Design a 15 dB Symmetrical Attenuator with 600Ω Characteristic Impedance

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Discussion Overview

The discussion revolves around designing a symmetrical π section attenuator that provides a voltage attenuation of 15 dB with a characteristic impedance of 600 Ω. Participants explore various approaches to the calculations involved in the design, including the use of specific formulas and the interpretation of decibel values.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution using specific formulas for R1, R2, and R3 based on the given parameters, calculating R1 and R2 values.
  • Another participant challenges the initial calculations, stating that their own equations yield different results, particularly for R2, which they claim is double the calculated value.
  • Several participants discuss the interpretation of decibels, noting that for voltage or current, the formula is dB = 20 log(ratio), while for power, it is dB = 10 log(ratio).
  • One participant expresses confusion regarding the inclusion of a negative sign in the attenuation value, questioning its necessity in the context of their calculations.
  • Another participant suggests that the term "characteristic impedance" may include the load, indicating a potential misunderstanding of the problem's requirements.
  • There is a reference to an external resource for further clarification on the π pad attenuator design, with participants seeking confirmation on their interpretations and calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the interpretation of the attenuation value. Multiple competing views and approaches to the problem remain, with ongoing debate about the correct formulas and assumptions.

Contextual Notes

Participants express uncertainty regarding the definitions and implications of characteristic impedance and attenuation, as well as the correct application of formulas in their calculations. There are unresolved mathematical steps and differing interpretations of the problem's requirements.

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Homework Statement



Design a ∏ section symmetrical attenuator to provide a voltage attenuation of 15 dB and a characteristic impedance of 600 Ω.

Homework Equations



ZS = ZL = Z

R1 = R3 = Z (K + 1 / K -1)

R2 = Z (K2 - 1 / 2 K)

The Attempt at a Solution



Z = 600Ω
K = 15dB = 15(15/20) = 5.6234

R1 = R3 = 600 (5.6234 + 1 / 5.6234 - 1)

= 859.54 ΩR2 = 600 (5.62342 - 1 / 2 x 5.6234)

= 600 (30.6226 / 11.2468)

= 1633.67 Ω

Can someone please confirm if I am on the right lines? Thanks
 

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Not what I got. I don't know where you got your formulas. I just wrote a set of equations to force Zin = 600 ohms from both sides, and an attenuation of 15 dB = gain of 0.1778. My R1 was not too much below yours but my R2 was exactly double yours. Interesting coincidence!

Maybe I'm misunderstanding what the problem asks for. My pi network looks like 600 ohms from both directions and the attenuation is 15 dB. I'm not assuming a source nor a load impedance.
 
Hi, when talking about current or voltage, it is db = 20log(ratio).

If you are taking the ratio of powers, then it is dB = 10log(ratio).

We are using first one, which gives us 15db= 20 log (n),

10 (15/20) = n

10 ( 3/4) = n

4√ 10 (3) = 5.6234 = k

I have found similar example but still want to get confirmation is it right or not
 
agata78 said:
Hi, when talking about current or voltage, it is db = 20log(ratio).

If you are taking the ratio of powers, then it is dB = 10log(ratio).

We are using first one, which gives us 15db= 20 log (n),

10 (15/20) = n

10 ( 3/4) = n

4√ 10 (3) = 5.6234 = k

EDIT: I think I misinterpreted the meaning of "characteristic impedance". Aparently it includes the load.

Use this:
http://www.electronics-tutorials.ws/attenuators/Pi-pad-attenuator.html


I have found similar example but still want to get confirmation is it right or not

If it's attenuation, it's -15dB, not 15 dB, which is 10^(-15/20) = 01778 = k.
 
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I know that attenuation means the reduction of signal strength during transmission. But how you can tell its with this example?
Your calculations would be right, but i don't know why is minus there.
 
Thanks but it is actually the same example that I was using, but I still can't find a minus there in that example!
 
Last edited:
agata78 said:
Thanks but it is actually the same example that I was using, but I still can't find a minus there in that example!

If you go by the link, attenuation is considered a positive quantity.

Follow their example.
 

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