Hey! I need to solve the following problem:(adsbygoogle = window.adsbygoogle || []).push({});

a) Show the trap filter below acts to reject signals at a frequency

[tex] \omega = \frac{1}{\sqrt{LC}} [/tex]

http://img418.imageshack.us/img418/8168/rlc2wf.gif [Broken]

b) How does the width of the frequency band rejected depend on the resistance R?

For a) I thought that at resonance (when

[tex] \omega = \frac{1}{\sqrt{LC}} [/tex] )

the reactance of L & C = 0 and therefore Z= R. In a series circuit the current is the same everywhere, but the voltage divides itself according to V= IR. Therefore the voltage on L & C is 0 (R=0) and the voltage on R= IR. But is that correct? I think Vout is measured on R and NOT L & C, but how would I know? Btw what is the function of that grounded symbol, what does it mean and is it relevant for this question?

As for b) I know that

[tex] Q= \frac{\omega_0}{\Delta \omega} [/tex]

so

[tex] \Delta \omega = \frac{\omega_0}{Q} [/tex]

with

[tex] Q= \frac{\omega_0 L}{R} [/tex]

Substitution gives:

[tex] \Delta \omega = \frac{\omega_0}{\frac{\omega_0 L}{R}} [/tex]

[tex] =\frac{R}{L}[/tex]

[tex] \Delta f= \frac{\Delta \omega}{2 \pi}[/tex]

[tex] = \frac{\frac{R}{L}}{2 \pi} [/tex]

[tex]= \frac{R}{2 \pi L}[/tex]

Though my textbook says that

[tex] \Delta \omega = \frac{R}{2L} [/tex]

Could anybody please tell me what I'm doing wrong?!

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# Homework Help: Series RLC filter (trap filter)

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