Laplace Transform

Laplace transform is an operation which maps time function \(f(t)\) to a new function \(F(s)\) of a complex variable \(s\): \[ L\{f\}(s) = \int_{0}^{\infty} e^{-st} f(t) dt \tag{1} \] In this equation, \(L\{\cdot\}\) is the Laplace operator.

  • (Remark 1) Strictly speaking, the integration domain of Laplace transform is from \((-\infty, \infty)\) rather than \((0, \infty)\), and the latter one is called the unilateral Laplace transform . However, since we usually analyze functions starting from initial time \(t_0=0\), the unilateral Laplace transform will be used throughout this semester.
  • (Remark 2) Laplace transform is one of the integral transforms which maps a function from its original function space into another function space. \(e^{-st}\) in equation 1 is called the ``kernel function’’ of the transform. Myriad of integral Transforms with different kernel functions exist.
  • (Remark 3) If you substitute \(s\) to \(j\omega\) in equation 1, then the equation simply becomes the Fourier transform. This implies that Laplace transform is a generalization of Fourier transform to the whole complex domain.


Region of Convergence

It is clear that equation (1) is not always defined for all \(s\), e.g., for constant function $f(t) = C $, \(C\in\mathbb{R}\), equation (1) becomes unbounded for any negative \(s\) values. Hence, for every function \(f(t)\) there exists an associated region of \(s\) on which \(F(s)\) can be defined.

In general, consider function \(|f(t)|\le e^{at}\), where \(a\in\mathbb{R}\) is a constant value. For equation (1) to converge, the real part of \(s\) should be bigger than \(a\). This implies that \(F(s)\) is only defined on region \(Re(s)>a\) of the complex plane, where \(Re(\cdot)\) stands for the real part of the complex variable \(s\) FIGURE_FILL_IN. This region is called the region of convergence of \(f(t)\).

As mentioned, Laplace transform is a generalization of Fourier transform that exists for a much broader range of functions. The reason becomes clear via the example described on FIGURE_FILL_IN — Fourier transform \(F\{\cdot\}\) requires the function \(f(t)\) to be absolutely integrable: \[ \int_{-\infty}^{\infty} |f(t)|dt < \infty \] whereas Laplace transform exist for non-integrable functions, e.g., \(f(t)=e^{2t}\).


Useful Properties

Laplace Transform of Exponential Functions

Consider exponential function \(f(t)=e^{at}\). The Laplace transform of this exponential function is: \[ L\{e^{at}\} = \int_{0}^{\infty} e^{-st} \cdot e^{at} dt = \int_{0}^{\infty} e^{(a-s)t} dt = \frac{1}{s-a}, \;\;\; Re(s) > a \]


Laplace Transform of Derivatives

Consider function \(f(t)\) and its Laplace transform \(F(s)\). The Laplace transform of \(f'(t)\) (the time derivative of \(f(t)\)) is given as: \[ L\{f'\}(s) = \int_{0}^{\infty} e^{-st} f'(t) dt = e^{-st}f(t)\biggr\rvert^{\infty}_{t=0} + s \int_{0}^{\infty}e^{-st} f(t) dt \] If \(s\) is in the region of convergence of \(f(t)\), then \(\lim_{t\rightarrow\infty}e^{-st}f(t)=0\). Hence, \[ L\{f'\}(s) = sF(s) - f(0) \] This can be generalized to \(f^{(n)}(t)\), the \(n\)-th time derivative of \(f(t)\): \[ L\{f^{(n)}\}(s) = s^nF(s) - s^{n-1}f'(0) - \cdots - f^{(n-1)}(0) \]

If \(f(0), f'(0), \cdots f^{(n-1)}(0)\) are all zeros, then \[ L\{f^{(n)}\}(s) = s^nF(s) \] Which means, the number of \(s\) multiplied on function \(F(s)\) shows the number of time derivatives of function \(f(t)\). This is the reason why \(s\) corresponds to a derivative operator.


Laplace Transform of Integrals

Consider function \(f(t)\) and its Laplace Transform \(F(s)\). Then, the Laplace Transform of \(\int_{0}^{t}f(\tau)d\tau=I_f(t)\), which is the integral of \(f(t)\), is given as: \[ L\{I_f\}(s) = \int_{0}^{\infty} e^{-st} \bigg(\int_{0}^{t}f(\tau)d\tau\bigg) dt = \int_{0}^{t}f(\tau)d\tau \cdot \frac{e^{-st}}{s} \biggr\rvert^{\infty}_{t=0} + \frac{1}{s} \int_{0}^{\infty}e^{-st} f(t) dt \] Again, if \(s\) is in the region of convergence of \(f(t)\), the first term on the right-hand side is zero. This implies that \[ L\{I_f\}(s) = \frac{1}{s}F(s) \] And as with (Eq. \(\ref{eq:derivative_integration_n}\)) the number of \(1/s\) multiplied on function \(F(s)\) shows the number of time integrals of function \(f(t)\). This is the reason why \(1/s\) corresponds to an integrator.


Laplace Transform of Convolution

Without proof, the Laplace transform of the convolution of two functions \(f(t)\) and \(g(t)\) is: \[ L\{f*g\}(s) = F(s)\cdot G(s) \] Meaning, the convolution operator of two functions \(f(t)\) and \(g(t)\) in the time domain is a simple multiplication of \(F(s)\) and \(G(s)\) in the complex frequency domain \(s\).