# RLC bandpass filter

I am working on a circuit:

simulate this circuit – Schematic created using CircuitLab

My goal is to find $$\ \omega_0, \beta, \$$ and $$\\varrho \$$

The way I approached this problem was to first find the transfer function $$\\frac{V_{out}}{V_{in}}\$$.

KCL

• $$\I_1=I_2+I_3+I_4\$$

EFC

• $$\I_1=\frac{V_{in}-V{out}}{100000}\$$

• $$\I_2=\frac{V_{out}}{\frac{1}{s\cdot c}}\$$

• $$\I_3=\frac{V_{out}}{s \cdot l}\$$

• $$\I_4=\frac{V_{out}}{400000}\$$

• $$\H(s)=\frac{V_{out}}{V_{in}}\$$

My result for the transfer function is as follows:

$$\\mathcal{H}(s)=\frac{50000 \cdot s}{s^2+62500 \cdot s + 1000000000000}\tag1\$$

My approach from here is to 1. Take the magnitude of this function, then find $$\\omega_c\$$ by substituting $$\j\omega\$$ in for s. From here set the magnitude of the transfer function equal to $$\\frac{\mathcal{H}_{max}}{\sqrt{2}}\$$ Once I have the cutoff equations then I can find $$\\beta\$$

I've tried the following

$$\| \mathcal{H}(s)| = \frac{\sqrt{(50000 \cdot w)^2}}{\sqrt{(62500 \cdot w)^2 + (-w^2 + 1000000000000)^2}}\tag2\$$

$$\mathcal{H}_{max} = \frac{R_L}{R+R_L}$$

Therefore,

$$\frac{\sqrt{(50000 \cdot w)^2}}{\sqrt{(62500 \cdot w)^2 + (-w^2 + 1000000000000)^2}}=\frac{\frac{R_L}{R+R_L}}{\sqrt{2}}$$

$$\omega = 31250 \cdot (5 \cdot \sqrt{41}\pm1)\tag3$$

I can find $$\\omega_0\$$ by taking the derivative of the magnitude =0 with respect to $$\\omega\$$

• Your transfer function is not correct. You have a parallel RLC circuit (band-reject filter), but you have the transfer function of a series RLC circuit (bandpass filter). Nov 12, 2018 at 20:57
• @sean M Strange I found the transfer function by solving the circuit, then dividing $\frac{V_{out}}{V_{in}}$. I edited my original question to show this. I am not sure how to add current to my circuit through the simulator Nov 12, 2018 at 21:06
• @Sean M When I use that method, I get $\frac{20000000000 \cdot s}{s^2+20080000000 \cdot s + 1000000000000}$ as my transfer function Nov 12, 2018 at 21:12
• I've moved my comment to the answers. Nov 12, 2018 at 21:15
• You seem to be trying to find $\omega_0$ in the longest way possible. Remember that the critical frequency is where the magnitude of the impedances of the capacitor and inductor are equal to each other, and then you should be able to get the critical frequency using a single formula. Nov 13, 2018 at 1:26

There is one way to get this transfer function right without pain and multiple iterations: use the fast analytical circuits techniques or FACTs as described in my book. By determining the time constants of this circuits with a zeroed excitation when the energy-storing elements are set in their dc or high-frequency states, the transfer function comes easily as shown below:

The formula below the sketch represents the brute-force approach while the final expression $$\H_{10}\$$ is the formula you want. In this low-entropy form, you see the peaking magnitude at nearly -2 dB and only by rearranging under this form you can access this peak. You can certainly determine this expression using nodal approach as you've tried but the FACTs as shown in the picture lead me to the factored form without writing a single line of algebra, just inspecting the circuit.

The easiest way to get the transfer function is realizing that the circuit is essentially a voltage divider, $$V_{out} = V_{in}\frac{Z}{R+Z}$$ and just solving for $$\frac{V_{out}}{V_{in}}=\frac{Z}{R+Z}$$

where Z is the impedance of your RLC combination. Once you have that completed, you would plug in your "cutoff" frequency, and reduce your transfer function to a manageable form of $$H(s)=\alpha+s\beta$$ which is possible.

Hint: $$s^2=-\omega^2$$

• Thank you. I have gotten a new transfer function using the voltage divider $\frac{20000000000 \cdot s}{s^2+20080000000 \cdot s + 1000000000000}$ Nov 12, 2018 at 21:15
• I see, now I should plug $\omega$ in for s and $$s^2=-\omega^2$$ Nov 12, 2018 at 21:23
• That's still not correct. I suspect you're summing the RLC impedances as if they are in series, but you need to sum them in parallel. And $s = j\omega$. Nov 12, 2018 at 21:44
• my original transfer function is correct. Nov 12, 2018 at 22:16
• You're right, I was mixing up my transfer functions. That's what I get for trying to answer questions and being at my job at the same time I suppose. Nov 12, 2018 at 22:50