This question is related to my last post. To avoid confusion, I opened another question.

From a website that John shared in my last post, this is a plot that is a 3D picture combining the Bode and the pole-zero diagram:

enter image description here

Thanks to @user287001, I learned that when we think about the frequency response, we only think about the situation when s = jω and ω>0. Although ω could be <0, we ignore this situation because we cannot see what it means.)

When we think about effect of the pole and zero of transfer function on the frequency response, we project these poles and zeros on the +jω axis. Like the poles in the plot shown below, the point \$ ωn*sqrt(1-\zeta ^2) \$ means the frequency response starts to decline.

When I think about it again, for a one order system, such as \$\frac{1}{s+2}\$ the pole is -2, and it is projected on origin of the jω axis. It isn't projected to the corner frequency like a second-order system. I think this 3D image does not work with other transfer functions. What could I do to connect the Bode plot with the poles and zeros?

Now I know the meaning of the point in the +jω axis, but what does the point \$ -\zeta ωn \$ in the σ axes mean? When does it mean when pole is projected on the σ axis?

  • \$\begingroup\$ The image in your question is of a 2nd order low-pass filter (this is stated on the website under the main heading). Any pole can be projected onto the jω axis. \$\endgroup\$
    – Andy aka
    Sep 25, 2022 at 7:59
  • \$\begingroup\$ TL, there's no \$\zeta\$ in a 1st order. The concept isn't there. Your proposed 1st order can be constructed with a pair of 1 Ohm resistors set up as a resistor divider with the grounded resistor paralleled by a 1 F capacitor (input and output at obvious places.) The pole will sit right at \$j\,\omega=0\$ and \$\sigma=-2\$ with a relatively obvious-looking shape where the \$\sigma=0\$ plane cuts through the 3D pole surface. Given \$\sigma=-2\$ is further away from this cutting plane (than -1), the curve's height will be 1/2 at \$j\,\omega=0\$. \$\endgroup\$
    – jonk
    Sep 25, 2022 at 8:36
  • \$\begingroup\$ @Andy Thanks for help! My question statement is wrong. I have changed it. What I really want to say the pole in the plot is projected on the \$ ωn*sqrt(1-\zeta ^2) \$, which is the corner frequency of the 2nd order system. But as to the one order system I mention, the pole is projected on the origin rather than the corner frequency 2. So what does it means when the pole is projected on the 0? I can't connect the pole with the Bode just like the 2nd order system do. \$\endgroup\$
    – T L
    Sep 25, 2022 at 11:54
  • \$\begingroup\$ For a TF of \$\frac{1}{s+2}\$, the magnitude at the origin in 0.5 <-- it's just simple math, make s = 0 (i.e. the origin) and the TF = \$\frac{1}{0+2} = 0.5\$. \$\endgroup\$
    – Andy aka
    Sep 25, 2022 at 12:09

3 Answers 3


That 3 dimensional diagram is only relevant to a second order system. The concepts of zeta and wn are also only relevant to a second order system.

That 3 dimensional diagram illustrates that the peaking frequency in the frequency domain (bode plot) is a bit lower than the frequency which the system will oscillate at in the time domain when subjected to a step input.

A first order system is not oscillatory and so the jw axis is not relevant to a first order system. Nor is there a peaking frequency for a first order system instead there is a pole frequency where the magnitude response is down 3dB from its DC value.

So to make the diagram relevant to a first order system you would re-label the first order pole position on the real axis as -1/Tau, (-zeta*wn is not relevant to a first order system), or for your 1/(s+2) example label it as -2.

Next you would draw a frequency plot (bode plot) above and parallel to the real axis. For your example first order transfer function, this frequency plot would start at a magnitude of 0.5 vertically above the origin, it would be -3dB lower at the pole position and then reduce away to zero as the real axis extends to infinity.

In this way we can relate the frequency response of the Bode plot to the pole position of the same first order system on the real axis on the s-plane.

