SC · Laplace

The Laplace Transform

The Laplace transform multiplies the Fourier transform's integrand by a convergence factor e^(−σt), widening the frequency jω into the complex number s = σ + jω. See for yourself that because σ makes growing signals die away, signals Fourier could not handle can now be transformed.

Does the weighted signal die down

In the upper figure the faint dashed curve is the growing signal e^(at) (a = 1), and the blue curve is the weighted signal, multiplied by the convergence factor e^(−σt). The lower figure is the running integral of that weighted signal from 0. Raise σ until the once-growing signal dies away to 0 and the integral reaches a ceiling and stops: at that point the Laplace transform exists.

growing signal e^(at)weighted e^((a−σ)t)

running integral

convergence factor σ0.50

Transform and ROC

X(s) = 1/(s − 1)   ROC: Re(s) > 1  · ∞

σ is smaller than the growth rate a = 1, so the weighted signal e^((a−σ)t) still grows. The integral diverges, so the transform does not exist, and this is exactly where Fourier (σ = 0) fails too.

Diverges

Multiplying Fourier by a convergence factor

The Fourier transform multiplied the signal by e^(−jωt) and integrated, and the Laplace transform multiplies by one more layer, the real exponential e^(−σt). Combining the two factors gives e^(−st), where s = σ + jω is the complex frequency. The jω part still picks out oscillations, but when σ > 0, e^(−σt) shrinks over time, so the σ part forces the signal to die away. That is why even a signal that grows without end, making the Fourier integral diverge, has a transform once σ is made large enough for the integral to converge.

ObserveX(s) = ∫ x(t) e^−st dt

The transform multiplies by e^−st and integrates.

Chooses = ?

The complex frequency is s = σ + jω.

Fill ine^atu(t): ROC Re(s) ?

e^atu(t) converges for Re(s) > a.

On your ownσ = 0: s = ? (Fourier)

With σ = 0, Laplace becomes Fourier.

The region of convergence

The Laplace integral does not converge for every σ. For the growing signal e^(at)u(t), the weighted signal is e^((a−σ)t), so it dies away to 0 and the integral converges only when σ > a, and the range of σ for which the integral converges is called the region of convergence, or ROC. This signal's transform is X(s) = 1/(s − a) with ROC Re(s) > a. The same formula with a different ROC means a different signal, so a Laplace transform is complete only as the pair of X(s) and its ROC.

Fourier is one slice of Laplace

s = σ + jω is a point on a plane. Setting σ to 0 gives s = jω, the vertical imaginary axis, and there the Laplace transform becomes exactly the Fourier transform, but only when that imaginary axis lies inside the ROC. So Fourier is a cross-section, the single line of the imaginary axis cut out of the whole s-plane that Laplace sees. Being free to choose σ is what lets Laplace handle transient response and stability, and the next unit reads systems by marking poles and zeros on this s-plane.

Back to the first screen

When σ was small, the weighted signal still grew and the integral diverged, and that is exactly where the Fourier transform fails. Raising σ beyond the growth rate a made e^((a−σ)t) die away to 0, and as the integral settled at the ceiling 1/(σ − a), the Laplace transform began to exist. What Laplace added to Fourier is exactly this choosable σ, a real exponential that forces convergence. The range of convergence that σ creates is the ROC, and a transform is the pair of X(s) and its ROC.

The Laplace transform X(s) = ∫₀^∞ x(t) e^−st dt uses the complex frequency s = σ + jω. Of the factor e^−st = e^−σt e^−jωt, e^−σt is the convergence factor: it makes growing signals die away, so signals Fourier could not handle can be transformed too. The range of σ for which the integral converges is the region of convergence (ROC), and for e^atu(t), X(s) = 1/(s − a), ROC: Re(s) > a. With σ = 0 (s = jω) and the imaginary axis inside the ROC, Laplace reduces to Fourier.

On to the next unit

If s is a point on a plane, the points where the transform X(s) becomes infinite or 0 hold almost everything about the system's character. In the next unit, the s-plane, poles and zeros, the pole s = a, which makes the denominator of X(s) = 1/(s − a) equal to 0, is marked on the plane. A pole to the left of the imaginary axis (Re < 0) means a response that dies away, one to the right a response that blows up, and the a < 0 stability condition seen in B4 reappears as exactly this pole location. That is where reading a system from a few points on a plane begins.