Step Response

a=[0.2,0.4,0.5,0.7,2,1]
g = (tf(1, [1 2*a 1]));
S= stepinfo(g)

This is a step response problem where as you can see 'a' is the variable to be multiplied. I know that we can solve this by using the stepinfo(g)and the step information with display the results automatically.

My question is how does matlab calculate the results?

% RiseTime: 1.1308
% SettlingTime: 38.7087
% SettlingMin: 0.4685
% SettlingMax: 1.7292
% Overshoot: 72.1625
% Undershoot: 0
% Peak: 1.7292
% PeakTime: 3.1416
I got these numbers by doing this:
a= 0.1
g= tf(1,[1 2*a, 1])
[ y, t ] =step(g) % default of 60 seconds
S = stepinfo(y,t)
plot ( t, y )

So you can solve this for any values, how can use indexing within the transfer function?

Now I need to just label each, do we need to input calculations just use

function S=stepinfo(y,t)

Frequency response:
a=[0.2,0.4,0.5,0.7,2,1]
g = (tf(1, [1 2*a 1]));
a=[0.2;0.3;0.4;0.7;1;2]'
H =(tf(1, [1 2*a(index) 1]));
bode(H,{0.1,10})

My goal here is to graph the freq response for all values so I am thinking to subplot and index also. I am not sure how this can be done with Bode.

This works, I double checked but can I index and subplot for 6 values?

a=0.1
H =(tf(1, [1 2*a 1]));
bode(H,{0.1,10})

Any information will help, I am just going in circles, mean while I will be reading the book again and again.

Gabriel

Answer 1

[Arnaud Miege]

First, a comment: it would help if your question was properly formatted. As it is, it's very difficult to read.

If I understand your question correctly, the best way is probably to use a for loop:

a=[0.2,0.4,0.5,0.7,2,1];
for k=1:length(a)
  g(k) = tf(1,[1 2*a 1]);
  [y(k,:), t(k,:)] = step(g(k));
  S(k) = stepinfo(y(k,:),t(k,:));
end

This grows the variables in a loop, so is not exactly best practice, but it should work:

>> S(1).RiseTime
ans =
  0.8600

You can then pass the whole g to the bode command:

bode(g,{0.1,100})

or again, do this in a loop if you want to do them one by one.

HTH,

Arnaud

[Gabriel]

I am grateful for your help and yes I do understand what you are talking about.

I thought that calculations were used in order to solve the characteristics of the step response.

Example POS= percentage overshoot = Max(y)-ss/ss

Tp = Peaktime = y(t)

Ts = SettlingTime = ?

Tr = RestingTime = ?

The settling time at 10% instead of the default 2% values.

Maybe I should create a function m file instead of a script file.

Please excuse me of my inexperience in matlab I am just starting out.

That indexing looks right,I am working out matlab code by hand first then on the program in order to understand the code. Any suggestions or tips please forward if you have time.

Thank You!

[Arnaud Miege]

Looking at the documentation for the stepinfo function (http://www.mathworks.com/help/releases/R2011a/toolbox/control/ref/stepinfo.html), you can specify a different settling time ratio from the default 2%.

[Arnaud Miege]

S = stepinfo(...,'SettlingTimeThreshold',ST) lets you specify the threshold ST used in the settling time calculation. The response has settled when the error |y(t) - yfinal| becomes smaller than a fraction ST of its peak value. The default value is ST=0.02 (2%).

[Gabriel]

Understood on the stepinfo, I have read that information already thanks for clearing that up.

In (bode(g,{0.1,100})

or again, do this in a loop if you want to do them one by one )

I was just thinking instead of using bode can this be used also

b=[0 1 -1];

>> a=[1 8 8];

>> w=[0:.1:2*pi];

>> freqs(b,a,w)

What is the difference of Bode and freqs, does it depend on the applications used.I have the textbook definition I just want to know real world examples.

Thanks for your help

[Arnaud Miege]

bode is from the Control System Toolbox, whereas freqs is from the Signal Processing Toolbox. bode allows you to compute/display the Bode plot (magnitude & phase) of any LTI system (Linear-Time Invariant), transfer function, zero-pole gain or state-space system. freqs, on the other hand, only displays/computes the frequency response of an analog filter.