# Analog filter -> cutoff frequency question

My "teacher" gave me the following filter circuit:

Then, asked me to find the cutoff frequency, when $$\R_1=1000,R_2=250,C=100\cdot10^{-9}\$$.

I did find the poles and zeros and they gave me $$\\omega=8000\$$ and $$\\omega=40000\$$,

• but the $$\-3\$$dB point gave me $$\\omega=\frac{40000}{\sqrt{23}}\approx8340\$$

• and $$\-3\$$dB above the limit when $$\\omega\to\infty\$$

gave me: $$\\omega=8000\sqrt{23}\approx38367~~~\$$.

So who is wrong and where?

• If the pole is at 8000, it is very reasonable for the -3dB cutoff frequency to be very near. 8340 looks correct. Where do you see a problem? Commented May 16, 2017 at 17:56
• Why do you think cut-off frequency should be found my multiplying or dividing a pole or zero frequency by $\sqrt{23}$? Commented May 16, 2017 at 18:12
• I think I see your problem: Is the problem that you don't know why don't give the same value? @ThePhoton of course that is not the case, but the OP wants to know why the math is giving him/her that Commented May 16, 2017 at 18:22

The simplest approach here are the fast analytical circuit techniques or FACTs. They are based on the Extra-Element Theorem (EET) forged my Dr. Middlebrook a while ago. I am going to show you how you can derive this transfer function and determine the pole and zero position just by inspection. Have a look a the below picture:

On a simple circuit like this, you start with $s=0$. It means in your circuit, you replace the capacitor by an open circuit. As nothing loads $R_1$, you have

$H_0$=1.

Then, you need to determine the natural time constant of this circuit. You determine the time constant of any circuit by reducing the stimulus value to 0, 0 V for a voltage source or 0 A for a current source. Here, reducing $V_{in}$ to 0 V means replace it by a short circuit. Now, if you disconnect the capacitor $C_1$, in your head, connect an ohm-meter at its connecting terminals in the circuit and measure the resistance. You "see" the series combination of $R_1$ and $R_2$. The time constant of this circuit is thus $\tau_1=C_1(R_1+R_2)$. We can show that for a first-order circuit, the pole is the inverse of the natural time constant. In other words:

$\omega_p=\frac{1}{C_1(R_1+R_2)}$

For the zero, consider the circuit redrawn while $C_1$ is replaced by its impedance expression. This is what is called the transformed circuit. A zero in a function $y=f(x)$ is the value of $x$ which brings $y=0$. In a transfer function, the output or $y$ is the response, $V_{out}$ in the drawing. What specific impedance combination in this transformed circuit observed at $s=s_z$, the zero frequency, could make $V_{out}=0\;V$? Could $R_1$ become an infinite value? No. Could the series combination of $R_2$ and $C_1$ become a transformed short circuit? Maybe:

$R_2+\frac{1}{s_zC_1}=0$ implies that $s_z=-\frac{1}{R_2C_1}$ or

$\omega_z=\frac{1}{R_2C_1}$

and this is your zero frequency. The final transfer function is thus expressed in the so-called low-entropy format

$H(s)=\frac{1+sR_2C_2}{1+sC_1(R_1+R_2)}=\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}$

Of course, you can determine the transfer function using the raw approach in which you consider the impedance $Z_1(s)=R_2+\frac{1}{s_zC_1}$ loading $R_1$ and you have

$H(s)=\frac{Z_1(s)}{Z_1(s)+R_1}$

Develop and rearrange to fit the low-entropy format and you will also find your pole and zero. However:

1. you can make mistake while expanding this expression (simple here I agree)
2. you need to factor and rearrange the result so that a pole and a zero appear.
3. there is no fun in doing that! : )

With the FACTs approach, you draw small sketches and obtain a transfer function by inspection without writing a line of algebra. You can also consider the gain for $s$ approaching infinity instead and use inverted poles and zeros as in the below picture:

Best of all, the transfer function is already arranged properly. See if you can figure this out yourself: take the same circuit and now load $V_{out}$ with a resistor $R_3$. If you calculate the new time constant, it becomes

$\tau_1=C_1(R_2+R_1||R_3)$ meaning that $\omega_p=\frac{1}{C_1(R_2+R_1||R_3)}$

and the dc gain for $s=0$ becomes

$H_0=\frac{R_3}{R_3+R_1}$

The zero remains unchanged and the new transfer function is

$H(s)=\frac{R_3}{R_3+R_1}\frac{1+sR_2C_2}{1+sC_1(R_2+R_1||R_3)}=\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}$

Again, no equations, just in your head with a little bit of habit.

As student, if you are designing circuits (passive or active) and need to determine transfer functions, I encourage you to acquire this skill because once you master the technique, you won't go back to the classical approach. If you start slowly step by step, it is quite simple actually. Complicate expressions when you understand the approach with 1st-order circuits.

You can discover FACTs further here

and also through examples published in the introductory book

Good luck!

If you draw a bode plot with your pole, and overlay a bode plot with the zero. You'll see that because your pole/zero aren't well seperated ~1/2 decade the zero is adding a small amount of magnitude near the pole. Yeilding the -3dB point to be slightly further out.

• I was writing the same answer, but I could find a nice plot, can you maybe? But +1 absolutely Commented May 16, 2017 at 18:42
• intuitive sanity check on math

• with cap C shorted, transfer function H(s)= 0.2 or ~ -14dB gain=R2/(R1+R2)
• when C is open H(s) = 0dB
• thus with -6dB/octave that's more than 2 octaves ( but less than a decade)
• Since R2 is smaller, when |Zc|=R1

it's almost 0dB ____        so 1/R1C=1e4  near 1st break ω1
14dB     \_____     1/R2C=4e4  near 2nd break ω2