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Drawing bode plot

  • #1
Femme_physics
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Homework Statement



Given this function in an open loop control system

http://img255.imageshack.us/img255/3686/700qa.jpg [Broken]

Draw a bode plot of the transfer function: The upper graph will present the lines for each of the amplification factors for each of the system components, the lower graph will sum them all.

Homework Equations



http://img189.imageshack.us/img189/693/400al.jpg [Broken]


The Attempt at a Solution



I haven't drawn the second graph they asked for (that sums all the components) , only the first...what do you think?

http://img521.imageshack.us/img521/1471/7001w.jpg [Broken]

http://img69.imageshack.us/img69/2880/ithink.jpg [Broken]
 
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Answers and Replies

  • #2
I like Serena
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  • #3
Femme_physics
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Hey Fp!

Mmh, I showed you WolframAlpha before, didn't I?

Check this out! :smile:
http://m.wolframalpha.com/input/?i=bode+plot&x=0&y=0&f1=400/((s+1)*(s+4))&assumption=*F.BodePlotCalculator.transferfunc-_400/((s+1)*(s+4))

Does the first plot look what you have?

Btw, how did you get your 0.25 and 1.25?
And what is the reason you think it's going up from 1.25?
Well, sadly I can't use Mr. WolframAlpha-solve-everything in the test, just a basic scientific calculator... :( Regardless, the answer doesn't help me see how I get there... *scratches head*

Do you see how I got to the characteristic formula?

So I end up with one component that is

1/0.25S^2

And another compnent that is

5/4S

If you look at the formulas I posted, I just took the figure associated with each S component
and started drawing from there. Isn't that what those numbers indicate?
 
  • #4
rude man
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There are no upward-sloping asymptotes in your final diagram, just flat one(s) and downward-sloping ones.

To get an upward-sloping one you would need an (s+a) term in the numerator. All you have in the numerator is a constant.
 
  • #5
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Yes, that is about right.

Hmm, you can't use mr. Wolfram-solve-everything can you?
What a bummer!
Ah well, it's a good reason to learn some more math! :wink:
Do you know the rules how to deal with ##\log##?

Basically, you get 3 straight lines.
These lines are given by:

$$[1]\quad dB = 20 \log 100 = 40$$
$$[2]\quad dB = 20 \log({100 \over \frac 5 4 s}) = 20 \log({80 \over s}) = 20 \log 80 - 20 \log s$$
$$[3]\quad?$$

Can you tell what the third line is?

You are supposed to put this in a log-log graph, that has the ##dB## on the vertical axis and ##\log s## on the horizontal axis.
The first line is a horizontal line that intersects the y-axis at 40 dB.
The second line is a line that slopes down; it intersects the y-axis at ##20 \log 80##, and it has a (downward) slope of ##-20##.

Can you draw those lines, including the third?
 
  • #6
Femme_physics
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  • #7
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Good!
You have the right formula for the 3rd line!

Even better for the 3rd line is: ##dB = 20 \log 400 - 40 \log s##
That way you can see that the downward slope is ##-40##.

But I wrote question marks where it's unclear to me what the figures are there
The first and second question mark should be zero.
That is, the second line should go down starting from the interection with the y-axis.
The same for the third line.

Furthermore the 3rd line should go down twice as steep as the 2nd line.

The actual Bode plot is:
along the 1st line until it intersects the 2nd line (this is actually left of the y-axis).
along the 2nd line until it intersects the 3rd line.
along the 3rd line until you go off the paper.

Now you should be able to calculate the 3rd and 4th question mark...


Next post I'll include a reference to a Wolfram plot to show you what I mean... (unless of course you already got it . ;p)
 
  • #8
Femme_physics
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The first and second question mark should be zero.
What made you decide it?

Now you should be able to calculate the 3rd and 4th question mark...
I don't see how... do I just approximate based on the graph? In that case, how do I know the angle of 20 db/dec and 40 db/dec of how I draw the lines exactly?
 
  • #9
gneill
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Your original formulation for the transfer function gives you everything you need to know :smile:

http://img255.imageshack.us/img255/3686/700qa.jpg [Broken]

If you take the limit as s goes to zero you'll get the gain when frequency goes to zero. Convert to db. That's where your plot starts for very low frequencies, and begins as a straight horizontal line (sketch one in lightly on your plot).

