why discrete transfer function result is different the math function equantion ???

hi everyone!!

i tested some simple equation and trans function.

1) 60S/S+1 (it is simple filtered derivative trans function) wth ramp(0.25/sec) input

h =
 
   60 s
  -------
  s + 1

2)i think that next is the same math equation

t=[0:1:10];
y=0.25*60*(1-exp(-t));
plot(t,y)

Y'(t)+Y(t)=60*X'(t)
here x'(t) is 0.25 (this is ramp function with 0.25/sec)

i think they must make the same results in matlab and simulink

but they show the different results. actualy final value is the same but time constant is not match.

for reach to the final value, it takes almost 1000sec, but math function time constant is very shot

why results are diffrent. is there any missing my test???

[Bjorn Gustavsson]

What is the time-constant of your exponential...

[jingon lim]

i am sorry .time constant is 1

[Bjorn Gustavsson]

Yes, then your second plot should be an OK response to X(t) constant from t = 0, shouldn't it, if X is 1/4 the steady state value for Y should be at a quarter of 60, right?

What you've plotted in the upper figure I cannot resolve, have you checked the scaling of the horizontal axis for that one?

[Aquatris]

I got the exact same responses. So I think you have a mistake somewhere in your code.

[Sai Sri Pathuri]

I cannot reproduce the issue. I could get similar response with both matlab and simulink. Can you attach the model you used?

[jingon lim]

i am so sorry for waiting..because of internet excess not good.

i attached my file.

actualy i am in matlab beginner..

i simulate my file with changing some condition and i found that model setting ->solver detail condtion makes affect the result .

first time i did't change solver selection(varialbe step and auto) and then that result is diffrent with math equation.

second solver selection(fixed and step size 1) two results are not same.

after changing the solver selection(fixed and step size 0.1), both are almost same.

can you let me know why solver selction make the results diffrent??

what is the right setting in this case?

at second time, the reason why i choosed step time to 1sec is that math equation have also 1sec time step.

[Aquatris]

In simulink, you are solving the differential equation numerically. So the smaller step size you use, the closer you get to the actual solution. Solving the equation analytically does not care about the step size so you get same result regardless of the step size you use in the time vector (just more points with smaller step size). There is no one right setting for this. It depends on how accurate you want your solution to be. Also the solver you choose effects things as well (i.e., Euler integration, Runge-Kutta, stiff solvers) For your system, any step size less than 0.1 with Euler integration should work fine.

[jingon lim]

Aquatris. Thank you. your reply.

before using simulik, i have to study about soving method...

[Walter Roberson]

You ask about 60S/2S+1 but your h shows 60*s/(s+1) ? That is a difference of 2 on the s?

[jingon lim]

sorry. i write wrong number. i fixed 1 from 2