Inverse Laplace Transform and Court

m_celikok
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Homework Statement


I had a question in my midterm, it was to find inverse laplace tansform of:
(4s+5) / (s^2 + 5s + 18.5)

Where ^ denotes power.


Homework Equations





The Attempt at a Solution


My answer was to find the complex roots of equation (s^2 + 5s + 18.5) , by them transform it to form (s - w0)(s - w1) where w0 and w1 are complex roots of original equation. Then I divided the total rational (4s+5) / (s^2 + 5s + 18.5) into sum of two different rationals in form of (A)/(s -w0) + (B)/(s-w1). Found A and B and then solved each sum as different inverse laplaces. My solution was,
33jh576.gif



Yet my teacher insists this is not the solution and gave me a 0 from that question. I am going to take this to Court. Can you confirm my solution please?
 
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m_celikok said:

Homework Statement


I had a question in my midterm, it was to find inverse laplace tansform of:
(4s+5) / (s^2 + 5s + 18.5)

Where ^ denotes power.


Homework Equations





The Attempt at a Solution


My answer was to find the complex roots of equation (s^2 + 5s + 18.5) , by them transform it to form (s - w0)(s - w1) where w0 and w1 are complex roots of original equation. Then I divided the total rational (4s+5) / (s^2 + 5s + 18.5) into sum of two different rationals in form of (A)/(s -w0) + (B)/(s-w1). Found A and B and then solved each sum as different inverse laplaces. My solution was,
33jh576.gif



Yet my teacher insists this is not the solution and gave me a 0 from that question. I am going to take this to Court. Can you confirm my solution please?

Your solution is approximately equal to <br /> 4e^{-2.5t}\cos(3.5 t) - \tfrac{10}{7} e^{-2.5t}\sin(3.5 t)<br /> which can be obtained by completing the square in the denominator of the transform and using a table of transforms.

You should get some credit for having found a substantially correct answer by a correct method, but you really need to express the coefficients as exact fractions, not decimal approximations, and express the complex exponentials in terms of cosines and sines.
 
fractions

the coefficients are expressed as fractions in my paper i just calculated them from wolfram alpha to easily post here. also i noted at the end of my solution that e's in this solution can be expressed as cosine and sines through euler identity but i rathered not which is because i find this form clearer. this was an computer engineering , systems and signals course so no such rigor was required. my teacher is quite hostile towards me. thanks for your reply sir.
 
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I would NOT try to factor the denominator. Instead, I would complete the square:
s^2+ 5s+ 18.5= s^2+ 5s+ \frac{25}{4}- \frac{25}{4}+ \frac{74}{5}= (s+ \frac{5}{2})^2+ \frac{49}{4}

So \frac{4s+ 5}{s^2+ 5s+ 18.5}= 4\frac{s+ 5/2}{(s+ 5/2)^2+ (7/2)^2}+ \frac{10}{7}\frac{7/2}{(s+ 5/2)^2+ (7/2)^2}
Which can be looked up in a table of transforms.
 
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HallsofIvy said:
I would NOT try to factor the denominator. Instead, I would complete the square:
s^2+ 5s+ 18.5= s^2+ 5s+ \frac{25}{4}- \frac{25}{4}+ \frac{74}{5}= (s+ \frac{5}{2})^2+ \frac{49}{4}

So \frac{4s+ 5}{s^2+ 5s+ 18.5}= 4\frac{s+ 5/2}{(s+ 5/2)^2+ (7/2)^2}+ \frac{10}{7}\frac{7/2}{(s+ 5/2)^2+ (7/2)^2}
Which can be looked up in a table of transforms.

Thanks for the reply. On the other hand the question here is not a "wouldn't". It is a "can/can't" question. Table way might be easier and cleaner which i agree too. But, does that invalidates my solution?
 
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