Is there a problem with this Source Transformation?

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Discussion Overview

The discussion revolves around a source transformation method used to find the transfer function of a circuit. Participants are examining the differences between the user's result and an expected solution, focusing on the implications of Thévenin equivalents and impedance considerations.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant, danie9, presents a transfer function derived from a source transformation and notes a discrepancy with a provided solution.
  • Another participant suggests that danie9's analysis is correct but points out a potential oversight regarding the Thévenin impedance, indicating that it should not be assumed to be zero ohms.
  • A third participant offers a suggestion for clarity in notation, recommending the use of a cursive lower-case "s" to avoid confusion with the numeral "5" in complex algebra.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correctness of the transfer function, as there is a disagreement regarding the treatment of the Thévenin impedance in the analysis.

Contextual Notes

The discussion highlights the importance of considering loading effects in circuit analysis, particularly when using Thévenin equivalents, but does not resolve the specific mathematical discrepancies presented.

Danie9
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ssadas.jpg

Hey guys, i did this source transformation as an alternate method to find the transfer function of a circuit, however I am getting a different transfer function of 2/(2s+(s+3)(s^2+1)) to the solution in the following image. Any help would be really appreciated :)
solution2.JPG
 
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Hi danie9! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Your analysis using the first method looks right.

The mistake you're making in the second is that while representing the first stage by its Thévenin voltage source you are forgetting about its Thévenin impedance. The result is as though there exists a unity gain buffer amplifier between the two stages because you are, in effect, assuming/using a Thévenin impedance of zero ohms. But there is no such isolating amplifier. So as well as redrawing with a voltage source you need to include the impedance of that voltage source to properly account for the loading effect of stage II on stage I.
 
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To avoid the likelihood of sooner or later mistaking your lower-case "s" for the numeral "5" somewhere in the middle of many pages of complex algebra, consider adopting a cursive lower-case letter "s" for your electronics work. Writing it like this will obviate one glaring opportunity for going astray in your work. (Leave off the lower-right upward flourish, that stem is needed only when joining to a following alphabet letter.)

http://www.kbteachers.com/alphabet/cursive/lowercase-s.gif ;);)
 
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NascentOxygen said:
Hi danie9! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Your analysis using the first method looks right.

The mistake you're making in the second is that while representing the first stage by its Thévenin voltage source you are forgetting about its Thévenin impedance. The result is as though there exists a unity gain buffer amplifier between the two stages because you are, in effect, assuming/using a Thévenin impedance of zero ohms. But there is no such isolating amplifier. So as well as redrawing with a voltage source you need to include the impedance of that voltage source to properly account for the loading effect of stage II on stage I.
Yes thank you so much, i will definitely start using a cursive 's' :)
 
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