deriving the transfer function given bode plot

I have been given the following question on finding the transfer function from the following bode plot

I know that the transfer function should look like this $$H\left(jw\right)\:=\:\frac{k}{\left(jw+10\right)\left(jw+100\right)\left(jw+1000\right)}$$ but I don't know if am doing this right, also how to find the value for k. Thank you for your help.

• Your assumed TF is wrong. What does an initial slope of -20dB/dec tell you?
– Chu
Commented Nov 23, 2017 at 6:30
• It is not possible to uniquely determine the transfer function through the amplitude response alone. Hence all of the answers below that suggest a transfer function (without stating their implicit assumptions) are incomplete or just blatantly wrong. Unfortunately OP has selected an answer that contains false claims.
– Carl
Commented May 21 at 14:50

The magnitude of a transfer function in dB is $$Magnitude=20log_{10}|H(jw)|$$ where H(jw) is the transfer function. Seeing the slopes in the graph shown above, there are poles at 100 and two poles at 1000Hz frequencies. So the transfer function would be $$H(jw)=\frac{k}{(jw+100)(jw+1000)^2}$$ Observe that there are two poles at 1000Hz.

Now at dc frequency/near dc (0.1 rad/s), the gain is 20dB. Gain at 0.1 rad/s is similar to dc frequency considering the pole magnitudes.

Therefore, $$20dB=20log_{10}|H(jw)|$$ $$log_{10}k-log_{10}[100*1000^2]=1$$ Solving we get $$k=10^{9}$$

• @Raykh The slope changes from -40 dB/dec to -80 dB/dec. One pole results in -20 dB/dec. So if it changes from -40 to -80 then it has to be two on top of each other. Commented Nov 23, 2017 at 4:42
• There is not a break freq at 10 rad/sec - Bode plots always show the LF characteristic - you can't assume something that isn't shown.
– Chu
Commented Nov 23, 2017 at 6:32
• @AdityaMadhusudhan, This is nonsense. The gain at $\omega=0$ is infinite.
– Chu
Commented Nov 24, 2017 at 22:05
• @AdityaMadhusudhan; the transfer function has an integrator, hence the -20dB/dec low frequency gain. The gain of an an integrator at w=0 is infinite. 0.1 rad/sec is not even on the graph, its two decades below 10 rad/sec
– Chu
Commented Nov 25, 2017 at 16:35
• @AdityaMadhusudhan, it doesn't matter how far below 0.1 rad/sec is, the point is that you cannot just assume a break point exists somewhere outside the range of a graph. There is no breakpoint below 100 rad/sec.
– Chu
Commented Nov 25, 2017 at 20:53

Your assumed transfer function is wrong. A Bode plot MUST show the LF(low frequency) and HF(high frequency) asymptotes, otherwise it's not giving the full picture. Hence, we must assume the Bode plot presented contains all the information - there are no surprises above or below the frequency range shown.

In this case the LF asymptote is a slope of -20 dB/dec.

There are two break frequencies: one pole at 100 rad/sec, and a double pole at 1000 rad/sec. There is not a break frequency at 10 rad/sec.

You've made a good start, the changes in slope of the bode plot will occur at the poles of the transfer function as you have noted. All you need to do now is find an expression for the magnitude of the transfer function in terms of w and k, then choose some (frequency, magnitude) point on the plot and solve for k.

Without discounting what has been written in the answers, the answer in my opinion is

$$H(s)=\frac{1\times10^{11}}{s\, (s+100) \, (s+1000)^2}$$

There should be a pole at zero frequency as indicated in the given Bode magnitude plot. We cannot ignore this entry of -20 dB/dec.

Nevertheless, we do not have the complete picture as @Chu and @a concerned citizen point out. So my answer while can be correct for the given Bode magnitude plot should not be interpolated outside the given frequency range and assumed still correct.

• @Chu's answer is correct: there is not enough information for below 10Hz and above ~5kHz to determine whether this filter has a lowpass behaviour, or integrator, or even highpass content. The only thing you can know for sure is the location of the two visible poles. Commented Jul 3, 2018 at 5:26
• @a concerned citizen. Thank you for this observation. We can assume that there is a pole at zero. I will remove the word "correct" in my post so that one may judge for himself/herself. The transfer function which I gave fit the given magnitude plot for the frequencies shown. Commented Jul 3, 2018 at 5:37
• We could assume, but we really shouldn't. :-) What if the lowest pole is at 10Hz? Or at 1Hz? It will not be shown and it would change the transfer function. Commented Jul 3, 2018 at 5:51
• @a concerned citizen. Yes I agree that we do not have the complete picture. But for the given magnitude plot, the transfer function which I povided fits. Now OP should take this answer with a grain a salt as you and Chu are pointing out. I will add this in my answer. Commented Jul 3, 2018 at 5:58

In my opinion the answer will be 10^11 / (s+10) (s+ 100) (s+1000)

First change is occurring at frequency 10 .

See where the transfer function is bending. You can see that it has poles at 10 rad/s, 100 rad/sec, 1000 rad /s and its DC gain is at 40 dB which is 100. At first pole the transfer function falls at -20 db/dec slope, at the second pole, it falls at -40 db/decade slope and at the third pole, the slope is -60 db /decade. The gain becomes one at 1200 rad/sec which is your unity gain frequency. Just know that at each pole the transfer function bends with a fixed negative slope in an ideal case although the practical result may vary slightly.

So your ideal transfer function may look like:

$$H(s) = \frac {100}{ (\frac{s}{10} + 1)(\frac{s}{100} + 1) ( \frac{s}{1000} + 1) }$$

In frequency domain:

$$H(jw) = \frac {100}{ (\frac{jw}{10} + 1)(\frac{jw}{100} + 1) ( \frac{jw}{1000} + 1) }$$

The reason to write each poles and zeroes as $$\ \frac{jw}{p} or \frac{jw}{z} \$$ instead of $$\ (jw + p)\$$ or $$\(jw + z)\$$ is that the numerator K in your transfer function should be low frequency gain also called DC gain which can be easily seen without any calculation.