The two terms in the denominator tell you that there will be a couple of -20 db/decade slopes to add in. Further, the numerical constants 1 and 4 tell you what the corner frequencies are (essentially where the new slopes will begin). These frequencies are natural frequencies: radians per second.

You can determine what a slope of -20 db/dec looks like by placing a straightedge on your graph paper so that it spans a frequency range of a decade and drops 20 db over that span.

These slopes are additive, so as you add terms starting at their corner frequencies, you 'bend" the plot line so that it achieves the required total slope.
 
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  • #10
Femme_physics
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If you take the limit as s goes to zero you'll get the gain when frequency goes to zero.
But that's exactly the problem - we haven't studied how to take the limit as S goes to zero--we only studied to solve bode plot through laplace plane... although I don't recall a question that asks drawing every component separately yet
 
  • #13
Femme_physics
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What the fajebus! My manual's answer is different than yours!


EDIT: could we have picked up on a mistake?
 
  • #14
gneill
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But that's exactly the problem - we haven't studied how to take the limit as S goes to zero--we only studied to solve bode plot through laplace plane... although I don't recall a question that asks drawing every component separately yet
I imagine that you've done limits before. For the present case the limit is trivial; just set s to zero. What's the result?

Sketching an approximation of the bode magnitude plot for a given transfer function is a matter of identifying the "corner frequencies" where slope changes (where a new term of the function starts to manifest its effect) and approximating that effect by a change in slope for the line.
 
  • #15
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I do not understand the top diagram. It seems to show something different.
Perhaps it belongs to a different problem?
Or it is an example of something?

The second diagram looks like mine.
The slopes of the lines are the same.
And the locations of the connections are almost the same.

Note that 0 on the x-axis in my diagram corresponds to ##10^0## in yours.
And 1 corresponds to ##10^1## in yours.
Furtermore my x-axis is at a different dB level. Mine is at dB=0.

Can't explain the slight difference yet, though.
 
  • #17
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It looks to me as if in your plot the 2nd line is slightly off.
The 2nd line intersects the y-axis at 40 dB, while it should be slightly lower (at 20 log 80).
That would explain the slight differences.
They made an approximation for the 2nd line.
 
  • #18
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fajebus?
 
  • #19
Femme_physics
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I'll go with your logic, ILS, the manual doesn't make any sense to me! How do you exactly decide where they cross the X axis, is the one thing I haven't quite gotten yet.

I imagine that you've done limits before. For the present case the limit is trivial; just set s to zero. What's the result?
We really haven't done limits! I'd have remembered that.
Settings S -> 0 for the original Gs I get

Gs = 400 / 4 = 100

Sketching an approximation of the bode magnitude plot for a given transfer function is a matter of identifying the "corner frequencies" where slope changes (where a new term of the function starts to manifest its effect) and approximating that effect by a change in slope for the line.
I'm still not sure what angle do I give -20 db/dec and -40 db/dec

I'll have to figure it out tomorrow, bedtime here. Be back in the morning. Thank you :)
 
  • #20
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I'll go with your logic, ILS, the manual doesn't make any sense to me! How do you exactly decide where they cross the X axis, is the one thing I haven't quite gotten yet.
You have the equations (I have calculated the values and replaced "log s" by dec):
$$[1]\quad dB = 40$$
$$[2]\quad dB = 38 - 20 \cdot dec$$
$$[3]\quad dB = 52 - 40 \cdot dec$$

The find the point where they cross the X axis, fill in dB=0 and solve for dec.


I'm still not sure what angle do I give -20 db/dec and -40 db/dec
A slope of -20 dB/dec means that the line goes down 20 dB if the dec increases by 1 point.

To make drawing them easier, you should draw for instance the 2nd line from ##(0,38)## to ##(\frac {38} {20},0)##.
If you do that, you will have the right slope.



And err... what is fajebus?
 
  • #21
Femme_physics
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Heh, fajebus is from The Simpsons :) Humor calls Jesus Jebus, and with a little more wordplay you end up with "what the fajebus!" :D


http://img441.imageshack.us/img441/2391/graphhhh.jpg [Broken]

How do I know how to turn it to one graph that combines all three?

