# bode plot, zero at 0Hz

I'm learning to draw magnitude bode plots from transfer functions:

$$H(s) = \frac{sRC}{sRC+1}$$

So I can see that I have a zero at s = 0, which corresponds to 0 Hz. I know this will add a +20dB/dec slope for all frequencies in a logarithmic plot. I'm struggling to figure where the gain will start at the left side of my plot. I can kind of cheat a little bit because I know that this is a passive high pass filter, so I can start at the right side of the plot and call that 0dB, move to the left, and then apply my 20dB/dec slope once I pass the break frequency of the pole. But if the transfer function was more complicated and I didn't already know what to expect, what would I do?

• Like Tim mentioned, you can start at a very small frequency value. Even industry standard applications draw the initial magnitude value at a frequency very close to 0 Hz. Unless you care about really granular values near 0 Hz, I wouldn't worry too much about what this value is since we usually look at the action as we move away from 0 Hz. Jan 9, 2020 at 22:35

$$H(s) = \frac{R C s}{R C s + 1}$$

If $$R C s$$ is much lower than 1 (i.e., $$R C s \ll 1$$, in Math), then you can make an approximation: $$H(s) \simeq \frac{R C s}{1}$$

Basically, the effect of the $$R C s$$ in the denominator becomes insignificant.

So, just pick a radian frequency well below $$1 / R C$$ and compute the magnitude of $$H(s)$$ at that point -- mark that on your plot, and draw your line rising at 20dB/decade from there.

Different approach: Let's start with rewriting using $$s = j \omega$$ and $$RC = 1/\omega_0$$. We get $$H(\omega) = \frac{j\omega}{j\omega+\omega_0}$$

Then we look at the most "relevant" frequency first. In this case that's $$\omega = \omega_0$$. At this frequency your magnitude is -3dB. For frequencies much larger then $$\omega_0$$ you simply have $$|H(\omega)|=1$$ and for frequencies much lower you get $$-20dB$$ per decade.

You can draw this as two straight lines than intersect at $$\omega_0$$ and than you round out the "knee" so that it goes through $$-3dB$$ at $$\omega_0$$.

You can't really draw anything at $$\omega = 0$$ since the log magnitude is minus infinity, so you are going to run out of paper or screen before you get there.

But if the transfer function was more complicated and I didn't already know what to expect, what would I do?