Quick math question

[TecHNooB]

Just need an explanation of this result.

x(t) = 2e^(-3t) u(t - 1)

dx(t)/dt = -6e^(-3t)u(t - 1) + 2 dirac(t - 1)

I don't get where the diract term comes from. If I do laplace transforms, the dirac term shows up. But I don't know how to get it in the time domain.

 

[2Xtreme21]

 

[Ticky]

Use multiplication rule for derivatives. d(u*v)=du*v+u*dv.

Edit: And remember what int(f(x)*dirac(x)) equals (f(0)).

 

[SmoochyTX]

Originally posted by: TecHNooB
Just need an explanation of this result.

x(t) = 2e^(-3t) u(t - 1)

dx(t)/dt = -6e^(-3t)u(t - 1) + 2 dirac(t - 1)

I don't get where the diract term comes from. If I do laplace transforms, the dirac term shows up. But I don't know how to get it in the time domain.

Instead of making multiple math threads each night, why don't you just create your own sole math thread????



ETA - If you're the same dude trying to do homework for your kid, then this applies. If it's you needing homework help for yourself try posting in the other guy's thread. LMAO

 

[TecHNooB]

Originally posted by: Ticky
Use multiplication rule for derivatives. d(u*v)=du*v+u*dv.

Edit: And remember what f(x)*dirac(x) equals.

So I would get..

-6e^(-3t) u(t - 1) + 2e^(-3t) dirac (t - 1) where 2e^(-3t) is evaluated at t = 1. I don't get the same result.

 

[Ticky]

Evaluate at t=0.

Edit: N/m that. One second.

 

[TecHNooB]

Originally posted by: SmoochyTX

Originally posted by: TecHNooB
Just need an explanation of this result.

x(t) = 2e^(-3t) u(t - 1)

dx(t)/dt = -6e^(-3t)u(t - 1) + 2 dirac(t - 1)

I don't get where the diract term comes from. If I do laplace transforms, the dirac term shows up. But I don't know how to get it in the time domain.

Instead of making multiple math threads each night, why don't you just create your own sole math thread????



ETA - If you're the same dude trying to do homework for your kid, then this applies. If it's you needing homework help for yourself try posting in the other guy's thread. LMAO

There's a dude doing homework for his kid? Thread?

 

[Ticky]

Originally posted by: TecHNooB

Originally posted by: SmoochyTX

Originally posted by: TecHNooB
Just need an explanation of this result.

x(t) = 2e^(-3t) u(t - 1)

dx(t)/dt = -6e^(-3t)u(t - 1) + 2 dirac(t - 1)

I don't get where the diract term comes from. If I do laplace transforms, the dirac term shows up. But I don't know how to get it in the time domain.

Instead of making multiple math threads each night, why don't you just create your own sole math thread????



ETA - If you're the same dude trying to do homework for your kid, then this applies. If it's you needing homework help for yourself try posting in the other guy's thread. LMAO

There's a dude doing homework for his kid? Thread?

And what kid would be taking a math class on this level?

 

[Ticky]

I can explain why the answer is that.

Think about it. One the right side, the derivative is just the derivative of the exponential. This is the first term. The second term represents the derivative right at the discontinuity. The derivative there is infinite, hence the dirac funcation. The 2 is just the overall coeficient. Now to explain that with math....

Edit: Oh, and I do think there should be a factor of e^(-3) in there.

 

[Renob]

Stay away from it its bad bad news, you get hooked fast!

Oh wait thought you said "meth" when you said "math" nevermind!!!

 

[TecHNooB]

Originally posted by: Ticky
I can explain why the answer is that.

Think about it. One the right side, the derivative is just the derivative of the exponential. This is the first term. The second term represents the derivative right at the discontinuity. The derivative there is infinite, hence the dirac funcation. The 2 is just the overall coeficient. Now to explain that with math....

Edit: Oh, and I do think there should be a factor of e^(-3) in there.

No, the laplace transform gives me the answer I stated originally. I just don't like having to perform laplace transforms in problems that deal with convolution. I should be able to get the answer while staying in the time domain.

 

[dighn]

just know that d/dt of u(t) = dirac_delta(t). the reason is that int(dirac_delta(t)*dt) around zero gives 1.

by chain rule you have

dx(t)/dt = -6e^(-3t)u(t - 1) + 2e^(-3t)* dirac(t - 1)
however since dirac delta is zero at every where except 0 (t = 1), you need only evaluate 2e^-3t at 1 = 2e^-3

are you sure it's 2*dirac(t-1)?

edit: actually I'm not sure about applying the chain rule around a discontinuity like that. but I agree with Ticky that the factor should be there. basically integrating around -1, +1, you should get e^-3 as that's what the orignal function evaluates to at t = 1 where it was 0 just an instant before.

 

[Ticky]

Alright, I just did it via Laplace, and I got a factor of e^-3. I can try to type up my Laplace work if you want....

