7 Laplace transform The Laplace transform is a generalised Fourier transform that can handle a larger class of signals. Instead of a real-valued frequency variable ω indexing the exponential component ejωt it uses a complex-valued variable s and the generalised exponential est. If all signals of interest are right-sided (zero for negative t) then a unilateral variant can be defined that is simple to use in practice. 7.1 Development The Fourier transform of a signal x(t) exists if it is absolutely integrable: ∞ x(t) dt < . | | ∞ Z−∞ While it’s possible that the transform might exist even if this condition isn’t satisfied, there are a whole class of signals of interest that do not have a Fourier transform. We still need to be able to work with them. Consider the signal x(t)= e2tu(t). For positive values of t this signal grows exponentially without bound, and the Fourier integral does not converge. However, we observe that the modified signal σt φ(t)= x(t)e− does have a Fourier transform if we choose σ > 2. Thus φ(t) can be expressed in terms of frequency components ejωt for <ω< . −∞ ∞ The bilateral Laplace transform of a signal x(t) is defined to be ∞ st X(s)= x(t)e− dt, Z−∞ where s is a complex variable. The set of values of s for which this transform exists is called the region of convergence, or ROC. Suppose the imaginary axis s = jω lies in the ROC. The values of the Laplace transform along this line are ∞ jωt X(jω)= x(t)e− dt, Z−∞ which are precisely the values of the Fourier transform. Writing s = σ + jω this integral can be expanded as ∞ σt jωt X(σ + jω)= x(t)e− e− dt, Z−∞ σt which is the Fourier transform of x(t)e− . σt Since x(t)e− = X(σ + jω) the Fourier reconstruction formula gives F{ } σt 1 ∞ jωt x(t)e− = X(σ + jω)e dω. 2π Z−∞ Multiplying both sides by eσt yields 1 ∞ x(t)= X(σ + jω)e(σ+jω)tdω. 2π Z−∞ With the change of variables s = σ + jω we have ds = jdω, and the inverse transform is σ+j 1 ∞ x(t)= X(s)estdω. 2πj σ j Z − ∞ 7-1 This is a line integral in the complex plane. In practice any value of σ can be used that corresponds to a line in the region of convergence. This integration in the complex plane requires knowledge in theory of functions of complex variables, but in practice we can avoid explicit integration by using tables of Laplace transform pairs. The Fourier synthesis formula reconstructs a signal using a set of scaled complex exponentials of the form ejωt for various values of ω. For a real-valued signal the positive and negative frequency components can be combined into sinusoids. The first plot below shows two such reconstruction basis functions for the Fourier transform. The basis functions for the Laplace transform include an exponential envelope that either increases or decreases with time, and two of these functions are shown in the second plot below. Since these functions grow with time they can represent signals that grow with time. This is the reason why the Laplace transform can represent a larger class of signals than the Fourier transform. 2 0 Fourier basis -2 -5 0 5 t 2 0 -2 Laplace basis -5 0 5 t The remainder of this section demonstrates how some bilateral Laplace transforms can be calcu- lated. To continue we will need the following result: With z = α + jβ we have zt (α+jβ)t αt jβt lim e− = lim e− = lim e− e . t t t →∞ →∞ →∞ αt jβt The quantity e− e is just a complex number expressed in polar form, with magnitude αt e− . For α> 0 this magnitude tends to zero as t , so in general → ∞ zt 0 if Re(z) > 0 lim e− = t if Re(z) < 0. →∞ (∞ at The Laplace transform of the signal x(t) = e− u(t) for a possibly complex value a can now be found: ∞ at st ∞ (s+a)t 1 (s+a)t ∞ x(t) = X(s)= e− u(t)e− dt = e− dt = − e− . L{ } 0 s + a t=0 Z−∞ Z h i Using the above result the transform is therefore 1 1 X(s)= − (0 1) = with ROC Re(s + a) > 0, s + a − s + a 7-2 leading to the Laplace transform pair at 1 e− u(t) Re(s) > Re(a). ⇐⇒ s + a − Note that this holds for all