# Interpretation of input impedance graph of a filter

I tried to simulate the input impudance of a multipass feedback low pass filter. I want to know if the graph I got is right, and if so, what does it show? Also, what is that dashed line about?

And here is the circuit itself:

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Yes, you have plotted the input impedance of your circuit.

The solid line is the magnitude, and the dashed line is the phase (relative to the right-hand scale) of the impedance. The phase might be 180 degrees off due to the definition of I(V1). If I(V1) is the current going in to the positive terminal of the supply you should use -I(V1) to get the current going out into the circuit.

Having the magnitude on a dB scale is not especially helpful -- better to figure out how to plot it on a linear scale. But this is not necessarily easy in LTSpice, it took me a while to figure out how to do it when I was doing something similar the other day, and I can't remember the correct incantations now.

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Thanks, is there any resonance in the plot? That is what I was looking for actually. –  Sean87 Feb 14 '12 at 23:22
@Sean87 There is: if you look at the magnitude curve, at 1 KHz the V/I gain is the maximum, which means that you have the highest input resistance, with ideally 0 reactance. –  clabacchio Feb 14 '12 at 23:32
@clabacchio, I mean not as big as Intel or TI, not as small as a garage start-up. –  The Photon Feb 15 '12 at 0:19
If you are working in materials other than silicon, you don't have to be quite so big to own your own fab... –  The Photon Feb 15 '12 at 17:46

If you look at the graph you can see that in the x-axis there are frequency values, and in the left and right y-axes, respectively, there are gain and phase values; so basically what you see is the representation in the frequency domain of the input impedance of your circuit.

The problem is that, plotting the ratio between a voltage and a current, in which the latter is in the sub-mA range (given the resistances), you end with a minimum gain of 60dB (=1000).

I'm not sure about what you expect to see, but you can see this as a transfer function that relates the input voltage to the input current. If you want to see a lowpass function you would better look at the transfer function (Vout/Vin).

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As said in the previous answers, this is the impedance indeed. The impedance has a magnitude and a phase. This enables one to represent impedances as complex numbers and their values are usually frequency-dependent, hence the representation in frequency domain.

The plots you have are so-called Bode plots. These characterize the amplification (or gain) and phase shift in the response of a system. So basically the magnitude plot (continuous line) is the ratio of the peak values of the output and input and the phase shift is the delay of the output compared to the input.

The Bode plot shows to what extent a filter can suppress different frequencies and whether it is going to introduce any delay.

As a closing remark, Bode plots are used for studying other systems than filters, too. They can be used to determine the stable operation regions of a given system.

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I think that Bode plot is something slightly different: AFAIK, it's the asymptotic approximation of those graph, that you can draw simply changing the inclination of the line of 20dB/decade when a pole or zero occurs. –  clabacchio Feb 15 '12 at 17:20
@clabacchio: That's right, you can draw them by hand, assembling them from straight lines (the said asymptotes), but numerically calculated plots will be smooth, just like the one from the OP. –  Count Zero Feb 15 '12 at 21:57
Yes, what I mean is that Bode diagram is the asymptotic approximation, the posted ones are only magnitude-phase diagrams, but not Bode. –  clabacchio Feb 15 '12 at 22:56
@clabacchio: No, it's still a Bode plot if it's an exact calculation or measurement. "Bode plot" does not imply asymptotic approximation. –  nibot Feb 16 '12 at 16:01