# Does frequency at zero matter?

Consider the following

simulate this circuit – Schematic created using CircuitLab

$$H(s) = \frac{sR_2C_1}{1+s(R_1+R_2)C_1}$$ $$f_z = \frac{1}{2\pi R_2C_1}$$ $$f_p = \frac{1}{2\pi (R_1+R_2)C_1}$$

I simulated the circuit in LTSPICE with $$\R_1=R_2=795\$$ and $$\C_1=1nF\$$ which should give me a $$\f_z = 200kHz\$$ and $$\f_p = 100kHz\$$.

Here is the AC anaylsis of the simulation.

Question: Given that there is a zero at the origin with a certain frequency - what should I see at that frequency ?

Additional information to add some clarity: I am reading this book about Linear Circuits and it says the following

$$H(s) = \frac{sR_2C_1}{1+s(R_1+R_2)C_1} = \frac{\frac{s}{w_{z0}}}{1+\frac{s}{w_{p1}}}$$

where $$\ w_{z0} = \frac{1}{R_2C_1} \$$ and $$\ w_{p1}= \frac{1}{(R_1+R_2)C_1} \$$

I believe the intent of this form is to visually be able to extract certain information about the system. I am confused on the significance of the $$\w_{z0} \$$ and what it means.

Traditionally I have seen forms such as $$H(s) = \frac{1+\frac{s}{w_z}}{1+\frac{s}{w_p}}$$ and I know that you have a flat line (magnitude plot) until the wz and then you have an +20db/dec line and then a -20db/dec at the wp. But this zero at the origin is messing me up.

• The "certain frequency" you see at the origin is $f = 0$. Why don't you substitute $s = 2 \pi f = 0$, plug it into $H(s)$, and see what you get? – TimWescott Mar 23 at 16:05
• Sure I get a H(s) = 0. I understand that a zero at an origin means that at DC, H(s) is zero, (because s = 0). But what about the RC for the numerator ? What does it do ? – Frankie Mar 23 at 16:11
• The coefficient in the numerator ($R_2 C_1$) establish the high-frequency gain, along with the coefficient in the denominator. – TimWescott Mar 23 at 16:15
• "...Given that there is a zero at the origin with a certain frequency - what should I see at that frequency ?": A plot in dB will not show the gain in $\omega = 0$, since $log (0)$ is undefined. – Dirceu Rodrigues Jr Mar 23 at 16:21
• You need to factor numerator and denominator. Your equation for fz is incorrect. fz = 0, as Tim says. – Spehro Pefhany Mar 23 at 16:29

## 1 Answer

The reason why you do not "see" the zero is because it lies at the origin, for $$\s=0\$$. You can see that when $$\s=0\$$, the numerator is 0 and so is the magnitude: the dc component is blocked by the capacitor.

There is actually a better way to write this transfer function because it does not provide insight on the high-frequency plateau. This form is called a low-entropy form, a term forged by the late Dr. Middlebrook some time ago. It implies that a transfer function linking a response ($$\V_{out}\$$) to a stimulus ($$\V_{in}\$$) must be preferably expressed in way where zeroes, poles and gains properly appear. In your case, you can advantageously factor $$\sR_2C_1\$$ in the numerator and $$\s(R_1+R_2)C_1\$$ in the denominator:

$$\H(s)=\frac{sR_2C_1}{s(R_1+R_2)C_1}\frac{1}{1+\frac{1}{s(R_1+R_2)C_1}}=H_{\infty}\frac{1}{1+\frac{\omega_p}{s}}\$$ in which:

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

In this expression, $$\\omega_p\$$ is an inverted pole and lets you shape the transfer function in this convenient way, with a leading term $$\H_{\infty}\$$ representing the plateau gain as $$\s\$$ approaches infinity.

This is the proper way of writing this transfer function. If you are interested by determining transfer functions in a quick and swift way, you can have a look at the fast analytical circuits techniques or FACTs for which an introduction is given here.

• I saw your answer in another post about FACTs. The book I am reading where the question was derived from was Linear Circuit Transfer Functions: An Introduction to Fast Analytical Techniques. I liked the link so much that I ended up buying a book and am working through it now but still have gaps - the notation / form in the question is actually taken from a FACTs book. – Frankie Mar 23 at 17:29
• Going through the FACTs and being fluent at them requires a little bit of time but once you master the technique, you won't go back to classical analysis! Feel free to post more questions if needed. – Verbal Kint Mar 23 at 21:57