We shouldn't put + C for many reasons 1°) it's a definite integral 2°) the inverse laplace transform always it's a one to one operator unique result 3°) we could think this problem like the L^-1[L{ -1/2(L {sin (t)} }'s ]= - t .f (t) then f=1/2 (sen (t)/t) ✔
i learn so much from your videos, not only do they help me with the subject you are teaching, but also with my math understanding in general, you are a great teacher.
I was struggling with a question in my textbook and decided to watch solution to another question, find this video and realized that you solved the same exact question that I was struggling with! I'm so happy thank you!
SUCH A GREAT EXAMPLE OF AN INVERSE LAPLACE TRANSFORMATION, I hope you can do examples of diferential ecuaions with the special functions like delta of dirac and others, i mean diferential ecuations where you have to use laplace transformation, keep doing those videos, they are so great
The plus C could be allowed, but might depend on the initial conditions. The problem is that how to make laplace of sin(x)+2 ? It doesn't work maybe. Then if inverse laplace should give sin(x)+2, how to? This method seems to work, but requires a lot of calculation. Basically signals can be transferred both on time and signal axes. sin(x) becoming sin(x-t0)+C if transferred in both axes. One could use sin(x)+2×u(t), but that wouldn't be same as sin(x)+2, except for t>=0. If Laplace valid for t>=0 only it seems maybe also reason why u(t) should be used instead. But one could add just 2 if the offset of 2 is certain. The calculation without offset is easier maybe and doing things around the 0 instead of 2 is more practical in this math. Like solving x^2 + 3x + 1 = 0 rather than x^2 + 3x + 3 = 2. You could tune the 2nd degree polynomial equation solving to work with right side =2 rather than =0, but it would get more complicated. Can you calculate inverse laplace transform of s^2/(s^2 + w^2) as an example case also?
@@eseranceese9305 A method I recommend, is to assume it is an arbitrary linear combination of t*sin(3*t), t*cos(3*t), sin(3*t), and cos(3*t). Then take the Laplace transforms of each of component function, using the s-derivative property of Laplace transforms. Set up the linear combination with undetermined coefficients, and use algebra to solve for them. L{cos(3*t)} = s/(s^2 + 9) L{sin(3*t)} = 3/(s^2 + 9) L{t*cos(3*t)} = -d/ds L{cos(3*t)}= (s^2 - 9)/(s^2 + 9)^2 L{t*sin(3*t)} = -d/ds L{sin(3*t)}= 6*s/(s^2 + 9)^2 Let the Laplace transform we're trying to invert, equal the following, and solve for A, B, C, and D: A*s/(s^2 + 9) + 3*B/(s^2 + 9) + C*(s^2 - 9)/(s^2 + 9)^2 + 6*D*s/(s^2 + 9)^2 For 1/(s^2 + 9)^2: 1/(s^2 + 9)^2 = A*s/(s^2 + 9) + 3*B/(s^2 + 9) + C*(s^2 - 9)/(s^2 + 9)^2 + 6*D*s/(s^2 + 9)^2 1 = A*s*(s^2 + 9) + 3*B*(s^2 + 9) + C*(s^2 - 9) + 6*D*s 1 = A*s^3 + 9*A*s + 3*B*s^2 + 27*B + C*s^2 - 9*C + 6*D*s A = 0 3*B + C = 0 27*B - 9*C = 1 D = 0 Solution for B&C: B = 1/54, C=-1/18 Thus: inverse Laplace of 1/(s^2 + 9)^2 = 1/54*sin(3*t) - 1/18*t*cos(3*t)
An easier way to evaluate the convolution let I = sin(t)*cos(t) = int 0 to t (sin(t-v)cos(v))dv since convolution is commutative I = int 0 to t (cos(t-v)sin(v))dv add the 2 together 2I = int 0 to t (sin(t-v)cos(v)+cos(t-v)sin(v))dv this becomes an angle sum formula for sin 2I = int 0 to t (sin(t-v+v))dv = int 0 to t (sin(t))dv = vsin(t) from 0 to t = tsin(t) - 0sin(t) = tsin(t) divide both sides by 2 I = tsin(t)/2
An easier way to evaluate it without using convolution. Use the s-derivative property of the Laplace transform, where L{t*f(t)} = -d/ds F(s). Take the s-derivative of sine's Laplace transform: d/ds 1/(s^2 + 1) = -2*s/(s^2 + 1)^2 Therefore: L{t*sin(t)} = 2*s/(s^2 + 1)^2 Multiply both sides by 1/2: 1/2*L{t*sin(t)} = s/(s^2 + 1)^2 Recognize the original transform we're trying to invert in the above. Thus: L-1 {s/(s^2 + 1)^2} = 1/2*t*sin(t)
you should probably add an argument after the laplace transform like L{f(t)}(s). otherwise one has always to remember which variables you use normally. And also: i am used to the convolution over the whole reals. is it actually the same thing as the integral from 0 to t?
blackpenredpen . Thank you sir , i got this question as well . Use convolution theorem to find the inverse laplace transform of the following. f(s) = 1/s+p)(s+q) Do you have an email ?
