How do I find gain values between decades on asymptotic Bode plots?

Question

I was doing some past exams for my control theory final and came across a question with the following asymptotic Bode plot given:

In order to solve for \$ \omega_c\$, I need to essentially find the equation of the line crossing 0 dB.

Using \$y=mx+c\$:

\begin{align*} \underbrace{\text{Gain @ } \omega = 50}_{y} = \underbrace{-20\log(\omega)}_{mx} + c \end{align*}

To solve for this \$y\$ I know that I cannot just take it as \$30\$ which would be half way between \$20 \text{ and } 40\$ as \$5\$ is halfway between the corresponding decade.

I already have the solution for it which involves first finding the line equation for the line between \$1 \text{ and } 5 \text{ rad/s}\$ but I don't really understand it.

What I am wondering if it is possible to find the gain using:

\begin{align*} \text{Difference between Gain @ $\omega = 10$ & @ $\omega = 5$} &= 20\log(\frac{5}{10})\\ \text{Difference between Gain @ $\omega = 10$ & @ $\omega = 5$} &= -20\log(2)\\ \text{Difference between Gain @ $\omega = 10$ & @ $\omega = 5$} &\approx -6\\ \end{align*}

This does give me the correct answer as 34 dB but it's just something I sort of figured out tinkering around. I am not exactly sure why this works. Is it a legitimate tool?

Sidenote: Calculators are not allowed which is why we have to delve into all of this hassle.

Answer 1

In this particular case finding the "y" value is not that difficult.

There is only one case that you need to keep in mind: $$ 20\log(\sqrt{2}) \approx 3 \text{dB}$$

As such,

$$10\log(2) \approx 3$$ $$10\log(2) \approx 3$$ $$\log(2) \approx \frac{3}{10}$$

Now acccording to the asymptotic bode plot, the gain at \$\omega = 50\$ is the same as the gain at \$\omega = 5\$.

To find the gain at \$\omega = 5\$:

$$20\log(5) = 20\log(\frac{10}{2})$$ $$20\log(10) - 20\log(2) = 20(1-0.3) = 14\text{dB}$$

This is the gain added by the zero @ \$\omega = 1\$ and the actual gain at this point is \$14+20 = 34\text{dB}\$.

Knowing the value of \$\log(2)\$ is quite important because this way you can get the values associated with the 3 dB cutoff frequency and since a lot of numbers are divisible by 2, you can use what I did to get the appropriate gain.

Answer 2

At 5 rad/s the level is 20 dB + (20 dB * log5) = 33.98 dB

Frequency at 0 dB = 10^(33.98 dB/20 dB) * 50 = 2500 rad/s

So it's basically a matter of using log base 10 to convert between the horizontal log scale and the linear vertical scale and then antilog to convert the other way, knowing that the 20 dB/decade and -20 dB/decade slopes are at 45 degrees to the horizontal.