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Bode Plot

  • Thread starter Duave
  • Start date
  • #1
81
0
Bode Plot It's The biggest one in the entire universe!

Homework Statement


The question is stated at the top of the image below.

Did I answer the TWO questions with this image?


https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-frc3/t1/1891009_10151901549150919_346181425_n.jpg



Homework Equations



The equation that I used is in the image

The Attempt at a Solution



The entire image is an attempt to solve the question shown at the very top of the image.

If I did not answer the TWO questions correctly, can you please point out where I made the error so that I can fix it.

Thank you

Regards,
Duave
 
Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,722
2,716
[STRIKE]You want to put some readable text in your posting template, not just a single link. Give the helper some idea of whether or not the question is one he's interested in or capable of answering without having to follow a link. When one is asking for help from volunteers, it pays to provide a clear information "bite" about the problem that can be assessed at a glance.[/STRIKE] -- EDIT: Poster corrected his "image problem" :smile:

Looking at the labels on your plot, it would appear that you've exchanged the order of some of the digits in your f3dB's. The values are okay later on, and the value that you get for the attenuation at 60Hz looks okay to me (I'm seeing an attenuation of -24.96 dB, so the difference is probably a matter of rounding/significant digits used in intermediate steps).
 
Last edited:
  • #3
81
0
[STRIKE]You want to put some readable text in your posting template, not just a single link. Give the helper some idea of whether or not the question is one he's interested in or capable of answering without having to follow a link. When one is asking for help from volunteers, it pays to provide a clear information "bite" about the problem that can be assessed at a glance.[/STRIKE] -- EDIT: Poster corrected his "image problem" :smile:

Looking at the labels on your plot, it would appear that you've exchanged the order of some of the digits in your f3dB's. The values are okay later on, and the value that you get for the attenuation at 60Hz looks okay to me (I'm seeing an attenuation of -24.96 dB, so the difference is probably a matter of rounding/significant digits used in intermediate steps).
gneill,

Thank you for your time. I made the corrections that you mentioned about. I had put numbers in the diagram that weren't right. Can you also look at this LOW - pass filter question that I answered. This time the high frequency is to the right. Is everything correct what I did?

https://scontent-b.xx.fbcdn.net/hphotos-ash3/t1/1661845_10151901762450919_830470428_n.jpg

Thank you
 
  • #4
gneill
Mentor
20,722
2,716
For the low pass case the numbers look okay. Again, I get something just a tad larger for the attenuation: -51.55 dB.

You could probably do to label the vertical axes on your plots and give an indication of scale. At least indicate the 0 dB reference.
 
  • #5
FOIWATER
Gold Member
434
12
I agree looks good, Have you ever derived these formulas for cut-off frequency yourself?

They can be found using the transfer functions of the circuit.

If you consider the high pass filter...

The Transfer function is the output voltage divided by the input voltage. The output voltage is IR
The input voltage is I(R+1/jwc)
So the transfer function is R/(R+1/jwc) If you find a common denominator for the bottom...

jwRC/(1+jf/(1/2piRC))

This is called the pole-zero form of the transfer function, the top gives the gain at zero frequency (DC) and the pole gives the corner (or -3db) frequency. This is useful to find the equation for -3dB yourself, try it out for the low-pass filter..
 
  • #6
81
0
I agree looks good, Have you ever derived these formulas for cut-off frequency yourself?

They can be found using the transfer functions of the circuit.

If you consider the high pass filter...

The Transfer function is the output voltage divided by the input voltage. The output voltage is IR
The input voltage is I(R+1/jwc)
So the transfer function is R/(R+1/jwc) If you find a common denominator for the bottom...

jwRC/(1+jf/(1/2piRC))

This is called the pole-zero form of the transfer function, the top gives the gain at zero frequency (DC) and the pole gives the corner (or -3db) frequency. This is useful to find the equation for -3dB yourself, try it out for the low-pass filter..

gneill & FOIWATER,

Thank you for your responses.
With respect to the transfer function I have a set of questions that are implanted within the illustrations below. It is another set of questions that I tried to answer for two RL circuits. High-Pass and Low-Pass. If I answered these questions correctly, then maybe I can derive the formula for cut-off frequency.

Did I answer all three questions correctly with both of the illustrations

P.S. The Questions are right there at the top of the picture

High Pass (RL-Circuit)

https://scontent-b.xx.fbcdn.net/hphotos-ash3/t1/1620934_10151902130255919_876213440_n.jpg


Low Pass (RL-Circuit)
https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn2/t1/1800438_10151902130275919_737264272_n.jpg
 
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