# 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: 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.

• Thanks, is there any resonance in the plot? That is what I was looking for actually. 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. Feb 14 '12 at 23:32
• @clabacchio, I mean not as big as Intel or TI, not as small as a garage start-up. 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... 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).

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.

• 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. 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. 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. 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. Feb 16 '12 at 16:01

Just throwing this in, because it took me sometime to figure it out.

If you use the above method of plotting V/I with AC analysis you get a dB and phase plot just like the image in the question.

If you click on the y axis, a pop up will appear and if you select linear, it will give you an impedance scale.

Now you have impedance vs frequency.

"I want to know if the graph I got is right, and if so, what does it show?"

In addition to the given answers (complex input impedance of the filter circuit), perhaps you are interested to learn if it is possible to derive some characteristic filter parameters from the given response?

In general, it can be shown that each valid transfer function that can be defined for such a circuit will have has the same denominator. This applies to all transfer functions to be defined between each grounded node (input) and each every other node within the circuit. And this also applies between these (grounded) inputs and the currents within the circuit. In particular, this applies, therefore, also to the input impedance of your circuit.

What does this mean: If the input impedance function Zin(s) and the classical voltage transfer function of the filter H(s) have the same denominator D(s), they will have the same pole frequency fp.

The input impedance as shown in your graph shows a typical bandpass response with a center frequency (identical to the pole frequency) at app. 800 Hz. Therefore, your active lowpass circuit will also have a pole frequency of fp=800 Hz.

For all second-order lowpass filters, the pole frequency has a direct relation to the well-known cut-off frequency (3 dB) - depending on the particular approximation (Butterworth, Chebyshev,...). For a Butterworth response, the pole frequency is identical to the 3dB-frequency.

Another example for equal pole frequencies: At the common node of R1, R2 and C2 you can observe a bandpass function with the same center frequency fp=800 Hz.