4.1 Properties of the laplace transform

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Derives basic properties of the Laplace transform.

Properties of the laplace transform

The properties associated with the Laplace transform are similar to those of the Fourier transform. First, let's set define some notation, we will use the notation $\mathfrak{L}\left\{\right\}$ to denote the Laplace transform operation. Therefore we can write $X\left(s\right)=\mathfrak{L}\left\{x,\left(,t,\right)\right\}$ and $x\left(t\right)={\mathfrak{L}}^{-1}\left\{X,\left(,s,\right)\right\}$ for the forward and inverse Laplace transforms, respectively. We can also use the transform pair notation used earlier:

$x\left(t\right)↔X\left(s\right)$

With this notation defined, lets now look at some properties.

Linearity

Given that ${x}_{1}\left(t\right)↔{X}_{1}\left(s\right)$ and ${x}_{2}\left(t\right)↔{X}_{2}\left(s\right)$ then for any constants $\alpha$ and $\beta$ , we have

$\alpha {x}_{1}\left(t\right)+\beta {x}_{2}\left(t\right)↔\alpha {X}_{1}\left(s\right)+\beta {X}_{2}\left(s\right)$

The linearity property follows easily using the definition of the Laplace transform.

Time delay

The reason we call this the time delay property rather than the time shift property is that the time shift must be positive, i.e. if $\tau >0$ , then $x\left(t-\tau \right)$ corresponds to a delay. If $\tau <0$ then we would not be able to use the single-sided Laplace transform because we would have a lower integration limit of $\tau$ , which is less than zero. To derive the property, lets evaluate the Laplace transform of the time-delayed signal

$\mathfrak{L}\left\{x,\left(,t,-,\tau ,\right)\right\}={\int }_{0}^{\infty }x\left(t-\tau \right){e}^{-st}dt$

Letting $\gamma =t-\tau$ leads to $t=\gamma +\tau$ and $dt=d\gamma$ . Substituting these quantities into [link] gives

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{x,\left(,t,-,\tau ,\right)\right\}& =& {\int }_{-\tau }^{\infty }x\left(\gamma \right){e}^{-s\left(\gamma +\tau \right)}d\gamma \hfill \\ & =& {e}^{-s\tau }{\int }_{-\tau }^{\infty }x\left(\gamma \right){e}^{-s\gamma }d\gamma \hfill \\ & =& {e}^{-s\tau }{\int }_{-\tau }^{0}x\left(\gamma \right){e}^{-s\gamma }d\gamma +{e}^{-s\tau }{\int }_{0}^{\infty }x\left(\gamma \right){e}^{-s\gamma }d\gamma \hfill \end{array}$

where we note that the first integral in the last line is zero since $x\left(t\right)=0,t<0$ . Therefore the time delay property is given by

$\mathfrak{L}\left\{x,\left(,t,-,\tau ,\right)\right\}={e}^{-s\tau }X\left(s\right)$

S-shift

This property is the Laplace transform corresponds to the frequency shift property of the Fourier transform. In fact, the derivation of the $s$ -shift property is virtually identical to that of the frequency shift property.

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{{e}^{-at},x,\left(t\right)\right\}& =& {\int }_{0}^{\infty }{e}^{-at}x\left(t\right){e}^{-st}dt\hfill \\ & =& {\int }_{0}^{\infty }x\left(t\right){e}^{-\left(a+s\right)t}dt\hfill \\ & =& {\int }_{0}^{\infty }x\left(t\right){e}^{-\left(a+\sigma +j\Omega \right)t}dt\hfill \\ & =& X\left(s+a\right)\hfill \end{array}$

The $s$ -shift property also alters the region of convergence of the Laplace transform. If the region of convergence for $X\left(s\right)$ is $\sigma >{\sigma }_{min}$ , then the region of convergence for $\mathfrak{L}\left\{{e}^{-at},x,\left(t\right)\right\}$ is $\sigma >{\sigma }_{min}-\text{Re}\left(a\right)$ .

Multiplication by $t$

Let's begin by taking the derivative of the Laplace transform:

$\begin{array}{ccc}\hfill \frac{dX\left(s\right)}{ds}& =& \frac{d}{ds}{\int }_{0}^{\infty }x\left(t\right){e}^{-st}dt\hfill \\ \hfill & =& {\int }_{0}^{\infty }x\left(t\right)\frac{d}{ds}{e}^{-st}dt\hfill \\ \hfill & =& -{\int }_{0}^{\infty }tx\left(t\right){e}^{-st}dt\hfill \end{array}$

So we can write

$\mathfrak{L}\left\{t,x,\left(,t,\right)\right\}=-\frac{dX\left(s\right)}{ds}$

This idea can be extended to multiplication by ${t}^{n}$ . Letting $y\left(t\right)=tx\left(t\right)$ , if follows that

$\begin{array}{ccc}\hfill ty\left(t\right)& ↔& -\frac{dY\left(s\right)}{ds}\hfill \\ \hfill & ↔& \frac{{d}^{2}X\left(s\right)}{d{s}^{2}}\hfill \end{array}$

Proceeding in this manner, we find that

${t}^{n}x\left(t\right)↔{\left(-1\right)}^{n}\frac{{d}^{n}X\left(s\right)}{d{s}^{n}}$

Time scaling

The time scaling property for the Laplace transform is similar to that of the Fourier transform:

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{x,\left(,\alpha ,t,\right)\right\}& =& {\int }_{0}^{\infty }x\left(\alpha t\right){e}^{-st}dt\hfill \\ \hfill & =& \frac{1}{\alpha }{\int }_{0}^{\infty }x\left(\gamma \right){e}^{-\frac{s}{\alpha }\gamma }d\gamma \hfill \\ \hfill & =& \frac{1}{\alpha }X\left(\frac{s}{\alpha }\right)\hfill \end{array}$

where in the second equality, we made the substitution $t=\frac{\gamma }{\alpha }$ and $dt=\frac{d\gamma }{\alpha }$ .

Convolution

The derivation of the convolution property for the Laplace transform is virtually identical to that of the Fourier transform. We begin with

$\begin{array}{ccc}\hfill \mathfrak{L}\left\{{\int }_{-\infty }^{\infty },x,\left(\tau \right),h,\left(t-\tau \right),d,\tau \right\}& =& {\int }_{-\infty }^{\infty }x\left(\tau \right)\mathfrak{L}\left\{h,\left(,t,-,\tau ,\right)\right\}d\tau \hfill \end{array}$

Applying the time-delay property of the Laplace transform gives

$\begin{array}{ccc}\hfill {\int }_{-\infty }^{\infty }x\left(\tau \right)\mathfrak{L}\left\{h,\left(,t,-,\tau ,\right)\right\}d\tau & =& H\left(s\right){\int }_{-\infty }^{\infty }x\left(\tau \right){e}^{-s\tau }d\tau \hfill \\ & =& H\left(s\right)X\left(s\right)\hfill \end{array}$

If $h\left(t\right)$ is the the impulse response of a linear time-invariant system, then we call $H\left(s\right)$ the system function of the system. The frequency response results by setting $s=j\Omega$ in $H\left(s\right)$ . The system function provides us with a very powerful means of determining the output of a linear time-invariant filter given the input signal. It will also enable us to determine a means of establishing the stability We will discuss stability shortly of a linear-time invariant filter, something which was not possible with the frequency response.

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