Electrical – Find quality factor from a bode plot

You've got to back up a step or three. The transfer function is complex valued so, to plot it, you need two plots, usually magnitude and phase. The magnitude plot is usually log-log but the phase plot is lin-log.

So, you need to find the magnitude and then take the log before plotting the Bode magnitude. To find the magnitude, multiply H by its conjugate and then take the root.

$$|H(j\omega)|^2 = \dfrac{100}{1 + \dfrac{10}{j \omega T}}\dfrac{100}{1 - \dfrac{10}{j \omega T}} = \dfrac{100^2}{1 + \dfrac{10^2}{(\omega T)^2}}$$

Also, omega is the radian frequency while f is the frequency. So, if omega = 1000, you don't multiply by 2 pi. However if f = 1000, you do.

UPDATE: fixed denominator of transfer function to match OP's original

UPDATE, PART DEUX:

We should try to put this transfer function in standard from so that can identify the asymptotic gain, the type, and the pole/zero frequency. Since the variable \$\omega\$ appears with highest exponent 1, it is a 1st order filter. There are only two types of 1st order filters: low-pass and high-pass. In standard form the OP's transfer function is:

\$H(j\omega) = 100 \dfrac{\frac{j\omega}{\omega_0}}{1 + \frac{j\omega}{\omega_0}}\$

\$ \omega_0 = \frac{10}{T}\$

Then:

\$ |H(j\omega)| = 100 \dfrac{\frac{\omega}{\omega_0}}{\sqrt{1 + (\frac{\omega}{\omega_0})^2}}\$

Now, if we stare at this a bit and ask it some questions, we can imagine exactly what this looks like.

When \$ \omega << \omega_0\$, the denominator is effectively "1" and so, the transfer function is decreasing by a factor of 10 as \$ \omega\$ decreases by a factor of 10. On a log-log scale, this is a line with a slope of +1.

When \$ \omega >> \omega_0\$, the denominator is effectively \$ \frac{\omega}{\omega_0} \$ and so the transfer function is effectively constant with a value of 100.

If we plot these two lines on a log-log plot and have them intersect at \$ \omega = \omega_0\$, we've created an asymptotic Bode magnitude plot. In fact, it's easy to see that when \$ \omega = \omega_0\$, the magnitude is \$ \frac{100}{\sqrt{2}}\$ so the lines we plotted are actually the asymptotes of the (magnitude) transfer function. That is, the function approaches these lines at the frequency extremes but never actually gets to them (on a log-log plot, \$ \omega = 0 \$ is at negative infinity)

A rough calculation for the first harmonic at 2kHz - assuming a gain decrease of 40dB/dec above 700 Hz - results in a damping factor of app. 9.44 (19.5dB). Hence, the first harmonioc at 2kHz should have an amplitude of 10/9.44=1.06 volts.

Visual justification: At f=1kHz we have a damping of 7.5dB (factor 2.37)

EDIT1: Due to some incorrect numbers (obtained from the diagram) I have corrected the damping values slightly.

EDIT2: Simulation of a corresponding 2nd-order lowpass with a squarewave input results in sinusoidal signal (good quality) with an amplitude of app. 1.3 V. This is more than roughly calculated above - however, the filter data couldn`t be quite exact because the evaluation of the given magnitude response is not error-free (pole frequency?, 0 dB crossing of the asymptote or the actual curve ?).