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The Laplace Transforms

Module by: Richard Baraniuk. E-mail the author

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Summary: Describes Laplace transforms.

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The Laplace transform is a generalization of the Continuous-Time Fourier Transform. However, instead of using complex sinusoids of the form ⅇⅈωt ω t , as the CTFT does, the Laplace transform uses the more general, ⅇst s t , where s=σ+ⅈω s σ ω .

Although Laplace transforms are rarely solved using integration (tables and computers (e.g. Matlab) are much more common), we will provide the bilateral Laplace transform pair here. These define the forward and inverse Laplace transformations. Notice the similarities between the forward and inverse transforms. This will give rise to many of the same symmetries found in Fourier analysis.

Laplace Transform

Fs=∫-∞∞ftⅇ-stdt F s t f t s t (1)

Inverse Laplace Transform

ft=12πⅈ∫c−ⅈ∞c+ⅈ∞Fsⅇstds f t 1 2 s c c F s s t (2)

Finding the Laplace and Inverse Laplace Transforms

Solving the Integral

Probably the most difficult and least used method for finding the Laplace transform of a signal is solving the integral. Although it is technically possible, it is extremely time consuming. Given how easy the next two methods are for finding it, we will not provide any more than this. The integrals are primarily there in order to understand where the following methods originate from.

Using a Computer

Using a computer to find Laplace transforms is relatively painless. Matlab has two functions, laplace and ilaplace, that are both part of the symbolic toolbox, and will find the Laplace and inverse Laplace transforms respectively. This method is generally preferred for more complicated functions. Simpler and more contrived functions are usually found easily enough by using tables.

Using Tables

When first learning about the Laplace transform, tables are the most common means for finding it. With enough practice, the tables themselves may become unnecessary, as the common transforms can become second nature. For the purpose of this section, we will focus on the inverse Laplace transform, since most design applications will begin in the Laplace domain and give rise to a result in the time domain. The method is as follows:

  1. Write the function you wish to transform, Hs H s , as a sum of other functions, Hs=∑i=1m H i s H s i 1 m H i s where each of the H i H i is known from a table.
  2. Invert each H i s H i s to get its h i t h i t .
  3. Sum up the h i t h i t to get ht=∑i=1m h i t h t i 1 m h i t

Example 1

Compute ht h t for ∀s,ℜs>-5:Hs=1s+5 s s -5 H s 1 s 5

This can be solved directly from the table to be ht=ⅇ-5t h t 5 t

Example 2

Find the time domain representation, ht h t , of ∀s,ℜs>-10:Hs=25s+10 s s -10 H s 25 s 10

To solve this, we first notice that Hs H s can also be written as 251s+10 25 1 s 10 . We can then go to the table to find ht=25ⅇ-10t h t 25 10 t

Example 3

We can now extend the two previous examples by finding ht h t for ∀s,ℜs>-5:Hs=1s+5+25s+10 s s 5 H s 1 s 5 25 s 10

To do this, we take advantage of the additive property of linearity and the three-step method described above to yield the result ht=ⅇ-5t+25ⅇ-10t h t 5 t 25 10 t

For more complicated examples, it may be more difficult to break up the transfer function into parts that exist in a table. In this case, it is often necessary to use partial fraction expansion to get the transfer function into a more usable form.

Visualizing the Laplace Transform

With the Fourier transform, we had a complex-valued function of a purely imaginary variable, Fⅈω F ω . This was something we could envision with two 2-dimensional plots (real and imaginary parts or magnitude and phase). However, with Laplace, we have a complex-valued function of a complex variable. In order to examine the magnitude and phase or real and imaginary parts of this function, we must examine 3-dimensional surface plots of each component.

Figure 1: Real and imaginary parts of Hs H s are now each 3-dimensional surfaces.
real and imaginary sample plots
(a) The Real part of Hs H s (b) The Imaginary part of Hs H s
Figure 1(a) (laplace1.png)Figure 1(b) (laplace2.png)
Figure 2: Magnitude and phase of Hs H s are also each 3-dimensional surfaces. This representation is more common than real and imaginary parts.
magnitude and phase sample plots
(a) The Magnitude of Hs H s (b) The Phase of Hs H s
Figure 2(a) (laplace3.png)Figure 2(b) (laplace4.png)

While these are legitimate ways of looking at a signal in the Laplace domain, it is quite difficult to draw and/or analyze. For this reason, a simpler method has been developed. Although it will not be discussed in detail here, the method of Poles and Zeros is much easier to understand and is the way both the Laplace transform and its discrete-time counterpart the Z-transform are represented graphically.

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