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In 'Franklin F kuo's book it is written that "The zeroes of transmission of the ladder occur at the poles of The series branch impendences or at the zeroes of shunt branch impendences" here is image of ladderenter image description here

But I don't understand why overall transfer function's zeroes are due to zeroes of shunt branches and Poles of series branch impendences? What is mathematical or intuitive explanation of it?

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  • \$\begingroup\$ Can you set up the transfer function of this circuit? \$\endgroup\$ Commented Aug 4, 2020 at 14:54
  • \$\begingroup\$ Yes, but that would be too complicated Rather I just wanted to understand intuitvily how it works for any other circuits i.e even More complex circuits \$\endgroup\$
    – user215805
    Commented Aug 4, 2020 at 15:00
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    \$\begingroup\$ Poles of series branch impedance -> infinite impedance is in series -> input signal doesn't reach output -> zero of the circuit. (my guess) \$\endgroup\$
    – AJN
    Commented Aug 4, 2020 at 15:13
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    \$\begingroup\$ Zeroes of shunt brand impedance -> short circuit to ground -> input gets shorted and doesn't reach output -> zero of circuit. (my guess). This seems to ignore the shift in frequencies due to loading of a branch from other parts of the ladder. May be there is more context in F.F.Kuo. \$\endgroup\$
    – AJN
    Commented Aug 4, 2020 at 15:14
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    \$\begingroup\$ @AJN, For ideal, lossless elements, if the impedance (for the series branch) goes to infinity, it doesn't really matter how it's loaded, still no signal is getting through. (Also, you should post your comments as an answer) \$\endgroup\$
    – The Photon
    Commented Aug 4, 2020 at 16:13

2 Answers 2

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Consider the ladder in the figure shown below. The ladder is shown in sub figure (A).

At the pole frequency of the series branch \$Z1\$, the equivalent circuit looks like (B) due to infinite impedance of \$Z1\$. There is no connection from input to output at the particular frequency. Hence a zero.

At the zero frequency of the shunt branch \$Z2\$, the equivalent circuit looks like (C) due to zero impedance of \$Z2\$. The voltage across \$Z2\$ is forced to zero by the short at that particular frequency. Here also the input signal cannot influence the output. Hence a zero.

impedance ladder

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Consider a simpler one-rung ladder with just L2/C4 and C3/L1/R3.

Let the impedance of L2/C4 be: $$\small Z_1=\frac{a+bs}{c+ds}$$ and the impedance of C3/L1/R3: $$\small Z_2=\frac{e+fs}{g+hs}$$ where \$\small s\$ is the Laplace operator.

The transfer function is $$\small G(s)=\frac{Z_2}{Z_1+Z_2} =\frac{(e+fs)(c+ds)}{(g+hs)(a+bs)+(e+fs)(c+ds)}$$

There will be transfer function zeros when the numerator is zero, i.e. when: $$\small (e+fs)=0 \:\: or \:\:(c+ds) =0$$

i.e., at the zero of \$\small Z_2\$, or the pole of \$\small Z_1\$.

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