00:06This video will cover the Laplace transform, and its analytical and MATLAB solutions.
00:11An improper integral is a definite integral where at least one of the limits of integration is incorrect.
00:16is infinite or the function to be integrated is not bounded on the integration interval.
00:21When the upper limit is lower, this is the general form that tells us, the definite integral
00:26at infinity and zero, when there exists a phi function, of x times the differential of x, is equal
00:31to the limit b, as it tends to infinity, of the definite integral at b, a, and analogously
00:37It is established when there are lower limits.
00:40Guys.
00:42Infinite integration limits, one or both integration limits tend to infinity.
00:47Unbounded function, the function tends to infinity at some point in the integration interval.
00:52Now, to do exercises related to integrals with infinite limits, the operators are established.
00:58with infinities, which establish rules, such as multiplying infinity by a constant
01:02is zero, or the division of a constant over infinity is zero, among other rules.
01:07Calculation for an improper integral.
01:10It is calculated using limits.
01:13The infinite limit is replaced by a variable, and the limit of the integral is calculated as the
01:18variable tends to infinity.
01:21Thus we have the following exercise: find the solution of the improper integral of
01:25integrating x squared, plus one, raised to the power of minus one, between, plus infinity and zero.
01:31This is equal to the integral of the upper infinite limit of the integrand, x squared, plus
01:36one, raised to the power of negative one, between, plus infinity and zero.
01:40And this is equal to one over the constant a, plus the inverse tangent of b over a, between the
01:45limits b, zero.
01:47Replacing b with infinite limits, and taking into account the limit operators,
01:52The division of a variable over a constant is equal to infinity, and also the division of a constant over
01:56The variable is equal to zero, we have the inverse tangent of infinity is 90 degrees, or π/2.
02:04The following exercise tells us to find the solution of the improper integral, x times the number raised
02:10negative x squared times the differential of x.
02:13This equates to the limit as b tends to infinity of the integrand raised to the power of -1
02:18two times the differential of x.
02:20First of all, because it was placed, integrating the number raised to negative x squared, and not
02:26x raised to the power of negative x squared, is because the substitution was performed.
02:31In the substitution we already designated a variable b, which is equal to e raised to the power of negative x squared,
02:36Taking the differential of b, and solving for b, we get negative one half times the differential of
02:41b equals x times the number raised to the power of negative x squared, times the differential of x, here
02:46It is observed that they have a similarity in the integral of the problem.
02:50And designating the limits and integrating, we have, the limit of b tends to infinity when, minus one
02:55half by the number raised to the power of minus x squared, between the limits
03:00b and zero.
03:02Expanding the limits, we obtain the number e raised to the power of negative infinity squared.
03:06negative e raised to the power of negative zero squared, to solve this you have that a number raised
03:11negative infinity is zero, and also that a constant over infinity is zero.
03:16Simplifying and multiplying, we get an answer of one half.
03:20The Laplace transform is a mathematical tool that allows us to transform a function
03:24from the time domain, where the independent variable is time, t, to a function
03:29of the frequency domain, where the independent variable is a complex frequency, s.
03:35That is, the function s is equal to the Laplace transform, equal to the definite integral from
03:40infinity to zero of the integrand of the function t, by the number e, raised to negative st, by
03:45the differential of t, where.
03:47S is a complex variable, t is the time variable, and e raised to the power of negative st is the function
03:56complex exponential.
03:58Main characteristics: linearity.
04:00The Laplace transform is a linear operator.
04:04This means that the transform of a sum of functions is equal to the sum of the transforms.
04:09of each function, and the transform of a function multiplied by a constant is equal to
04:13to the constant multiplied by the transform of the function.
04:16Transformation of derivatives.
04:18The Laplace transform converts the derivatives of a function into algebraic multiplications.
04:23by s.
04:24This property is fundamental to solving differential equations.
04:29Transformation of integrals.
04:30The Laplace transform also transforms the integrals of a function into divisions.
04:35algebraic by s.
04:36Initial and final value theorem.
04:38These theorems relate the value of the original function at t equal to zero and at infinity
04:42with the behavior of its Laplace transform in s.
04:46Solve a differential equation using the Laplace transform.
04:501.
04:51Take the Laplace transform of both sides of the differential equation.
04:54Use the properties of linearity and transformation of derivatives to obtain
04:58an algebraic equation in terms of f, s.
05:032.
05:04Solve the algebraic equation for f, s by clearing f, s from the algebraic equation
05:09obtained in the previous step.
05:113.
05:12Find the inverse Laplace transform.
05:15Use tables of Laplace transforms or partial fraction methods to find
05:19the function f, t, whose transform is f, s.
05:23This function f, t, a will be the solution to the original differential equation.
05:28It is typical to use Laplace transformations when referring to a table of pairs
05:32of transformations.
05:34Combining some of these simple Laplace transformations with the properties of the transformation
05:39From Laplace, we can discuss many applications of the Laplace transform.
05:43We also have some forms for derivatives in Laplace.
05:47The following exercise tells us to solve, using the Laplace transform, the following
05:51differential equation, second derivative of y, plus 7, by the first derivative, plus
05:5612y this equals 1, according to the conditions function 0 equals 0, and f' to 0.
06:02Here we first replace the known forms, for the derivatives according to Laplace, so we have
06:07For the first and second derivatives, we now perform the Laplace transform,
06:12We substitute into the differential equation, taking into account that the condition tells us,
06:16function 0 equals 0, and f' to 0, we have s squared, by function of s, plus, 7s, by
06:24function of s, plus 12 function of s, this equals 1 over s.
06:28By factoring out and solving we get, 1 over, s square, plus, 7s, plus 12.
06:37We obtain, 1 over, s, which is 1, over s square, plus, 7s, plus 12, factoring this
06:45We obtain two binomials, s plus 4, also s plus 3.
06:49Now we perform the exercise using partial fractions, that is, the fraction obtained
06:54We equate each binomial in its numerators with letters a, b.
06:58To find out the value of 'a', we multiply by the binomial 's', plus 4, and we get 1 over
07:03s, plus 3, this equals, plus b over s, plus 3, but the denominator is s equals, minus
07:104 according to the vertical line notation, and we replace in the denominator, therefore, a, is worth less
07:161.
07:17Now we perform the same method to find the value of b, in this case we multiply by
07:22s, plus 3, and we equate b, when its denominator is minus 3, b is 1.
07:28Once the values of a and b have been obtained, where a equals negative 1 and b equals 1,
07:34We replace in each numerator of each fraction, and by the inverse Laplace transform, and
07:39We get as an answer, number raised to the power of negative 3t, negative e raised to the power of negative 4t.
07:45Now to use MATLAB, we have the following function t, is equal to 10, e, raised to the power of minus
07:513t, plus, 2t times the number, e, raised to the power of negative t, plus 6, e, raised to the power of negative t, times
07:58cosine of 4t.
08:03In MATLAB, we created a comment with the percent symbol, called a transform
08:09Laplace as a function of t.
08:13Next, we type s, y, m, s, creating a symbolic variable x and another symbolic variable
08:19and, in this case s, t.
08:21And we type the function equal to, 10 raised to the power of minus 3t, plus, 2t times the number raised to the power of
08:27negative t, plus 6 raised to negative t, times cosine of 4t.
08:52We press enter and we see that MATLAB has entered the function correctly.
08:57Next, we type the laplace function of f in parentheses; it must always be like this.
09:08It gives us the correct answer; to make it look more formal, we type "pretty" between
09:13sans-serif parentheses.
09:26And it gives us another way of presenting it.
09:31That's all, if you found the video helpful, don't forget to subscribe.
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