- #1

- 2

- 0

how to get this equation:

Vout/Vin = [RL]/[-[ω][/2]+jω(R/L)+1/(LC)]

Vout/Vin = [RL]/[-[ω][/2]+jω(R/L)+1/(LC)]

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- Thread starter Ai En
- Start date

- #1

- 2

- 0

how to get this equation:

Vout/Vin = [RL]/[-[ω][/2]+jω(R/L)+1/(LC)]

Vout/Vin = [RL]/[-[ω][/2]+jω(R/L)+1/(LC)]

- #2

- 35

- 0

- Tell us how Vout is defined.

- Tell us what [omega][/2] stands for (use the boards LaTeX syntax).

- #3

- 2

- 0

http://upload.wikimedia.org/wikipedia/en/1/14/RLC_series_band-pass.svg

- #4

- 35

- 0

as a solution you could use the voltage divider formula to get a first equation:

[tex]\frac{V_{out}}{V_{in}} = \frac{R_L}{R_L + j \omega L + \frac{1}{j \omega C}}[/tex]

afterwards if you think thats useful you can try to manipulate this formula until you get the one given to you. Since i can not interpret your given equation (as i said use the LaTeX encoding to be more clear about your formula) i can only guess and did for example multiply the whole formula with [itex]\frac{j \omega L}{j \omega L}[/itex] and the denominator again with [itex]\frac{L}{L}[/itex] that results in:

[tex]\frac{V_{out}}{V_{in}} = \frac{j \omega R_L}{L \cdot \left( - \omega^2 + j \omega \frac{R_L}{L} + \frac{1}{C L}\right)}[/tex]

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