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This module describes the use of the Laplace transform in finding solutions to differential equations.

Differential equations

It is often useful to describe systems using equations involving the rate of change in some quantity through differential equations. Recall that one important subclass of differential equations, linear constant coefficient ordinary differential equations, takes the form

A y ( t ) = x ( t )

where A is a differential operator of the form

A = a n d n d t n + a n - 1 d n - 1 d t n - 1 + . . . + a 1 d d t + a 0 .

The differential equation in [link] would describe some system modeled by A with an input forcing function x ( t ) that produces an output solution signal y ( t ) . However, the unilateral Laplace transform permits a solution for initial value problems to be found in what is usually a much simpler method. Specifically, it greatly simplifies the procedure for nonhomogeneous differential equations.

General formulas for the differential equation

As stated briefly in the definition above, a differential equation is a very useful tool in describing and calculatingthe change in an output of a system described by the formula for a given input. The key property ofthe differential equation is its ability to help easily find the transform, H s , of a system. In the following two subsections, we will look atthe general form of the differential equation and the general conversion to a Laplace-transform directly from the differentialequation.

Conversion to laplace-transform

Using the definition, [link] , we can easily generalize the transfer function , H s , for any differential equation. Below are the steps taken to convert any differential equation into its transferfunction, i.e. Laplace-transform. The first step involves taking the Fourier Transform of all the terms in [link] . Then we use the linearity property to pull the transform inside thesummation and the time-shifting property of the Laplace-transform to change the time-shifting terms to exponentials. Oncethis is done, we arrive at the following equation: a 0 1 .

Y s k 1 N a k Y s s k k 0 M b k X s s k
H s Y s X s k 0 M b k s k 1 k 1 N a k s k

Conversion to frequency response

Once the Laplace-transform has been calculated from the differential equation, we can go one step further to define the frequencyresponse of the system, or filter, that is being represented by the differential equation.

Remember that the reason we are dealing with these formulas is to be able to aid us in filter design. A LCCDEis one of the easiest ways to represent FIR filters. By being able to find the frequency response, we will be able tolook at the basic properties of any filter represented by a simple LCCDE.
Below is the general formula for the frequency response of a Laplace-transform. The conversion is simply a matter of takingthe Laplace-transform formula, H s , and replacing every instance of s with w .
H w s w H s k 0 M b k w k k 0 N a k w k
Once you understand the derivation of this formula, look atthe module concerning Filter Design from the Laplace-Transform for a look into how all of these ideas of the Laplace-transform , Differential Equation, and Pole/Zero Plots play a role in filter design.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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