Transfer function of this simple circuit

  • #1
129
0

Homework Statement



The picture of the circuit is attached; I want to find [itex]|V_{A}/V_{J}|[/itex]. This seems really easy but I haven't done circuit analysis in forever.

Homework Equations



Complex impedances, [itex]Z_{C} = 1/i\omega C, Z_{R} = R[/itex].

The Attempt at a Solution



First [itex]R_{A}[/itex] and [itex]C_{A}[/itex] are in parallel, so the equivalent impedance is [itex]Z_{A} = Z_{C_{A}} Z_{R_{A}} /(Z_{C_{A}} + Z_{R_{A}})[/itex]. Then the circuit is a voltage divider, so the voltage across [itex]Z_{A}[/itex] is [itex]V_{A} = V_{J} Z_{A}/(Z_{A} + Z_{R_{J}})[/itex]. Therefore,

$$
\frac{V_{A}}{V_{J}} = \frac{Z_{A}}{Z_{A} + Z_{R_{J}}} = \frac{i R_{A}}{i(R_{A} + R_{J}) - C_{A} R_{A} R_{J} \omega} \implies \left| \frac{V_{A}}{V_{J}} \right| = \frac{R_{A}}{\sqrt{(R_{A} + R_{J})^2 + (C_{A} R_{A} R_{J} \omega)^2}}
$$
 

Attachments

  • circuit.png
    circuit.png
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Answers and Replies

  • #2
gneill
Mentor
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Looks good.
 
  • #3
129
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Okey dokey, thank you!
 

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