# Insertion loss in a T network

1. Feb 1, 2017

### David J

1. The problem statement, all variables and given/known data

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:-

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

2. Relevant equations

The example equation we are given is as shown below

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

3. The attempt at a solution

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

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

$A_{IL} = 9.70dB$

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

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2. Feb 1, 2017

### Staff: Mentor

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.

3. Feb 1, 2017

### The Electrician

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

4. Feb 1, 2017

### 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.

5. Feb 1, 2017

### 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:

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

6. Feb 2, 2017

### 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 dunno 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

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7. Feb 2, 2017

### Staff: Mentor

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 RA and RB are series connected to the input and output respectively, while RC is a shunt lying between them.

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

8. Feb 2, 2017

### David J

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

9. Feb 2, 2017

### 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

#### Attached Files:

• ###### Question 4.jpg
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10. Feb 2, 2017

### Staff: Mentor

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

11. Feb 2, 2017

### David J

Thanks again

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

much appreciated, thanks again for help and advice gents