4.2 Laplace transforms of common signals

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This module derives the Laplace transform of well-known signals.

Laplace transforms of some common signals

We'll next build up a collection of Laplace transform pairs which we will include in a table. It's important to keep in mind that once the transform pair has been derived, the focus should be on utilizing the transform pair found in the table rather than in recalculating the transform.

Exponential signal

Consider the Laplace transform of $x\left(t\right)={e}^{\alpha t}u\left(t\right)$ :

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{{e}^{-\alpha t},u,\left(t\right)\right\}& =& {\int }_{0}^{\infty }{e}^{-\alpha t}{e}^{-st}dt\hfill \\ & =& {\int }_{0}^{\infty }{e}^{-\left(\alpha +s\right)t}dt\hfill \\ & =& \frac{-1}{\alpha +s}{\left({e}^{-\left(\alpha +\sigma +j\Omega \right)t}|}_{0}^{\infty }\hfill \\ & =& \frac{1}{\alpha +s},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\sigma >-\alpha \hfill \end{array}$

where $\sigma >-\alpha$ defines the region of convergence. Notice also that if $\alpha <0$ , $X\left(s\right)$ still exists provided $\sigma >-\alpha$ . Therefore,

${e}^{-\alpha t}u\left(t\right)↔\frac{1}{\alpha +s}$

Unit step function

Recall that the Fourier transform of $u\left(t\right)$ does not converge for this signal (it contains a $\delta \left(\Omega \right)$ term, since $u\left(t\right)$ is not absolutely integrable). The Laplace transform on the other hand, converges. In fact, since $u\left(t\right)$ is a special case of the exponential function with $\alpha =0$ ,

$u\left(t\right)↔\frac{1}{s}$

The region of convergence is $\sigma >0$ , the right-half plane .

Ramp signal

This signal is given by $x\left(t\right)=tu\left(t\right)$ and also does not have a finite Fourier transform. The Laplace transform can be easily found using the multiplication by $t$ property,

$tu\left(t\right)↔\frac{1}{{s}^{2}}$

The region of convergence is once again the right-half plane, $\sigma >0$ .

Cosine signal

Even though we computed the Fourier transform of the cosine signal, $x\left(t\right)=cos\left({\Omega }_{0}t\right)$ , the Fourier transform technically does not converge for this signal. That is why $X\left(j\Omega \right)$ involves impulse functions. The Laplace transform produces quite a different result. First we use the fact that

$cos\left({\Omega }_{0}t\right)u\left(t\right)=\frac{{e}^{j{\Omega }_{0}t}u\left(t\right)+{e}^{-j{\Omega }_{0}t}u\left(t\right)}{2}$

Since each of the two terms is an exponential function we have

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{cos,\left({\Omega }_{0}t\right),u,\left(t\right)\right\}& =& \frac{\frac{1}{2}}{s-j{\Omega }_{0}}+\frac{\frac{1}{2}}{s+j{\Omega }_{0}}\hfill \\ & =& \frac{s}{{s}^{2}+{\Omega }_{0}^{2}}\hfill \end{array}$

Here, the region of convergence corresponds to the right-half plane, $\sigma >0$ .

More transform pairs

We can use the existing transform pairs along with the properties of the Laplace transform to derive many new transform pairs. Consider the exponentially weighted cosine signal. This signal is given by

$x\left(t\right)={e}^{-\alpha t}cos\left({\Omega }_{0}t\right)u\left(t\right)$

We can use the $s-$ shift property of the Laplace transform [link] along with the Laplace transform of the cosine signal [link] to get

${e}^{-\alpha t}cos\left({\Omega }_{0}t\right)u\left(t\right)↔\frac{s+\alpha }{{\left(s,+,\alpha \right)}^{2}+{\Omega }_{0}^{2}}$

Another common signal is

$x\left(t\right)=t{e}^{-\alpha t}u\left(t\right)$

We use the Laplace transform of the exponential signal "Exponential Signal" and the $t$ multiplication property to get

$t{e}^{-\alpha t}u\left(t\right)↔\frac{1}{{\left(s,+,\alpha \right)}^{2}}$

Extending this idea one step further, we have

$x\left(t\right)={t}^{2}{e}^{-\alpha t}u\left(t\right)$

Here, multiplication by ${t}^{2}$ applies, giving

${t}^{2}{e}^{-\alpha t}u\left(t\right)↔\frac{2}{{\left(s,+,\alpha \right)}^{3}}$

Example 3.1 Consider the signal $x\left(t\right)=t{e}^{-2t}u\left(t\right)$ . Therefore, we get

$t{e}^{-2t}u\left(t\right)↔\frac{1}{{\left(s,+,2\right)}^{2}}$

Example 3.2 Consider the signal $x\left(t\right)={e}^{-2t}cos\left(5t\right)u\left(t\right)$ . As seen in [link] ,

${e}^{-2t}cos\left(5t\right)u\left(t\right)↔\frac{s+2}{{\left(s+2\right)}^{2}+25}$
 $x\left(t\right)$ $X\left(s\right)$ ${e}^{-\alpha t}u\left(t\right)$ $\frac{1}{s+\alpha }$ $u\left(t\right)$ $\frac{1}{s}$ $\delta \left(t\right)$ $1$ $tu\left(t\right)$ $\frac{1}{{s}^{2}}$ $cos\left({\Omega }_{0}t\right)u\left(t\right)$ $\frac{s}{{s}^{2}+{\Omega }_{0}^{2}}$ $sin\left({\Omega }_{0}t\right)u\left(t\right)$ $\frac{{\Omega }_{0}}{{s}^{2}+{\Omega }_{0}^{2}}$ ${e}^{-\alpha t}cos\left({\Omega }_{0}t\right)u\left(t\right)$ $\frac{s+\alpha }{{\left(s,+,\alpha \right)}^{2}+{\Omega }_{0}^{2}}$ ${e}^{-\alpha t}sin\left({\Omega }_{0}t\right)u\left(t\right)$ $\frac{{\Omega }_{0}}{{\left(s,+,\alpha \right)}^{2}+{\Omega }_{0}^{2}}$ $t{e}^{-\alpha t}u\left(t\right)$ $\frac{1}{{\left(s,+,\alpha \right)}^{2}}$ ${t}^{2}{e}^{-\alpha t}u\left(t\right)$ $\frac{2}{{\left(s,+,\alpha \right)}^{3}}$ ${t}^{n}{e}^{-\alpha t}u\left(t\right)$ $\frac{n!}{{\left(s,+,\alpha \right)}^{n+1}}$

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