I have some struggles to find the transfer function of this plot. We got the hint, that the "breaks" of slope can be assumed to be on vertical and horizontal lines. Also we should pay attention on the phase in the higher frequencies.

Am I right, that the phase reserve (margin) is approximately zero and that the amplitude margin is 100 dB?

bode plot

  • \$\begingroup\$ Gain and phase margin look correct (of course the phase margin is awful), looks like a pole at origin, another pole, a zero then another pole. Approximate the frequencies by looking at the breakpoints where the slope changes \$\endgroup\$
    – John D
    Jun 22, 2018 at 20:15

2 Answers 2


Gain Margin

  1. Find the frequency where the PHASE becomes -180 degrees.
  2. Find the GAIN, G (in dB), at this SAME FREQUENCY (from the upper plot)
  3. Then, we define the GAIN MARGIN as: Gain Margin = 0 - G dB
  4. Gain Margin = 1/M if you are measuring Magnitude (M) as a ratio (not is dB).

(Note that G is in dB here... But you may want to convert between dB and magnitude as a ratio. To covert magnitude, M, to gain in decibels (dB), G, you use G=20*log10(M). To convert G to M, M=10^(G/20))

Am I right, ... that the amplitude margin is 100 dB?

Yes the gain margin is \$100dB\$.

Phase Margin

  1. Find the frequency where the GAIN is 0 dB. (This means the output and input amplitudes (magnitudes) are identical at this particular frequency; on the Bode plot, it's where the transfer function crosses 0 dB on the upper [magnitude] plot.)
  2. Find the PHASE, P (in degrees), at this SAME FREQUENCY (by now looking at the lower plot).
  3. Then, we define the PHASE MARGIN as: Phase Margin = +P + 180 degrees

Am I right, that the phase reserve (margin) is approximately zero... ?

Yes. Using the above technique: Phase Margin = -180 + 180 = Approx 0

(Link to above information)


The order of an open loop system depends on the number of reactive elements either L or C , or -C with negative feedback in a closed loop system or a high pass type passive filter.

We choose compensation to compromise for stability or or gain or bandwidth reasons towards making it dominated by a -1 order slope at 0dB for it’s closed loop gain. The gain margin is often less or unimportant.

Since each contributes a + or - 90 deg phase shift the break point of each occurs at |45 deg| then extends for +/-2 frequency decades towards its final contribution in phase shift.

Looking at the Bode Plot of amplitude we also know each n order of magnitude change in slope is 6dB per octave or 1/2 f to f to 2f .

Unfortunately using just amplitude margin at 180 deg is not enough to evaluate stability because you only have 5 deg of phase margin at 0dB, so your step response will ring , very-underdamped at that frequency of minimum phase margin at 0.1 rad/s.

There is a lead-lag filter in this curve at 20 rad/s but it is far too high in its R1:R2C break point to do any good at 0.1 rad/s

So you can define your design requirements for Gain BW and phase margin eg >30 deg or 45 deg and increase that lead-lag cap value by a factor of 20/0.1 =200x bigger or tell us what you want to do?

  • \$\begingroup\$ Thank you for your very detailed response. TBH I'm a newbie in this field. My main concern is the broad understanding of the topic. What are relations of poles/zeros on the gain and phase. When does the phase change direction and when does the db/dec values change. AFAIK poles change the gain from - 20dB/dec to - 40dB/dec and poles back to - 20 dB/dec. But I'm not sure how poles/zeros affect the phase. Is it correct to say that there is a pole at 10^-2; a zero at 10^1 and finally a pole at 10^2. Oh and a pole at 0. \$\endgroup\$
    – Kai
    Jun 22, 2018 at 22:39
  • \$\begingroup\$ 20dB/dec=6dB/octave @45 deg \$\endgroup\$ Jun 22, 2018 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.