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Electrician:
Yes, that's the image impedance, sorry i didnt look at your original link i was in a hurry.
You've calculated the same values i found using help from GIT, which Ratch claims won help for this
One thing i dont think you explained (or i missed something) is what you mean by 'iterative impedance'. the 'iterative impedance' value i get is the same as the square root equation.
Care to explain a little more about the method you mentioned?
Ratch:
You say that GIT wont help, but it does help, you just cant use it *once*, you have to use it *twice*. Alternately, you can use it iteratively until you find the right value,
but im not sure yet if this is the same 'iteratively' that Electrician is talking about. Notice the value for Rout that Electrician got is the same value i got previously. and that's not the same value that we get for the open circuit output impedance using GIT or current perturbation.
Confusion is arising here. The impedances known as "iterative impedances" have been known as such for about 100 years. You are describing the result of your square root equation as an "iterative impedance value" and it is not the same as the classical "iterative impedance". When I am discussing these things, I will use the word "iterative" only in connection with the classical concepts. I will use the phrase "repeated impedance value" to denote the number(s) you have calculated, and which are the same as the classical "image impedances". You can see the possibilities for confusion here, when the impedances that you are calculating, the "repeated impedances", are the same as the classical "image impedances", but not the same as the classical "iterative impedances" (explained further below).
The image impedances (Zi1 and Zi2) are these: when Zi2 is connected as a load to a 2-port, the impedance looking into the input of the 2-port will be Zi1; when Zi1 is connected to the input of a 2-port, the impedance looking into the output of the 2-port will be Zi2. The image impedances are the impedances which simultaneously match the input and output of the 2-port.
This page shows it well:
https://www.transtutors.com/homework-help/Networks-Systems/two-port-network/image-impedance/
The classical "iterative impedance" (not the same as the "image impedance") is the result of cascading a lot of identical 2-ports and calculating the input and output impedances of the cascade; this is why they called it "iterative impedance". Imagine the last 2-port in the cascade; its input impedance becomes the load for the next to last 2-port. The input impedance of the next to last 2-port is affected by the fact that its load is the input impedance of the last 2-port. As you work your way back the cascade, eventually the input impedances of each 2-port approaches a particular value. That value is the impedance (it is the output iterative impedance), which, if it is the load of the 2-port, will cause the input impedance of the 2-port to be the same as that load. For the transistor circuit under discussion, its value is 10316.2 ohms.
And, of course, it works the same in the other direction. The input iterative impedance is that impedance, which, when connected as the driving impedance for the 2-port, will cause the output impedance to have the same value as that driving impedance (the input iterative impedance).
From this thread:
https://www.electro-tech-online.com/threads/audio-transformers-and-two-ports.124261/
"Given a two port, measure the input impedance with the output port open-circuited; call that impedance Zoc. Then measure the input impedance with the output port short-circuited; call that impedance Zsc. Then Zi1 = SQRT(Zsc*Zoc); this value is the geometric mean of Zsc and Zoc. The same procedure can be followed to determine the output image impedance, Zi2.
This same procedure is well known as a method to determine the characteristic impedance of a transmission line. A length of transmission line is essentially a two-port; it has an input and output port.
"
I would use the word "repeated" where the red "iteratively" is above. It is being used to mean something other than what the classical "iterative impedance" means.
I used the word "iterative" in "iterative impedance"; I didn't use "iteratively".
You should realize that the short circuit and open circuit techniques require measurements
Looks simple right? But each of those impedance parameters has to be measured
You *really* felt that you had to clear up the fact that you used the word "iterative" rather that "iteratively"? Isnt that a bit much? I think we all here can read English good enough to tell the difference without someone pointing it out![]()
iPass Rout Rin
307.49 11832.23 (predicted with Sqrt formula)
1 355.67 11707.97
2 307.72 11831.57
3 307.49 11832.22
4 307.49 11832.23
5 307.49 11832.23
6 307.49 11832.23
7 307.49 11832.23
8 307.49 11832.23
9 307.49 11832.23
10 307.49 11832.23
GreatYou didnt specify what method(s) you used, but if you dont want to that's ok i guess, your input is still appreciated.
Hi again,
Electrician:
Ok you cleared up some things nicely, but i didnt say that you 'had' to measure, i said you can either 'measure' or 'calculate'.
You did a static 'calculation', which is fine, but that's still not what Ratch and I were after here. We were after a *single* formula, without any sub calculations, that would calculate the (open circuit) output impedance, and then later the 'image' impedances...again a formula not a static value. The static value comes at the end after the formula is developed.
In other words, we do want this:
356 or 307 ohms
but that's not all we were after. We also wanted this:
Rout_Image=function(R2,R3,R4,hie,hfe)
In other words, a formula with the resistors in it (not the static values but the variable names R2, R3, etc.) which usually comes out to some algebraic function.
This means if we want to 'calculate' the values with say the impedance parameters, we can not just say:
z11, z12, etc.,
we have to say:
z11(R2,R3,R4,hie,hfe), z12(R2,R3,R4,hie,hfe), etc.,
so each parameter becomes a function rather than a static value like 1.2345 or 56.321 or what have you.
You did however clear some things up very nicely so you did help a lot here with or without that kind of formula, so your input was greatly appreciated.
So you were using the word 'iteratively' as a placeholder for a more elaborate idea, that's completely understandable..
Here is the result of my iterated (or repeated) calculation:
Code:iPass Rout Rin 307.49 11832.23 (predicted with Sqrt formula) 1 355.67 11707.97 2 307.72 11831.57 3 307.49 11832.22 4 307.49 11832.23 5 307.49 11832.23 6 307.49 11832.23 7 307.49 11832.23 8 307.49 11832.23 9 307.49 11832.23 10 307.49 11832.23
As can be seen above, it only takes 4 iterations (repetitions) to converge. This speed of convergence may not be typical however.