laplace transform problems

[steveB]

Q1:



Could you please check if I have it correct conceptually?

That doesn't look correct to me. It looks like you make two mistakes that offset each other. When applying the derivative rule, we use the initial value at 0-, and the initial value of cos(t) u(t) is 0 at t=0-. This is exactly what I want you to see and understand. Then when you transform cos(t) δ(t) you try to integrate from 0- to 0+, which typically we dont' try to do. There may be a way to do this (although I've never tried) using special rules, but we don't even try because we can use the sifting property of the delta function to get the answer more directly. Whenever you integrate a function with the δ(t), you simple extract the value of that function at t=0. Remember that the 0- and 0+ indicate the instruction to include or exclude the weird behavior of certain functions at t=0. Hence, the integral of a any function with δ(t) from -∞ to +∞ is the same as from 0- to 0+ , both of which will include the impulse function. The answer just amounts to plugging t=0 into the function you are trying to integrate. Hence the Laplace transform of cos(t) δ(t) is simply cos(0) exp(0)=1.

So, you had the initial value of 1 and the transform of cos(t) δ(t) as 0, but it should be the initial value is 0 and the transform of cos(t) δ(t) is 1.

Q2:
Could you please help me with this query too?

Remember what I said before that the initial value at t=0- is zero for most functions that we deal with, which includes all derivatives of δ(t) and you are correct that the n'th derivative of δ(t) transforms to s^n.

Q3:
Kindly help me with this query? Thank you very much.

I'll have to scan through this later to see if I can spot the discrepancy.

 

[PG1995]

Many thanks for the help.

Q1:

That doesn't look correct to me. It looks like you make two mistakes that offset each other. When applying the derivative rule, we use the initial value at 0-, and the initial value of cos(t) u(t) is 0 at t=0-. This is exactly what I want you to see and understand. Then when you transform cos(t) δ(t) you try to integrate from 0- to 0+, which typically we dont' try to do...

I have tried to correct it. I hope it's okay now.

Q3:

I'll have to scan through this later to see if I can spot the discrepancy.

No problem.

Best regards
PG

 

[steveB]

That looks correct now, but I'm confused by what you wrote after the "because" symbol. You say cos(0-)=0, which is not true because cos(0)=1. While it is correct that u(0-)=0, it's not clear to me why that is relevant. If you delete these two statements, which don't seem relevant, it all looks good.

 

[steveB]

Q3:
Kindly help me with this query?

It looks like you just made an algebra mistake when you calculated A1 and B1.

You end up with a condition that the numerator must be equated which gives: 2 B1 + A1 (s+1) = 2 s + 12

This leads to A1 s + (2 B1 + A1) = 2 s + 12 from which it is clear that A1=2 and B1=5.

 

[PG1995]

Thank you.

It looks like you just made an algebra mistake when you calculated A1 and B1.

You end up with a condition that the numerator must be equated which gives: 2 B1 + A1 (s+1) = 2 s + 12

This leads to A1 s + (2 B1 + A1) = 2 s + 12 from which it is clear that A1=2 and B1=5.

Seriously I have been through it almost five times but I couldn't trace any mistake although you are correct (A1*s) + A2 = A1(s+ alpha) + (B1*beta) isn't satisfied. Heaven knows where that nasty mistake occurred!

Regards
PG

 

[steveB]

Heaven knows where that nasty mistake occurred!

In the second to the last line or your work, you say: 2s+12=2(s+1)+6(2)

This is wrong. It should be 2s+12=2(s+1)+5(2)

 

[PG1995]

My sincerest apologies.

But this is what I was doing. I was thinking that, 2s + 12 = 2(s+1) + 5(2), corresponding terms should agree with each other, and this made me to equate "12" with "6(2)". Sorry.

Many thanks.

Regards
PG

 

[steveB]

My sincerest apologies.

But this is what I was doing. I was thinking that, 2s + 12 = 2(s+1) + 5(2), corresponding terms should agree with each other, and this made me to equate "12" with "6(2)". Sorry.

Many thanks.

Regards
PG

No problem. That's the difficulty with mistakes. Once we make one, it's then hard to see the error right away. Often it takes a fresh eye to catch it. Hence, either having a friend look or taking time away from the problem and going back later usually reveals the mistake much faster than just staring at it.

The good thing here is that you knew there might be a mistake in your work and your were intent on tracking it down.

I wish I could tell you that this gets better as you get more experienced, but my experience is that it does not get better, and then you get old and it gets worse. Age gives you experience that keeps you from making amateurish mistakes, but it degrades your faculties which can allow you make more stupid mistakes, if you don't keep your guard up.

The inevitability of invisible mistakes is why engineering always includes the "due-dilligence" effort of verification, which includes theoretical double- triple- or quadruple-checking, experimental verifications under temperature, vibration and other extremes and field testing under worst case conditions, if possible.

The consequences of mistakes also gets worse as you get more experienced. A mistake on a test might hurt your grade, but a mistake on a real-world design might hurt a person.

 

[PG1995]

Hi

Could you please help me with this query? Thank you.

Regards
PG

PS: Perhaps using L'Hopital's rue might help here.

 

[steveB]

Hi

Could you please help me with this query? Thank you.

Regards
PG

PS: Perhaps using L'Hopital's rue might help here.

Even though ∞/∞ is undefined, you have a very specific limit that can be evaluation by various techniques. Actually, one technique (which is what I would use) is "inspection/experience". You don't need to do very many limit problems before you realize that exp(x) increases faster than any fixed power x^n, where n is a finite integer. Hence, the denominator goes to infinity so much faster than the numerator, that the limit does evaluate to zero as you expected.

If you don't have the experience to be sure of this, then you can do a power series of the numerator and denominator to get insight, as follows.

[latex]\frac{x^2}{\exp (x)}=\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{6} + ...}=\frac{1}{\frac{1}{x^2}+\frac{1}{x}+1+\frac{x}{6}+ ...}[/latex]

You should then see that, in the limit as x goes to infinity, the denominator will go to infinity, while the numerator is finite, and hence the total ratio must go to zero.