Hello Steve, my name is Teo and I come all the way from Greece. I was trying to solve an inverse Laplace and googled for help so I stumbled upon this video.The problem is I can't exactly understand how to use this method on my problem. The problem is inverse Laplace of (s+3)/(s^2 + 4)^2 . I tried the partial fraction method as well but it seems that it can't be divided. I'd love to hear back from you with some help because I'm sure it will take you 5 minutes to solve it. Thank you for your time anyway.
The first thing that comes to mind is separating it. L^-1{s/(s^2 + 4)^2} + 3L^-1{1/(s^2 + 4)^2}. The first part is pretty much the same as in the video, but you only have s^2 + 2^2 in the denominator, so that results in tsin(2t)/4. There's an extra factor of 1/2, because you have to match the 2 on the top for the sine part. For the other fraction you will have to calculate a convolution of two sines: 3(L^-1{1/(s^2 + 4)} * L^-1{1/(s^2 + 4)}) = 3/4(L^-1{2/(s^2 + 2^2)} * L^-1{2/(s^2 + 2^2}) = 3/4(sin(2t) * sin(2t)).
Just watching this for fun, seems pretty cool. can someone explain the step from sint * cost? why does the argument of sin become "t-v", whereas for cos it becomes simply "v"? and perhaps i should know what is being convoluted in this convolution 😂
Broo, I got a big question, how to know, where put, t-v, for expmle, cos(t-v)sin(v), ¿Cómo sabes donde poner (t-v), ?¿ Podría haber sido cos(t-v)sin(v)? Disculpa el inglés, no soy nativo.
Partial fractions won't help you here, because you already have the denominator as reduced as possible. Unless you explore complex roots of the denominator.
Given: s/(s + 4)^2 Add zero in a fancy way, to form a term we can cancel: (s + 4 - 4)/(s + 4)^2 = (s + 4)/(s + 4)^2 - 4/(s + 4)^2 = 1/(s + 4) - 4/(s + 4)^2 The first term inverts as e^(-4*t). The second term requires us to use the s-derivative property to unpack it. In general, L{t*f(t)} = -d/ds L{f(t)}. The expression we have to unpack, is related to the s-derivative of 1/(s + 4), so differentiate this: -d/ds 1/(s + 4) = 1/(s + 4)^2 Thus: L-1{-4/(s+ 4)^2} = -4*t*e^(-4*t) Thus, the solution is: e^(-4*t) - 4*t*e^(-4*t)
This the worst way to do this. The whole purpose of laplace transform is to avoid convolution. S/(s^2+1)^2 is the derivative of (1/2)/(s^2+1). The ILT of this expression is (1/2)sint. therefore the ILT of the original expression will be (1/2)tsintn
There is a way in general to avoid convolution, to carry out the inverse Laplace of: p(s)/(s^2 + w^2)^n Where p(s) is any polynomial, w is any real constant, and n is any integer constant. Assume the solution is a linear combination of t^k*sin(w*t) + t^k*cos(w*t), with all possible k-values from 0 to (n-1). Then construct a linear combination of corresponding Laplace transforms, each with an undetermined coefficient, to solve for algebraically. If p(s) only contains odd powers of s, then your solution will be an even function. Likewise, if p(s) only contains even powers of s, then your solution will be an odd function. These inspections allow you to eliminate half your terms, and simplify the number of coefficients to solve for.
It's called the transfer function. The function that relates the output to the input of a linear time independent dynamic system, assuming X(s) represents the Laplace transform of the input, and Y(s) represents the Laplace transform of the output.
@@vasilisa4723 You can take s-derivatives of the Laplace of sine and cosine, and use the s-derivative property to find the Laplace transform of t*sin(t) and t*cos(t). Then set up a linear combination of A*cos(t) + B*sin(t) + C*t*cos(t) + D*t*sin(t). Solve for A, B, C, and D.