The standard form transfer function of a first order system would be written as:-

k/(Tau*s + 1) where k is the dc gain.

To get your 1/(s+2) transfer function into this form divide each term by 2.

This gives 0.5/(0.5*s +1) yielding up both the value of Tau and the value of the dc gain which is the gain at the origin when w = 0.


The diagram is a plot for a 2nd order system. The standard transfer function for a 2nd order low pass filter is: $$G(s) = \frac{\omega_{N}^2}{s_{2}+2\zeta\omega_{N}+\omega_{N}^2}$$

When the poles are factored from the denominator the are: $$p_{1},p_{2} = -\zeta\omega_{N}\pm\sqrt{\omega_{N}^2-1}$$

If \$\omega_{N}^2<1\$ then the result is complex: $$p_{1},p_{2} = -\zeta\omega_{N}\pm j\sqrt{1-\omega_{N}^2}$$

It is seen that in both cases, the real part is \$\sigma=-\zeta\omega_{N}\$

The significance of the real part is called damping. From a hardware point of view it is the influence of resistance or friction.

From a signal point of view the the response of an underdamped 2nd order system to a step input is a damped sinusoid as shown below. The further the pole is from the origin, the faster the sinusoid damps out. enter image description here

  • \$\begingroup\$ Thanks RussellH! But what if it is a system of order n? I need a more general answer that applies to systems of order n. Qualitative explanations are fine, no quantitative explanations like expressions are needed. \$\endgroup\$
    – T L
    Sep 25, 2022 at 5:08
  • \$\begingroup\$ @TL In general, for order \$n\$ there are \$q\$ pairs of complex conjugate poles and \$r\$ real poles. I won't belabor how to get the partial fractions but if you want you can see why the standard form has no zeros in it. The bottom line is that you can decompose order \$n\$ into a sum of 2nd order and 1st order partial fractions in standard form. \$\endgroup\$
    – jonk
    Sep 25, 2022 at 9:22

A pole which has a negative real part represents an exponentially decaying component in the natural response. The further to the left the pole is on the s-plane then the higher the rate of decay will be.

For a second order system with a damping ratio less than one, there will be a pair of complex conjugate poles. The higher (more negative) the sigma values of those poles then the faster will be the decay of the envelope of the oscillatory response to a step input. That is to say the further to the left on the s-plane the pole positions are then the more stable will be the system.

For a first order (1 pole) system (or indeed for the two real poles of a second order overdamped system) the value of the pole is equal to -1/Tau where Tau is the time constant of the pole. So for a pole value equal to -2 we have a time constant of 1/2 = 0.5. For a pole value of -3 we have a time constant of 1/3 = 0.3333. This illustrates that the more negative the pole value (the further to the left the pole is on the s-plane) then the smaller the time constant of the pole and the higher the rate of decay.

For an overdamped second order system, which will have two poles at different positions on the real axis, the more dominant of the two poles will be the pole nearest the origin which has the largest time constant.

For systems with more poles than two, lets consider a 5th order 5 pole system with no zeros. If 3 of the poles are way to the left of the other two then we can consider the system to be 2nd order dominant, ignore the 3 left most poles and approximate the behaviour of the system by just considering the two rightmost dominant poles. Now we can get values for zeta and wn and determine the behaviour of the system in terms of rise time, overshoot etc by applying the usual equations.

Alternatively if we are designing a system to have a particular rise time and overshoot (perhaps specified by a customer) and we have a fifth order system then we would place 3 of the poles way out to the left of the other two and treat the system as 2nd order dominant. With 3 of the poles placed way out to the left, we can place the two dominant poles with the particular values of zeta and wn in mind. The closed loop system will now approximate the required 2nd order behaviour. The behaviour of a 2 pole, 2nd order system with no zeros is determined by the values of zeta and wn because the values of zeta and wn define the 2 pole positions.


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