4.1.2. Properties of the Unilateral Laplace transform

These are from the Wikipedia page on the Laplace transform

Table 4.2 Unilateral Laplace Transform Properties

	Time Domain	\(s\)-Domain
Linearity	\(a\,x(t)+b\,y(t)\)	\(a\,X(s)+b\,Y(s)\)
Complex Conjugation	\(x^*(t)\)	\(X^*(s^*)\)
Real Signals	\(x(t)\in\setR\)	\(X(s^*) = X^*(s)\)
Time Shifting	\(x(t-a)u(t-a)\quad a>0\)	\(e^{-as} X(s)\)
Time Scaling	\(x(at) \quad a>0\)	\(\frac{1}{a}X\left(\frac{s}{a}\right)\)
Differentiation	\(x'(t)\)	\(sX(s)\) assuming \(x(0)=0\)
\(n\)-th Order Differentiation	\(\frac{d^n}{dt^n} x(t)\)	\(s^n X(s)\) assuming \(x(0)=x^{(1)}(0) = \ldots = x^{(n-1)}(0) = 0\)
Integration	\(\int_0^t f(u)du\)	\(\frac{1}{s}X(s)\)
Convolution	\(x(t)\ast y(t)\)	\(X(s) Y(s)\)

Time Shifting

Note that \(x(t)u(t)\) is shifted to \(x(t-a)u(t-a)\) and not \(x(t)\) alone. The proof is a straightforward use of the definition.

Time Scaling

The Laplace transform of the time scaled function can be calculated as:

\[\begin{split}\LT\left\{ x(at) \right\} &= \int_0^\infty x(at) e^{st} dt\\ &= \int_0^\infty x(u) e^{su/a} \frac{1}{a} du\\ &= \frac{1}{a} \int_0^\infty x(u) e^{su/a} du\\ &= \frac{1}{a} X\left( \frac{s}{a} \right)\\\end{split}\]

Differentiation in the Time Domain

This is going to be the proof were we need integration by parts. So let me remind you:

\[\int_a^b u dv = \left[u v\right]_a^b - \int_a^b v du\]

Now let us look at the Laplace transform of \(x'(t)\) given that the Laplace transform of \(x(t)\) is \(X(s)\).

\[\LT\left\{ x'(t) \right\} = \int_0^\infty x'(t) e^{-st} dt\]

Set \(u = e^{-st}\) and \(v = x(t)\) then \(dv = x'(t) dt\) so we have:

\[\begin{split}\int_0^\infty x'(t) e^{-st} dt &= \left[ x(t)e^{-st} \right]_0^\infty + s\int_0^\infty x(t) e^{-st} dt\\ &= -x(0) + s \LT\left\{ x(t) \right\}\\ &= s X(s) - x(0)\end{split}\]

In practice (of control theory for instance) it is often easier to assume that \(x(0)=0\) and this then requires some scaling and shifting of the variables involved, e.g. working with \(x(t)-x_0\) if \(x_0\) is the true value of \(x\) at \(t=0\).

For higher order derivatives the reasoning above can be easily extended. Note that for \(d^n(x)/dt^n\) we have to correct with the sum of not only \(x(0)\) but also of \(x^{(k)}(0)\) for \(k=0,\ldots,n-1\).

Integration in the Time Domain

Consider the signal \(y(t)\) obtained from \(x(t)\) by integration:

\[y(t) = \int_0^t x(u)du\]

In we then differentiate \(y(t)\) once we get the original function \(x(t)\) back. In the s-domain that would mean that \(s Y(s) = X(s)\) and thus

\[Y(s) = \frac{1}{s} X(s)\]

Of course this reasoning seems a bit sloppy and of course you should be able to prove it from the definition. Again you will need integration by parts.

Convolution in the Time Domain

We have established that for the eigenfunctions \(e^{st}\) the Laplace transform value \(H(s)\) acts as the eigenvalue (\(e^{st}\ast h(t) = H(s)e^{st}\)). Therefore it should not come as a surprise by now that convolution in the time domain (the ‘action’ of any LTI system in de time domain) corresponds with multiplication in the s-domain.

The proof is quite the analog of what we did to prove the convolution property for the Fourier transform. Use the integral form of the convolution and substitute it into the definition of the Laplace transform. After some change of variables the result will follow.