I am known with the basics of what poles and zeros of transfer function are and how to determine them, but I certainly don't know how to apply such poles and zeros to the gain curve, which describes frequency response of an amplifier. Also, this video explains the basics of poles and zeros, and I understand it completely.

This is the picture of Bode plot of an amplifier from some research paper that can be downloaded here:

enter image description here

The steeper curve shows two pole compensation frequency response, while the other one is for dominant pole/ single pole compensation of an amplifier. With the red dot I marked a pole, with green mark I marked a zero (which are located there, accordingly to that research paper). The pole is defined as the point in an imaginary plane to which function approaches.

How can the red dot be a pole according to the upper definition of it? Zero is defined as the point in a real plane in which transfer function that is being observed reaches value of zero. According to this research paper, the zero of two pole compensation is located at 320 kHz (that is somehow close to green mark).

How can according to upper definition of zero a zero be located, where the curve is marked with green dot? I would say that a zero is located at the frequency where a both functions "cross" 0 dB of gain.

  • \$\begingroup\$ A bode plot looks at the transfer function for each frequency on the jw axis (i.e. s=jw) which means that while the transfer function at a zero may have a zero response at some complex s it isn't necessarily zero for just the imaginary s=jw. \$\endgroup\$
    – John D
    May 16, 2018 at 15:01
  • \$\begingroup\$ From the article you cited: "A solution is the use of two-pole compensation. This consists of introducing an additional low-frequency pole, followed by a zero at a higher frequency, placed to bring the gain slope back to -20 dB/decade before the loop gain crosses 0 dB. In this way, as frequency falls from the unity-loop-gain point, the loop gain initially increases at 20 dB per decade, as in the single dominant pole case. However, at the zero frequency, this rate increases to 40 dB per decade, resulting in much higher loop gains in the audio band." Emphasis mine. \$\endgroup\$
    – jonk
    May 16, 2018 at 15:17
  • \$\begingroup\$ web.njit.edu/~levkov/classes_files/ECE232/Handouts/… page (4) or this ece.utah.edu/~ee3110/bodeplot.pdf \$\endgroup\$
    – G36
    May 16, 2018 at 15:23
  • \$\begingroup\$ I think you are confusing: a pole-zero plot, a root locus plot, and a Bode plot (or 'frequency response'). \$\endgroup\$
    – Chu
    May 16, 2018 at 16:09

2 Answers 2


I don't understand the question nor do I understand why the bode plot is marked with a pole and a zero and here's why: -

A typical pole zero diagram for a 2nd order low pass filter: -

enter image description here

The vertical axis is jw and this is also the base frequency axis of a bode plot like this one (also a 2nd order low pass filter): -

enter image description here

Behind the resonant peak in the bode plot above is lurking a pole but you cannot draw it on the bode plot because neither of the pole's co-ordinates are exactly along the frequency axis.

And together, in 3 dimensions they look like this: -

enter image description here

As you can see, you cannot mark a pole on a bode plot unless it occurs right on the jw axis but, then the bode plot peak would rise to infinity and it doesn't on your bode plot.

You also cannot mark a zero on a bode plot for the same reasons namely the bode plot HAS to be zero if there is a zero with a co-ordinate along the jw axis. If it's not there then you can't mark it on the bode plot because it makes no sense.

Pictures from here.

  • \$\begingroup\$ Lacking analysis, this is how I find poles and zeros on Bode plots (when that's all I have to work with.) \$\endgroup\$
    – jonk
    May 16, 2018 at 16:19
  • \$\begingroup\$ @jonk a good method. \$\endgroup\$
    – Andy aka
    May 16, 2018 at 16:21
  • \$\begingroup\$ @Andyaka Could you say that these things you wrote in your answer can be learned from some filter design book (next to mathematics)? Or are poles and zeros a topic of some other area of electronics? \$\endgroup\$
    – MucaGinger
    May 16, 2018 at 16:25
  • 1
    \$\begingroup\$ @Keno I learned them back in the 1980s at college and luckily have never had to look much in books since. Poles and zeros are part of 2nd order differential equations so a decent math book that is engineering biased would help. I have never seen the 3D view drawn anywhere before I started drawing it for SE answers but memories fade.... \$\endgroup\$
    – Andy aka
    May 16, 2018 at 17:00

Since you say you understand poels and zeroes, then you understand that they are nothing but roots of the characteristic polynomials of the numerator and denominator, and they don't have to be purely real (s=1), or purely imaginary (s=i), they can be complex (s=-1\$\pm\$i).

In the same manner, the poles and zeroes in your case are complex, which means they are damped, which results in their projection on the j\$\omega\$ axis to not be a peak to infinity, or a valley to zero -- they are damped.

For example, a 2nd order transfer function with a pole at 1Hz, and a zero at 10Hz. The transfer function is:

$$H(s)=\frac{a_2 s^2+a_1 s+a_0}{s^2+b_1 s+b_0}=\frac{s^2+4s+100}{s^2+0.4s+1}$$

and the plot:


In this case, the roots are z=-2\$\pm\$9.798i and p=-0.2\$\pm\$0.978i. They are not purely real, or purely imaginary, therefore their projection on the frequency axis appears as damped. If the terms b1 and a1 were zero, then the roots would be purely imaginary:


Looking at your picture, the transfer function has a second order denominator and a first order numerator, because of the slope after the zero. The pole looks to be slightly underdamped. If I'd try to obtain a similar result, I'd try with b1=1.2, a2=0 (need 1st order), a1=1 and a0=10:


Notice how the pole peaks very slightly, and the zero is smooth due to the numerator being a 1st order polynomial, thus the root (zero) is purely real. If it were to be a funnel-like magnitude, it would have been a complex pole without a real part, only imaginary. But like this, the slope from the pole until the zero is -40dB/dec, then, after the zero, -20dB/dec. Also look at the phase, how it goes up by 90o.

  • \$\begingroup\$ What means "damping"? Also, I don't understand imaginary numbers and complex ones either much so I think I am not able to fully understand your answer. And from the first equation, how did you get those values for a's and b's?... \$\endgroup\$
    – MucaGinger
    May 16, 2018 at 16:37
  • \$\begingroup\$ Could you maybe recommend a book to read for a "greeny" like me to learn these things properly? \$\endgroup\$
    – MucaGinger
    May 16, 2018 at 16:38
  • \$\begingroup\$ @Keno On the most neutral tone possible, you might want to understand those, first, the complex numbers, then trigonometry, because no matter what book I'd recommend they'd all be based on that math. "Opamps for everyone" might be a good one for you, or, if you think you have the stomach, "Analog Electronic Filters" by Hercules Dimopoulos. \$\endgroup\$ May 16, 2018 at 16:47

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