I want to print the value of t when value of vc1 is 90.5931 but I am not able to do so. please help

for t1=0:0.00001:0.15   
  vc1 = 100-100*exp(-5000*t1)*(cosh(2000*6^(1/2)*t1) + ...
                   (5*6^(1/2)*sinh(2000*6^(1/2)*t1))/12);
  while(vc1 == 90.5931)
    display(t1)
  end
  plot(t1,vc1)
  hold on
end

[Guillaume]

Note that you're actually lucky that your comparison didn't work as if it did, you would have entered a never ending loop

while vc1 == 90.5931
   display(t1)
end

means: enter the loop if vc1 is equal to 90.5931, then display t1 and try the loop again. Since vc1 has not change, it will display t1 again and try again, and again...

Answer 1

[Walter Roberson]

t1 =0:0.00001:0.15;
vc1 = 100 - 100*exp(-5000*t1) .* (cosh(2000*6^(1/2)*t1) + ...
                 (5*6^(1/2)*sinh(2000*6^(1/2)*t1))/12);
plot(t1, vc1);

tidx = find(vc1 >= 90.5931, 1, 'first');
t1(tidx)

[Guillaume]

I'm not convinced your tidx is the one the OP wants as this returns the t1 for ~90.603 which is the next one after ~90.5931.

Answer 2

[Guillaume]

As per Dr Siva's answer, do not use == to compare floating points numbers unless the two numbers have been obtained exactly the same way (both results of the same calculation or both typed in the code), due to floating point accuracy (the way numbers are actually stored and the round-off error of calculation).

Always compare the absolute difference of the number relative to an arbitrary small number. Some multiple of eps may be a good idea (the actual function not some made up variable with an unrealistic value), or just an arbitray small number. In your case, the actual value is around 2*1e-5 away from 90.5931, so I suggest using 1e-4

if abs(vc1 - 90.5931) < 1e-4  %with no eps variable existing

To answer your second problem, the slow plotting, the answer is not to use a loop:

t1 = 0 : 1e-5 : 0.15;
vc1 = 100 - 100*exp(-5000*t1) .* (cosh(2000*6^(1/2)*t1) + ...
                                 (5*6^(1/2)*sinh(2000*6^(1/2)*t1))/12);
plot(t1, vc1, '.r');
%and to find for which t1, vc1 is near 90.5953:
t90 = t1(abs(vc1 - 90.5931) < 1e-4)

Answer 3

[KSSV]

Dont try to equate floating point numbers. Try the following:

eps = 10^-100 ;  % Set a small number epsilon for equating 
for t1 =0:0.00001:0.15
  vc1 = 100-100*exp(-5000*t1)*(cosh(2000*6^(1/2)*t1) + ...
                   (5*6^(1/2)*sinh(2000*6^(1/2)*t1))/12);
   if ((vc1 - 90.5931)< eps)
     display(t1)
   end
  plot(t1,vc1,'.r')
  drawnow
  hold on
end

[sayan chakraborty]

thank you. It works but takes a lot of time to plot. Is there a way to tackle this problem?

[Guillaume]

I'm sorry to say that Dr. Siva's answer is wrong on several levels. He's got the right idea that you cannot compare floating point number for equality using == due to floating point accuracy. The example code is totally wrong though.

For a start, do not call a variable eps as it is a the name of the matlab built-in function, which gives you the distance to the next floating point number.

Secondly, if you actually use that function you'll see that eps(90.5931) is around 1e-14, a much larger value than 1e-100. What that means in practice is that the if test will only be true if and only if vc1 is exactly 90.5931 (or whatever the internal representation of that number actually is). As it is the test is exactly the same as doing the equality test that was advised against.

Finally, it also the absolute of the difference that you should compare. As it is the test will be true with any vc1 smaller than 90.5931. That later point is actually the reason that Dr. Siva's code actually works (in that it displays something, but that something is wrong).

[Walter Roberson]

1e-14 is a much larger value than 1e-100 ;-)

[Guillaume]

Doh! That's what I meant. Fixed. My point still stand in that 1e-100 is completely useless in this case.