# Finding upper cutoff and lower cutoff frequencies on notch circuit

I need to find upper cutoff (UCF) and lower cutoff (LCF) frequencies based on formulas of a notch circuit, but I'm getting a quadratic equation (as expected) with an imaginary number (IN; not expected.)

I tried to solve this equation transforming the IN to polar coordinates then cutting off the angle (which gave me f'=2MHz and f''=2,57MHz, values very far from what was expected - something around fo.)

How can I find these frequencies?

This is the circuit I've been working on.

This is the formula I used to find UCF and LCF. (Which are represented by f in the formula.)

Some values I have already found:

• Zp:21256ohms
• Rp:123000ohm
• Qp:70ohm
• fo=2,269MHz

The roots of the formula fo will be the lower cutoff frequency and the upper cutoff frequency.

I also tried to solve this by calculating the bandwidth (BW), then

• UFC=fo+BW/2
• LFC=fo+BW/2 Other formulas I used can be found on this photo:

• You need to ask a question and not repeat what you might write in your diary. This is a Q and A site and you need to concentrate on the Q bit. Commented Oct 30, 2020 at 23:05
• It looks ok but far better to make a 4th order notch at this Q. falstad.com/afilter/… Commented Oct 30, 2020 at 23:14
• Change any parameter you wish. Http://www.falstad.com/afilter/… Commented Oct 30, 2020 at 23:20
• Keep in mind Q>=100 makes LC tolerances impossible to achieve in 2nd order, and easier in higher order filters, also keep in mind the Falstad screen resolution is a clue to change spectrum BW and filter design sensitivity, these are emitter follower or Op Amp driven with any load.you cannot buy /afford 0.1% tolerance LC parts Commented Oct 30, 2020 at 23:23
• Do you know to find the upper and lower cutoff frequencies with the formula given? Commented Oct 31, 2020 at 0:10

If it's about calculating the bandwidth, it would be better to calculate the transfer function instead of relying on pre-made formulas:

\begin{align} Z_{ser}(s)&=R_{ser}+sL\tag{1} \\ Z_{par}(s)&=\frac{1}{\frac{1}{Z_{ser}}+sC}\tag{2} \\ H(s)&=\frac{R}{R+Z_{par}(s)} \\ &\Rightarrow \\ H(s)&=\frac{RLCs^2+RR_{ser}Cs+R}{RLCs^2+(RR_{ser}C+L)s+R_{ser}+R} \\ {}&=\frac{s^2+\frac{R_{ser}}{L}s+\frac{1}{LC}}{s^2+\left(\frac{R_{ser}}{L}+\frac{1}{RC}\right)s+\frac{R_{ser}+R}{R}\frac{1}{LC}}\tag{3} \end{align}

This gives a transfer function in the generic form:

$$G(s)=K\frac{s^2+\frac{\omega_z}{Q_z}s+\omega_z^2}{s^2+\frac{\omega_p}{Q_p}s+\omega_p^2}=K\frac{s^2+BW_zs+\omega_z^2}{s^2+BW_ps+\omega_p^2}\tag{4}$$

From $$\(3)\$$ and $$\(4)\$$ the center frequency and the bandwidths can be extracted:

\begin{align} K=H(0)&=\frac{R}{R+R_{ser}}\tag{5} \\ \omega_p=\omega_z&=\frac{1}{LC}\tag{6} \\ BW_z&=\frac{R_{ser}}{L}\tag{7} \\ BW_p&=\frac{R_{ser}}{L}+\frac{1}{RC}\tag{8} \\ \end{align}

Which shows that the formulas in your post discard the $$\R_{ser}\$$ element and, for small enough values it can be discarded. In this case: $$\25\;\Omega\$$ and $$\15\;\text{k}\Omega\$$, so it's fine. To verify use a calculator or simulator to compare the circuit with the Laplace expression:

The RLC version overlaps with the Laplace expression, so it's a match. The measured quantities are the frequencies, f1 and f2, the passband bandwidth bw, and the center frequency, fc, against the calculated bwp and f0. The numbers come very close, save precision, parasitics, etc.

Given your formulation I'm not sure how to interpret your question, but I'll add this, to be sure. In case what you want is to find independent formulas for $$\f_L\$$ and $$\f_H\$$, then all you have to do is use the following system of equations:

\left\{ \begin{aligned} f_Hf_L&=\frac{1}{LC} \\ f_H-f_L&=\frac{R_{ser}}{L}+\frac{1}{RC} \\ \end{aligned} \right.

which results in a 2nd order equations with the two valid roots:

$$f_L=\frac{\sqrt{(RR_{ser}C)^2+2RLC(R_{ser}+2R)+L^2}-RR_{ser}C-L}{2RLC} \\ f_H=\frac{\sqrt{(RR_{ser}C)^2+2RLC(R_{ser}+2R)+L^2}+RR_{ser}C+L}{2RLC}$$

And the result of the .meas scripts are:

f1: freq=(2.36937e+006,0) at 2.36937e+006
f2: freq=(2.66826e+006,0) at 2.66826e+006
f_lo: (sqrt(c**2*r**2*rs**2+2*c*l*r*rs+4*c*l*r**2+l**2)-c*r*rs-l)/(2*c*l*r)/2/pi=(2.36856e+006,0)
f_hi: (sqrt(c**2*r**2*rs**2+2*c*l*r*rs+4*c*l*r**2+l**2)+c*r*rs+l)/(2*c*l*r)/2/pi=(2.6736e+006,0)