I have a TF I wanted to plot using a step input. However, I have more zeros than poles and MATLAB won't plot that. So I divided the num by the den to get a remainder and a quotient.(adsbygoogle = window.adsbygoogle || []).push({});

Original TF: (more zeros than poles)

0.1741 s^7 + 137.8 s^6 + 511.8 s^5 + 888.8 s^4 + 662.1 s^3 + 166.9 s^2

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0.002721 s^7 + 2.147 s^6 + 33.96 s^5 + 75.43 s^4 + 89.24 s^3

I then plotted the remainder over the original denominator and used the quotient to shift the graph up. (code not shown) I get a linear ramp function. My advisor expects the result to be an "s-shaped" curve.

Transfer function: (remainder/original den)

0.4 s^6 - 1661 s^5 - 3938 s^4 - 5048 s^3 + 166.9 s^2 (this is the remainder)

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0.002721 s^7 + 2.147 s^6 + 33.96 s^5 + 75.43 s^4 + 89.24 s^3 (this is the orig den)

So...I played around with the remainder and the denominator. I noticed that when I reduce the system and added the quotient as the constant in the denominator, I get an "s-shaped" curve!!

Reduced the system by s^2 with quotient as constant in den:

0.4 s^4 - 1661 s^3 - 3938 s^2 - 5048 s + 166.9

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0.002721 s^5 + 2.147 s^4 + 33.96 s^3 + 75.43 s^2 + 89.24 s + (quotient = 64)

This gives me an"s-shaped" curve. Is there anyway in which this can make sense...or be correct?

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# Transfer function manipulation is this right?

Can you offer guidance or do you also need help?

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