01 The tenth homework

Introduction to Exercises

In the tenth homework, there is an exercise problem, using the Laplace property to solve the Laplace transform of the signal. There are two types of questions included. The first three signals are given mathematical expressions. The next two exercises are two-period half-edge signals. Let's take a look at the solution to this problem.

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problem solving

The first sub-problem is about multiplying t by 1 minus e to the negative at power. This needs to be applied to the Laplace transform s-domain differential properties. The signal is multiplied by the independent variable t, and the corresponding Laplace transform expression is differentiated for s, and then multiplied by minus one. Therefore, first, for the negative at of 1 minus e, write its Laplace transform, which is a second-order rational fraction. Consider multiplying by t and differentiating the result just now. Finally get the answer to the first question. If it is solved directly according to the definition of Laplace transform, integration by parts is required, and the solution process is relatively complicated.

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The second sub-problem is to solve the Laplace transform of the product of sine, cosine and exponential signal. This needs to be applied to the Laplace transform s-domain translation properties. The time-domain signal is multiplied by the exponential signal, and the corresponding Laplace transform translates on the s-plane. Using this property, first write the Laplace transform corresponding to the sine(t) plus 2 times the cosine(t) signal, and then consider the Laplace transform multiplied by the negative t power of e, on the basis of the previous results, all the Replace s with s+1, and you can get the final answer of this small question.

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The third sub-problem actually has two ways of analysis. It can be regarded as u(t-2), multiplied by the negative t power of e, and solved by applying the Laplace transform s-domain translation property. It can also be seen as e raised to the negative t power, delayed by 2 to the right to form. The second way of thinking is applied below, using the Laplace transform time-shift theorem to solve. The Laplace transform of e to the negative power of t is 1/1 of s+1. Translate to the right by 2, and the corresponding Laplace transform is multiplied by the negative 2s power of e. Since the original question corresponds to the negative t power of e, it can be split into the negative 2 power of e and multiplied to the right delay signal. Therefore, the corresponding Laplace transform needs to be multiplied by the negative 2 power of e. Simplify the answer to this small question. The same result can be obtained by applying the first idea.

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The fourth sub-problem is a periodic triangular pulse signal with a period of 1. Since it is a unilateral Laplace transform, this signal only retains the part after t is greater than or equal to 0. The Laplace transform of the signal in one period can be solved first, and then the time-shift characteristic of the Laplace transform can be applied to solve the problem. For the convenience of solution, the differential property of Laplace transform is applied here. Here the differential signal of a single periodic signal is given. It contains four parts. Write down the corresponding Laplace transforms respectively. For the shock signal at the origin, the corresponding Laplace transform is 1. For a rectangular signal between 0 and 0.5, the corresponding Laplace transform is 2 divided by s, multiplied by 1 and minus 0.5 t power of e. Similarly, the Laplace transform of a rectangular pulse signal between 0.5 and 1 can be written. The last one is the shock signal at 1, and the corresponding Laplace transform is negative e negative s power. Next, they are combined, simplified, and finally divided by 2 to obtain the Laplace transform of a single periodic signal. This form still looks more complicated. Applying the time-shift characteristic of Laplace transform, the Laplace transform of a half-periodic signal can be obtained by shifting a single periodic signal to the right by an integer multiple of the period, which is finally equal to dividing F0(s) by 1 minus the negative s power of e. According to the Laplace transform result of the single periodic signal obtained above. After the simplification is substituted, the Laplace transform of the half-periodic signal can be obtained.
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With the idea of the fourth sub-problem, the fifth sub-problem is even simpler. Its single periodic signal is two impulse signals, and it is easy to write the corresponding Laplace transform. According to the previous ideas, the Laplace transform of a half-periodic signal is equal to the transform of a single periodic signal divided by 1 minus the negative s power of e. This is the result of the fifth subsection.

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※ Summary ※

This article gives the Laplace transform exercises in the tenth homework = solving ideas, and the rational application of the properties of Laplace transform can greatly simplify the difficulty of solving Lapalce transform.

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■ Links to related literature:

Homework Tutorial Video SS2023-HW10: Laplace Transformation Properties Exercises

01 The tenth homework

Introduction to Exercises

problem solving

※ Summary ※

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