I have this trigonometric equation;
cos(2*pi*50*t)+cos(2*pi*100*t)
I want to graphic of equation and I want to find field for a period. How can I do?
Asked
Active
Viewed 42 times
-4
-
What does "find field" mean in this context? – rayryeng Aug 09 '17 at 15:08
-
Example; I have rectangle. `Short side lenght: 10 cm` and `Long side lenght: 20 cm` `Rectangle field = 10 cm x 20 cm = 200 cm^2` – Tuna EREN Aug 09 '17 at 16:22
-
So you mean [area](https://en.wikipedia.org/wiki/Area)? – Wolfie Aug 09 '17 at 16:27
-
Yes. Are area and field different? – Tuna EREN Aug 09 '17 at 20:06
-
Not different. I've just never heard field used in that context before. Why do you need to do this graphically when you can do it analytically with integrals? – rayryeng Aug 09 '17 at 23:55
-
I need both. Graphic and area. – Tuna EREN Aug 10 '17 at 06:58
1 Answers
1
Graph:
>> f = @(t) cos(2*pi*50*t) + cos(2*pi*100*t);
>> x = linspace(0, 1/50, 100);
>> y = f(x);
>> plot(x,y)
Area over 1 period:
>> integral(f, 0, 1/50)
or just do it manually:
∫ ( cos(2π·50t) + cos(2π·100t) ) dt =
-1/2π·( 1/50·sin(2π·50t) + 1/100·sin(2π·100t) )
which, evaluated between 0 and 1/50, equals 0.
Rody Oldenhuis
- 37,726
- 7
- 50
- 96
-
Good answer but I have problem. Yes, I find equals 0. Area below zero is negative. I want this to be positive too. I shared picture. I want to find the scanned area. – Tuna EREN Aug 10 '17 at 08:07
-