Laplace Transform Properties


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Laplace Transform Properties
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1 Properties and theorems

Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):

LaTeX:  f(t) = \mathcal{L}^{-1} \{  F(s) \}
LaTeX:  g(t) = \mathcal{L}^{-1} \{  G(s) \}

the following table is a list of properties of unilateral Laplace transform:

Properties of the unilateral Laplace transform
Time domain Frequency domain Comment
Linearity LaTeX: a f(t) + b g(t) \ LaTeX: a F(s)  + b G(s) \ Can be proved using basic rules of integration.
Frequency differentiation LaTeX:  t f(t) \ LaTeX:  -F'(s) \
Frequency differentiation LaTeX:  t^{n} f(t) \ LaTeX:   (-1)^{n} F^{(n)}(s) \ More general form
Differentiation LaTeX:  f'(t) \ LaTeX:   s F(s) - f(0^-) \ Write the exact integral form of the given function, and add another integral to complement the former to deduce the sum to indefinite integration of a differential. Next few steps are simple.
Second Differentiation LaTeX:  f''(t) \ LaTeX:   s^2 F(s) - s f(0^-) - f'(0^-) \ Apply the Differentiation property to LaTeX:  f'(t) .
General Differentiation LaTeX:  f^{(n)}(t)  \ LaTeX:   s^n F(s) - s^{n - 1} f(0^-) - \cdots - f^{(n - 1)}(0^-) \ Follow the process briefed for the Second Differentiation.
Frequency integration LaTeX:  \frac{f(t)}{t}  \ LaTeX:   \int_s^\infty F(\sigma)\, d\sigma \
Integration LaTeX:  \int_0^t f(\tau)\, d\tau  =  u(t) * f(t) LaTeX:   {1 \over s} F(s) LaTeX: u(t) is the Heaviside step function.
Scaling LaTeX:  f(at) \ LaTeX:   {1 \over |a|} F \left ( {s \over a} \right )
Frequency shifting LaTeX:  e^{at} f(t)  \ LaTeX:  F(s - a) \
Time shifting LaTeX:  f(t - a) u(t - a) \ LaTeX:   e^{-as} F(s) \ LaTeX: u(t) is the Heaviside step function
Convolution LaTeX:  (f * g)(t) \ LaTeX:  F(s) \cdot G(s) \
Periodic Function LaTeX:  f(t) \ LaTeX: {1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt LaTeX: f(t) is a periodic function of period LaTeX: T so that LaTeX: f(t) = f(t + T), \; \forall t
  • Initial value theorem:
LaTeX: f(0^+)=\lim_{s\to \infty}{sF(s)}
  • Final value theorem:
LaTeX: f(\infty)=\lim_{s\to 0}{sF(s)}, all poles in left-hand plane.
The final value theorem is useful because it gives the long-term behaviour without having to perform partial fraction decompositions or other difficult algebra. If a function's poles are in the right hand plane (e.g. LaTeX: e^t or LaTeX: \sin(t)) the behaviour of this formula is undefined.

1.1 Proof of the Laplace transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find the transform of a function's derivative. This can be derived from the basic expression for a Laplace transform as follows:

LaTeX: \mathcal{L} \left\{f(t)\right\}  =\int_{0^-}^{+\infty} e^{-st} f(t)\,dt
LaTeX:  ~~ = \left[ \frac{f(t)e^{-st}}{-s} \right]_{0^-}^{+\infty} - </dd></dl>
<p>\int_{0^-}^{+\infty} \frac{e^{-st}}{-s} f'(t)dt (by parts)

LaTeX:  ~~ = \left[-\frac{f(0)}{-s}\right] + </dd></dl>


LaTeX: \mathcal{L}\left\{\frac{df}{dt}\right\} = s\cdot\mathcal{L} \left\{ f(t) \right\}-f(0),

and in the bilateral case, we have

LaTeX:  \mathcal{L}\left\{ { df \over dt }  \right\}</dd></dl>
<pre> = s \int_{-\infty}^{+\infty} e^{-st} f(t)\,dt  = s \cdot \mathcal{L} \{ f(t) \}.

2 See Also