Help with bode plots

  • Thread starter rusty009
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  • #1
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Main Question or Discussion Point

Hey, could some one please help me with bode plots. I have a transfer function,lets say

G(s)=0.16s/(1+0.16s)

if I want to find the gain at 10 rads/s, do I subsitute s for 10 and then take 20log( G(10) )? Thats what I thought the method was, but now in my notes it says,

Gain = 0.16ω/√(1 + (0.16ω)^2)

then 20log(Gain)

I have a bode plot for this transfer function, which is in the same notes, and it agrees with the first method. Thanks in advance.
 

Answers and Replies

  • #2
MATLABdude
Science Advisor
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In actuality, it's [tex]s=\sigma+j\omega[/tex], so you should be substituting in s = 10j, and finding the frequency response using the transfer function. If you substituted that in directly, you'd have the complex gain.

But since you usually want to find the magnitude of the gain (and separately find the phase difference), you actually want to find [tex]\left|G(j\omega)\right|[/tex] using the following (note that the asterisk denotes complex conjugate):

[tex]\left|G(j\omega)\right|^{2}=G(j\omega)G^{\ast}(j\omega)
\left|G(j\omega)\right|=\sqrt{\left|G(j\omega)\right|^{2}}[/tex]

EDIT: 20log10(gain) just gives you the gain in terms of dB. LaTeX seems to be broken at the moment, so hopefully the equations show up in a few hours. If not, you should be able to click on the red "LaTex broken" placeholder to bring up the code snippet, and then paste it into say, the following (after you remove the tex tags from both ends of the snippet):
http://www.codecogs.com/components/equationeditor/equationeditor.php [Broken]
 
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