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

In summary: Yes, you are correct. The first line would have a constant amplitude of 20log(0.5/3145), the second would have a slope of +20dB/decade, and the third would have a slope of -20dB/decade after 3145 rad/s. These three segments would be added together to get the final Bode plot.
  • #1
TheRedDevil18
408
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 ?
 
Physics news on Phys.org
  • #2
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
TheRedDevil18 said:

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
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
 
  • #5
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
 

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

What is a Bode plot?

A Bode plot is a graph that depicts the frequency response of a system, showing how the magnitude and phase of a signal are affected by different frequencies. It is commonly used in engineering and science to analyze the behavior of a system.

How do I sketch a Bode plot?

To sketch a Bode plot, you first need to determine the transfer function of the system. This can be done by taking the ratio of the output signal to the input signal. Then, plot the magnitude and phase of the transfer function against frequency on a logarithmic scale. The magnitude is typically plotted in decibels, and the phase is plotted in degrees.

What is the transfer function for the given equation?

The transfer function for the given equation is H(jw) = 0.5 / (1 - 3145j/w). This represents a low-pass filter, as the output signal is inversely proportional to the input signal at higher frequencies.

What does the j symbol represent in the equation?

In this equation, the j symbol represents the imaginary unit, which is used to denote the presence of complex numbers. In the context of a Bode plot, it represents the phase shift of the signal at different frequencies.

How can a Bode plot be used to analyze a system?

A Bode plot can provide important information about the behavior of a system, such as its stability, frequency response, and gain margin. By analyzing the magnitude and phase plots, engineers can identify the dominant frequencies and the overall response of the system. This can be helpful in designing and optimizing systems to meet specific requirements.

Similar threads

Replies
3
Views
2K
Replies
4
Views
3K
Replies
5
Views
2K
Replies
5
Views
4K
Replies
2
Views
2K
Back
Top