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It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.NaS4 said:Thx man. I’ll look up dimensional analysis
It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.NaS4 said:Thx man. I’ll look up dimensional analysis
DaveE said:It can get scary. Look for the useful bits. If they get too abstract (math, philosophy and such) that may a good place to stop, or not, if you like that stuff.
Yeah the w = 2(pi)f but since the question tells us that w is 200Master1022 said:Just a quick update, I have checked my expression:
$$ Z = \left( \frac{\omega^4 R L^2 C^2}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) + j \left( \frac{\omega L (1 - \omega^2 LC + (\omega R C)^2)}{(1-\omega^2 LC)^2 + (\omega CR)^2} \right) $$
and think it is equal to yours (thus we are either both correct... or both wrong!)
Quick question about the problem statement: are the units of ## \omega ## a typo when it says Hz? Usually we expect ## \omega ## to be in rad/s (a conversion factor of ## 2 \pi ## is needed).
Am doing the calculations in a Python notebook, so will post pictures here to show how my calculations are resulting
NaS4 said:Yeah the w = 2(pi)f but since the question tells us that w is 200
Master1022 I got nothing but respect for you man. Really makes me happy to see people like you helping out people like meMaster1022 said:Ah okay, that explains the discrepancy I had earlier. After setting ## \omega = 200 ##, I did get the same value of the gain and the final voltage as you (give or take some rounding errors) - the computer output ## 11551.24... ## (V), so this does agree with your result.
Hope that is of some help.
It's very common for people to be sloppy about that. I guess they assume you'll know about the 2π correction.Master1022 said:Quick question about the problem statement: are the units of ω a typo when it says Hz? Usually we expect ω to be in rad/s (a conversion factor of 2π is needed).
Happy to help, hopefully that answer is correct. One other sanity check is to see whether you can find a formula for the damping ratio of an RLC circuit with this setup and see whether it is very small (which is the only way such a large value of the gain is possible). I did have a quick glance on wikipedia, but I cannot see any similar variation. Regardless of the numerical values of the question, you did get the overall method correct (finding the transfer function and then calculating the gain).NaS4 said:Master1022 I got nothing but respect for you man. Really makes me happy to see people like you helping out people like me
Agreed! (I also prefer working in the ##s##-domain)DaveE said:The canonical form for this particular LCR impedance is:
Z(s) = sL⋅(1+sRC)/(1+sRC+s2LC), where s=jω (sorry, I have habits that are hard to break!)
Yes, that is correct. This leads toDaveE said:Then if you define some common features: ω=1/sqrt(LC), Zo=sqrt(L/C), Q=1/(2ζ)=Zo/R
The quadratic term looks like 1+(1/Q)(s/ω)+(s/ω)2.