Electronic – How to determine the phase margin and transfer function of this bode plot

bode plotcontrol systemphase margin

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

Best Answer

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)