Suppose we have a transfer function H(s) = 1/s+1. Is there a general method by which using electric components we can construct such a system?
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\$\begingroup\$ Possible duplicate: Implementing an op-amp based circuit that has a given transfer function \$\endgroup\$– GrapefruitIsAwesomeCommented May 8, 2022 at 10:38
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1\$\begingroup\$ (1/s)+1 or 1/(s+1) ? \$\endgroup\$– LvWCommented May 8, 2022 at 14:50
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\$\begingroup\$ LvW H(s) = 1/(s+1) \$\endgroup\$– Jun Seo-HeCommented May 8, 2022 at 15:52
1 Answer
It is not so common passing from transfer functions to circuits. Maybe it is more common han usual in time discrete circuits. I guess you are referring to time continues circuits.
However, It depends if your transfer function will amplify the signal or just attenuate.Active components for the former, passive ones for the latter.
At that point you have to decompose your transfer functions as product of monomial terms and use the well-known circuits to realize your transfer function. During the design you have to keep in mind this rule: The product of transfer function is the cascade interconnection of electrical circuits only if the output impedance of the first filter is lower than the input impedance of the second filter and so on. That's the only way to avoid unwished interface effects between two circuits.
For example, you would find a configuration like that:
\$H(s)=\frac{1}{(s^2a^2+2sa+1)}=\frac{1}{(1+sa)(1+sa)}\$
It is the product of two RC filter.
Let's consider basics configurations: 1)RC filter with \$H(s)=\frac{1}{1+sRC}\$ 2)RL filter with \$H(s)=\frac{sL}{R+sL}\$ 3)RLC, LR, CR etc
Does it work?
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\$\begingroup\$ "It is not so common passing from transfer functions to circuits" -- it's done all the time when going from the drawing board to the design. The product of cascaded transfer functions happens whether you want it or not, it's called convolution. The loading effect is what you're referring to, and it should only be avoided if it's not needed, otherwise that's how classical passive filter design works, for all N>2. You can attenuate with active devices, too. In this case H(0)=1, so it can be done either active or passive. And even passive networks can amplify. \$\endgroup\$ Commented May 8, 2022 at 20:15