Laplace Transform Question

Hi all,

Currently stuck on this question. v(0) is given to be v(subscript)0 (the last term in the numerator of my solution below.)

I have done it, but not sure whether I have the right solution.

I got:

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Reply 1

ghostwalker

Original post by random_duck

...

One of the many things I've not studied, but as no one else has replied (where are you all?), from just checking transform tables, I can't see any problem with that.

Reply 2

random_duck

Original post by ghostwalker

One of the many things I've not studied, but as no one else has replied (where are you all?), from just checking transform tables, I can't see any problem with that.

Hi ghostwalker, thanks for the reply.

So my answer is correct then?

I also need help then finding the inverse laplace transform of my solution to convert it back into a function of v(t) - would you be able to help me with that?

Thanks.

Reply 3

ghostwalker

Original post by random_duck

Hi ghostwalker, thanks for the reply.

So my answer is correct then?

As best I can tell.

I also need help then finding the inverse laplace transform of my solution to convert it back into a function of v(t) - would you be able to help me with that?

Thanks.

I doubt if I can do more than you can do yourself. I'd only check it against tables, and where necessary see if I can get it into a form I can see in tables.

Anything else would need research on my part, which I've not time for at present.

Reply 4

random_duck

Original post by ghostwalker

As best I can tell.

I doubt if I can do more than you can do yourself. I'd only check it against tables, and where necessary see if I can get it into a form I can see in tables.

Anything else would need research on my part, which I've not time for at present.

Ok, thanks; bump for more help. (Is anybody familiar with Laplace?)

Reply 5

davros

Original post by random_duck

Ok, thanks; bump for more help. (Is anybody familiar with Laplace?)

I'm very rusty on it too, but the general method is to get your final transform into partial fractions - lots of separate terms like

\dfrac{A}{s + a}

and

\dfrac{B}{s^2 + b^2}

and then look up the results in a table of (inverse) transforms and add the different bits together.

I haven't checked your initial result, but it looks like what you've written will separate fairly nicely into reciprocals of linear factors.

Reply 6

RichE

Yes, your transform V(s) is correct.

And if you put it into partial fractions of the form

A/s + B/(k+s) + C/(1+RCs)

those are all standard.

PS do you know that k does not equal 1/(RC)? That would be a special case if they were equal.

Reply 7

random_duck

Thanks for the additional help guys.

Original post by RichE

Thanks, glad to hear its correct. What exactly did you mean in your last sentence? I didn't understand...

Since you probably know about this more than I do, would you please be able to help check my answer for v(t)?

(I.e finding the inverse laplace transform of my result.) My answer seems incredibly long, and I am not sure if its correct.

Many thanks. :smile:

Reply 8

RichE

Original post by random_duck

Thanks for the additional help guys.

Thanks, glad to hear its correct. What exactly did you mean in your last sentence? I didn't understand...

Since you probably know about this more than I do, would you please be able to help check my answer for v(t)?

(I.e finding the inverse laplace transform of my result.) My answer seems incredibly long, and I am not sure if its correct.

Many thanks. :smile:

If kRC = 1 then k+s and 1+RCs are repeated factors, so the method of partial fractions works differently. (I presume your answer would also go wrong with kRC equalling 1 - it would mean dividing by 0 somewhere or something like that).

Reply 9

random_duck

Original post by RichE

I *think* I know what you mean now.

Are you able to help me find the inverse Laplace Transform of my answer V(s)?

Reply 10

RichE

Original post by random_duck

I *think* I know what you mean now.

Are you able to help me find the inverse Laplace Transform of my answer V(s)?

You've been told everything you need to know provided:

(i) you know how to do partial fractions
(ii) you know what the inverse Laplace of 1/(s-a) is

Reply 11

random_duck

Original post by RichE

You've been told everything you need to know provided:

(i) you know how to do partial fractions
(ii) you know what the inverse Laplace of 1/(s-a) is

Well, could you check my answer please? I have done it, it's just I am not sure if it is correct...

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Reply 12

steve44

If you post it I'll see if it is the same as what I got. (I've done it for k=1/RC as well).

Reply 13

random_duck

Original post by steve44

If you post it I'll see if it is the same as what I got. (I've done it for k=1/RC as well).

I think my answer is wrong, so if you could guide me in the right direction, I'd appreciate it.

Nonetheless, here it is:

(edited 10 years ago)

Reply 14

steve44

Original post by random_duck

I think my answer is wrong, so if you could guide me in the right direction, I'd appreciate it.

Nonetheless, here it is:

Attachment not found

Correct, top marks (except the sign in front of v_0 should be + not -, as you stated at the beginning).

When you get to considering the case k=1/RC you will need the inverse Laplace transform of 1/(s+k)^2... You can get this by doing

- d/dk \int_0^infty e^{-(k+s)t} dt = -d/dk (1/(k+s))

which gives

\int_0^infty t e^{-(k+s)t} dt = (1/(k+s)^2) :smile:

(edited 10 years ago)

Reply 15

steve44

In the case when the exponents k and 1/RC are equal in differential equations, it is usual to get a term of the form te^{-kt}.

(edited 10 years ago)

Reply 16

random_duck

Original post by steve44

Correct? Incredible, it took me at least 3 goes, and a few wasted pages filled with errors. :smile:

Yes, I spotted the sign error, thanks for the heads-up.

Though what do you mean exactly by the case k=1/RC? I don't think k=1/RC in this question?

Reply 17

steve44

Original post by random_duck

Correct? Incredible, it took me at least 3 goes, and a few wasted pages filled with errors. :smile:

Yes, I spotted the sign error, thanks for the heads-up.

Though what do you mean exactly by the case k=1/RC? I don't think k=1/RC in this question?

It may not be in the question but strictly speaking you should consider this case as RichE mentioned k=1/RC. This is because the partial fraction expansion will look different, it will be of the form

{1 \over s(s+k)^2} = {A \over s} + {B \over s+k} + {D \over (s+k)^2}

(edited 10 years ago)

Reply 18

random_duck

Original post by steve44

It may not be in the question but strictly speaking you should consider this case as RichE mentioned k=1/RC. This is because the partial fraction expansion will look different, it will be of the form

{1 \over s(s+k)^2} = {A \over s} + {B \over s+k} + {D \over (s+k)^2}

Hmm, yes, good point. But I think the constant k is undefined in the question, unless I have missed it.

Here is the full question:

Capture.JPG20.1KB

Reply 19

steve44

Original post by random_duck

Hmm, yes, good point. But I think the constant k is undefined in the question, unless I have missed it.

Here is the full question:

Yes k is undefined. But what you have gone and done is raised a subtlety of inverse Laplace transforms, which I was trying to avoid, which is the dreaded `time step function'. Some books say the inverse transform is what you expect and others give what you expect times the time step function. This requires me going back and looking at this stuff again...