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Correct understanding of Wolfram Alpha continued fraction notation

Posted 8 months ago

Hello, I am new to Wolfram Alpha and this forum. I am now trying to synthesize Low pass Chebyshev filter from a transfer function which is verified to be correctly calculated. Based on this transfer function I am trying to count input impedance of the filter, which should be

Divide[4.4392Power[s,3]+1.58559Square[s]+3.20288s+0.49693,12.4Power[s,4]+4.4392Power[s,3]+13.9856Square[s]+3.20288s+2.04693]

I am trying to calculate the continuous fraction from the above formula:

https://www.wolframalpha.com/input?i2d=true&i=continued+fraction+Divide%5B4.4392Power%5Bs%2C3%5D%2B1.58559Square%5Bs%5D%2B3.20288s%2B0.49693%2C12.4Power%5Bs%2C4%5D%2B4.4392Power%5Bs%2C3%5D%2B13.9856Square%5Bs%5D%2B3.20288s%2B2.04693%5D

I transcribed the above form of output to this fraction:

0,242768+Divide[1\(44)18486s,-2\(44)81876*Power[10,16]s-Divide[7\(44)94538*Power[10,32]Square[s],1-Divide[2\(44)81876*Power[10,16]s,1+Divide[2.22045*Power[10,-16]s,0\(44)986281-Divide[0\(44)735921s,1+Divide[1,0\(44)51732s]]]]]]

Did I do it correctly? I am not sure if I understand the notation of the calculation of continued fraction correctly.

Thank you very much in advance for reply!

Vaclav

POSTED BY: Vaclav Zajic
2 Replies

You probably should write s^2 instead of Square[s].

I suspect that Wolfram Alpha knows continued fractions only for numbers, not for rational functions.

POSTED BY: Gianluca Gorni
Posted 8 months ago

Thank you for reply! Meanwhile I managed to compute the continued fractions manually, it required correction of some rounding errors and then I got quite good looking continued fractions.

POSTED BY: Vaclav Zajic
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