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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 \begin{equation} H\left(jw\right)\:=\:\frac{k}{\left(jw+10\right)\left(jw+100\right)\left(jw+1000\right)} \end{equation} but I don't know if am doing this right, also how to find the value for k. Thank you for your help.
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 $$k=10^{9}$$
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.
