Electronic – Finiding composition of circuit from frequency response and bode plot

analogbode plotfilterpassive-networks

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?

Best Answer

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.

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