Transfer function and normalized transfer function!

Vidya Kantamneni

2012 10 24

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Hallo everybody,

I encountered a small issue today, though quite basic. I hope you can help me regarding the same.

I have designed a filter using some user defined toolbox. I got a discrete transfer function with four zeros and four poles, and like this:

Transfer function:

0.0002606 z^3 + 0.0002606 z^2 + 0.0002606 z + 0.000260

-------------------------------------------------------

z^3 - 2.823 z^2 + 2.667 z - 0.8425

I wanted to check the step response of the above transfer function and so I have entered the block as is in Simulink(Discrete transfer function block).

When I give step input, I get a beautiful step response but the amplitude is somewhere around 60. Why do I get the step response so amplified?

Is my transfer function already normalized? Does the step response of a normalized transfer function and not normalized transfer function makes any difference?

Thanks for your time and effort. Best regards, Vidya

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Vidya Kantamneni 2012-10-24

Hallo Rajiv,

Thanks for writing. The initial value is 0 and final value is 1. For numerator, the coefficients are [0.0002606 0.05788 0.0002606 0.0002606] and for denominator [1 -2.8231 2.6667 -0.8425]. (as is copy pasted!)