Question

RX(τ)=exp(−a∣τ∣) where a is a positive real number. Find the PSD of X(t), given by S(f)=F{RX

​ [infinity]
(τ)}=∫ Rₓ (τ)e⁻ʲ²πᶠτ dτ
−[infinity]

Answer 1

To find the power spectral density (PSD) of X(t), given by S(f) = F{RX[infinity](tau)} = integral RX(tau)e^(-j2pif tau) dtau where RX(tau) = exp(-a|tau|), you need to evaluate the integral.

Step-by-step explanation:

To find the power spectral density (PSD) of $X(t)$, given by $S(f) = F\{R_X[\infty](\tau)\} = \int R_X(\tau)e^{-j2\pi f \tau} d\tau$ where $R_X(\tau) = e^-a$, you need to evaluate the integral. Let's start by substituting $R_X(\tau)$ into the integral:

$S(f) = \int e^-ae^{-j2\pi f \tau} d\tau$

Next, we can use the properties of exponentials to simplify the expression:

$S(f) = \int e^{-(a + j2\pi f)\tau} d\tau$

Now, we can integrate the exponential function:

$S(f) = \frac{1}{-(a + j2\pi f)}e^{-(a + j2\pi f)\tau}+C$

Finally, taking the limit of $\tau$ as $\infty$ and setting $C = 0$ since this term will approach zero, we get the PSD of $X(t)$:

$S(f) = \frac{1}{a + j2\pi f}$