Answer:
[tex]\mathbf{R(\tau) = \dfrac{A^2}{2} cos (\omega \tau)}[/tex]
Step-by-step explanation:
To find Autocorrelation function of the following periodic function
Given that:
X(t) = A sin(wt +θ)
with the period T=2π/w , A, θ, and w are constants.
The autocorrelation function of periodic function with period and phase θ can be expressed as:
[tex]R(\tau) = \dfrac{1}{T} \int \limits ^{T/2}_{-T/2} x(t) *x(t - \tau) \ dt[/tex]
[tex]R(\tau) = \dfrac{A^2}{T} \int \limits ^{T/2}_{-T/2} \ A sin ( \omega t + \theta)*A sin [ \omega (t- \tau ) + \theta] \ dt[/tex]
where;
[tex]sinAsin B = \dfrac{1}{2}[cos (A-B) -cos (A+B)][/tex]
Then;
[tex]R(\tau) = \dfrac{A^2}{2T} \int \limits ^{T/2}_{-T/2} \ cos ( \omega t- \omega \tau + \theta - \omega t - \theta) - cos (\omega t - \omega \tau + \theta + \omega t + \theta) \ dt[/tex]
[tex]R(\tau) = \dfrac{A^2}{2T} \int \limits ^{T/2}_{-T/2} \ cos ( - \omega \tau ) - cos (2 \omega t - \omega \tau + 2 \theta) \ dt[/tex]
[tex]R(\tau) = \dfrac{A^2}{2T} \int \limits ^{T/2}_{-T/2} \ cos ( - \omega \tau ) \ dt - \dfrac{1}{2T} \int \limits ^{T/2}_{-T/2} cos (2 \omega t - \omega \tau + 2 \theta) \ dt[/tex]
The term 2 is the cosine wave of frequency and the phase = [tex]- w \tau + 2 \theta[/tex]
if we integrate that, the second term in the expansion for R(t) = zero
As such,
[tex]R(\tau) = \dfrac{A^2}{2T} \int \limits^{T/2}_{-T/2} \ cos ( - \omega \tau ) dt[/tex]
where ;
[tex]cos (-\omega \tau )[/tex]is constant
Then :
[tex]R(\tau) = \dfrac{A^2}{2T} cos (-\omega \tau) [t]^{T/2}_{-T/2}[/tex]
[tex]R(\tau) = \dfrac{A^2}{2T} cos (-\omega \tau) \times [\dfrac{T}{2}+ \dfrac{T}{2}][/tex]
[tex]R(\tau) = \dfrac{A^2}{2T} cos (-\omega \tau) \times [\dfrac{2T}{2}][/tex]
[tex]R(\tau) = \dfrac{A^2}{2T} cos (-\omega \tau) \times T[/tex]
[tex]R(\tau) = \dfrac{A^2}{2} cos (-\omega \tau)[/tex]
since [tex]cos (-\omega \tau) = cos (\omega \tau)[/tex]
[tex]\mathbf{R(\tau) = \dfrac{A^2}{2} cos (\omega \tau)}[/tex]
