Math Is Fun Forum

yscheok

Member

Registered: 2009-06-18

Posts: 7

Please help...

hi, any one can help me to solve below equation? i would like to know that what is the derivation for a and b.

y = a + jb = √[(R+jwL)(G+jwC)]

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Please help...

I'm not really sure what you mean. What does "derivation" mean in this context? What do each of the letters represent? Is j the imaginary unit? Is w complex or purely real? Do you want to find a and b in terms of the other letters?

Last edited by Daniel123 (2009-06-19 06:15:12)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi yscheok;

Does j = √ (-1) = i  that notation is used in electronics.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yscheok

Member

Registered: 2009-06-18

Posts: 7

Re: Please help...

Hi all,

thanks  for reply.
For J = i that notation is used in electronics. actually this is the transmission line equation and i would like to find the a and b equation. below is some step to get the a and b answer, i not sure whether is correct or not? can you all help me?

y = a + jb = √[(R+jwL)(G+jwC)]

y = √[RG + jwRC + jwLG + j²w²LC]

when G=0,

y = √[jwRC + j²w²LC]
  = √[j²w²LC  (R/jwL + 1)]
  = jw√LC *(R/2jwL + 1)
  = R√LC/2L + jw√LC

a= R√LC/2L  ,     b= w√LC

I have another question would like to check with you all regarding the Modular arithmetic. may i know that is the name for the "inverse Modular arithmetic" and what is the formula for it?

Last edited by yscheok (2009-06-24 00:21:55)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi yscheok;

Everything is OK up to here
=jw√LC *(R/2jwL + 1).

Please bracket what is in the square root. Do you just have √L ?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yscheok

Member

Registered: 2009-06-18

Posts: 7

Re: Please help...

hi bobbym,

sory for that, is √LC.

=jw√(LC) *(R/2jwL + 1)
= R√(LC)/2L + jw√LC

Last edited by yscheok (2009-06-24 13:52:43)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi

This mistake is here y =jw√(LC) *(R/2jwL + 1) is not true, it doesn't equal your y. Just substitute some numbers for R, C, L, w, j and you will agree.

As for your inverse modular arithmetic (modular inverse) go here first.
http://mathreference.com/num-mod,minv.html
It is done by computer.

Last edited by bobbym (2009-06-24 21:00:17)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yscheok

Member

Registered: 2009-06-18

Posts: 7

Re: Please help...

Hi, i think i make another mistake again,
how about as below?

y =jw√(LC) *(R/(2jwL) + 1)
y = R√(LC)/2L + jw√LC

if my equation is wrong, can you help me to solve this equation?

Last edited by yscheok (2009-06-24 23:28:27)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi yscheok;

I already tried that too, it is also wrong. You are making a mistake by pulling the j and w out of the radical.

Below is the last true statement. Work from here 1 step at at a time so I can see where you are going. I will assist with each step you make.

I would start by squaring the extreme left hand side and the right hand side to eliminate the radical.

Last edited by bobbym (2009-06-25 09:48:05)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yscheok

Member

Registered: 2009-06-18

Posts: 7

Re: Please help...

a + jb = √[j*w*R*C + j²*w²*L*C]
          = √[j²*w²*L*C*(R/(jwL)+1)]
          = √j²*√w²*√(L*C)* √[R/(jwL)+1]
          = j*w*√(L*C) * √[R/(jwL)+1]
          = after here i dont have any idea for next step

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi;

I worked on this for a while and this is the best I could do with it.

Square the right hand side.

Now you can say.

provided that you say

instead of 

This concession was the only way to obtain a and b by themselves on the left hand side.

Last edited by bobbym (2009-06-25 20:56:08)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

yscheok

Member

Registered: 2009-06-18

Posts: 7

Re: Please help...

hi bobbym, thanks for the answer.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Please help...

Hi yscheok;

  Remember that the solution is not unique, there are probably other forms. I  just thought that this one is the simplest.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.