Bode plot question...

[CraKaJaX]

Studying for my LAST SEMESTER OF THE SEMESTER!!! But I'm a bit confuzzled. This is one of the practice exam questions (with solution) ... but I'm stumped on the magnitude part. For 40 =< omega =< 400... how does he get 800/omega? Wouldn't it be an exact value? Am I missing something? Don't you just plug in a value of omega and then take the greater of the 2 (1 or the ratio of omega/whatever?

 

[RESmonkey]

last semester of the semester?

 

[CraKaJaX]

Originally posted by: RESmonkey
last semester of the semester?

Sorry, I'm burnt out. That would be last exam of the semester

 

[RESmonkey]

At least you made it this far. I died when I had to do assembly/machine programming.

 

[drinkmorejava]

Isn't that funny; I have a modeling and controls test on thursday with bode plots

 

[CraKaJaX]

Originally posted by: drinkmorejava
Isn't that funny; I have a modeling and controls test on thursday with bode plots

That's great, so you can answer it right? :laugh:

 

[krotchy]

Heres what I see,
0-P1 - Rising with value = 20log(5*w) (0<w<4)
P1-P1 - Flat response, so it is constant@a=20log(20)=26dB (4w<40)
P2-Z - Falling response with value = 20log(800/w) (40<w<400)
Z - Infinity - Flat again@B=2, constant 20log(2) = 6dB


at exactly 40, 800/40 = 20, which is the same as the flat response at the knee
at exactly 400, 800/400 = 2 which is the same as flat response at the final value

between then its a constant drop of 20db/dec

 

[drinkmorejava]

Originally posted by: CraKaJaX

Originally posted by: drinkmorejava
Isn't that funny; I have a modeling and controls test on thursday with bode plots

That's great, so you can answer it right? :laugh:

Nope, haven't studied for it yet, and I have an aeroelasticity exam tomorrow

 

[SparkyJJO]

I survived bode plots.

I don't think I really fully understood them. Probably because of the use of logarithms, which still confuse me. Calculus, makes sense, logarithms, nope.

 

[frostedflakes]

Originally posted by: RESmonkey
At least you made it this far. I died when I had to do assembly/machine programming.

You should have stuck with it. Assembly was very weird at first (for me at least), but I'm sure you would have picked it up quick enough.

OP, looks like he just derived the magnitude as a function of ? from the bode plot. For example you can use y=mx+b (where y is |H(?)| and x is ?). Solve for b using known values to obtain the general equation for the line.

Between 40 and 400 I got |H(?)|=(-1/20)?+22, which gives the same values as ?/800. Your instructor must have used some method to simplify the equation that isn't obvious to me.

 

[Born2bwire]

Is it pronounced: "bode" or "bode - eeeee"?

I've only had one professor say the word verbally and he pronounced it as "bode." But I have heard a few students tack on a "ey" sound and say "bode-ey." What's the pronounciation you guys have heard?

 

[krotchy]

double response..... oops

 

[krotchy]

Originally posted by: krotchy

Originally posted by: Born2bwire
Is it pronounced: "bode" or "bode - eeeee"?

I've only had one professor say the word verbally and he pronounced it as "bode." But I have heard a few students tack on a "ey" sound and say "bode-ey." What's the pronounciation you guys have heard?

Bow - dee

As for their usefulness, by hand they are virtually worthless but apply pspice and you can optimize a filter pretty easily, especially when you need to draw an AC signal off a DC voltage (for data and power on the same line)

 

[Leros]

We touched on bode plots in systems and signals. I didn't really get it. Analog ftl.

 

[blahblah99]

The area between 40 rad/s and 400 rad/s is approximately 1/40 * / (1+s/40) for unity gain. Since the gain is actually about 20, you have a coefficient of 800 ( 20 / (1/40) = 800). Because it's a 20dB drop per decade between those two frequencies, the magnitude is about 800/omega.

Looks confusing at first, but after you work with bode plots for a while you can do it all in your head.

 

[esun]

Originally posted by: Leros
We touched on bode plots in systems and signals. I didn't really get it. Analog ftl.

Originally posted by: SparkyJJO
I survived bode plots.

I don't think I really fully understood them. Probably because of the use of logarithms, which still confuse me. Calculus, makes sense, logarithms, nope.

Sadly, the real issue isn't that logarithms or Bode plots are inherently difficult. It's that so many professors suck at explaining them. It's a very simple mathematical formula to find magnitude and phase, and it's a very simple approximation to use poles and zeros for transfer functions. It's just that for some reason professors don't explain it well.

 

[CraKaJaX]

Originally posted by: krotchy
Heres what I see,
0-P1 - Rising with value = 20log(5*w) (0<w<4)
P1-P1 - Flat response, so it is constant@a=20log(20)=26dB (4w<40)
P2-Z - Falling response with value = 20log(800/w) (40<w<400)
Z - Infinity - Flat again@B=2, constant 20log(2) = 6dB


at exactly 40, 800/40 = 20, which is the same as the flat response at the knee
at exactly 400, 800/400 = 2 which is the same as flat response at the final value

between then its a constant drop of 20db/dec

My confused part was for P1-P2... you got 800/jw by multiplying out the 4 and 40 to the numerator to get 800/jw... is that the standard form or something? Does it always have to be in that form? Or is that your way of doing it?

Thanks for the help though guys

 

[chorb]

I somehow got a B+ in that class, I dont remember a dam thing. good luck.