# Finiding composition of circuit from frequency response and bode plot

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

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?

• 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.) – George Herold Nov 7 '14 at 0:13
• @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 :( – Sean87 Nov 7 '14 at 1:23
• Check again - and you will see that diverger`s answer is correct. – LvW Nov 7 '14 at 8:45

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.

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