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#1 Re: Help Me ! » Please help... » 2009-06-28 16:42:43
hi bobbym, thanks for the answer.
#2 Re: Help Me ! » Please help... » 2009-06-25 18:10:56
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 
#3 Re: Help Me ! » Please help... » 2009-06-24 23:26:02
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?
#4 Help Me ! » Inverse modular arithmetic? » 2009-06-24 15:51:57
- yscheok
- Replies: 1
Hi all,
normally dθ/dt is the inverse of ∫θ dt,
what is the inverse name for modular arithmetic?
#5 Re: Help Me ! » Please help... » 2009-06-24 13:50:30
hi bobbym,
sory for that, is √LC. 
=jw√(LC) *(R/2jwL + 1)
= R√(LC)/2L + jw√LC
#6 Re: Help Me ! » Please help... » 2009-06-24 00:09:24
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?
#7 Help Me ! » Please help... » 2009-06-18 21:03:18
- yscheok
- Replies: 12
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)]
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