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CivilSigma
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Homework Statement
For a constant power signal x(t) = c, determine the auto correlation function and the spectral density function.
Homework Equations
The auto correlation function is:
$$R_x (\tau) = \int_{-\infty}^{\infty} E(x(t) \cdot x(t+\tau)) d\tau$$
To my understanding, here to find the expected value of the signal we must also multiply the function x(t)x(t+tau) by the probability function which is taken to be 1/period. Then, we change the integral limits to one period and get:
$$R_x (\tau) = \int_{-T/2}^{T/2} E(x(t) \cdot x(t+\tau)) \cdot \frac{1}{T} d\tau$$
Is my understanding of the auto-correlational function correct?
For the spectral density function :
$$S_x (\omega) = \int_{-\infty}^{\infty} R_x(\tau) \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$
Here, we don't change the integral limits and consider all the possible values of tau.
The Attempt at a Solution
The auto-correlation is:
$$R_x (\tau) = \int_{-T/2}^{T/2} E(x(t) \cdot x(t+\tau)) \cdot \frac{1}{T} d\tau$$
$$R_x(\tau) = \int_{-T/2}^{T/2} c^2 \cdot \frac{1}{T} d\tau$$
$$R_x(\tau)=c^2$$
The spectral density is then:
$$S_x (\omega) = \int_{-\infty}^{\infty} R_x(\tau) \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$
$$S_x (\omega) = \int_{-\infty}^{\infty} c^2 \cdot e^{-i\omega \tau} \cdot \frac{1}{2\pi}d\tau$$
$$S_x (\omega) = \frac{-c^2}{2\pi \cdot i \omega} \times (e^{-i \omega \tau})|_{-\infty}^{\infty}$$
$$S_x (\omega) = \frac{-c^2}{2\pi \cdot i \omega} \times (0 - \infty)=?$$
However, my lecture notes suggest that the answer is :
$$S_x(\omega) = c^2 \cdot \delta(t)$$How did they get to here? How is the Dirac function obtained from evaluating the integral? Moreover, what happened to the other constants such as 2pi, i and omega from my original solution? If someone can shed some light on this mystery I would really appreciate it.
