Winterstone
Banned
PG1995,
I don´t understand your question.
The equation as derived by you includes 3 resistors and 2 capacitors (including the element I call Y5 which in your drawing is Y5=sC1).
That means: All 5 elements are present.
After element allocation for lowpass behaviour (proper selection of R and C) you will notice that there is an expression (Y1+Y5), which simply means: Y1=1/R1 in parallel to Y5=1/R5 .
Thus: Rp=R1||R5.
Thus, to realize a particular lowpass transfer function the element Rp must have a certain value which can be realized by R1 only (why use two R in parallel when one R can do the job?).
Please note the word "CAN". Of course, if you like you can design also with a resistor R5.
That´s what I mean with "redundant".
Perhaps you have dropped R5 from the beginning (because somebody or Wikipedia told you this).
But how could you know? When you like to understand a circuit and the process of proper dimensioning you need the equation including all elements. Only then you can decide if one part can be dropped or not.
I hope I was able to explain it clearly.
REMARK: By the way, the fact that - after proper grouping - you see that Y5 appears only tgether with Y1 (in a sum) is one of the advantages of using admittances Y rather than impedances Z.
In case of Z it would me not easy to see that R1||R5.
I don´t understand your question.
The equation as derived by you includes 3 resistors and 2 capacitors (including the element I call Y5 which in your drawing is Y5=sC1).
That means: All 5 elements are present.
After element allocation for lowpass behaviour (proper selection of R and C) you will notice that there is an expression (Y1+Y5), which simply means: Y1=1/R1 in parallel to Y5=1/R5 .
Thus: Rp=R1||R5.
Thus, to realize a particular lowpass transfer function the element Rp must have a certain value which can be realized by R1 only (why use two R in parallel when one R can do the job?).
Please note the word "CAN". Of course, if you like you can design also with a resistor R5.
That´s what I mean with "redundant".
Perhaps you have dropped R5 from the beginning (because somebody or Wikipedia told you this).
But how could you know? When you like to understand a circuit and the process of proper dimensioning you need the equation including all elements. Only then you can decide if one part can be dropped or not.
I hope I was able to explain it clearly.
REMARK: By the way, the fact that - after proper grouping - you see that Y5 appears only tgether with Y1 (in a sum) is one of the advantages of using admittances Y rather than impedances Z.
In case of Z it would me not easy to see that R1||R5.
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