4th Assignment: Laplace Transform
Fourth 3 Week Blog: Laplace Transform
In the fourth 3 week block we were introduced to Laplace transform. For majority of the class professor Davis lectured to us about the Laplace transform. Laquita and I compromised that we would be working closer together. Laquita took notes during lecturers while I took notes on the board and summarized our days work. We both contributed in completing our matlab and other diagrams and etc…
In order to run laplace transformers on Matlab we have to use ” syms s t; ” in Matlab. Below are the problems:
>> syms s t;laplace(1,t,s)
ans =
1/s
>> syms s t;laplace(exp(t),t,s)
ans =
1/(s – 1)
>> syms s t;laplace(exp(t^2),t,s)
ans =
laplace(exp(t^2), t, s)
>>
As you can see we do not get a solution when we do laplace transform equations in that form.
So now we will go into Matlab and type this formulated equation to find a more precise result.
If we can find any function that has this form then that is the laplace transform.
The inverse laplace ” ilaplace(….,s,t) ” gives the solution. Inverse calculates the function to whom laplace transform belongs to.
The Matlab code for inverse laplace is as follows:
>> syms A B s t; ilaplace(A*(2+s)/(s^2 +2*s-1) +B/(s^2+2*s-1)+1/(s*(s^2+2*s-1)))
ans =
((cosh(2^(1/2)*t) + (2^(1/2)*sinh(2^(1/2)*t)*((2*A + B + 2)/(A + 1) – 1))/2)*(A + 1))/exp(t) – 1
>> syms A B s t; x=ilaplace(A*(2+s)/(s^2 +2*s-1) +B/(s^2+2*s-1)+1/(s*(s^2+2*s-1)),s,t)
x =
((cosh(2^(1/2)*t) + (2^(1/2)*sinh(2^(1/2)*t)*((2*A + B + 2)/(A + 1) – 1))/2)*(A + 1))/exp(t) – 1
>>
Now we will take or function and give A B values to make our picture more informative.
This technique can only be applied to linear formulas because the laplace transform is linear.
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(1) Some specific first and second-order liner differential equations:
1.
2.
3.
4.
The First equation that I am assigned is:
The Second equation that I am assigned is:
