Implementing an op-amp based circuit that has a given transfer function

Question

I have this question but I am really confused about how to start, can you please give me some helpful hints.

Let us assume we have a DC voltage measured in a circuit, called \$V_{\text{in}}\$. We intend to build a circuit such that it has a transfer function defined as

$$H(s) = \frac{V_{\text{out}}(s)}{V_{\text{in}}(s)} = \frac{s^2+s+1}{s}$$

Draw this electrical circuit. The circuit can be composed of resistors, inductors, capacitors and op amps only.

I tried op amps with resistors and capacitors in a few different forms, but I didn't get the same \$H(s)\$.

Also I tried a few filters but non of them gave me the same answers.

Answer 1

\$\dfrac{V_{OUT}}{V_{IN}}\$ = \$\dfrac{s^2 + s + 1}{s}\$

Divide thru by s to get \$s+1+\dfrac{1}{s}\$

Now you have three terms that are summed together and one of them is based around a resistor, one a capacitor and one an inductor. Does this help? I'm not going to do the full job because it sounds like homework.

Can you take it from here?

Answer 2

First thing- since the vin is DC, s=0 and H(0)=infinity, Second- for the case where both input and output are voltages, the degree of the numerator can't exceed the degree of the denominator which is being violated here, so it's impossible to build the circuit in real world.

Answer 3

Can you please verify that this circuit is what the posts are about?

Answer 4

This is a improper transfer function: The system has more zeros than poles; it is not causal, cannot be implemented, has a strictly proper inverse and has infinite high-frequency gain.

$$\frac{Y(s)}{X(s)}=H(s)=\frac{s^2+s+1}{s}$$ with impulse response: $$y_i(t)=\delta(t)' + \delta(t) + u(t)$$ with step response (also with a discontinuity in origin): $$Y_s(s)=1+\frac{s+1}{s^2}$$ or, in time: $$y_s(t)=\delta(t) + u(t)+ t$$

Figure 1

Additional comment on realizable x unrealizable TFs:

Figure 2

UPDATE:

Seen as a PID controller,

$$H(s)=\frac{s^2+s+1}{s}$$ has the form

$$\frac{K_Ds^2+K_Ps+K_I}{s}=K_p(1+\frac{1}{T_Is}+T_Ds)$$

Clearly, this transfer function is improper and practically it can not be used as frequency increases, its gain also increases In consequence, practical applications apply a low-pass filter in the D-term, of the form \$1/(T_fs+1)\$. This low-pass filter has the effect of attenuating high-frequency signals. What lead us to a practical PID controller:

$$H(s)=K_p(1+\frac{1}{T_Is}+\frac{T_Ds}{T_fs+1})$$

Note that, for the modified D-term, the high frequency gain is given by \$\frac{T_d}{T_f}\$ (where \$T_d > T_f\$), instead of \$\infty\$, for the pure differentiator \$T_ds\$. Compare with Figure 2 above (ideal x real).