You have the equations (I have calculated the values and replaced "log s" by dec):
0 = 38 - 20dec
dec = 38/20 = 1.9

0 = 52 - 40dec
dec = 52/40 = 1.3

If that' the case it doesn't match the graph, as the lines reach beyond "2"
 
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  • #22
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Heh, fajebus is from The Simpsons :) Humor calls Jesus Jebus, and with a little more wordplay you end up with "what the fajebus!" :D
Aha! I keep learning new things!



How do I know how to turn it to one graph that combines all three?
That's how a Bode plot works.

You have the formula
$$G(s) = {100 \over {1 + \frac 5 4 s + \frac 1 4 s^2}}$$

For small values of s, say s=0.001, or s=0.0001, the denominator ##1 + \frac 5 4 s + \frac 1 4 s^2## is almost 1.
Certainly if you round to a couple of digits.
The terms ##\frac 5 4 s## and ##\frac 1 4 s^2## can be neglected relative to 1.
(Calculate them if you don't believe me.)

For very big values of s, say s=100, or s=1000, it's the other way around.
The third term is the biggest by far, and the terms ##1## and ##\frac 5 4 s## can be neglected.

Somewhere in the middle the second term dominates the others.
But you'll have to draw the lines to find out where he is on top.




0 = 38 - 20dec
dec = 38/20 = 1.9

0 = 52 - 40dec
dec = 52/40 = 1.3

If that' the case it doesn't match the graph, as the lines reach beyond "2"
Oops. I made a mistake! :surprised
Can you spot it?

EDIT: No I didn't.
Check it on this plot:

http://m.wolframalpha.com/input/?i=plot[{20*log_10(100),20*log_10(80)-20x,20*log_10(400)-40x},+{x,-1,3}]&x=0&y=0
 
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  • #23
NascentOxygen
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For the record I do have the solution from the manual, I just don't know what to make of it
http://img802.imageshack.us/img802/29/bode1.jpg [Broken]
Hi Fp! The transfer function you are dealing with is:

G(s) = 400/❨(1+s) (4) (1+s/4)❩

Notice how I have converted each term involving s into the form (1+s/Ѡ) because when this is plotted piecewise linear it is a horizontal line up until s=Ѡ at which point the graph changes to have a slope of 20dB/decade. For s « Ѡ you can see that (1+s/Ѡ) is approximately 1 and that's a horizontal line of 0dB; for s » Ѡ the 1 term is insignificant and (1+s/Ѡ) is approximated by s/Ѡ, and a plot of s/Ѡ is a straight line of slope 20dB/dec.

So your transfer function comprises the sum of 3 plots:
▸ 100 ....which on a dB scale is a constant 40dB
▸ 1/(1+s) ....which is a horizontal line of 0dB until s=1 then it dives at -20dB/dec
▸ 1/(1+s/4) .... which is a horiz line of 0dB until s=4 then it dives at -20dB/dec

These are shown on the graph you provided above.. (Remember, 100 = 1)
The product of terms in G(s) become added when drawn on a dB scale (since dBs are logarithms and the logarithm of X.Y is logX + logY)

I do understand how this graph is achieved, however:

http://img834.imageshack.us/img834/6669/bode2.jpg [Broken]

http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon2.gif [Broken] Don't forget that the Bode Plot requires a plot of dB vs. frequency together with a plot of angle vs. frequency.
 
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  • #24
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Check this out.

attachment.php?attachmentid=54964&stc=1&d=1359029940.jpg

Does this look more like your graph?
(Courtesy of Excel.)


To quickly find the corner points there is another trick:

Your denominator is ##1+\frac 5 4 s + \frac 1 4 s^2##.

The first corner point occurs when the first 2 terms are equal.
That is when ##1 = \frac 5 4 s##.
That is the point where the domination of ##1## ends and the domination of ##\frac 5 4 s## begins.
(These domination games should appeal to you, don't they?)
Calculate ##s##?

The second corner point occurs when the last 2 terms are equal: ##\frac 5 4 s = \frac 1 4 s^2##. What is ##s##?

Do those points match the graph?
 

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  • #25
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Suppose you take the transfer function:
$$G(s)={400 \over (s+2)^2}$$

Can you make a Bode plot for that one now?
(You should be able to recognize the result. ;))
 

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