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Properties and rules of Laplace transformation

1.) Laplace transformation of addition operation can be executed by element due to the linear property of Laplace transformation,

\mathcal{L}\left\{\sum_{k=1}^{n}c_k f_k (t) \right\} = \sum_{k=1}^{n}c_k \mathcal{L}\left\{f_k (t)\right\} = \sum_{k=1}^{n}c_k F_k (s)

2.) Laplace transformation of derivative

Let $                         f=f(t) function a general step function, where its Laplace transformation is $  F(s)=\mathcal{L} \left \{ f(t) \right \}.The question is: How is possible to derive the Laplace transformation of $          \frac{\mathrm{d}f(t)}{ \mathrm{d}t} derived function.

$                    \mathcal{L} \left \{ \frac{\mathrm{d}f(t)}{ \mathrm{d}t}   \right \} = \int_0^{\infty}\frac{\mathrm{d}f(t)}{ \mathrm{d}t}e^{-st}dt where the improper integral has to carry out.

Let’s use the partial integral rule:
$                \left.\begin{matrix}
\frac{\mathrm{d} f(t)}{\mathrm{d} t}={v}'\Rightarrow v=f(t) \\ 
e^{-st}=u\Rightarrow {u}'=-se^{-st}
\end{matrix}\right\} , where ’ denotes the derivative by „t”.

$          \mathcal{L} \left \{ \frac{\mathrm{d}f(t)}{ \mathrm{d}t}   \right \} = \int_0^{\infty}\frac{\mathrm{d}f(t)}{ \mathrm{d}t}e^{-st}dt= \left [ f(t)e^{-st} \right ]_0^{\infty} - \int_0^{\infty}(-se^{-st})f(t)dt

$                  \mathcal{L} \left \{ \frac{\mathrm{d}f(t)}{ \mathrm{d}t}   \right \} = s\int_0^{\infty}e^{-st}f(t)dt-f(0)=sF(s)-f(0)

Then apply the rule by inductive way for higher (i.e. “n” order) derivative, by using the conditions, that $                         f=f(t) general step function, and “n” ordered continuously differentiable in the $                  \left [ 0;\infty \right ) interval.

$            \mathcal{L} \left \{ \frac{\mathrm{d}^nf(t)}{ \mathrm{d}t^n}   \right \} =s^nF(s)-s^{n-1}f(0)-s^{n-2}\left.\begin{matrix}
\frac{\mathrm{d} f(t)}{\mathrm{d} t}
\end{matrix}\right|_{t=0}-...\frac{\mathrm{d}^{n-1} f(t)}{\mathrm{d} t^{n-1}}=s^nF(s)-\sum_{k=1}^ns^{n-k}}}\left.\begin{matrix}
\frac{\mathrm{d}^{k-1} f(t)}{\mathrm{d} t^{k-1}}
\end{matrix}\right|_{t=0}, where $            f(0); \quad \left.\begin{matrix}
\frac{\mathrm{d}f(t)}{\mathrm{d} t}
\end{matrix}\right|_{t=0} ;\cdots; \left.\begin{matrix}
\frac{\mathrm{d}^{k-1} f(t)}{\mathrm{d} t^{k-1}}
\end{matrix}\right|_{t=0} are the higher (max. “n-1” order) ordered derivative value at $t=0 point, i.e. at the entry point /these are the initial conditions/.

3.) Laplace transformation of integrals

Let $                         f=f(t) be a general step function and $                         F(s)=\mathcal{L} \left \{ f(t) \right \} .

