How Do You Calculate Insertion Loss in a T Network?

[David J]

Homework Statement

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I am trying to follow a given example as part of my notes. The example is described below

Using the equation below calculate the insertion loss of the T network of figure 20 (figure 20 is in the attachment) given the values in the table.

I am given a transmission matrix for the T network and values as follows:-

$R_S = 300\Omega$
$R_L = 300\Omega$
$R_A = 200\Omega$
$R_B = 200\Omega$
$R_C = 300\Omega$

Homework Equations

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The example equation we are given is as shown below

$A_{IL} = 20\log\left[\frac {AR_L + B +(CR_L + D)R_S)^2 }{R_S +R_L}\right]$

The Attempt at a Solution

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The example solution I am given in my notes is shown below in its entirety :-

$A_{IL} = 20\log\left[\frac {AR_L + B +(CR_L + D)R_S)^2 }{R_S +R_L}\right]$

So $A_{IL} = 20\log\left[\frac{ 1 + \left(\frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C} R_L+\left(1+\frac{R_A}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

So $A_{IL} = 20\log\left[\frac{ 1 + \left(\frac{200}{300}\right)300+200+200+\frac{200\times200}{300}+\left(\frac{1}{300} 300+\left(1+\frac{200}{300}\right)\right)300}{300+300}\right]$

The example then moves straight to

$A_{IL} = 20\log\left(3.056\right)$

and then an answer of

$A_{IL} = 9.70dB$

The above is what I am given as an example. My problem is that even when I follow each step as shown above I cannot arrive at $3.056$

I end up with $3.648055556$ which gives an answer of $11.241dB$ which is not correct.

My working out is shown below.

Starting with

$= 20\log\left[\frac{ 1 + \left(\frac{200}{300}\right)300+200+200+\frac{200\times200}{300}+\left(\frac{1}{300} 300+\left(1+\frac{200}{300}\right)\right)300}{300+300}\right]$

$= 20\log\left[\frac{\left(1.666666667\right)\times833.3+\left(1+1.666666667\right)\times300}{600}\right]$

$= 20\log\left[\frac{\left(1.666666667\right)\times833.3+800}{600}\right]$

$= 20\log\left[\frac{2,188.833334}{600}\right]$

$= 20\log\left(3.648055556\right)$

So using my working out I get $A_{IL} = 11.241dB$ which is in correct

As I said in the beginning of this post this is an example question I am given and I am trying to work my way through it. I just need any advice as to why my final result does not match the example. I am obviously doing something wrong somewhere, I just cannot see where. I am thinking its to do with rounding off numbers etc and I am not able to program recurring numbers into my calculator (casio fx-100MS) so that may have something to do with it. Maybe I have messed up the calculations out with the brackets or something along those lines. Any advice would, as always be most appreciated.

Thanks

 

[gneill]

In the attached image it shows the D term of the matrix being given by

$D = 1 + \frac{R_A}{R_B}$

But in their solution they appear to have used $R_C$ in place of $R_B$. That is, they've employed the term A rather than D. So either the attachment is incorrect or they've used the wrong term.

 

[The Electrician]

I think the D element of the given transmission matrix is wrong. It should be 1+RB/RC, not 1+RA/RB.

 

[The Electrician]

When you went from here:
$A_{IL} = 20\log\left[\frac {AR_L + B +(CR_L + D)R_S)^2 }{R_S +R_L}\right]$

To here:
So $A_{IL} = 20\log\left[\frac{ 1 + \left(\frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C} R_L+\left(1+\frac{R_A}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

You messed up your parentheses in the first term after the left [. You should have:

$A_{IL} = 20\log\left[\frac{\left( 1 + \frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C} R_L+\left(1+\frac{R_B}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

And, as I said in the previous post, the D term of the given transmission matrix is wrong, but it doesn't matter because RA = RB.

 

[The Electrician]

You made a couple of mistakes going from here:
$= 20\log\left[\frac{ 1 + \left(\frac{200}{300}\right)300+200+200+\frac{200\times200}{300}+\left(\frac{1}{300} 300+\left(1+\frac{200}{300}\right)\right)300}{300+300}\right]$

to here:
$= 20\log\left[\frac{\left(1.666666667\right)\times833.3+\left(1+1.666666667\right)\times300}{600}\right]$

You should have gotten:

$= 20\log\left[\frac{\left(1.666666667\right)\times 300+533.3+\left(1+1.666666667\right)\times300}{600}\right]$

 

[David J]

Good morning and thanks for the information. This gives me something to work on today. I can see where the D term is wrong and I think that is typo from the university question paper. I have attached the full example question and answer for your info.

