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Drawing bode plot

  • #27
NascentOxygen
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This for ILS's new problem? The break points (corner frequencies) will not be different when both factors are identically 1/(s+2). The break points must coincide.

The -40dB/dec slope is correct. :smile:
 
  • #28
Femme_physics
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Well, I followed the same logic as in the first exercise, didn't I?

I don't see why would they coincide if that same logic indicates differently.
 
  • #29
NascentOxygen
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I haven't been able to follow ILS's "shortcut" method, but here's the amplitude plot with double poles at s=-2. So whatever method you use, to be valid it must produce this answer for 400/((s+2)(s+2)).

Note: horiz scale is Hz, not rad/sec, and Ѡ=2 rad/sec corresponds to f≃⅓ Hz.
 

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  • #30
gneill
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I think you'll find that the factored version of the transfer function spells out the poles conveniently in terms of the natural angular frequency (rad/sec).

attachment.php?attachmentid=55070&stc=1&d=1359268363.gif


It's as easy as that.

For single poles the slope increases by 20 db/decade, for double poles like ##(s + 2)^2##, the slope increases by 40 db/decade (that's still 20 db/decade per pole).
 

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  • #31
Femme_physics
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I think you'll find that the factored version of the transfer function spells out the poles conveniently in terms of the natural angular frequency (rad/sec).
Then this logic does not match Nascent's graph who posted right before you. The break point seems to be at 0.1

the slope increases by 40 db/decade
Increase or decrease - that depends whether this expression is in the nominator or denominator
 
  • #32
gneill
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Then this logic does not match Nascent's graph who posted right before you. The break point seems to be at 0.1
Right. His graph has frequency f in Hz along the x-axis. Natural frequency ω is in rad/sec, where ##ω = 2\pi f##.
Increase or decrease - that depends whether this expression is in the nominator or denominator
Heh. Yes, by "increase" I meant steeper descent for each pole. Zeros in the numerator have the opposite effect, tilting the slope upwards by 20 db/decade for each zero.
 
  • #33
NascentOxygen
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Then this logic does not match Nascent's graph who posted right before you. The break point seems to be at 0.1
That's because gneill helpfully threw in a new transfer function, for clarity.
Increase or decrease - that depends whether this expression is in the nominator or denominator
Certainly.
 
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  • #34
gneill
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That's because gneill helpfully invented another transfer function, for clarity.
Ah yes, er, that too :smile:

Actually I went back to the original transfer function for the thread, since I though it might be helpful to compare with the related plots.
 
  • #35
NascentOxygen
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I went back to the original transfer function for the thread, since I though it might be helpful to compare with the related plots.
The original TF had poles at -1 and -4. Though easy to get confused here, I know. :wink:
 
  • #36
gneill
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The original TF had poles at -1 and -4. Though easy to get confused here, I know. :wink:
Aurrgh! So much for short term memory. Heck, it was only three pages ago, too!
 
  • #37
NascentOxygen
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  • #38
I like Serena
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The way I described it, is how I learned it some time ago.

However, after gneill's and NascO's comments, I've looked it up on wiki, and I found this section on how to draw Bode plots by hand:
http://en.wikipedia.org/wiki/Bode_plot#Rules_for_handmade_Bode_plot
This is basically what they already said.

In the end I believe it is somewhat arbitrary, since the real transfer function plot does not consist of straight lines.
But I certainly concur that we should follow the standard methods.

Here's the graph I've made with the "actual" transfer function for ##G(s)={400 \over (s+1)(s+4)}##, my version, and NascO/gneill's version.

attachment.php?attachmentid=55077&stc=1&d=1359286223.jpg


As you can see, it doesn't really matter (my version is actually closer), but I do propose to follow NascO/gneill's suggestions.
 

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  • #39
NascentOxygen
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Well, now that ILS has cleared that up.... :wink:

there is no stopping yet, because as I pointed out earlier, we are only halfway there!

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon2.gif [Broken] Don't forget that the Bode Plot requires a plot of angle vs. frequency as well as dB vs. frequency.

BTW, I think we have left FemmePhysics some miles back along the way. Should we turn back to look for her?
 
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  • #40
gneill
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BTW, I think we have left FemmePhysics some miles back along the way. Should we turn back to look for her?
I'm sure that FP will track us down in due course. Probably just waiting for the dust to settle :smile:
 
  • #41
Femme_physics
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I'm here! Except, I already had the Control Systems semester test yesterday. The sad part is that everything BUT the Bode Plot I'm sure I got right ;) But at least I didn't have to DRAW a graph, the graphs were given to me, I just had to find the function that describes it. So far I'm able to keep a 90+ average (out of a 100) with the help of physicsforums, or most specific you guys.
I had some solved examples to work with...hopefully I got it right. I might approach this thread before the final test of my degree (May-July), provided you'll still be around :)

I really appreciate your help, especially ILS that always goes above and beyond (amazing). If I didn't get the bode plots right I'll come here to whine, rest be assured!

Now I have machine parts test to take in 5 days so I better get in shape. :-) I appreciate your help, guys.
 

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