# Extracting Transfer function from Bode's gain plot

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

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!

• What is the x-axis? $\omega = 2\pi f$?
– Jdip
Commented Mar 13 at 18:45
• @Jdip yess exactly Commented Mar 13 at 18:47
• Take a look at this answer and come back if you have questions!
– Jdip
Commented Mar 13 at 19:06
• What are the axes? It looks like dB vs. linear frequency. Commented Mar 14 at 19:43
• Sorry, I should have mentioned this at first: please edit your question with this information. That makes it complete, and people do not have to look into the comments to figure out what the question means. Commented Mar 15 at 1:45

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.

• so how would it be interpreted mathematically, that exactly my problem that i fail to tame... thanks again Commented Mar 13 at 20:17
• See my update; I hope that clears it up for you. Commented Mar 13 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 $H(s) = 200 \frac{s-j\omega_0}{(s-j\omega_1)(s-j\omega_2)}$ Commented Mar 13 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. Commented Mar 13 at 22:34
• Is it better now? $H(s) = 200 \frac{s+j\omega_0}{(s+j\omega_1)(s+j\omega_2)}$ Commented Mar 13 at 22:39