# Two-port network models

[/SUB]

## 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 Ω

R2 = 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

#### Attachments

• P2.2(a).png
18.9 KB · Views: 411
Last edited:

rude man
Homework Helper
Gold Member
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

rude man
Homework Helper
Gold Member
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:

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.

Last edited:
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 dont 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:
rude man
Homework Helper
Gold Member
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.