- #1
Rectifier
Gold Member
- 313
- 4
Hello!
This is problem is a part of a bigger problem which I solved and came to a formula which is correct.
This is the equation for one transfer function. The next thing I would like to find out here is when arg(H(jw)) is -90 or 90 degrees but I get stuck.
Transfer function:
## H(jw) = \frac{R}{R(1-w^2LC)+jwL} ##This is how I proceeded till I got stuck.
## H(jw) = \frac{R}{R(1-w^2LC)+jwL} \\ H(jw) = \frac{R}{\sqrt{(R(1-w^2LC))^2+(wL)^2}e^{jarctan( \frac{wL}{R(1-w^2LC)})}} \\ H(jw) = \frac{ R }{ \sqrt{ (R(1-w^2LC))^2+(wL)^2} } e^{-jarctan( \frac{wL}{R(1-w^2LC)})} \\ ##
Then we want to know where the argument is -90 or 90 degrees.
## 90=-jarctan( \frac{wL}{R(1-w^2LC)}) ##
Here is the step where I get stuck. Could you please help me out?
Thanks in advance!
This is problem is a part of a bigger problem which I solved and came to a formula which is correct.
This is the equation for one transfer function. The next thing I would like to find out here is when arg(H(jw)) is -90 or 90 degrees but I get stuck.
Transfer function:
## H(jw) = \frac{R}{R(1-w^2LC)+jwL} ##This is how I proceeded till I got stuck.
## H(jw) = \frac{R}{R(1-w^2LC)+jwL} \\ H(jw) = \frac{R}{\sqrt{(R(1-w^2LC))^2+(wL)^2}e^{jarctan( \frac{wL}{R(1-w^2LC)})}} \\ H(jw) = \frac{ R }{ \sqrt{ (R(1-w^2LC))^2+(wL)^2} } e^{-jarctan( \frac{wL}{R(1-w^2LC)})} \\ ##
Then we want to know where the argument is -90 or 90 degrees.
## 90=-jarctan( \frac{wL}{R(1-w^2LC)}) ##
Here is the step where I get stuck. Could you please help me out?
Thanks in advance!