Derive x(t) = Acos(Bt+C) for a Periodic Function with Given Properties

Since x(t) is periodic with period T=6, we can write:

x(t) = x(t + 6)

Using the fact that x(t) = -x(t-3), we can write:

x(t+6) = -x((t+6)-3) = -x(t+3)

So we have:

x(t) = -x(t+3)

Now, we can write x(t) as a Fourier series:

x(t) = a0 + ∑(n=1 to ∞) [an cos(nωt) + bn sin(nωt)]

where ω = 2π/T = π/3 is the fundamental frequency.

Using the fact that ak = 0 for k > 2, we have:

a0 = (1/T) ∫(0 to T) x(t) dt = (1/6) ∫(0 to 6) x(t) dt = (1/6) x(t) dt

a1 = (1/T) ∫(0 to T) x(t) cos(ωt) dt = (1/6) ∫(0 to 6) x(t) cos(πt/3) dt

a2 = (1/T) ∫(0 to T) x(t) cos(2ωt) dt = (1/6) ∫(0 to 6) x(t) cos(2πt/3) dt

Since x(t) is an odd function (i.e., x(-t) = -x(t)), we have:

a1 = 0

Using the fact that x(t) = -x(t+3), we have:

a2 = (1/6) ∫(0 to 3) x(t) cos(2πt/3) dt - (1/6) ∫(3 to 6) x(t) cos(2πt/3) dt

= (1/6) ∫(0 to 3) x(t) cos(2πt/3) dt + (1/6) ∫(0 to 3) x(t+3) cos(2π(t+3)/3) dt

= (1/6) ∫(0 to 3) x(t) cos(2πt/3) dt - (1/6) ∫(0 to 3) x(t) cos(2πt/3) dt

= 0

Therefore, the Fourier series of x(t) is:

x(t) = a0 + a2 cos(2ωt)

= (1/6) x(t) + 0 cos(2πt/3)

Solving for x(t), we get:

x(t) = (6/5) a2 cos(2πt/3)

= (6/5) (1/6) ∫(0 to 3) x(t) cos(2πt/3) dt cos(2πt/3)

= (1/5) ∫(0 to 3) x(t) cos(2πt/3) dt cos(2πt/3)

Comparing this with the form x(t) = Acos(Bt+C), we see that:

A = (1/5) ∫(0 to 3) x(t) cos(2πt/3) dt

B = 2π/3

C = 0

Therefore, x(t) = (1/5) ∫(0 to 3) x(t) cos(2πt/3) dt cos(2πt/3)