How Do Imaginary Numbers Affect the Inverse Laplace Transform Calculation?

[grandpa2390]

Homework Statement

find the inverse laplace transform.

1/(s^3 + 7s)

Homework Equations

sin(kt) = k/(s^2 + k^2)
cos(kt) = s/(s^2 + k^2)

The Attempt at a Solution

so this one is different from all the others I have done because it involves an imaginary number and I am not sure what the rule is that changes the problem.
somehow the answer becomes 1/7 - cos(sqrt(7)*t ) / 7

but I am getting positive sine instead of negative cosine and 7i in the denominator instead of 7. does the i reverse sine to cosine or something?

 

[Simon Bridge]

somehow the answer becomes 1/7 - cos(sqrt(7)*t ) / 7

... try taking the Laplace transform of the answer and see if it matches the question.

 

[grandpa2390]

... try taking the Laplace transform of the answer and see if it matches the question.

Well... considering it is the right answer... it should match the question. What is the method for getting that answer.

 

[Simon Bridge]

It can be instructive to check that the "correct answer" matches the question ... did you try? Did you try the transform on your answer?

Going forward:
Did you try completing the square in s^2 in the denominator.

 

[grandpa2390]

hint: complete the square in s^2 in the denominator.

I don't know what that means.

I set up my partial fraction. A(s^2+7) + B(s) = 1
s=0 and sqrt(-7)
A= 1/7 and B= 1/sqrt(-7)

1/(7s) + 1/(sqrt(-7)*(s^2+7))

1/7 * 1/s = 1/7
1/sqrt(-7)sqrt(7) * sqrt(7)/(s^2 + 7) = 1/(7i) * sin(sqrt(7)t)

1/7 + sin(sqrt(7)t)/7i
but that is wrong I should get 1/7 - cos(sqrt(7)t)/7

 

[Simon Bridge]

Try getting the denominator in form: $(s^2 + a^2)^2$

Put it another way: please show your working.

 

[grandpa2390]

Try getting the denominator in form: $(s^2 + a^2)^2$

Put it another way: please show your working.

there I posted my work.

 

[grandpa2390]

Try getting the denominator in form: $(s^2 + a^2)^2$

so you are saying it is ok to multiply the initial problem by s/s and then I won't have to worry about i.

edit: no that didn't work

I do not have a clue.

 

[Simon Bridge]

I set up my partial fraction. A(s^2+7) + B(s) = 1
s=0 and sqrt(-7)
A= 1/7 and B= 1/sqrt(-7)

1/(7s) + 1/(sqrt(-7)*(s^2+7))

1/7 * 1/s = 1/7
1/sqrt(-7)sqrt(7) * sqrt(7)/(s^2 + 7) = 1/(7i) * sin(sqrt(7)t)

1/7 + sin(sqrt(7)t)/7i
but that is wrong I should get 1/7 - cos(sqrt(7)t)/7

... In the partial fractions decomposition: don't you need a Bs+C in the numerator of the quadratic denominator?
I found it helped to put a^2=7 and just write everything out in terms of a.

 

[grandpa2390]

I'll try yahoo answers.

... In the partial fractions decomposition: don't you need a Bs+C in the numerator of the quadratic denominator?
I found it helped to put a^2=7 and just write everything out in terms of a.

that might be it then. I should have done Bs+C. and that would put the s in the numerator required for the cosine function. but will it get rid of the i?

 

[Simon Bridge]

If I read you correctly: the "i" comes from the imaginary root of the quadratic - nothing to do with the decomposition.
In general, the polynomial in the numerator should be, at most, one order less than the polynomial in the denominator.

 

[Ray Vickson]

Homework Statement

find the inverse laplace transform.

1/(s^3 + 7s)

Homework Equations

sin(kt) = k/(s^2 + k^2)
cos(kt) = s/(s^2 + k^2)

The Attempt at a Solution

so this one is different from all the others I have done because it involves an imaginary number and I am not sure what the rule is that changes the problem.
somehow the answer becomes 1/7 - cos(sqrt(7)*t ) / 7

but I am getting positive sine instead of negative cosine and 7i in the denominator instead of 7. does the i reverse sine to cosine or something?

Use either (1) If $f(t) \leftrightarrow g(s)$ then $\int_0^t f(\tau) \, d\tau \leftrightarrow \frac{g(s)}{s}$; or (2) a partial-fraction expansion of $\frac{1}{s^3 + 7s}$.

 

[Simon Bridge]

... will it get rid of the i?

I had a think further - you tried a "partial fractions" of form:
$$\frac{A}{s}+\frac{B}{s^2+7}$$... and discovered the relations for A and B were: $$As^2 + 7A +Bs = 1\\ \implies As^2=0,\; 7A=1,\; Bs=0$$ ... the first tells you that A=0 while the second tells you that A=1/7 ... which is an inconsistency, which should have been a hint that you'd done something wrong.
By working out the roots of s^2+7 you were trying to find the values of s that would make the relations true(?) ... which is an OK strategy except that the decomposition is supposed to work for all values of s not just a few of them, so the strategy failed. How you incorporated the roots into the partial fractions is still beyond me - but I suspect it is enough to notice that you had not understood what the partial fractions method is supposed to do.