Help Finding Transfer Function Vo/Vs

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1. Nov 30, 2014

Captain1024

1. The problem statement, all variables and given/known data

Find $a)\ H(s)=\frac{V_o}{V_s}$ of the filter in the circuit below.
$b)\$The center frequency $\omega_0$
$c)\$The band-width B

$a)\ H(s)=\frac{6s}{12s^2+11s+1}$
$b)\ \omega_0=0.28\frac{rad}{s}$
$c)\ B=0.71\frac{rad}{s}$

2. Relevant equations

Ohm's law: $V=IR$
$s=j\omega$
For series resonance:
$\omega_0=\sqrt{\omega_1\omega_2}$
$B=\omega_1-\omega_2$
$\omega_1=-\frac{R}{2L}+\sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}\$, Where R is in Ohm's, L in Henry's, C in Farads
$\omega_2=-\omega_1$

3. The attempt at a solution

Part $a)\$I used KVL to find currents $I_1\$(Left loop) and $I_2\$(Right loop)
$C_2=2F\ \Rightarrow\ \frac{1}{2s}$
$C_1=3F\ \Rightarrow\ \frac{1}{3s}$
Left loop: $V_s=(1+\frac{1}{2s})I_1-\frac{1}{2s}I_2\$ (Eqn. 1)
Right loop: $\frac{-1}{3s}I_2-2I_2+\frac{1}{2s}I_1=0$
$\Rightarrow\ -(\frac{1}{3s}+2)I_2=\frac{-1}{2s}I_1$
$\Rightarrow\ I_1=2s(\frac{1}{3s}+2)I_2$
$\Rightarrow\ I_1=(\frac{2}{3}+4s)I_2\$ (Eqn. 2)
Plugging (Eqn. 2) into (Eqn. 1): $V_s=(1+\frac{1}{2s})(\frac{2}{3}+4s)I_2-\frac{1}{2s}I_2$

Distributing and solving for $I_2$:
$I_2=\frac{V_s}{\frac{8}{3}+4s+\frac{1}{3s}-\frac{1}{2s}}$

Now, $V_0$ using Ohm's Law is $V_0=2I_2$
Therefore:
$H(s)=\frac{V_o}{V_s}=\frac{2}{\frac{8}{3}+4s+\frac{1}{3s}-\frac{1}{2s}}$

Multiplying by $\frac{s}{s}$:
$H(S)=\frac{2s}{\frac{8s}{3}+4s^2+\frac{1}{3}-\frac{1}{2}}$

Simplifying:
$H(s)=\frac{2s}{4s^2+\frac{8s}{3}-\frac{1}{6}}$

Parts $b)\$& $c)$
Two issues: 1) Circuit is not solely series or parallel 2) There is no inductor
1) I can handle series and parallel RLC circuits, but this circuit is mixed. Can I turn this into a series or parallel circuit?
2) My equations for center frequency and band-width involve an inductor value and I'm assuming I shouldn't just ignore it.

Any assistance is greatly appreciated.

2. Nov 30, 2014

Staff: Mentor

For your right loop equation, note that I2 also flows through the 2F capacitor. Your equation doesn't reflect this, it only shows I1 flowing through it.

Note that your circuit comprises two elementary RC filters, one after the other: A low pass filter followed by a high pass filter. There are no inductors in the circuit so you can throw away any LC formulas for this one.

Since the filters have different component values they will have different corner frequencies, so a bode plot should show a "hump" where the gain will be maximum (or rather the attenuation the least), and heading to large negative dB values on either side. Note that the filters are not isolated from each other so there will be interaction of the component values and these corner frequencies will be displaced from their "unloaded" positions.

I'd start by finding the frequency where the gain is maximum (attenuation least). That should address part (b). Then contemplate what "band width" means in the context of the overall filter, and how you might determine it.

Last edited: Dec 1, 2014
3. Dec 1, 2014

Captain1024

I reworked Part a) with the following loop equations:
Left loop: $V_s=(1+\frac{1}{2s})I_1-\frac{1}{2s}I_2$
Right loop: $(\frac{-1}{2s}-\frac{1}{3s}-2)I_2+\frac{1}{2s}I_1=0$
$\Rightarrow\ I_1=(4s+\frac{5}{3})I_2$
Then $I_2=\frac{V_s}{4s+\frac{5}{6s}+\frac{11}{3}}$

So $H(s)=\frac{2s}{4s^2+\frac{11s}{13}+\frac{5}{6}}$

Still not correct, although I think I'm close. Where am I going wrong?

4. Dec 1, 2014

Staff: Mentor

Something went wrong in your working out of I2. Your expression for I1 looks fine.

5. Dec 1, 2014

Captain1024

Don't know how I missed it.

$I_2=\frac{V_s}{4s+\frac{1}{3s}+\frac{11}{3}}$

That led me to the correct answer for part a)

Should I use the transfer function to accomplish this?

6. Dec 1, 2014

Staff: Mentor

Yes. More specifically, you want to find the frequency which maximizes the magnitude of the transfer function.

7. Dec 1, 2014

Captain1024

H(s) would approach infinity as s approaches the poles, right? Poles $P=\frac{1}{24}(-11\pm\sqrt{73})$

8. Dec 1, 2014

Staff: Mentor

Those are both negative values, so the transfer function would go to negative infinity (due to the s in the numerator). But that's in the s domain.

You're looking for a value of $\omega$. So substitute $j \omega$ for s. Then find the value of $\omega$ that maximizes it.

9. Dec 1, 2014

Captain1024

$H(\omega)=\frac{2j\omega}{-4\omega^2+\frac{11j\omega}{3}+\frac{1}{3}}$

Should I then multiply by the conjugate to have no imaginary terms in the denominator? If so, would the conjugate be:
$-4\omega^2-\frac{11j\omega}{3}+\frac{1}{3}$

10. Dec 1, 2014

Staff: Mentor

Sure.

11. Dec 1, 2014

Captain1024

$H(\omega)=\frac{-8j\omega^3+\frac{22}{3}\omega^2+\frac{2}{3}j\omega}{16\omega^4+\frac{97}{9}\omega^2+\frac{1}{9}}$

WolframAlpha tells me that the poles $P=\pm[\frac{1}{12}i\sqrt{\frac{1}{2}(97\pm11\sqrt{73})}\ ]$

This yields two positive imaginary poles: 0.1023i & 0.8143i (approx.)
Are one of these my center frequency?

12. Dec 2, 2014

Staff: Mentor

ω is a positive real number (and angular frequency). So imaginary values should tell you that something's not right. The Bode plot is going to rise to a maximum value and then descend again (it's a hump-shaped curve). So there should be a single maximum in the middle of the hump.

In your first post you listed the "Correct answers", so you should be able to tell when you get it right.

$$H(s) = \frac{6 s}{12 s^2 + 11 s + 1}$$