As George noted in his comments, if you label the nodes 1-5 from left to right, nodes 1,3 & 5 are all the same voltage due to the op-amps forcing enough current until those nodes are equal.

Node 1 would also be known as the input voltage. Notice that the output current is Vin / Z5 then. Which is also equal to the current through Z4.

Note that the current through Z2 and Z3 have to be equal to eachother. From the information so far, you can determine the voltages at nodes 2 and 4 with respect to Vin. Once you know node 2 voltage, you know the input current (Vin-V2)/Z1. The output current is in terms of Vin. If you divide the output current by the input current, you'll know the difference in impedance because the equations will be true over all Vin. That is, Vin/Iin / Vout/Iout = impedance_conversion. Vin=Vout in this case, so Iout/Iin = impedance=conversion.

Working through what I just described you should end up with this:

$$ Iout=Vin/Z5 $$
$$ V4=Vin+Vin/Z5*Z4 $$
$$ V2=Vin + (Vin-V4)/Z3*Z2$$
$$ (Vin-V2)/Z1=Iin$$

$$ \frac{Iout}{Iin} = \frac{\frac{Vin}{Z5}}{\frac{Vin-V2}{Z1}} = \frac{Z1}{Z5}\frac{Vin}{Vin-V2}$$
$$ = \frac{Z1}{Z5}\frac{Vin}{\frac{-Vin+V4}{Z3}Z2} = \frac{Z1Z3}{Z5Z2}\frac{Vin}{-Vin+V4}$$
$$ = \frac{Z1Z3}{Z5Z2} \frac{Vin}{\frac{Vin}{Z5}Z4} = \frac{Z1Z3}{Z2Z4} $$

$$ \frac{Zin}{Zout} = \frac{Iout}{Iin} = \frac{Z1Z3}{Z2Z4} $$

If the impedances are just resistances I can't imagine it being terribly interesting as it's just a resistor made up by 4 resistors.

If the impedances are capacitive and resistive and you place them right, you could potentially get an impedance that looks like an inductor. This has an interesting potential usage. In integrated circuits, capacitors are much prefered to inductors due to their small size. Now you can make a circuit that acts like an inductor without an inductor. It can also be tuned and altered using the resistors and capacitors so that sizing can be optimal for a chosen inductance value. This has applications in integrated filters and probably other integrated circuits.

This should be enough to get you started on this topic. You should probably read up on gyrators if you're interested in this more. Hopefully I didn't make any typos in all of my equations.

Another way of looking at the result is this as Z5 = Zout:
$$ Zin = \frac{Z1Z3}{Z2Z4}Z5 $$
So Z5 can be used as just another knob to adjust the impedance you're creating.

The answer, before the interesting diversion:

A PID + a second derivative will do the trick:

^{simulate this circuit – Schematic created using CircuitLab}

With transfer function:

$$
H(s) = -\frac{R_{11}}{R_6} C_1 R_1 C_2 R_2 s^2 - \frac{R_{11}}{R_7} C_3 R_3 s + \frac{R_{11}}{R_{10}}\frac{R_6}{R_5} + \frac{R_{11}}{R_9} \frac{1}{R_4 C_4 s}
$$

Which matches your transfer function of:

$$
H(s) = As^2 + Bs + C + \frac{D}{s}
$$

You didn't mention the signs of \$A\$, \$B\$, \$C\$ and \$D\$.

If you would like to reduce the number of op amps, you can combine the PID op amps into one using the information from this article.

^{simulate this circuit}

With transfer function:

$$
H(s) = K \frac{(s/z_1 + 1)(s/z_2 + 1)}{s (s/p_1 + 1)(s/p_2 + 1)}
$$

Here, \$p_1\$ and \$p_2\$ are extra zeros. Ideal PID controllers don't have either, "filtered derivative" PID controllers only have \$p_1\$, and "type 3" PID have both. See the article for more information.

