Help with Bode Plots | Gain Calculation

[rusty009]

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.

 

[MATLABdude]

In actuality, it's $$s=\sigma+j\omega$$, 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 $$\left|G(j\omega)\right|$$ using the following (note that the asterisk denotes complex conjugate):

$$\left|G(j\omega)\right|^{2}=G(j\omega)G^{\ast}(j\omega) \left|G(j\omega)\right|=\sqrt{\left|G(j\omega)\right|^{2}}$$

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

 

[oswaler]

Hello,

Thank you for reaching out for help with Bode plots and gain calculation. It seems like you are on the right track with your understanding. Let me explain the two methods you mentioned in more detail.

The first method you mentioned is called the "direct substitution" method. It involves substituting the desired frequency (ω) for s in the transfer function and then taking the logarithm of the magnitude of the resulting expression. In your case, the desired frequency is 10 rads/s. So, you would substitute 10 for s and then take the logarithm of the magnitude of G(10) (which is 0.16*10/(1+0.16*10) = 1.6/2.6 = 0.615). This would give you 20log(0.615) = -2.5 dB. This method is commonly used to quickly estimate the gain at a specific frequency.

The second method you mentioned is called the "magnitude formula" method. It involves using a formula to calculate the magnitude of the transfer function at a given frequency. In your case, the formula is Gain = 0.16ω/√(1 + (0.16ω)^2). Using this formula, you would plug in 10 for ω and then take the logarithm of the resulting value. This would also give you -2.5 dB, as the two methods should agree with each other.

It's great that you have a Bode plot for this transfer function, as it can serve as a visual representation of the gain and phase at different frequencies. The Bode plot should also show a gain of -2.5 dB at 10 rads/s, confirming your calculations.

In conclusion, both methods are valid and should give you the same result. The direct substitution method is quicker and can be used for estimation, while the magnitude formula method is more accurate. I hope this helps clarify your understanding of Bode plots and gain calculation. Let me know if you have any further questions. Best of luck with your studies!