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From: Jonathan on 23 Mar 2010 14:37
G'day,

I have the following matrix...

Beta = [52 52 52 52 60 52 40 24 32 44];
yB = zeros(1,length(Beta)).';
for i = 1:length(Beta)
if Beta>0 & Beta<10
yB(i)=1;
elseif Beta>10 & Beta<63
yB(i)=cos((Beta-10)*pi()/180);
elseif Beta>63
yB(i)=0.63;
end
end

For starters, the if loop doesn't do anything (other than creating the initial matrix) and it is not clear to me why. I guess one approach to vectorizing the code is to eliminate the first and third if statements and use the find function instead. However, is there also something wrong with the second if statement?

Thanks for any help on this.
Jon
From: dpb on 23 Mar 2010 14:54
Jonathan wrote:
> G'day,
>
> I have the following matrix...
>
> Beta = [52 52 52 52 60 52 40 24 32 44];
> yB = zeros(1,length(Beta)).';
> for i = 1:length(Beta)
> if Beta>0 & Beta<10
> yB(i)=1;
> elseif Beta>10 & Beta<63
> yB(i)=cos((Beta-10)*pi()/180);
> elseif Beta>63
> yB(i)=0.63;
> end
> end
>
> For starters, the if loop doesn't do anything (other than creating the
> initial matrix) and it is not clear to me why. I guess one approach to
> vectorizing the code is to eliminate the first and third if statements
> and use the find function instead. However, is there also something
> wrong with the second if statement?

Well, there is no such thing as an "if loop" for starters. If you
really meant that as written that may be your conceptual problem.

The if statement(s) _DO_ do, something, just not what you (apparently)
think.

Type

doc if

for details of what, precisely.

> yB = zeros(1,length(Beta)).';

yB = zeros(length(Beta),1);
for i = 1:length(Beta)
if Beta(i)>0 & Beta(i)<10

% and etc., will fix your problem using the for...end

Alternatively, study the following

Beta(Beta>10 & Beta<63) = cos((Beta-10)*pi/180);

and salt to suit...

Also, another way to apply limits is something like

x = max(min(x,MaxVal),MinVal); % Study this, too...

It would be applied before the conversion above altho your sample vector
is all within the limits given.

--
From: Matt Fig on 23 Mar 2010 15:03
Look at the value produced by Beta>0. Now how does an IF statement deal with non-scalar arguments? (HINT: read the help for IF)

Beta = [52 52 52 52 60 52 40 24 32 44];
yB = zeros(length(Beta),1);


for ii = 1:length(Beta)
if Beta(ii)>0 & Beta(ii)<10
yB(ii)=1;
elseif Beta(ii)>10 & Beta(ii)<63
yB(ii)=cos((Beta(ii)-10)*pi/180);
elseif Beta(ii)>63
yB(ii)=0.63;
end
end




Is there a FAQ on this one yet? There should be!
From: Jan Simon on 23 Mar 2010 15:18
Dear Jonathan!

> yB(i)=cos((Beta-10)*pi()/180);

BTW. "pi" does not need brackets.

Jan
From: Bruno Luong on 23 Mar 2010 15:27
"Jan Simon" <matlab.THIS_YEAR(a)nMINUSsimon.de> wrote in message <hob45a$8p6$1(a)fred.mathworks.com>...
> Dear Jonathan!
>
> > yB(i)=cos((Beta-10)*pi()/180);
>
> BTW. "pi" does not need brackets.
>

The bracket should be here: cos((Beta-10)*(pi/180))
so that the vector is scanned one time less when multiplying.

Bruno
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