How can I create a step response with the following unit step function as input?

input : F

when (0<t), F = 3

when (0.98<t) = 6

I want to make step response with that unit step function as input in matlab...

please help me..

[madhan ravi]

And how is the transfer function defined?

[Si eun Jung]

TF is

1/(s^2+2*zeta*w_n*s+w_n^2)!!

w_n = 10, zeta = 0.2

so, I made that code

clc;

clear all;

close all;

w_n = 10;

zeta = 0.2;

t=0:0.01:10;

F(0<t) = 3;

F(0.98<t) = 6;

X = tf([F.*w_n^2],[1 2*zeta*w_n w_n^2])

step(X,t)

but I finally failed...

Answer 1

[madhan ravi]

F = @(t) (t < .98) * 3 +(t > .98) * 6
fplot(F,[0,2])

[Gyanu Gautam]

You will get even more refined plot of step function if you use the following code:

F = @(t) (t < .98) * 3 +(t >= .98) * 6
fplot(F,[0,2])

I just tweaked a liitle bit of your code.

madhan ravi: Thank you ;)

[Si eun Jung]

TF is

1/(s^2+2*zeta*w_n*s+w_n^2)!!

w_n = 10, zeta = 0.2

so, I made that code...

clc;
clear all;
close all;
w_n = 10; 
zeta = 0.2;
t=0:0.01:10;
F(0<t) = 3;
F(0.98<t) = 6;
X = tf([F.*w_n^2],[1 2*zeta*w_n w_n^2]) 
step(X,t)

but I finally failed...

[madhan ravi]

w_n = 10; 
zeta = 0.2;
t = 0:0.01:10;
X = tf(w_n^2,[1 2*zeta*w_n w_n^2]);
F = @(t) (t < .98) * 3 +(t >= .98) * 6
subplot(2,1,1)
plot(t,F(t))
title('Step Input')
subplot(2,1,2)
lsim(X,F(t),t)

[Si eun Jung]

oh my god.... youre genius....

Thank you soooooo much...

[Paul]

Madhan's code gives the expected result. However, keep in mind that it only worked because lsim was able to detect the jump at t = 0.98 and use an appropriate approximation in its conversion of X to discrete time. So be careful using lsim. doc lsim for more information.

You can generate the step response for X using:

y = step(X,t)

Then you can use y, and y only, to compute the final result that you need. If all you wanted was the plot of the final result, then madhan's approach is fine. But if you're doing this for a class where you need to understand the fundametals of LTI systems, you should try it this way as well.

[Gyanu Gautam]

I have a doubt here.

How can we plot step response for the "specified step input" from MATLAB-inbuilt 'step()' function.

Here the step input specified is:

F = 3 when (0<t)

and F = 6 when (0.98<t).

While the default step() function will plot the step response of unit step function:

U = 1 when (0<t)

U = 0 otherwise.

Also, doc lsim says it simulates time response of both continuous and discrete systems.

BTW, Beautiful solution Madhan Ravi :)

[Paul]

The original definition of the input F was ill posed. What is F(2) supposed to be?

Clearly 0 < 2, so F = 3.

But 0.98 < 2, so F = 6.

Because madhan's solution was accepted by the OP, we see that the input F is really supposed to be:

F(t) = 3 (0 <= t < 0.98)

F(t) = 6 (0.98 <= t).

Now we can define the unit step function U(t) as:

U(t) = 0 (t < 0)

U(t) = 1 (t >= 0) % note the >=

You are correct that the command

Y = step(X,t)

returns the output of the linear system with transfer function X in response to the unit step input U.

However, the actual input we care about, F, can be written in terms of U:

F(t) = 3*U(t) + 3*U(t - 0.98)

Therefore, the output that we care about, let's call it Z(t) is:

Z(t) = 3*Y(t) + 3*Y(t - 0.98).

So you only really need to compute Y once using the step command.

You're correct that lsim can accept either continous and discrete systems. Is that relevant to this discussion?