Integral Transforms and Complex Variable Functions
5.1 Laplace Transforms and Their Inverses
Integral transform introduced by the French mathematician Pierre-Simon Laplace (1749–1827) can be used to map the ordinary differential equations into algebraic equations. Thus it established the foundation for many re-search areas. For instance, the Laplace transform established the basis for the modeling, analysis and synthesis of control systems. In this section,
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the definition and basic properties of Laplace transform and inverse Laplace transform are summarized first. Then we focus on the solutions to Laplace transform problems and their applications using MATLAB.
5.1.1 Definitions and properties
The one-sided Laplace transform of a time-domain function f (t) is defined as
L [f (t)] = Z ∞
0
f (t)e−stdt = F (s) (5.1) where L [f (t)] is the notation of Laplace transform.
The properties of Laplace transform is summarized below without proofs.
(i) Linear property L [af (t)± bg(t)] = aL [f(t)] ± bL [g(t)] for scalars a and b.
(ii) Time-domain shift L [f (t − a)] = e−asF (s).
(iii) s-domain property L [e−atf (t)] = F (s + a).
(iv) Differentiation property L [df (t)/dt] = sF (s)− f(0+). Generally, the nth order derivative can be obtained from
L
dn dtnf (t)
= snF (s)−sn−1f (0+)−sn−2df (0+)
dt −· · ·−dn−1f (0+) dtn−1 . (5.2) If the initial values of f (t) and the other derivatives are all zero, (5.2) can be simplified as
L [dnf (t)/dtn] = snF (s) (5.3) and this property is the crucial formula to map differential equations into algebraic equations.
(v) Integration property If zero conditions are assumed, L [Rt
0f (τ ) dτ ] = F (s)/s. Generally, the Laplace transform of the multiple integral of f (t) can be obtained from
L
Z t 0 · · ·
Z t 0
f (τ ) dτn
= F (s)
sn . (5.4)
(vi) Initial value property lim
t→0f (t) = lim
s→∞sF (s).
(vii) Final value property If F (s) has no pole with non-negative real part, i.e., Re(s) > 0, then
tlim→∞f (t) = lim
s→0sF (s).
(viii) Convolution property L [f (t) ∗ g(t)] = L [f(t)]L [g(t)], where the convolution operator ∗ is defined as
f (t) ∗ g(t) = Z t
0 f (τ )g(t − τ) dτ = Z t
0 f (t − τ)g(τ) dτ. (5.5)
(ix) Other properties
L [tnf (t)] = (−1)ndnF (s) dsn , L
f (t) tn
= Z ∞
s · · · Z ∞
s
F (s) dsn. (5.6) If the Laplace transform of a signal f (t) is F (s), the inverse Laplace trans-form of F (s) is defined as
f (t) = L−1[F (s)] = 1 j2π
Z σ+j∞ σ−j∞
F (s)estds (5.7)
where σ is greater than the real part of the poles of function F (s). The definitions of poles will be given later.
5.1.2 Computer solution to Laplace transform problems
It is hard if not impossible to write numerical programs to solve Laplace transform problems. Computer algebra systems should be used instead. For instance, the Symbolic Math Toolbox of MATLAB can be used to solve the problems easily and analytically. The procedures for solving such problems are summarized as follows:(i) The symbolic variables such as t should be declared using the syms command. Then the time-domain function f (t) should be defined in variable fun.
(ii) Call the laplace() function to solve the problem. Thus the Laplace transform can be obtained with the following function call
F=laplace(fun) % the time variable is given in t F=laplace(fun,v,u) % with domain variables v, u specified
and the function simple() can be used to simply the obtained symbolic result.
(iii) For complicated problems, the returned results are difficult to read. The function pretty() can be used to better display the results. Also the latex()function can be used to convert the results into LATEX string.
(iv) If the Laplace transform function F (s) is known, it can also be described in the variable fun. Then the MATLAB function ilaplace() can be used to calculate the inverse Laplace transform of the given function.
The syntaxes of the function are
f=ilaplace(fun) % default variable is s
f=ilaplace(fun,u,v) % specify the domain variables v and u
Example 5.1 For a given time domain function f (t) = t2e−2tsin(t + π), compute its Laplace transform function F (s).
Solution From the original problem, it can be seen that the time domain variable t should be declared first. With the MATLAB statements, the function f (t) can be specified. Then the laplace() function can be used to derive the Laplace transform of the original function
>> syms t; f=t^2*exp(-2*t)*sin(t+pi); F=laplace(f) and the result is as follows:
F (s) = −8 (s + 2)2
(s + 2)2+ 13 + 2 1
(s + 2)2+ 12.
