To differentiate we use the diff function variable command For integration we

# To differentiate we use the diff function variable

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Maple also has built-in functions for integration and differentiation. To differentiate, we use the diff( [function], [variable] ) command. For integration, we can use int( [function], [variable][=lower..upper]). The lower and upper limits are optional. > laplace( diff(f(t),t), t, s); > laplace( int(f(tau),tau=0..t), t, s); addtable fourier fouriercos fouriersin hankel hilbert invfourier invhilbert , , , , , , , , [ invlaplace invmellin laplace mellin savetable , , , , ] 1 s 2 := f t 2 ( ) sin 10 t 5 e ( ) 10 t 10 15 20 s 2 100 5 s 10 10 s 5 ( ) 24 s 2 340 s 3 s 3 2000 ( ) s 2 100 ( ) s 10 s 20 s s 2 100 50 s 10 5 ( ) 24 s 2 340 s 3 s 3 2000 ( ) s 2 100 ( ) s 10 s 2

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30 Lecture Slides : MIE346 Lecture Slides 1.pptx Current Topics : Review of Common Laplace Transforms, Laplace Properties Next Topics : Inverse Laplace, Applications to Circuits (Review) Slide 5,6 Complex Plane Notes: Class Notes
31 Slide 5 Notes Continued:

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32 Slide 8 Example of Poles and Zeros Number of Poles: Number of Zeros: List of Poles: List of Zeros:
33 In the last section, we saw the basics of the Laplace Transform. In order for this to be a useful tool, we also require an inverse of the transform to allow us to move back from the frequency domain to the time domain. The basic Inverse Laplace Transform is mathematically defined as follows: { } Evaluation of this integral requires careful consideration of the ROC for the function , among other things. In general, we will almost never apply this mathematical definition directly when dealing with simple circuits and their signals. Instead, we will typically use an approach that makes use of the tables we constructed in the previous section. General Procedure Partial Fraction Expansion To begin, we will be restricting the calculation of the inverse Laplace Transform to functions which can be written in fraction form with specific polynomial numerators and denominators. Such functions are typical of those we will see shortly when dealing with the transfer functions and responses of circuits; we will be able to calculate the majority of required transforms with this method. To begin, we assume that the function, , is of the following form: With the above form, we specifically restrict ourselves to cases where . In other words, we will never have cases where the numerator contains a higher power of than the denominator. Under these restrictions, we can specifically apply the partial fraction expansion method to the above . For functions not meeting the above criteria, we must either construct the inverse in an ad-hoc manner using the tables, or (as a last resort) apply the inverse integral directly. Our procedure for partial fraction expansion is as follows: Step 1 Calculate the expansion of the fraction above using one of four distinct cases, based on the poles and zeros and any cases of repetition. For example, if we have the following function, : Pre-Reading: Inverse Laplace Transform

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34 By applying a partial fraction expansion, we break the fraction into multiple terms, each of which has an easy-to-calculate inverse: Step 2 Take the inverse of the expanded fraction components. Each will typically be easy to calculate, as mentioned above. For the example above:
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