a. However, if Re(a) is negative then the transform does not converge on the line s = jω, and the Fourier transform does not exist. The region of convergence is a critical component of the Laplace transform. Using a derivation analogous to the one above we determine that the following is a valid Laplace transform pair: at 1 e− u( t) Re(s) < Re(a). − − ⇐⇒ s + a − at at The signals e− u(t) and e− u( t) therefore have the same Laplace transform — only the ROCs are different. Thus there− is no one-to-one− correspondence between x(t) and X(s) unless the ROC is specified, which makes the transform tricky to use. This ambiguity vanishes if we restrict all our signals to be causal or right-sided. For example, the Laplace transform X(s)=1/(s + a) then at has only one inverse, namely e− u(t), and we are not required to know the ROC. This leads to the unilateral Laplace transform. 7.2 Unilateral Laplace transform When people talk about the Laplace transform they usually mean the unilateral variant ∞ st X(s)= x(t)e− dt. − Z0 This is essentially just the bilateral Laplace transform applied to a signal that is known to be zero for negative time, also called a right-sided signal. The lower integration limit is taken to be 0−, infinitesimally before t = 0. This allows for impulses at the origin, and permits a natural specification of initial conditions when using the transform to solve differential equations. The unilateral Laplace transform cannot handle signals that are not right-sided, or systems that are not causal. There is essentially no difference between the unilateral and the bilateral Laplace transform, except that the former deals with the subclass of signals that start at t = 0. The inverse transform therefore remains unchanged. By way of example consider firstly the transform of δ(t): ∞ st ∞ s(0) s(0) ∞ δ(t) = δ(t)e− dt = δ(t)e− dt = e− δ(t)dt =1. L{ } − − − Z0 Z0 Z0 Thus the following Laplace transform pair is obtained: δ(t) 1 for all s. ⇐⇒ Similarly jω t jω t 1/2 1/2 cos(ω t)u(t) = 1/2e 0 u(t)+1/2e− 0 u(t) = + L{ 0 } L{ } s jω s + jω − 0 0 for Re(s) > Re( jω0) and Re(s) > Re(jω0), or just Re(s) > 0. Multiplying out yields the Laplace transform− − pair − s cos(ω0t)u(t) 2 2 for Re(s) > 0. ⇐⇒ s + ω0 7-3 Since for unilateral Laplace transforms any F (s) has a unique inverse, we generally ignore any reference to the ROC. Note however that finding a Fourier transform by evaluating the Laplace transform at s = jω is only valid if the imaginary axis lies in the ROC. Finding the inverse Laplace transform using the synthesis formula requires integration in the complex plane, which is a subject in its own right. In practice for the signals of interest we can simply find inverse transforms using tables. This is because most transforms F (s) that we care about are rational functions of the form F (s)= P (s)/Q(s), where P (s) and Q(s) are polynomials in s. The values of s for which F (s) = 0 are called the zeros of F (s), and the values where F (s) are the poles of F (s). For rational functions the zeros satisfy P (s) = 0 and the poles satisfy→Q ∞(s) = 0. The table below contains some common unilateral Laplace transform pairs: c+j 1 ∞ st ∞ st x(t)= 2πj c j X(s)e ds X(s)= 0− x(t)e− dt − ∞ δ(t)R 1 R 1 u(t) s 1 tu(t) s2 n n! t u(t) sn+1 λt 1 e u(t) s λ − λt 1 te u(t) (s λ)2 − n λt n! t e u(t) (s λ)n+1 − s cos(bt)u(t) s2+b2 b sin(bt)u(t) s2+b2 at s+a e− cos(bt)u(t) (s+a)2+b2 at b e− sin(bt)u(t) (s+a)2+b2 at (r cos θ)s+(ar cos θ br sin θ) re− cos(bt + θ)u(t) s2+2as+(a2+−b2) − at 0.5rejθ 0.5re jθ re− cos(bt + θ)u(t) s+a jb + s+a+jb − at As+B re− cos(bt + θ)u(t) s2+2as+c r = A2c+B2 2ABa , b = √c a2, θ = tan 1 Aa b c a−2 − A√c−a2 − − − q Suppose for example we wanted to find the right-sided inverse transform of 8s + 10 F (s)= . (s + 1)(s + 2)3 This has a zero at s = 10/8= 5/4, a pole at s = 1, and a pole of order three (or three poles) at s = 2. − − − − Using partial fractions we can write 8s + 10 2 6 2 2 = + .