The result can also be simplified with the statement simple(F) and the simplified result is F (s) = −23s2+ 12s + 11
(s2+ 4s + 5)3.
Example 5.2 Assume that the original function is given by f (x) = x2e−2xsin(x+
π), compute the Laplace transform and then take inverse Laplace transform and see whether the original function can be recovered.
Solution Similarly, the laplace() function can still be used
>> syms x w; f=x^2*exp(-2*x)*sin(x+pi); F=laplace(f,x,w)
and the result is the same as the one obtained in the previous example, albeit the name of the variables used are different.
If the command f1=ilaplace(F) is used, the result f1(t) = −t2e−2tsin t can be obtained. Since the equation sin(t + π) = − sin t holds, the original function is actually restored.
Example 5.3 Compute the inverse Laplace transform for the following complex variable function
G(x) = −17x5− 7x4+ 2x3+ x2− x + 1 x6+ 11x5+ 48x4+ 106x3+ 125x2+ 75x + 17. Solution The following statements can be used
>> syms x t; G=(-17*x^5-7*x^4+2*x^3+x^2-x+1)...
/(x^6+11*x^5+48*x^4+106*x^3+125*x^2+75*x+17); f=ilaplace(G,x,t) However, the result is displayed as
-1/31709*sum((39275165+45806941*_alpha^4+156459285*_alpha+5086418*
_alpha^5+149142273*_alpha^3+221566216*_alpha^2)*exp(_alpha*t), _alpha = RootOf(_Z^6+11*_Z^5+48*_Z^4+106*_Z^3+125*_Z^2+75*_Z+17)) and it can be seen that the RootOf function is returned, since the denominator polynomial function does not have analytical solutions. Thus the original problem has no symbolic solution. With the use of vpa(f,16) function, high-precision numerical solutions can be obtained.
Example 5.4 For the function f (t) given in Example 5.1, explore the relationship between L [d5f (t)/dt5] and s5L [f (t)].
Solution To solve the problem, the fifth-order derivative to the given function f (t) can be obtained by function diff(). Then the Laplace transform can be obtained
>> syms t s; f=t^2*exp(-2*t)*sin(t+pi);
F=simple(laplace(diff(f,t,5)))
which yields F (s) = −23000 + 6825s + 6660s2+ 960s4+ 3471s3+ 110s5
(s2+ 4s + 5)3 .
Taking Laplace transform to function f (t), multiplying the result by s5, one may then subtract the above result to find the difference
>> F0=laplace(f); simple(F-s^5*F0) and the difference obtained is 6s − 48.
It is obvious that the difference of the two terms is not zero, which seems not to agree with (5.3). This is because the initial conditions here are non-zero. It can easily be found that f (0) = f′(0) = f′′(0) = 0, while f(3)(0) = −6, and f(4)(0) = 48.
Hence the difference equals 6s − 48.
Example 5.5 Display the differentiation property of Laplace transform L
d2f (t) dt2
. Solution Some of the properties of Laplace transform can be displayed with the use of the Symbolic Math Toolbox of MATLAB. For this problem, the function f (t) should be declared first, then the second-order derivative can be obtained with the use of function diff(). The Laplace transform of the second-order derivative can then be obtained
>> syms t; y=sym(’f(t)’) % declare the original function laplace(diff(y,t,2))
with the display s*(s*laplace(f(t),t,s)-f(0))-D(f)(0). Also the formula for the Laplace transform of the eighth-order derivative can be derived
>> laplace(diff(y,t,8)) and the formula is
s*(s*(s*(s*(s*(s*(s*(s*laplace(f(t),t,s)-f(0))-D(f)(0))-@@(D,2)(f)(0))-@@(D,3)(f)(0))-@@(D,4)(f)(0))-@@(D,5)(f)(0))-@@(D,6)(f)(0))-@@(D,7)(f)(0)
Example 5.6 For the function f (t) = e−5tcos(2t+1)+5, compute L [d5f (t)/dt5].
Solution For the given function f (t), the Laplace and then the inverse Laplace transform can be performed with the following MATLAB statements
>> syms t; f=exp(-5*t)*cos(2*t+1)+5;
F=laplace(diff(f,t,5)); F=simple(F)
the result obtained is F (s) = 1475 cos 1s − 1189 cos 1 − 24360 sin 1 − 4282 sin 1s
s2+ 10s + 29 . In
fact, a simplified result can further be obtained if needed. For instance, collecting the terms in the numerator by using the following statements
>> syms s; collect(F)
the result can be obtained as (1475 cos 1 − 4282 sin 1) s − 1189 cos 1 − 24360 sin 1
s2+ 10s + 29 .