# Transfer Function From Bode Plot

I have been doing some work on this for a while now and I am confused by the very first point.

Looking at the bode plot it starts off at 50dB and has a slope of -20d/dec. I am not sure how to handle this first section.

My transfer function so far is: I know how to find K and zeta but in my equation the value for a alludes me. My initial assumption is to take a=.00001, something close to 0.

As in class we have only ever gotten bode plots with an initial section which is a constant, if anyone can tell me a way to reasonably handle the first second I would be appreciative.

• Re the first section, what function has a constant phase angle of -90deg?
– Chu
Dec 11, 2015 at 15:39
• That is just jw there is no real part. Now that you say that I realize I ignored the phase plot. Dec 11, 2015 at 16:11
• K/jw to be exact!
– Chu
Dec 11, 2015 at 17:22

The piece that you're looking for is $$T(j\omega) = \frac{K}{j\omega}$$ This has a denominator that looks like $1 / \omega$, so the slope of the gain is -20 dB / decade. It also has a phase of -90$^\circ$, which matches your phase plot at low frequencies.
You can find $K$ by looking at any point on the transfer function. For instance, at $\omega = 0.01$, you see a gain of 10 dB, so $$|T(j0.01)| = \frac{K}{0.01} = 100 K = 10^{10\text{ dB} / 20\text{ dB}} \approx 3.16$$ so $K \approx 3.16 \times 10^{-2}$.