Help with TF and STEP function

Hi, I have a transfer function considering as input the function "STEP". I need to know whats is the value of this function when the time is Inf. How can I do that in a simple way. I have it:

   TF1 = tf([3 7],[1 2 5])
    h2 = stepplot(FT) 
    grid on

Thanks

Answer 1

[Star Strider]

See if this does what you want:

TF1 = tf([3 7],[1 2 5])

figure
h2 = step(TF1, Inf);

figure
stepplot(TF1)
grid on

SteadyStateValue = h2(end)

SteadyStateValue =
   1.397791589266900

The additional step call is necessary because stepplot does not give up its secrets willingly.

[Bruno Souza]

thank you!

[Star Strider]

As always, my pleasure!

Answer 2

[Walter Roberson]

I do not know what the real way of handling this is, but:

In any case in which the degree of the denominator is not equal to the degree of the numerator, the limit towards infinity is going to be ((sign of leading coefficient of numerator) / (sign of leading coefficient of denominator)) /(infinity to the (degree of denominator minus degree of numerator)) .

In your case that would be ((1)/(1)) / (infinity^(3-2)) which would be 0.

tf() arranges so that leading negative coefficients in the denominator are transferred to the numerator, so for example tf([3 7],[-1 2 5]) is treated as tf([-3 -7], [1 -2 -5]) . So if you extract the coefficients from the tf, you do not need to examine the sign of the denominator, as you can assume it will be positive.

Short summary: if degree of numerator is greater than degree of denominator, then you will go to +/- infinity at infinity; if the degree of denominator is greater than degree of numerator, then you will go to +/- 0 at infinity. The analysis for equal degree takes more work.