1

Let the figure of the Bode's Gain plot of a certain transfer function, estimate what could this transfer function be: enter image description here

Here is what I tried to do: since at $\omega = 0 , |G(j \omega)|_{db} = 20\log(K) = 20$ I see that it goes up to $35$, I dont have an idea to translate it to something mathematical, because I think this plot isn't the asymptotical plot, instead its the real gain's plot, any hints on how to approach this problem is appreciated thanks!

Jdip
  • 5,980
  • 3
  • 7
  • 29
Knowledge Seeker
  • 11
  • 1

1 Answers1

1

A common source of peaking is complex conjugate poles (assuming a real impulse response). The sharper (tighter) the peaking, the closer the poles are to the $j\omega$ axis at the frequency where the peaking occurs (but in the left half plane assuming a stable causal system. If in discrete time, replace “$j\omega$ axis” with “unit circle” and “left half plane” with inside the unit circle.

Here we see two peaks, one near $\omega=1$ which is sharper and another near $\omega=10$.

The poles and zeros for a transfer function are expressed mathematically as

$$H(s)=K\frac{(s-z_1)(s-z_2)\ldots}{(s-p_1)(s-p_2)\ldots}$$

Where $K$ is a real gain scaling and $z_1, z_2, \ldots$ are the complex zeros and $p_1, p_2, \ldots$ are the complex poles. In this case it appears that there are no finite zeros and two pairs of complex conjugate poles as previously explained.

Dan Boschen
  • 50,942
  • 2
  • 57
  • 135
  • so how would it be interpreted mathematically, that exactly my problem that i fail to tame... thanks again – Knowledge Seeker Mar 13 '24 at 20:17
  • See my update; I hope that clears it up for you. – Dan Boschen Mar 13 '24 at 20:33
  • Ah okey, but i think we do have a zero at somewhere near $\omega_0 = 3.85$ and one question since we have two peaks at respectively $\omega_{1} = 1, \omega_{2} = 10$ then $H(s) = 200 \frac{s-j\omega_0}{(s-j\omega_1)(s-j\omega_2)}$ – Knowledge Seeker Mar 13 '24 at 20:39
  • No your poles and zeros are imaginary which would me they are right on the $j\omega$ axis which is not the case (otherwise the response would go to infinity and zero). So add a real component to push them into the left half plane, the zero if there would be further into the left half plane and the poles and zeros (if there is a zero) would likely be complex conjugate pairs if the frequency response is symmetric about zero (or shows the positive freq axis only). Adjust how far into the left half plane and the gain to match the response. – Dan Boschen Mar 13 '24 at 22:34
  • Is it better now? $H(s) = 200 \frac{s+j\omega_0}{(s+j\omega_1)(s+j\omega_2)}$ – Knowledge Seeker Mar 13 '24 at 22:39
  • No - the terms would be $(s-\sigma \pm j\omega)$ – Dan Boschen Mar 13 '24 at 22:51
  • and the sigma/omega terms are available in the graph or do i have to do some extra work? – Knowledge Seeker Mar 13 '24 at 22:52
  • Plot the magnitude of the response and adjust the real term as well as the overall gain $K$ and I think it will then make more sense to you. Try the pair of poles at $\omega=1$ first by themselves -they will be closer to the $j\omega$ axis (smaller $\sigma$) so are more dominant and then add the higher frequency poles. – Dan Boschen Mar 13 '24 at 23:05
  • First i have no idea how to determine K, second when you tell me there is a peak at w = 1, so that means there is a conjugate poles which imaginary part is w=1, and how can i determine its real part? – Knowledge Seeker Mar 14 '24 at 02:53
  • Plot the magnitude of the response and adjust K and adjust the real part so that they match the target response. You don’t know what either is so you start with a value and adjust. If you move in the right direction you will minimize the error. Start with the poles that have an imaginary component $j\omega = \pm 1$. Please try and I hope it will be clearer for you – Dan Boschen Mar 14 '24 at 03:36
  • Ah i see the methodology, but i think you dont understand that i dont have any source or tools to manipulate the graph beside my eyes – Knowledge Seeker Mar 14 '24 at 04:32