Bode plot doesn't look as expected for circuit

Hi - I was trying to analyze a band pass filter, but the results I'm getting don't make sense to me.

I've attached the simulation.

I first solved for the transfer function of it. It looks like a simple voltage divider - so:

Vout/Vin = R / (R + SL + 1/(SC)) = RSC / (RSC + S^2LC + 1)

So I have a zero at the origin, and poles at (R/(2L)) +/- (1/(2L)) * (R^2 - 4L/C)^0.5. I chose values for R, L, and C so that both the poles are at R/(2L), which in this case is 796Hz.

My understanding is that the zero at the origin gives me +20db/decade starting at w = 1. The two poles at 796 hz should give me -40db/decade starting at 796Hz, before 796Hz it should give me 0db/decade.

But that isn't what I'm seeing. What I'm seeing looks more like I have a zero just before my poles. In fact, the zero at the origin looks a whole lot like the high pass filter that is produced if I short out the inductor (zero at 400Hz).

I just can't seem to figure out what is going on here. My math says one thing but the graph says another.

Hi!

I don't see any problem with the simulator. With the zero @ the origin, you are getting +20 dB/dec starting from -oo in log w because the zero is @ w=0. A zero @ w=0 in TF does not imply 0 dB gain or flat gain, for that matter. Only after the poles @ w = 2*pi*796. you are supposed to be getting +20-40 = -40 dB/dec. This makes sense because what u have there is an idealized 2nd order bandpass filter.

My math says one thing but the graph says another.

1.) I think "your math" tells you something wrong. A 2nd order bandpass never has a gain drop of 40dB/dec.
2.) I don't think that your values arrive at a double pole at 796 Hz. Recalculate!
3.) Correction to 993066935: +20-40 = -20 dB/dec

LvW said:

My math says one thing but the graph says another.

1.) I think "your math" tells you something wrong. A 2nd order bandpass never has a gain drop of 40dB/dec.
2.) I don't think that your values arrive at a double pole at 796 Hz. Recalculate!
3.) Correction to 993066935: +20-40 = -20 dB/dec

Click to expand...

1. I agree - I'm expecting +20dB from the zero and -40dB from the poles.
2. R/(2L) = 1K/ (2 * 100m) = 5000 rad/sec = 796Hz
3. Agreed.

---------- Post added at 20:45 ---------- Previous post was at 20:44 ----------

993066935 said:

Hi!

I don't see any problem with the simulator. With the zero @ the origin, you are getting +20 dB/dec starting from -oo in log w because the zero is @ w=0. A zero @ w=0 in TF does not imply 0 dB gain or flat gain, for that matter. Only after the poles @ w = 2*pi*796. you are supposed to be getting +20-40 = -40 dB/dec. This makes sense because what u have there is an idealized 2nd order bandpass filter.

Click to expand...

What do you mean I get +20dB/dec starting at -infinity? I mean, how does one draw that? I thought it was +20db starting at 0?

Thanks, typo.
log(0) -> -infinity You don't draw w=0 in bode plot.

I think I have found my mistake. I was forgetting to calculate in the constant gain of the circuit. Since I had a "RSC" term in the numerator, that contributes a (20*log(R*C)) = -68db constant term. The zero at the origin contributes a +20db/dec term that crosses 0db at w = 1.

The bode plot now matches what I expect!

2. R/(2L) = 1K/ (2 * 100m) = 5000 rad/sec = 796Hz

That's only the REAL part of the pole (however, you neglected the minus sign); what about the imaginary part?