Find Fourier Series of f(t)=2u(t)-2u(t-2) with T=4s | Step Function Question

[th3plan]

consider f(t)=2u(t)-2u(t-2) Volts between t=0 and t=4 seconds. So i have to find the Fourier Series Representation of this given that T=4 secounds. I am not worried about the Fourier part. If i draw the f(t) function out on graph its 2 Volts from t=0 till t=2 seconds and then wehn it gets to 2 it goes down -2 volts back to Zero Volts , then when it gets to 4 seconds it goes up back to 2 volts correct ?

Is what i stated correct ?

 

[Redbelly98]

Yes.

 

[oswaler]

Yes, your understanding of the function f(t) and its graph is correct. To find the Fourier Series representation of this function, we would need to use the formula:
f(t) = a0/2 + Σ(an*cos(nωt) + bn*sin(nωt))
where ω = 2π/T and T is the period (in this case, T=4 seconds).

First, we need to calculate the coefficients an and bn. Since f(t) is an even function, we only need to calculate the coefficients for cos(nωt). We can write f(t) as:
f(t) = 2u(t) - 2u(t-2) = 2u(t) - 2u(t)u(t-2)

We can see that the function is 2 for t between 0 and 2 seconds, and 0 for t between 2 and 4 seconds. Therefore, we can write the function as:
f(t) = 2u(t) - 2u(t)u(t-2) = 2u(t) - 2u(t-2)u(t)

Using the formula for the Fourier Series coefficients of an even function, we can calculate:
an = (2/T)∫f(t)*cos(nωt)dt = (2/4)∫(2u(t) - 2u(t-2)u(t))*cos(nωt)dt = (1/2)∫(2*cos(nωt) - 2*cos(nωt)*u(t-2))dt
= (1/2)*(2*sin(nωt)/nω - 2*sin(nωt)*u(t-2)/nω) evaluated from 0 to 4 seconds
= (1/2)*(2*sin(nω*4)/nω - 2*sin(nω*4)*u(4-2)/nω - 2*sin(nω*0)/nω + 2*sin(nω*0)*u(0-2)/nω)
= sin(nπ)/nπ

Similarly, we can calculate bn:
bn = (2/T)∫f(t)*sin(nωt)dt = (2/4)∫(2u(t) - 2u(t-2)u(t))*sin(nωt)dt = (