Amplifier gain in resonant RLC circuit

[Meow12]

Homework Statement

A resonant RLC circuit can be used as an amplifier for a certain band of frequencies around the resonant frequency. Consider a series RLC circuit as an audio band amplifier with an AC voltage source as the input, and the voltage across the 8.0 Ω resistor as the output. The amplifier should have a gain (=output/input) of 0.5 at 200Hz and 4000Hz. What is the required value of the inductor in Henry's? What is the required value of the capacitor in Farad's?

Relevant Equations

At resonance, $\displaystyle\omega_0=\frac{1}{\sqrt{LC}}$
$\omega=2\pi f$
$\displaystyle A_V=\frac{V_{output}}{V_{input}}$

Amplifier gain $A_V$ is defined as the ratio of an amplifier's output voltage to its input voltage,
i.e. $\displaystyle\frac{V_R}{V}=\frac{IR}{IZ}=\frac{R}{R}=0.5$ at 200 Hz.

But this is absurd. Where have I gone wrong? Please nudge me in the right direction.

 

[Meow12]

I think I got it!

$R/Z_1=0.5$ at $\omega=2\pi\cdot 200$

$R/Z_2=0.5$ at $\omega=2\pi\cdot 4000$

I have 2 equations and 2 unknowns L and C. (R is given.)

 

[Meow12]

Hmm....solving those 2 equations is harder than I expected.

The first equation says that $Z_1=R/0.5=2R$ at $\omega=400\pi$ rad/s

$R^2+(X_L-X_C)^2=4R^2$ at $\omega=400\pi$ rad/s

$(X_L-X_C)^2=3R^2=192$ at $\omega=400\pi$ rad/s

$\displaystyle\left(\omega L-\frac{1}{\omega C}\right)^2=192$ at $\omega=400\pi$ rad/s

$\displaystyle\left(400\pi L-\frac{1}{400\pi C}\right)^2=192$

$\displaystyle 400\pi L-\frac{1}{400\pi C}=\pm 13.856$

Similarly, from the second equation,

$\displaystyle 8000\pi L-\frac{1}{8000\pi C}=\pm 13.856$

But I can't decide which is + and which is -.

Please help!

 

[BvU]

So you get four answers ? All with positive $\omega$?

[edit] I mean sensible $L$, $C$ ?

 

[Meow12]

$\displaystyle 8000\pi L-\frac{1}{8000\pi C}=\pm 13.856$

Let us multiply the above equation by 20. We get

$\displaystyle 160000\pi L-\frac{1}{400\pi C}=\pm 13.856\times 20$ ----------- (1)

The other equation is

$\displaystyle 400\pi L-\frac{1}{400\pi C}=\pm 13.856$ ----------- (2)

Subtracting equation (2) from (1), we get $159600\pi L$ on the left-hand side.

Note that since the left-hand side is positive, the right-hand side also must be positive. Thus we may pick the positive sign in equation (1). $400\pi C$ is going to be extremely small. So, $\displaystyle\frac{1}{400\pi C}$ is going to be extremely large--larger than $400\pi L$. So, we pick the negative sign in equation (2).

$159600\pi L=13.856\times 20-(-13.856)=13.856\times 21$

$L=5.80E-4$ H

Substituting this value of $L$ in one of our equations, we get $C=5.46E-5$ F

Both these values match the ones given by our prof. :)

 

[BvU]

There is so much good stuff in this exercise, I can't help but add a few comments

The 'gain' expression for this circuit $$A_V = {R\over \sqrt{R^2 +\left (\omega L-{1\over \omega C}\right)^2}}$$simplifies to $$\begin{align*}
A_V &= \omega RC \quad &\text{for}\quad \omega << \omega_0 \\
A_V &= R/(\omega L) \quad &\text{for}\quad \omega >> \omega_0
\end{align*} $$as a plot of $A$ vs frequency shows (log-log plot so that approximations show up as straight lines; also: $\omega_0 = \sqrt{\omega_2\omega_1} \Rightarrow f_0= 894 Hz$).

(Blue horizontal line for $A_V = 0.5$)

In the exercise circuit, damping is considerable ($\omega_2-\omega_1 > \omega_0$ -- I get $\alpha = 7332\ (1167 \ {\sf\text{Hz}}), \ \zeta = 1.3 )$ for the points where $A={1\over 2}\sqrt 2$.

$\$