Yes I have, didn't workIt's a very open question.....just showing a network. Do you have any practical context? A circuit or a function that your working on?
Have you tried flipping the circuit over?
Edit: Sorry Alec t.......didn't see your post.
didn't seem to workTry swapping L for C and C for L.
Thanks for your reply... but I believe you are plotting the reactance in dB ....while I'm plotting it in Ohm...
hmm..now I'm confused..I verified using circuit simulator and formula and just couldn't seem to get your response.....
I don't think you are right, the series resonant network will look like a capacitor below resonance and inductor above resonance, locally at the resonant frequency, the reactance is almost a straight line. Left side <0, crosses 0, then right side > 0.Don't know what you are using to do these plots, but try using more frequencies. You've got a fairly high Q network with rapid changes in impedance. Checking it only every 0.01 MHz may not be enough.
As I say, I don't know what you are using to do the plotting, but your original plot seems wrong. You have a basic series resonant circuit, which will have a low impedance at resonance. A series resonant network will look like an inductance above it's resonant frequency and C2 will parallel resonate with that inductance. So the impedance plot *should* have an impedance minimum at a lower frequency that the impedance maximum.
That series resonant network will look like a capacitance below resonance. By replacing C2 with L2, you are now resonating that apparent C with L2 producing a parallel resonant circuit -below- the series resonant frequency.
Look closer in with you plotting app, use more frequencies.
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