Generating a single pulse in time?

Hi I would like to write a simple code for a single pulse.So for example: if I would like to see pulse in the time period: 400<t<500

Why does If (t>400&&t<500) s=2.0 else s=0.0 end

When I plot this I get: a result where s=0 for all time and no pulse.

but

if (t>400&&t<500) s=2.0; else s=0.01; end

a pulse with a baseline of 0.01 (and maximum value 2.0)?

Thank you for your help.

Answer 1

[David Sanchez]

Your tspan is

tspan=[0,1000];

it should be:

tspan=0:1000;

You will get this plot:

[AA]

Thanks!

This is a bit worrying though (for me)- as for both 0:1000 and [0 1000] we are specifying the endpoints in the tspan. How can the result be so different?

[AA]

I meant the initial and endpoints of integration time.

Answer 2

[Wayne King]

You did not tell us the sampling frequency so I'll just assume it is 1 Hz in my example.

   t = 0:1:1000;
   s = 0.01*ones(length(t));
   s(t>400 & t<500) = 2;
   plot(t,s)

Note that in this case you could just do the simpler (assuming the first sample is t=0):

   s = 0.01*ones(10001,1);
   s(402:500) = 2;

but the example I have should work when the sampling frequency is not 1.

[AA]

Could you please shed some light on my comment above. I would be very grateful. What is the difference between tspan=0:1000 and [0,1000] Thanks!

[Talha Khalid]

@Wayne King

I just tested your answer. It doesn't generate the required result. See the attached picture

Answer 3

[David Sanchez]

It is easier to do:

   t = 0:1000;
   s = zeros(length(t));
   s(t>400 & t<500) = 2;
   plot(t,s)

[AA]

Thank you for your answer.

I'm trying to work out why I get the results that I do get? I cannot work it out. Why does having a baseline result in a pulse?

Thanks for your help.

Answer 4

[AA]

Thank you for your answer.

I'm trying to work out why I get the results that I do get? I cannot work it out. Why does having a baseline result in a pulse?

Thanks for your help.

Answer 5

[Wayne King]

See my earlier answer for how to do it.

Using your if statement approach, when t is a vector how do you expect the if statement to evaluate (t> 400 && t<500)?

To get that approach to work you'd have to loop through the t-values like this

s = zeros(1000,1);
 t = 1:1000;
 for nn = 1:length(t)
   tval = t(nn);
 if (tval > 400 && tval<500)
   s(nn) = 2;
 else
   s(nn) = 0.01;
 end
 end
 plot(t,s)

But that is not an efficient use of MATLAB's inherit array operations. Use the approach I gave you earlier.

Answer 6

[AA]

Thanks both. Heres the thing: I need a zero baseline: SO when I put in either David Sanchez or Wayne Kings' (modify 0.01 to 0 in the latter case) in my code I DO NOT get a pulse- just a flat 0 line. Here is my full code ( I know it seems really long- but its a simplified version of a larger code that I'm using and the setup needs to be preserved. I have included your suggestions):

function dy=testpulse(t,y);

N=100; dy=zeros(N,1); delta=2*pi/N;

s = zeros(length(t));

   s(t>400 & t<500) = 2;

    for k=1:N
        dy(k)=s-y(k);
    end

Calling routine: N=100; delta=2*pi/N;

y0=zeros(N,1);

for j=1:N

y0(j)=0.0;

end

x=zeros(N,1);

x=(1:N)*delta-delta;

tspan=[0,1000];

[t,y]=ode15s(@testpulse,tspan,y0);

a=y(:,1:N); mesh(x,t,a); xlabel('\Theta'),ylabel('t'), zlabel('[]'); title('X')

SO my question remains- even when using the codes that you both have suggested why don't I get a pulse when the baseline is 0?

I do know how to get a pulse without using the && (you can use the OR operator or just use a series of if and elseif commands to get it)- but this is really bugging me because it doesn't make sense to me!?