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Can you do explain how to draw a Bode plot with GH(s)?

How do I get the slope and intersection point?

This is one of the teacher's examples. He drew it but I don't know what it means. Can you explain it?

I only understood that the gain is 20 log lGHl.

enter image description here enter image description here

RNN master
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    Just a note: the final slope is labeled as -20dB/decade -- it should be -40dB/decade because there's three poles and one zero. So your instructor gets 9 out of 10 on that exam question, or maybe 9.5 out of 10, given that they've clearly demonstrated that they know what they're doing, by showing all their work. – TimWescott Dec 15 '20 at 17:08

1 Answers1

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How do I get the slope and intersection point?

Start with a very simple example: -

$$H(s) = \dfrac{1}{s}$$

When \$s = 1\$, the amplitude of the transfer function is clearly 1.

When \$s = 0.1\$, the amplitude is 10.

When \$s = 10\$, the amplitude is 0.1.

If you convert those amplitudes to dB you would have: -

  • When \$s = 0.1\$, gain = 20 dB
  • When \$s = 1\$, gain = 0 dB
  • When \$s = 10\$, gain = -20 dB

So, immediately you can see that if s increases by 10 the gain falls by 20 dB. This is why we call the slope -20 dB/decade: -

enter image description here

It's a little bit different with this: -

$$H(s) = \dfrac{1}{s + 1}$$

  • When \$s = 0\$, the gain is 1 (this is a DC gain of 1)
  • When \$s \neq 0\$ we have to remember that s has a complex value so, if \$s = 0.1\$, the transfer function becomes this: -

$$H(s) = \dfrac{1}{j0.1 + 1}$$

And, to solve that for gain magnitude we add the square of the terms in the denominator and take the square root like this: -

$$|H(s)| = \dfrac{1}{\sqrt{0.1^2 + 1^2}} \approx \dfrac{1}{1.005} \approx 1$$

Notice that we made an approximation here.

The impact of that approximation is that we say gain is flat from DC up to at least \$s = 0.1\$. If we then made \$s = 1\$ we get this: -

$$|H(s)| = \dfrac{1}{\sqrt{1^2 + 1^2}} = \dfrac{1}{\sqrt2} = 0.7071 = \text{-3 dB point}$$

For this particular equation (\$\frac{1}{s+1}\$), when \$s = 1\$ we are at the 3 dB point.

This is because 20 log(0.7071) = -3.01 dB. We call it the 3 dB point loosely but it's really the -3 dB point (get used to it!).

However, because we are wanting to sketch a frequency response from the H(s) formula, we don't mind a diagram error of 3 dB and so, we say when sketching it, that the transfer function is flat from DC all the way to \$s = 1\$.

We can draw a straight line on the bode plot at 0 dB from a very low frequency to the frequency where \$s = 1\$. Above that frequency at (say) \$s = 10\$, the gain magnitude is: -

$$|H(s)| = \dfrac{1}{\sqrt{10^2 + 1^2}} = \dfrac{1}{\sqrt{101}}$$

This is a gain magnitude of 0.0995 or about -20 db.

So then we draw another straight line falling from 0 dB at \$s = 1\$ to -20 dB at \$s = 10\$. The line (red) will clearly continue falling at 20 dB per decade hence it continues to -40 dB at s = 100: -

enter image description here

Image above from this website and you can put your own numbers into the page and get your own specific results. It even provides you the phase angle plots.

Here's what your actual transfer function is on the page above: -

enter image description here

And it results in this magnitude plot: -

enter image description here

I think you should be able to manage this from now?

Andy aka
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  • Thank you! but about the site I cant put number. i edited post and put about the site's problem, can you do add this number? i dont know what is the problem – RNN master Dec 15 '20 at 18:12
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    @RNNmaster firstly, no need to add these things to your question - it just confuses people coming along later and makes them think I'm not answering stuff correctly so please delete that question modification. If you can't add stuff to the linked site maybe it's temporarily offline or you've not entered a valid character or maybe you need to wait for 30 seconds? – Andy aka Dec 15 '20 at 18:15
  • Ohh I am sorry, I wanted put the Numerator Polynomial, N(s): s+5 and Numerator Polynomial, D(s): s^4+ 60 s^3+500 s^2 Multiplicative Constant, 50. but this cant plot so..asked to you again. thank you – RNN master Dec 15 '20 at 18:19
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    @RNNmaster I couldn't see anything wrong with what you entered. Maybe the gap between 60 and s^3 is the problem. Don't leave white spaces maybe? – Andy aka Dec 15 '20 at 18:20
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    Just mess around with it and take your time. I never used it until I searched for it 2 hours ago so I'm a beginner on it too. – Andy aka Dec 15 '20 at 18:23
  • Really thank you to help me.. but I dont know why that site cant change to plot window, anyway your pc can do it? – RNN master Dec 15 '20 at 18:24
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    I don't know - you've now probably used that site more than me so, you are now THE EXPERT - the baton is passed to you LOL. – Andy aka Dec 15 '20 at 18:26
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    Nitpick: "When s > 0 we have to remember that s has a complex value" complex numbers are not comparable, so saying s > 0 is meaningless when s can be complex. – Hearth Dec 16 '20 at 05:01
  • @Hearth maybe when |s| > 0 ?? Please do suggest a better way. – Andy aka Dec 16 '20 at 09:51
  • @Andyaka That would be how I'd write it. Either that or "when s ≠ 0". – Hearth Dec 16 '20 at 13:27
  • @Hearth you have persuaded me.... – Andy aka Dec 16 '20 at 13:29