Where:

$$ R_1 = Z_{in}\qquad R_2 = \frac{R_1 z_2}{p_1 + z_2}\qquad R_3 = \frac{R_1 p_2 K}{z_1 (p_2 - z_1)}$$

$$
C_1 = \frac{p_1-z_2}{R_1 z_2 K}\qquad C_2=\frac{p_2 - z_1}{R_1 p_2 K}\qquad C_3=\frac{z_1}{R_1p_2K}
$$

You can work out the necessary values.

## Now for the fun diversion:

So, I think that @Vladimir Cravero is correct, a transfer function with more zeros than poles is unphysical.

Physicists will think about this in terms of susceptibilities in the complex frequency domain (equivalent to what EEs refer to as transfer functions in the Laplace domain), and the Kramers-Kronig(KK) relations, however this can be extended to the [laplace domain](see appendix A) as well.

We know that the convolution theory allows us to take response functions in the time domain (\$h(t) \rightarrow H(s)\$) and turn them into transfer functions in the laplace domain:

$$ H(s)V(s) = \mathcal{L}\left\{ \int_{-\infty}^{t} h(t-\tau)v(\tau)d\tau \right\}$$

However, we require that \$h(>0)\$ is \$0\$, otherwise the transfer function is reacting to stimuli that hasn't happened yet. Making sure that this requirement is obeyed is done by making sure that such a function obeys the KK relations in the Laplace domain.

The KK relations have two requirements:

- Analyticity in the right half-plane of Laplace space. This means no poles in the right half-plane.
- \$\lim_{s\rightarrow \infty} H(s) = 0\$, and furthermore that it goes to zero at least as fast as \$1/|s|\$. (This can apparently be relaxed a bit, but I'm not sure how, or how much.)

These requirements make sense: we can't have any sort of gain for infinite frequency for any real system (energy conservation requires this), and any poles in the right half-plane would also lead to energy conservation violations, finite inputs would lead to infinite power eventually.

Given these requirements, the Kramers-Kronig relations give us a relationship between the real and imaginary parts of the transfer function:

$$ \Re\{H(s)\} ∝ PV \int_{-\infty}^{\infty}\frac{\Im\{H(s')\}}{s'-s}ds' $$

$$ \Im\{H(s)\} ∝ PV \int_{-\infty}^{\infty}\frac{\Re\{H(s')\}}{s'-s}ds' $$

Where \$PV\$ stands for Cauchy's Principle Value integral.

Ultimately, doing this integral isn't actually that important, but we need to make sure that the transfer functions obey the requirements of the KK relations.

For a system where there are more zeros than poles, it is pretty trivial to show that this doesn't hold.

But wait! While @Vladimir Cravero is ultimately correct, realizing a physical transfer function with more zeros than poles is not possible because we would break causality, @Chu is also correct, this is done all the time with PID controllers. What gives?

The answer is that PID controllers (and all real systems) have low pass filters that give the bandwidth of the system. The order of this low pass filter is determined by the order of the system. For PID controllers, this shows up in the \$K_d\$,\$K_i\$ and \$K_p\$ values. We don't actually have perfect Op amps where these are just numbers, they must always also include a low pass filter that gives the bandwidth of the op amp, and this adds an extra pole. Furthermore, the response of the thing we are driving has a low pass filter within it (as must any physically realizable system).

Notes:

The KK relations are also known as the Hilbert transform, which I found out while doing the research for this post. This might be a more well known name here in the EE community.

It is possible that there are typos here. It is also possible that there are (weird) real valued causal systems that don't obey the KK relationships, but are still causal. This would be an analog to non-hermitian hamiltonians which *happen* to have real valued eigenvalues and eigenvectors in quantum mechanics. I'm not sure about this. [edit: this paper should be looked at if you are interested in this question]

This actually continues to hold in the small signal limit for nonlinear systems. The nonlinear KK relations are still a thing, references available upon request.

## Best Answer

In theory, the

~~inverse of the inverse~~converse of the converse is the original impedance which was a filter with a bunch of RC's.NIC Negative Impedance Converters

give the negative of impedance, not the inversion.The performance as shown would be highly vulnerable with stray coupling. Yet the desired or expected performance was never specified.