Is there a problem with this Source Transformation?

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The discussion centers on a user's attempt to find the transfer function of a circuit using source transformation, resulting in a discrepancy with the expected solution. The error identified is the omission of the Thévenin impedance when representing the first stage as a voltage source, leading to an incorrect assumption of zero impedance. This oversight creates an illusion of a unity gain buffer amplifier between stages, which does not exist. To correct the analysis, the user is advised to include the Thévenin impedance to accurately reflect the loading effect. The user acknowledges the feedback and plans to adopt a cursive 's' in their work to avoid confusion in future calculations.
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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