I also attached the question I will shortly attempt (question 4) as this shows what I think is the "correct D term" of the matrix, as you have suggested

I can also see where i put the bracket in the wrong place on the first parenthesis. I don't know how I missed that because its written in the example answer in black and white exactly how to go about working out so I will put this down to an arithmetic mistake. I have spent quite a long time on this now and I simply did not see it

I will re attempt this example today to try and match what they say I should get.

Appreciated as always. thanks

 

[gneill]

I also attached the question I will shortly attempt (question 4) as this shows what I think is the "correct D term" of the matrix, as you have suggested

Yes, that shows the correct terms for the ABCD matrix for the given network.

The ABCD matrix for a 2-port network is fairly easy to find, and I really should have done it earlier when I suspected there was an issue with the D term used in the provided solution. Thankfully @The Electrician picked up the ball there.

When each component is either in series or a shunt, they have 2 x 2 matrix forms that can be multiplied in succession to arrive at the overall ABCD matrix. In this instance R_A and R_B are series connected to the input and output respectively, while R_C is a shunt lying between them.

Here's a quick derivation for the ABCD matrix for this network that I've done using MathCad:

 

[David J]

When you went from here:
$A_{IL} = 20\log\left[\frac {AR_L + B +(CR_L + D)R_S)^2 }{R_S +R_L}\right]$

To here:
So $A_{IL} = 20\log\left[\frac{ 1 + \left(\frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C} R_L+\left(1+\frac{R_A}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

You messed up your parentheses in the first term after the left [. You should have:

$A_{IL} = 20\log\left[\frac{\left( 1 + \frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C} R_L+\left(1+\frac{R_B}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

Hello again. Its a bit clearer now thanks. Looking at the first equation above i failed to recognize $AR_L$ as, effectively $\left(1 + \frac{R_A}{R_C}\right)R_L$
If I had noticed this then I would probably have gone with

$\left(1.66666667\right)R_L$ = $\left(1.66666667\right)300$ = $500$

I have managed to contact the university regarding the error in the example question matrix and they did confirm its wrong but as you have pointed out here, in this case, $R_A=R_B$ so it was not the reason I miscalculated. The reason was down to me failing to recognize $AR_L$

I will have a go at the question now

 

[David J]

So approaching the actual question (attached) i get the following

$A_{IL} = 20\log\left[\frac{\left( 1 + \frac{R_A}{R_C}\right)R_L+R_A+R_B+\frac{R_A R_B}{R_C}+\left(\frac{1}{R_C}R_L+\left(1+\frac{R_B}{R_C}\right)\right)R_S}{R_S+R_L}\right]$

Input the given data and simplify

$A_{IL} = 20\log\left[\frac{\left( 1 + \frac{13}{213}\right)\times100+13+13+\frac{13\times13}{213}+\left(\frac{1}{213}100+\left(1+\frac{13}{213}\right)\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{\left(1.06103\right)\times100+13+13+\frac{13\times13}{213}+\left(\frac{1}{213}100+\left(1+\frac{13}{213}\right)\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{\left(106.103\right)+26.793+\left(\frac{1}{213}100+\left(1+\frac{13}{213}\right)\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{\left(106.103\right)+26.793+\left(0.4695+\left(1+\frac{13}{213}\right)\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{\left(106.103\right)+26.793+\left(0.4695+1.06103\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{\left(106.103\right)+26.793+\left(1.53083\right)75}{175}\right]$

$A_{IL} = 20\log\left[\frac{106.103+26.793+114.81225}{175}\right]$

$A_{IL} = 20\log\left[\frac{247.70825}{175}\right]$

$A_{IL} = 20\log\left(1.415475714\right)$

So $A_{IL} = 3.01805dB$

I believe this to be correct. Any comments, as always much appreciated, thanks

 

[gneill]

Looks good. You'll want to round your result to a couple of decimal places.

 

[David J]

Thanks again

so the final result will be $A_{IL}=3.02 dB$

much appreciated, thanks again for help and advice gents