- When adding an extra element in the place of a short in the original system a voltage test source will enable you to find Zd. If you elect to use a current test source instead you will end up finding 1/Zd.
- When adding an extra element in the place of an open in the original system a current test source will enable you to find Zd. If you elect to use a voltage test source instead you will end up finding 1/Zd.
That said for problem 2a) if I wanted to treat the inductor as an extra element while creating a short in the original system then I should add a voltage test source to find Zn and Zd? By nulling vout does this effectively short the transformer leaving only the capacitor impedance for Zn?
Thanks,
Adam,
ReplyDeleteI think the results you described for the test sources mentioned is right.
In P2a, replacing the inductor EE with a test voltage source sounds reasonable. In finding Zn, nulling Vout is not the same as shorting anything. If you work backwards from the output, there will be 0 voltage across R and C and 0 current throughout the secondary. All that is left is the test voltage source. Because this is considered an ideal voltage source, its impedance is infinite so vtest/itest = inf.
- KT
Adam,
ReplyDeleteI don't agree with your first statement. This does not hold in the example given in class and HW problem 1.
Please see page 5 of the annotated notes: "L07_5807_out1"
Regards,
A-Dub
I also replaced the inductor with a voltage test source for Zn and Zd. However, the voltage across the capacitor is also nulled such that the voltage through the second winding, v_2 is equal to the voltage through the test voltage so that v = v_2 + v_test, 0 = v_2 + v_test.
ReplyDeleteAdam,
ReplyDeleteMy comment in response to your discussion of test sources was wrong. Zd is always vtest/itest (this was mentioned in class today actually). Zd and 1/Zd come into play when system transfer functions are considered.
- KT
Laura,
ReplyDeleteI agree with your result vtest = -V2. However, I'm struggling to understand how to computer Zn = v_test/i_test. If the secondary side has 0 current because the vout is nulled then doesn't that mean the primary side also has 0 current. If there is no current anywhere how can we find an expression for Zn? I've found expressions for V1, V2, Vg, and the associated voltage loops, but can't figure out how to include i_test in any equation. Any suggestions?
I agree with Keith's first statement when he said, "...vtest/itest = inf".
DeleteThe denominator (itest) goes to zero. Therefore Zn blows up to infinity.
Is it possible to push Vg and L through the transformers to simplify things and then go the rest of the way using EET? Or does this violate the statement regarding using the EET method to solve the problem? What does everyone think?
ReplyDeleteThat's what I did and I got an answer that matches with 7.15, so I think it is OK to do it.
ReplyDelete