The critical step is to remember you need f(t-a)u(t-a), yielding e^(-as)*F(s). So you need to add and subtract 3 in the exponential so you get the correct form.

 

[TecHNooB]

Originally posted by: Ticky
Alright, I just did it via Laplace, and I got a factor of e^-3. I can try to type up my Laplace work if you want....

The critical step is to remember you need f(t-a)u(t-a), yielding e^(-as)*F(s). So you need to add and subtract 3 in the exponential so you get the correct form.

Oh yea.. I forgot to time shift the laplace. In that case, the solution manual is wrong. Thanks Ticky

 

[Ticky]

Originally posted by: TecHNooB

Originally posted by: Ticky
Alright, I just did it via Laplace, and I got a factor of e^-3. I can try to type up my Laplace work if you want....

The critical step is to remember you need f(t-a)u(t-a), yielding e^(-as)*F(s). So you need to add and subtract 3 in the exponential so you get the correct form.

Oh yea.. I forgot to time shift the laplace. In that case, the solution manual is wrong. Thanks Ticky

Sure. I really FRACKING hate it when the solutions manual is wrong.

 

[TecHNooB]

Originally posted by: Ticky

Originally posted by: TecHNooB

Originally posted by: Ticky
Alright, I just did it via Laplace, and I got a factor of e^-3. I can try to type up my Laplace work if you want....

The critical step is to remember you need f(t-a)u(t-a), yielding e^(-as)*F(s). So you need to add and subtract 3 in the exponential so you get the correct form.

Oh yea.. I forgot to time shift the laplace. In that case, the solution manual is wrong. Thanks Ticky

Sure. I really FRACKING hate it when the solutions manual is wrong.

Actually, I may be wrong again.. 1 sec.

OK

X(S) = 2 e^-s / (s + 3)

dx(t)/dt = sX(s) - x(0) = 2s e^-s / (s + 3) - 0 = [2 - 6 / (s + 3)] e^-s
= 2 e^-s - 6 e^-s / (s + 3)

If there is no / then assume multiplication, because * is now reserved for convolution

 

[Ticky]

Originally posted by: TecHNooB

Originally posted by: Ticky

Originally posted by: TecHNooB

Originally posted by: Ticky
Alright, I just did it via Laplace, and I got a factor of e^-3. I can try to type up my Laplace work if you want....

The critical step is to remember you need f(t-a)u(t-a), yielding e^(-as)*F(s). So you need to add and subtract 3 in the exponential so you get the correct form.

Oh yea.. I forgot to time shift the laplace. In that case, the solution manual is wrong. Thanks Ticky

Sure. I really FRACKING hate it when the solutions manual is wrong.

Actually, I may be wrong again.. 1 sec.

OK

X(S) = 2 e^-s / (s + 3)

dx(t)/dt = sX(s) - x(0) = 2s e^-s / (s + 3) - 0 = [2 - 6 / (s + 3)] e^-s
= 2 e^-s - 6 e^-s / (s + 3)

If there is no / then assume multiplication, because * is now reserved for convolution

You lost your e^(-3). Remember, you need to rewrite 2e^(-3) u(t-1) as 2e^(-3) e^(-3(t-1)) u(t-1) as the very first step.


Edit: Convolution is for sissies.

 

[Ticky]

Alright, hold on, I'm installing the symbolic math toolbar in Matlab, and then I can double check the answer.

 

[Born2bwire]

Originally posted by: Ticky
Alright, hold on, I'm installing the symbolic math toolbar in Matlab, and then I can double check the answer.

Maple FTW! I'm still pissed that the 64 bit version doesn't have the symbolic toolkit. I have to ssh into our group's servers back in the US to use them.

 

[Ticky]

Ok.

Matlab says the answer is:

(2*dirac(x - 1))/exp(3*x) - (6*heaviside(x - 1))/exp(3*x)

However, I imagine that matlab doesn't know how to handle that first term. I think we're agreed that that simplifies to 2 dirac(x-1) e^(-3)

 

[Ticky]

Originally posted by: Born2bwire

Originally posted by: Ticky
Alright, hold on, I'm installing the symbolic math toolbar in Matlab, and then I can double check the answer.

Maple FTW! I'm still pissed that the 64 bit version doesn't have the symbolic toolkit. I have to ssh into our group's servers back in the US to use them.

Yeah, I had to install x32 to use it.

 

[Born2bwire]

Originally posted by: Ticky
Ok.

Matlab says the answer is:

(2*dirac(x - 1))/exp(3*x) - (6*heaviside(x - 1))/exp(3*x)

However, I imagine that matlab doesn't know how to handle that first term. I think we're agreed that that simplifies to 2 dirac(x-1) e^(-3)

That answer looks ok to me.

 

[sao123]

just buy a TI 89 titanium...

 

[Ticky]

Originally posted by: sao123
just buy a TI 89 titanium...

Is there a way to enter heaviside in a TI?