Sketch Bode Plot for $\frac{0.5}{1 - \frac{3145j}{w}}

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Discussion Overview

The discussion revolves around sketching the Bode plot for the transfer function \(\frac{0.5}{1 - \frac{3145j}{w}}\). Participants explore the mathematical manipulation of the function and its implications for the Bode plot, including the identification of different components of the function and their contributions to the overall plot.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a step-by-step transformation of the function into a more convenient form for sketching the Bode plot, suggesting it can be expressed as \((\frac{0.5}{3145})(s)(\frac{1}{(\frac{s}{3145}+1)})\).
  • Another participant identifies two separate functions to be combined for the Bode plot: a differentiating function \(H1=0.5*s\) with a slope of +20 dB/dec and a first-order lowpass \(H2=1/(s+3145)\) with a horizontal line below the pole frequency and a -20 dB/dec slope above it.
  • Some participants clarify that the function can be viewed as a single function \(F(s) = \frac{ks}{Ts+1}\), emphasizing the identification of constants \(k\) and \(T\) for the Bode plot.
  • There is a discussion about the approach to sketching the Bode plot, with one participant suggesting that they would draw the three separate functions and then combine them, while others focus on the overall function and its segments.

Areas of Agreement / Disagreement

Participants generally agree that the function can be represented in different ways for the purpose of sketching the Bode plot. However, there is some disagreement regarding whether to treat it as a single function or as a combination of separate functions, indicating multiple competing views on the approach to the Bode plot.

Contextual Notes

Some participants express uncertainty about the form required for the lowpass function, noting that they were taught to express it as \(1/(s+1)\), which may influence their approach to the problem.

TheRedDevil18
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Homework Statement



\frac{0.5}{1 - \frac{3145j}{w}}, sketch the bode plot

Homework Equations

The Attempt at a Solution



Moving the jw to the bottom
\frac{0.5}{1+ \frac{3145}{jw}}

Let s = jw

\frac{0.5}{1+ \frac{3145}{s}}

= \frac{0.5}{\frac{s+3145}{s}}

= (0.5) \frac{s}{s+3145}

= (0.5)(s) \frac{1}{3145(\frac{s}{3145}+1)}

= (\frac{0.5}{3145})(s) (\frac{1}{(\frac{s}{3145}+1)})

So are those the three separate bode functions that I have to sketch ?
 
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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.
 
TheRedDevil18 said:

Homework Statement



\frac{0.5}{1 - \frac{3145j}{w}}, sketch the bode plot

Homework Equations

The Attempt at a Solution



Moving the jw to the bottom
\frac{0.5}{1+ \frac{3145}{jw}}

Let s = jw

\frac{0.5}{1+ \frac{3145}{s}}

= \frac{0.5}{\frac{s+3145}{s}}

= (0.5) \frac{s}{s+3145}

= (0.5)(s) \frac{1}{3145(\frac{s}{3145}+1)}

= (\frac{0.5}{3145})(s) (\frac{1}{(\frac{s}{3145}+1)})

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.
 
LvW said:
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
 
rude man said:
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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