Hello,
In HW5, problem 1, I cannot seem to determine the s^2 numerator zero contributed by L2,C1. For determining the first order zero contributed by C1, I assumed a samll resistance 'r' in series with L2 and the result was: v^test/i^test = r. Now, to determine the zero contributed by L2C1 together, I have set C1 as short (abnormal) and put the test source across L2 (with shorted L1 represented by a small resistance 'r') . I was expecting the resistance see would be 1/r....but I cannot get that.
Any hints would be useful.
Regards,
nitish
I'm not sure I know what you're trying to do but I'll take a crack at it.
ReplyDeleteThe resulting a2 term is s^2*C1*(L1+L2). So (s/w)^2 = s^2*C1*(L1+L2) and 1/wo = sqrt(C1*(L1+L2)). fz is then 1/(2piC1(L1+L2)).
Are you trying to simulate this in Spice?
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ReplyDeleteThanks for responding wrich. I was trying to understand how the L2C1 part of the s^2*C1(L1+L2) was obtained. I was trying to explain the procedure I was following to determine that term. Do you think I am on the right track?
ReplyDeleteI was not trying to simulate it in Spice.
I remember this one now. There's a little trick to it involving the canceling of infinities. You use the L2 term you found in a1, make L2 abnormal, place the source in the C1 location and multiply it by the resulting impedance.
ReplyDeleteFor C1, you get r. For L2 (with C1 s/c), you'll get similar like 1/(1/r-D^2/D'^2/R). So for C1*L2, you'll have 1-r*(D^2/D'^2/R). r-->0, then you'll have C1*L2. This is the hard way.
ReplyDeleteEasy way: I don't think r has to be placed inside the loop of the d^ controlled voltage/current sources. It can be placed outside the loop, then the sources can be s/c because of L1. Hence you'll get r*(1/r)=1.