Bode Plot

  • #1

Homework Statement



[itex] \frac{0.5}{1 - \frac{3145j}{w}} [/itex], sketch the bode plot

Homework Equations




The Attempt at a Solution



Moving the jw to the bottom
[itex] \frac{0.5}{1+ \frac{3145}{jw}} [/itex]

Let s = jw

[itex] \frac{0.5}{1+ \frac{3145}{s}} [/itex]

[itex] = \frac{0.5}{\frac{s+3145}{s}} [/itex]

[itex] = (0.5) \frac{s}{s+3145} [/itex]

[itex] = (0.5)(s) \frac{1}{3145(\frac{s}{3145}+1)} [/itex]

[itex] = (\frac{0.5}{3145})(s) (\frac{1}{(\frac{s}{3145}+1)}) [/itex]

So are those the three separate bode functions that I have to sketch ?
 

Answers and Replies

  • #2
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I would start somewhat earlier (3rd line from the bottom formula): Two separate functions to be combined:
1.) Differentiating function : H1=0.5*s : Straight line with a slope of +20dB/dec crossing the f-axis at w=2)
2.) First order lowpass H2=1/(s+3145) : Horizontal line (1/3145) below the pole frequency ; -20dB/dec above the pole frequency.
 
  • #3
rude man
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Homework Statement



[itex] \frac{0.5}{1 - \frac{3145j}{w}} [/itex], sketch the bode plot

Homework Equations




The Attempt at a Solution



Moving the jw to the bottom
[itex] \frac{0.5}{1+ \frac{3145}{jw}} [/itex]

Let s = jw

[itex] \frac{0.5}{1+ \frac{3145}{s}} [/itex]

[itex] = \frac{0.5}{\frac{s+3145}{s}} [/itex]

[itex] = (0.5) \frac{s}{s+3145} [/itex]

[itex] = (0.5)(s) \frac{1}{3145(\frac{s}{3145}+1)} [/itex]

[itex] = (\frac{0.5}{3145})(s) (\frac{1}{(\frac{s}{3145}+1)}) [/itex]

So are those the three separate bode functions that I have to sketch ?
What? It's just one function: F(s) = ks/(Ts+1)
where you can easily identify k and T, right?
You did just fine, your last expression is in the most convenient form for drawing the Bode plot.
There are two segments, one corresponding to the zero at s=0 and one to the pole at s = -1/T.
 
  • #4
I would start somewhat earlier (3rd line from the bottom formula): Two separate functions to be combined:
1.) Differentiating function : H1=0.5*s : Straight line with a slope of +20dB/dec crossing the f-axis at w=2)
2.) First order lowpass H2=1/(s+3145) : Horizontal line (1/3145) below the pole frequency ; -20dB/dec above the pole frequency.
For number 2, I was taught that it has to be in the form 1/(s+1), which is why I took the 3145 out
 
  • #5
What? It's just one function: F(s) = ks/(Ts+1)
where you can easily identify k and T, right?
You did just fine, your last expression is in the most convenient form for drawing the Bode plot.
There are two segments, one corresponding to the zero at s=0 and one to the pole at s = -1/T.
Yes, It's one function but I meant that I would draw those three separate functions and then add them together

So the first one would be a constant line 20log(0.5/3145), the second one, s, would be a differentiation with +20 db/dec slope and the last one would be zero gain from 0 to 3145 rad/s and then -20db/dec slope
 

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