SteveR
Active member
I'm having trouble re-arranging equations for RLC circuits into a form typically used for Bode plots.
For example, see NCEES practice exam problem #114. A series RLC circuit is given. I have no problem writing the I(s)= equation for this circuit:
I(s) =(50/s)/(10 + S*.005 + 5000/s)
But this form does not lend itself to simple interpretation for a Bode plot. The final form is:
I(s) = 10,000/(S+1000)^2
My problem is knowing what algebraic transformations I need to do to get from this:
I(s) = (50/s)/(10 + S*.005 + 5000/s)
To this:
I(s) = 10,000/(S+1000)^2
Reading their solution I am clear on what they did, however, that does not help me with other problems. When looking at the raw equation written directly fromt he schematic, what are the steps you should typically go through to get the equation into the more simplified form? I"m guessing it is something like "remove fraction from denominator/numerator, change decimals to whole numbers, factor, etc., but when I try actually doing it, I don't get the equations where they need to be to write a Bode plot.
For example, see NCEES practice exam problem #114. A series RLC circuit is given. I have no problem writing the I(s)= equation for this circuit:
I(s) =(50/s)/(10 + S*.005 + 5000/s)
But this form does not lend itself to simple interpretation for a Bode plot. The final form is:
I(s) = 10,000/(S+1000)^2
My problem is knowing what algebraic transformations I need to do to get from this:
I(s) = (50/s)/(10 + S*.005 + 5000/s)
To this:
I(s) = 10,000/(S+1000)^2
Reading their solution I am clear on what they did, however, that does not help me with other problems. When looking at the raw equation written directly fromt he schematic, what are the steps you should typically go through to get the equation into the more simplified form? I"m guessing it is something like "remove fraction from denominator/numerator, change decimals to whole numbers, factor, etc., but when I try actually doing it, I don't get the equations where they need to be to write a Bode plot.