Let’s determine the following Laplace transformation $               \mathcal{L} \left \{\int_0^t f(\tau)d\tau \right \}.
Definition of the Laplace transformation will be used as previously:
$                \mathcal{L} \left \{\int_0^t f(\tau)d\tau \right \}   = \int_0^{\infty} \left [  \int_0^t f(\tau)d\tau \right ] e^{-st}dt

Let’s use the partial integral rule as happened above at the derivative case:

$           \left.\begin{matrix}
\int_0^t f(\tau)d\tau=u  \Rightarrow {u}'=f(t)\\ 
e^{-st}={v}'\Rightarrow v=-\frac{1}{s}e^{-st}

$                 \mathcal{L} \left \{\int_0^t f(\tau)d\tau \right \}   = \int_0^{\infty} \left [  \int_0^t f(\tau)d\tau \right ] e^{-st}dt=\left [-\frac{1}{s}e^{-st} \int_0^t f(\tau)d\tau \right ]_0^{\infty} - \int_0^{\infty} \left ( -\frac{1}{s} \right ) e^{-st}f(t)dt

$                 \mathcal{L} \left \{\int_0^t f(\tau)d\tau \right \}   =\frac{1}{s}  \int_0^{\infty}  e^{-st}f(t)dt=\frac{F(s)}{s}

Similar to the “n” order derivative determination can be driven the Laplace transformation of “n” ordered integral.
$                \mathcal{L} \left \{ \int_0^t \left (\int_0^t \text{...} \left (\int_0^tf(\tau)d\tau \right ) d\tau \text{...} \right )d\tau \right \} =\frac{F(s)}{s^n}

4.) Theorem of similar

Let $                         f=f(t) be a general step function and $                  \lambda an optional real number.

$                   \mathcal{L} \left \{ f(\lambda t)\right \} =\frac{1}{\lambda}F \left ( \frac{s}{\lambda} \right )

$                    \mathcal{L} \left \{ f(\lambda t)\right \} = \int_0^{\infty}  e^{-st}f(\lambda t)dt=\frac{1}{\lambda} \int_0^{\infty}  e^{-\frac{s}{\lambda}(\lambda t)}f(\lambda t)d(\lambda t)

$                \mathcal{L} \left \{ f(\lambda t)\right \} =\frac{1}{\lambda}F\left (\frac{s}{\lambda} \right )

This property of Laplace transformation is according to the property of linearity (see above the linear properties of Fourier and Laplace transformation).

5.) Attenuation theorem

Let $                         f=f(t) be a general step function and $                \gamma an optional real number.

$          \mathcal{L} \left \{ f(t)e^{-\gamma t}\right \} =F(s+\gamma)

It is known, if $    f(t) general step function, then $   f_1(t)=f(t) e^{-\gamma t} is the same.

Consequently, $                \mathcal{L} \left \{ f_1(t) \right \} =\int_0^{\infty} f(t)e^{-\gamma t} e^{-st}dt=\int_0^{\infty} f(t)e^{-(s+\gamma) t}dt=F(s+\gamma)

6.) Shift theorem

Figure 27. and 28. show the graphs of general step function and shifted general step function.

Figure 27.


Figure 28.

If Laplace transformation of $ f=f(t) is $      \mathcal{L} \left \{ f(t) \right \} = F(s) , then Laplace transformation of $  f(t-\tau) is: $    \mathcal{L} \left \{ f(t-\tau) \right \} = F(s) e^{-s\tau}

In order to proof the shift theorem, apply the original definition of Laplace transformation:
$  \mathcal{L} \left \{ f(t-\tau) \right \} = \int_0^{\infty}f(t-\tau)e^{-st}dt
Apply the following new variable: $          \xi=t-\tau \quad \Rightarrow  \quad t=\xi+\tau  \\  \quad \Rightarrow \quad dt=d\xi
Consequently the new integral bounds are: $   t:(0;\infty) \rightarrow \xi:(-\tau;\infty)

$                    \mathcal{L} \left \{ f(t-\tau) \right \} = \int_0^{\infty}f(t-\tau)e^{-st}dt=\int_{-\tau}^{\infty}f(\xi)e^{-s(\xi+\tau)}d\xi =\int_{-\tau}^{\infty}f(\xi)e^{-s\xi}e^{-s\tau}d\xi= e^{-s\tau}\int_{-\tau}^{\infty}f(\xi)e^{-s\xi}d\xi

$               \mathcal{L} \left \{ f(t-\tau) \right \}= e^{-s\tau}F(s)

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