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I have the following bode plot and nyquist plot result of an unknown DUT (it is a filter I asume!).

enter image description here

I wonder what kind of arrangement it has in terms of being a series or parallel RC or RLC network.

Since the top left plot starts at 500 and to the middle of the frequency it reaches the minimum, I assume there is a resistor involved.

Also because there is no so called half-circle in the nyquist diagram I assume there is no capacitor in the circuit.

The phase shift from -90 to +90 in the lower left graph makes my head scratchy!

What could be the circuit for this graph?

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  • \$\begingroup\$ The impedance rises at low frequency. (what could that be?) And then rises again at high frequency (What could that be?) In the middle it's pretty flat so a 100 ohm resistor is a good guess. The phase shift tells you something about the order, number of poles. (And I think) each pole gives 90 degrees. (A single RC give 90 degrees of phase shift.) \$\endgroup\$ – George Herold Nov 7 '14 at 0:13
  • \$\begingroup\$ @GeorgeHerold thanks for the hints. I get from what you said there is a cap, a resistor and an inductor involved in the circuit? I checked plots for both series and paralel rlc circuits but none looked like this :( \$\endgroup\$ – Sean87 Nov 7 '14 at 1:23
  • \$\begingroup\$ Check again - and you will see that diverger`s answer is correct. \$\endgroup\$ – LvW Nov 7 '14 at 8:45
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It's a \$RLC\$ circuit, maybe.

First, from the right graph, when the real part is \$100\Omega\$ or so, the image part range from \$-500\$ to \$500\Omega\$, so i guess it has a \$R\$ in series with a reactive part. And from the the phase graph, it apparently capacitive at low frequency, inductive at high frequency, so it maybe has a \$C\$ and \$L\$ in series. Now the whole circuit should be a \$RLC\$ in series.

Omit the \$L\$ part at low frequency, and omit the \$C\$ part at high frequency, then

$$ R=100\Omega\\ |Z_{x}|=\sqrt{R^2+\frac{1}{w^2C^2}} = 500 \quad \text{when}\quad w=2\pi \times 1Hz\\ |Z_{x}|=\sqrt{R^2+w^2L^2}=500 \quad \text{when}\quad w=2\pi \times 10^3Hz\\ $$

Solve it, we get $$ C=2.69 \times 10^{-4}\text{F},L=7.8 \times 10^{-2}\text{H} $$

Because we omitted \$C\$ at high frequency, and omitted \$L\$ at low frequency, there should some error. So, we adjust the value of \$C\$ and \$L\$, finally get your graphs.

$$ C=3 \times 10^{-4}\text{F},L=8 \times 10^{-2}\text{H},R=100\Omega $$

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