of f is required, because in the example I calculated, some kind of union of ROCs was required. If s>0, union of ROCs of f and g works.

In your example since Re(s) >-3, the numerator of sigma1 has a positive imaginary part and is in the upper half plane. To get to -sigma1 and avoid the branch cut along the negative x axis,

-sigma1 = sigma1*exp(-i*pi) 

Not counting 1/(s+3) the expression is

(-i/pi)*( log(sigma1)-log(sigma1*exp(-i*pi)) )
= (-i/pi)*( log(sigma1) -(log(sigma1)-i*pi) ) = (-i/pi)*i*pi = 1

[Paul]

We might be talking about two different things, I'm not sure. Also, I mistakenly used the word "union," which further confused things. To clarify ....

We have f(t) with Laplace transform F(s) with ROC Re(s) > a_f (assuming f(t) is causal)

We have g(t) with Laplace transform G(s) with ROC Re(s) > a_g (assuming g(t) is causal).

h(t) = f(t)g(t) has Laplace transform H(s) with ROC Re(s) > a_f + a_g, which is the union of the ROCs of F(s) and G(s)

Edit (11 Nov 2024). The preceding statement is incorrect. It should just simply say:

h(t) = f(t)g(t) has Laplace transform H(s) with ROC Re(s) > a_f + a_g

In computing the integral

syms H(s) F(s) G(s) sigma c T
H(s) = int(F(sigma)*G(s-sigma),sigma,c-1j*inf,c+1j*inf,'Hold',true)/2/sym(pi)/1j

the only requirement on c is that the vertical line defined by c has to lie within the ROC of F(s). But apparently it doesn't have to also lie within the ROC of G(s) according to the tables in the Wikipedia links (and one of my bookshelf text books).

Let's modify the example to try this.

syms t real
f(t) = exp(-4*t); F(s) = laplace(f(t)); % ROC: Re(s) > -4
g(t) = exp(-2*t); G(s) = laplace(g(t)); % ROC: Re(s) > -2
H(s) = int(F(sigma)*G(s-sigma),sigma,c-1j*inf,c+1j*inf,'Hold',true)/2/sym(pi)/1j
H(s) = subs(H(s),c,-3)  % value of c so that the integration is in ROC of F(s) but not G(s)
H(s) = release(H(s))
assume(real(s) > -6);  % ROC of H(s)
H(s) = simplify(H(s))

Verify as above

sval = [-4, -3, -2+1j, 0, 1-1j];
[double(H(sval)); 1./(sval + 6)]

ans = 
   0.5000 + 0.0000i   0.3333 + 0.0000i   0.2353 - 0.0588i   0.1667 + 0.0000i   0.1400 + 0.0200i
   0.5000 + 0.0000i   0.3333 + 0.0000i   0.2353 - 0.0588i   0.1667 + 0.0000i   0.1400 + 0.0200i

So, in this case, the vertical line is in the ROC of F(s), but it's not in the ROC of G(s), which is what I was surprised about, i.e., the line of integration doesn't have to lie in the intersection of the ROCs of F(s) and G(s). But it is in the ROC of H(s), which is the union of the ROCs of F(s) and G(s).

Edit (11 Nov 2024): The preceding sentence is incorrect. It should just say:

But it is in the ROC of H(s).

Interestingly, this use of simplify yields an incorrect result

simplify(H(s),'IgnoreAnalyticConstraints',true)

[David Goodmanson]

< modified Nov 14 2024. Slight change in notation, comments added, basic content is unchanged. >

Hi Paul, when you said you made a mistake when you said 'union, I think also I made a mistake when I also said 'union'. Your mention of ROCs helped quite a bit. It appears that (at least for the simplest example) the requirement is an intersection of two ROCs, one of which is flipped around.

f(t),g(t),h(t),F(s),G(s),H(s) have been defined already. Assume we are headed toward the convolution integral

H(s) = (1/2pi i) Int{C-i*inf, C+i*inf}  F(sigma)G(s-sigma) dsigma      (1)
where C = Re(sigma) on a vertical path of integration

Keeping to the simple example, suppose a,b real, since making them complex does not really change anything, and

f(t) = exp(-a*t);  F(sigma) = 1/(a+sigma)
     % ROC: Re(a+sigma) > 0 -->  -a < Re(sigma)
g(t) = exp(-b*t);  G(s-sigma) = 1/(b+s-sigma)
     % ROC: Re(b+s-sigma) > 0 --> Re(sigma) < Re(s)+b

Because in G(s-sigma) the argument of sigma is negative, the ROC is rotated by 180 degrees about an axis perpendicular to the page. Re(sigma) is less than some value, rather that greater than some value as with F(sigma). The total requirement is

-a < Re(sigma) < Re(s)+b                                     (2)

Both conditions must be met, so which puts Re(sigma) into the intersection of the two ROCs. And that condition implies a ROC condition on s:

-a < Re(sigma) < Re(s)+b   -->   -a < Re(s)+b   -->   -(a+b) < Re(s) 

which is the ROC that is needed for H(s) and its transform pair f(t)g(t)

h(t) = f(t)g(t) = exp(-(a+b)*t);    H(s) = 1/(a+b+s)
     % ROC: Re(a+b+s) > 0 --> -(a+b) < Re(s)

For the example, the integration (1) is

H(s) = (1/2pi i) Int{C-i*inf, C+i*inf}  1/((a+sigma)(b+s-sigma)) dsigma

Note that by (2), the path of integration Re(sigma) = C is forced to run between the pole at -a and the pole at (b+s). That works, since running in between is the only way to get a nonzero result for the integral.

[Alex]

To find the Laplace transform of a second-order nonlinear time-delayed term, you need to apply the delay property in the Laplace domain. If the term includes a time delay τ\tauτ, the Laplace transform of the delayed function f(t−τ)f(t-\tau)f(t−τ) is given by:L{f(t−τ)}=e−τsL{f(t)}\mathcal{L}\{f(t-\tau)\} = e^{-\tau s} \mathcal{L}\{f(t)\}L{f(t−τ)}=e−τsL{f(t)}

For nonlinear terms, you may need to linearize the function before applying the Laplace transform, depending on the nature of the nonlinearity. For more detailed guidance, you can refer to specialized resources on control systems or Laplace transforms on the relevant website.

Make sure to check available resources on the Website for specific examples related to your nonlinear time-delayed term.