I'm reading a book (see figure 5-12) about (among other things) pole zero plots and there is an example I don't understand. For the schematics and formula below:

enter image description here

it adds the following zero pole plots:

enter image description here

and says:

The individual pole zero plots show the dc gain of 1/2 plotting as a straight line from the –6 dB intercept. The two zeros occur at the same break frequency, thus they add to a 40-dB/decade slope. The two poles are plotted at their breakpoints of ω = 0.44/τ and ω = 4.56/τ. The combined amplitude plot intercepts the amplitude axis at –6 dB because of the dc gain, and then breaks down at the first pole.

I understand why the breakpoints are located at the mentioned points and why the slopes go up/down where they do. However, I don't understand why the individual plot starts at 0 dB while the combined one starts at -6 dB.

I have noticed this:

[...] The combined amplitude plot intercepts the amplitude axis at –6 dB because of the dc gain, and then breaks down at the first pole.

First, I don't see why the dc gain is -6 dB for all \$\omega < \frac{0.44}\tau \$. Second, I don't see why this dc gain is applied only to the combined plot and not to the individual one.


In the equation 5-10 put s=0 to get the dc-gain. It is ½ = -6dB. this fits perfect with figure 5-12

Figure 5-11 shows 4 different bode plots in the same diagram - one for each factor. I.e. one for x½, two for the two poles in the denominator, and one for the double zero in the nominator. The latter has slope of 40dB/dec which actually is the same as two single zeros added up.

Figure 5-12 shows the sum of all the these individual plots. It's the sum - because in logarithmic scale summing correspond to multiplication as seen in equation 5-10. There, all the 4 frequency dependent factors (poles and zeros) are multiplied together. And you may only multiply each factor once - of course. The same goes for the ½ which is the same as: only add -6dB once.

Note however that there is a minor peculiarity in the figure. It should have been drawn like this: enter image description here It is customary to draw the x-axis with Y=0 as the x-axis. Maybe this explains your confusion

  • \$\begingroup\$ "In the equation 5-10 put s=0 to get the dc-gain. It is ½ = -6dB. this fits perfect with the last figure." How does it fit in the first one? \$\endgroup\$
    – Martel
    Feb 19 at 16:43
  • 1
    \$\begingroup\$ the horizontal line at -6dB is the factor ½. All the other factors have a dc gain of 1 equal to 0dB. In reference to the circuit diagram: How to find the dc gain? Remove all capacitors and replace all inductors by short circuits, then do the calculation of the gain. Here it is R/(R+R). Lower left R is irrelevant for DC gain. And along the same lines to find the high frequency gain, do the opposite replacements C=short and L=open. Just to round it off: The high frequency gain is seen to be 1 because the capacitors short input to output. Voila. \$\endgroup\$ Feb 19 at 20:43
  • \$\begingroup\$ does this answer your questions? \$\endgroup\$ Feb 20 at 16:08
  • \$\begingroup\$ I understand the explanation for the second figure (the -6dB line). However I still don't see why that is different in the first one, which starts at 0 dB. When you say "All the other factors have a dc gain of 1 equal to 0dB", "other factors" mean each of the factors multiplying in the numerator and denominator respectively? \$\endgroup\$
    – Martel
    Feb 20 at 19:33
  • \$\begingroup\$ i tried to clarify my answer a bit - does this help? \$\endgroup\$ Feb 21 at 10:24

For readers who would like to determine the transfer function of this filter but also for the sake of a simple personal exercise, I have applied the fast analytical circuit techniques or FACTs described in my last book. The principle is simple and consists of determining the time constants of this 2nd-order circuit in two conditions: when the excitation is zeroed (the input source is replaced by a short circuit) and when the output is a null (no ac response) despite the excitation being back in place.

We start with \$s=0\$ and open-circuit all capacitors, exactly as SPICE does when determining a dc operating point. When you do this, the circuit becomes a simple resistive divider and if all resistors are equal, the gain is 0.5 or -6 dB. Then you carry as shown in the below drawings:

enter image description here

What is cool is that you determine all time constants by inspection without the need to resort to algebra: less mistakes and you can easily come back to a faulty sketch and fix it individually leaving the rest intact. If all goes well, you have the following expressions:

enter image description here

I have derived the expression with different values for the resistors and capacitors but if you consider all resistances equal to \$R\$ and the capacitors equal to \$C\$, then you can simplify the expression and factor the denominator as the product of two separated poles after solving a simple two-unknown equation and confirm the coefficients given in the text:

enter image description here

We can now plot and compare the reference response given by the brute-force approach and that of the FACTs. As you can see, they are rigorously identical:

enter image description here

  • \$\begingroup\$ Are you really Christophe P. Basso the author of that book? Interesting. EET could morph into GFT and vice versa. Reading this post, reminds me of the late R. D. Middlebrook, I love his paradigm of D-OA. authors.library.caltech.edu/4537/1/MIDieeemm06.pdf \$\endgroup\$
    – Unknown123
    Feb 21 at 12:00
  • \$\begingroup\$ Oui, I confess this is me : ) I have met Monsieur Middlebrook several times and he was a really kind person. The EET extended into higher order is really an efficient tool and you can acquire the skill quickly. You can see it working here in a 6th-order transfer function. \$\endgroup\$ Feb 21 at 16:11

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