# Drawing a bode plot of transfer function with unknown gain

I want to draw bode plot of this transfer function:

$$G(p) = {K \over p \space (1+0.1p) \space (1+0.05p)}$$

But I don't know what to do with that K (static gain) -- I've only drawn TF with known gain.

• Just make your y axis G/K in dB – Scott Seidman Jun 25 '16 at 2:18
• Thats sound good, but if i want to measure wc0 from the graph what to do ? because it will be changing with K. – saad abouzahir Jun 25 '16 at 2:21
• No, it won't change with K – Scott Seidman Jun 25 '16 at 2:22
• Assume K is 1. Then include the change in gain as a linear vertical shift in the final bode plot. For this (homework?) question, the K is relatively inconsequential versus working with the second-order system. – user2943160 Jun 25 '16 at 2:24
• pulse at 0dB, it did change, i choosed k=1 and it was about wc0 = 1, and for K=10 it becomes wc0 = 10 – saad abouzahir Jun 25 '16 at 2:25

The shape of the function is exactly the same for all values of K (assuming you're drawing a Bode plot). Different values of K just mean a translation of the graph upwards for higher values or downwars for lower values.

Ok, let's do some math to explain more explicitly what has been said in some comments to your question.

Let's rewrite G as another TF multiplied by K:

$$G(p) = K \cdot G_n(p)$$

where

$$G_n(p) = \dfrac{1} {p \cdot (1+0.1p) \cdot (1+0.05p)}$$

is the normalized (with respect to K) TF.

Let's define the logarithmic (dB) amplitude response of the system this way:

$$A_{(dB)}(\omega) = 20 \log_{10} \left| G(j\omega) \right|$$

We see easily that:

$$A_{(dB)}(\omega) = \\[1em] = 20 \log_{10} \left| K \cdot G_n(j\omega) \right| = \\[1em] = 20 \log_{10} \left| K \right| + 20 \log_{10} \left| G_n(j\omega) \right| = \\[1em] = K_{(dB)} + A_{n(dB)}(\omega)$$

Where $A_{n(dB)}$ is the amplitude response relative to the normalized TF and $K_{(dB)}$ is the constant K expressed in dB:

\begin{align*} A_{n(dB)}(\omega) &= 20 \log_{10} \left| G_n(j\omega) \right| \\[1em] K_{(dB)} &= 20 \log_{10} \left| K \right| \end{align*}

From that you can see that the only difference in the amplitude Bode plot between the original and the normalized TF is just a vertical shift, so the corner frequencies of both plots will remain the same.

Here is an LTspice simulation that shows practically the situation:

Of course I had to choose a value for K (100 = 40dB), but you can easily see that any change to K will just change the amount of the vertical shift.