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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.

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    \$\begingroup\$ Just make your y axis G/K in dB \$\endgroup\$ – Scott Seidman Jun 25 '16 at 2:18
  • \$\begingroup\$ Thats sound good, but if i want to measure wc0 from the graph what to do ? because it will be changing with K. \$\endgroup\$ – saad abouzahir Jun 25 '16 at 2:21
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    \$\begingroup\$ No, it won't change with K \$\endgroup\$ – Scott Seidman Jun 25 '16 at 2:22
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    \$\begingroup\$ 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. \$\endgroup\$ – user2943160 Jun 25 '16 at 2:24
  • \$\begingroup\$ pulse at 0dB, it did change, i choosed k=1 and it was about wc0 = 1, and for K=10 it becomes wc0 = 10 \$\endgroup\$ – saad abouzahir Jun 25 '16 at 2:25
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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.

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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:

enter image description here

enter image description here

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.

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