Transfer function and normalized transfer function!

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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
  3 个评论
Vidya Kantamneni
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!)
Vidya Kantamneni
Vidya Kantamneni 2012-10-24
Oopps, How did 0.05788 come in the middle! :(
That is the mistake. Thanks Rajiv for point it out! :)

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