Question

Find the Autocorrelation function of the following periodic function: X(t) = A sin(wt +θ) 21 With T=2π/w the period, A, θ, and w are constants.

Answer 1

`\mathbf{R(\tau) = (A^2)/(2) cos (\omega \tau)}`

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:

`R(\tau) = (1)/(T) \int \limits ^(T/2)_(-T/2) x(t) *x(t - \tau) \ dt`

`R(\tau) = (A^2)/(T) \int \limits ^(T/2)_(-T/2) \ A sin ( \omega t + \theta)*A sin [ \omega (t- \tau ) + \theta] \ dt`

where;

`sinAsin B = (1)/(2)[cos (A-B) -cos (A+B)]`

Then;

`R(\tau) = (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`

`R(\tau) = (A^2)/(2T) \int \limits ^(T/2)_(-T/2) \ cos ( - \omega \tau ) - cos (2 \omega t - \omega \tau + 2 \theta) \ dt`

`R(\tau) = (A^2)/(2T) \int \limits ^(T/2)_(-T/2) \ cos ( - \omega \tau ) \ dt - (1)/(2T) \int \limits ^(T/2)_(-T/2) cos (2 \omega t - \omega \tau + 2 \theta) \ dt`

The term 2 is the cosine wave of frequency and the phase =
`- w \tau + 2 \theta`

if we integrate that, the second term in the expansion for R(t) = zero

As such,

`R(\tau) = (A^2)/(2T) \int \limits^(T/2)_(-T/2) \ cos ( - \omega \tau ) dt`

where ;

`cos (-\omega \tau )`is constant

Then :

`R(\tau) = (A^2)/(2T) cos (-\omega \tau) [t]^(T/2)_(-T/2)`

`R(\tau) = (A^2)/(2T) cos (-\omega \tau) * [(T)/(2)+ (T)/(2)]`

`R(\tau) = (A^2)/(2T) cos (-\omega \tau) * [(2T)/(2)]`

`R(\tau) = (A^2)/(2T) cos (-\omega \tau) * T`

`R(\tau) = (A^2)/(2) cos (-\omega \tau)`

since
`cos (-\omega \tau) = cos (\omega \tau)`

`\mathbf{R(\tau) = (A^2)/(2) cos (\omega \tau)}`