Laplace of a Periodic Function

In summary, the Laplace of a periodic function is a mathematical tool used to analyze the behavior of a function that repeats itself at regular intervals. It is a complex integral that transforms a function from the time domain to the frequency domain, allowing for the identification of its fundamental frequency and harmonics. This can be useful in various applications, such as signal processing, control systems, and differential equations. The Laplace of a periodic function is calculated using a special formula that takes into account the periodicity of the function. It is a powerful tool in the study of periodic systems and can provide valuable insights into their behavior.
  • #1
Brian Moughtin
5
0
I need to plot several cycles of the output response.

The waveform is sawtooth, starting at 0 and taking 1 second to peak at 10, returning to 0 in 1 second.

The transfer function is S/S+1

So far I've got

10/1-e^-2s ((1/s^2) - ( e^-s/s^2) - (e^-2s/s^2) as the Laplace Transform of the waveform but now I'm stuck !


Any pointers appreciated !
 
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  • #2
Brian Moughtin said:
I need to plot several cycles of the output response.

The waveform is sawtooth, starting at 0 and taking 1 second to peak at 10, returning to 0 in 1 second.

The transfer function is S/S+1

So far I've got

10/1-e^-2s ((1/s^2) - ( e^-s/s^2) - (e^-2s/s^2) as the Laplace Transform of the waveform but now I'm stuck !


Any pointers appreciated !

Something is not right here. There are some parentheses missing in your expression for the Laplace Transform. If I just add one at the end, it looks like

[tex]\frac{{10}}{1} - e^{ - 2s} \left[ {\frac{1}{{s^2 }} - \frac{{e^{ - s} }}{{s^2 }} - \frac{{e^{ - 2s} }}{{s^2 }}} \right][/tex]

Now you say that you have a sawtooth waveform, but what you describe is not a sawtooth. A sawtooth would return abruptly to zero at t = 1 second. If it takes a second to return to zero you have a triangular waveform.

I cannot find a way to interpret what you have written for the transform that looks like the correct result for either of these waveforms. Both cases are done in this document

http://mmweb.cis.nctu.edu.tw/course/AM/chap5.pdf

As for plotting the output, if you already know the waveform, why are you taking the transform? I assume it is because this is an exercise and you are supposed to be showing that the inverse transform of your Laplace Transform gets you back to the correct waveform. How do you go about doing the inverse tansform?
 
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  • #3


The Laplace transform of a periodic function is a useful tool in analyzing the behavior of a system over time. In this case, the given waveform is a sawtooth function that starts at 0, peaks at 10 in 1 second, and returns to 0 in another 1 second. The transfer function is given as S/S+1.

To plot several cycles of the output response, we can use the inverse Laplace transform to convert the Laplace transform back to the time domain. The inverse Laplace transform of 10/1-e^-2s ((1/s^2) - ( e^-s/s^2) - (e^-2s/s^2) can be simplified to 10(1-e^-s)/(s(s+1)^2). This can be further simplified to 10(1-e^-s)/(s^3+2s^2+s).

To plot the response, we can use a graphing software or manually plot points for different values of time. For example, if we choose time t=0, the output response will be 0. At t=1, the output will be 10, and at t=2, the output will be 0 again. This pattern will repeat for each cycle.

To plot multiple cycles, we can choose different values of time and plot the corresponding output response. For instance, at t=3, the output will be 10, at t=4, the output will be 0, and so on. By plotting multiple cycles, we can observe the periodic nature of the function and how it behaves over time.

In conclusion, the Laplace transform of a periodic function can be used to analyze and plot the response of a system over time. By using the inverse Laplace transform, we can convert the Laplace transform back to the time domain and plot the output response for multiple cycles.
 

1. What is the Laplace transform of a periodic function?

The Laplace transform of a periodic function is a mathematical technique that converts a function of time into a function of complex frequency, allowing for easier analysis and manipulation. It is denoted by the symbol ℓ, and is defined by the integral ℓ[f(t)] = ∫0 f(t)e-stdt, where s is a complex variable.

2. How is the Laplace transform used to solve differential equations involving periodic functions?

The Laplace transform can be used to solve differential equations involving periodic functions by transforming the differential equation into an algebraic equation, which is easier to solve. This is done by applying the Laplace transform to both sides of the equation, and then solving the resulting algebraic equation for the transformed function.

3. Can the Laplace transform be applied to all periodic functions?

No, the Laplace transform can only be applied to functions that satisfy certain conditions, such as being piecewise continuous and having a finite number of discontinuities. Additionally, the function must have a unique limit at each discontinuity.

4. How does the Laplace transform handle discontinuities in periodic functions?

The Laplace transform handles discontinuities in periodic functions by breaking the function into smaller, continuous pieces and transforming each piece separately. The transformed function will then have a series of terms, each representing the contribution of each continuous piece of the original function.

5. What are the advantages of using the Laplace transform for periodic functions?

The Laplace transform offers several advantages when dealing with periodic functions, such as simplifying the analysis of differential equations and allowing for easier manipulation of complex functions. It also provides a convenient way to handle discontinuities in functions, making it a useful tool in many areas of science and engineering.

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