1. **Problem Statement:** Find the power spectral density (PSD) of a random process whose autocorrelation function $R(\tau)$ is given by: $$ R(\tau) = \begin{cases} \lambda^2, & |\tau| > 3 \\ \lambda^2 + \lambda \left(1 - \frac{|\tau|}{3}\right), & |\tau| \leq 3 \end{cases} $$ 2. **Formula Used:** The power spectral density $S(f)$ is the Fourier transform of the autocorrelation function $R(\tau)$: $$ S(f) = \int_{-\infty}^{\infty} R(\tau) e^{-j 2 \pi f \tau} d\tau $$ 3. **Important Notes:** - Since $R(\tau)$ is real and even, $S(f)$ will be real and even. - The constant part $\lambda^2$ for $|\tau| > 3$ contributes a delta function at zero frequency, but since it is constant over infinite $\tau$ except a finite interval, we consider the piecewise structure carefully. 4. **Rewrite $R(\tau)$:** $$ R(\tau) = \lambda^2 + \lambda \left(1 - \frac{|\tau|}{3}\right) \mathbf{1}_{|\tau| \leq 3} $$ where $\mathbf{1}_{|\tau| \leq 3}$ is the indicator function equal to 1 when $|\tau| \leq 3$ and 0 otherwise. 5. **Fourier transform of $R(\tau)$:** $$ S(f) = \int_{-\infty}^{\infty} \left[ \lambda^2 + \lambda \left(1 - \frac{|\tau|}{3}\right) \mathbf{1}_{|\tau| \leq 3} \right] e^{-j 2 \pi f \tau} d\tau $$ Split the integral: $$ S(f) = \lambda^2 \int_{-\infty}^{\infty} e^{-j 2 \pi f \tau} d\tau + \lambda \int_{-3}^{3} \left(1 - \frac{|\tau|}{3}\right) e^{-j 2 \pi f \tau} d\tau $$ 6. **Evaluate the first integral:** $$ \int_{-\infty}^{\infty} e^{-j 2 \pi f \tau} d\tau = \delta(f) $$ where $\delta(f)$ is the Dirac delta function. So, $$ S(f) = \lambda^2 \delta(f) + \lambda \int_{-3}^{3} \left(1 - \frac{|\tau|}{3}\right) e^{-j 2 \pi f \tau} d\tau $$ 7. **Evaluate the second integral:** Since the function is even in $\tau$, and $e^{-j 2 \pi f \tau} = \cos(2 \pi f \tau) - j \sin(2 \pi f \tau)$, the integral over symmetric limits of an even function times an even function is twice the integral from 0 to 3: $$ \int_{-3}^{3} \left(1 - \frac{|\tau|}{3}\right) e^{-j 2 \pi f \tau} d\tau = 2 \int_0^3 \left(1 - \frac{\tau}{3}\right) \cos(2 \pi f \tau) d\tau $$ 8. **Compute:** Let $$ I = 2 \int_0^3 \left(1 - \frac{\tau}{3}\right) \cos(2 \pi f \tau) d\tau = 2 \left[ \int_0^3 \cos(2 \pi f \tau) d\tau - \frac{1}{3} \int_0^3 \tau \cos(2 \pi f \tau) d\tau \right] $$ 9. **Calculate each integral:** - First integral: $$ \int_0^3 \cos(2 \pi f \tau) d\tau = \frac{\sin(6 \pi f)}{2 \pi f} $$ - Second integral (by parts): Let $u=\tau$, $dv=\cos(2 \pi f \tau)d\tau$ then $du=d\tau$, $v=\frac{\sin(2 \pi f \tau)}{2 \pi f}$ $$ \int_0^3 \tau \cos(2 \pi f \tau) d\tau = \left. \tau \frac{\sin(2 \pi f \tau)}{2 \pi f} \right|_0^3 - \int_0^3 \frac{\sin(2 \pi f \tau)}{2 \pi f} d\tau $$ The second integral: $$ \int_0^3 \sin(2 \pi f \tau) d\tau = \frac{1 - \cos(6 \pi f)}{2 \pi f} $$ So, $$ \int_0^3 \tau \cos(2 \pi f \tau) d\tau = \frac{3 \sin(6 \pi f)}{2 \pi f} - \frac{1}{2 \pi f} \cdot \frac{1 - \cos(6 \pi f)}{2 \pi f} = \frac{3 \sin(6 \pi f)}{2 \pi f} - \frac{1 - \cos(6 \pi f)}{4 \pi^2 f^2} $$ 10. **Substitute back:** $$ I = 2 \left[ \frac{\sin(6 \pi f)}{2 \pi f} - \frac{1}{3} \left( \frac{3 \sin(6 \pi f)}{2 \pi f} - \frac{1 - \cos(6 \pi f)}{4 \pi^2 f^2} \right) \right] = 2 \left[ \frac{\sin(6 \pi f)}{2 \pi f} - \frac{\sin(6 \pi f)}{2 \pi f} + \frac{1 - \cos(6 \pi f)}{12 \pi^2 f^2} \right] = 2 \cdot \frac{1 - \cos(6 \pi f)}{12 \pi^2 f^2} = \frac{1 - \cos(6 \pi f)}{6 \pi^2 f^2} $$ 11. **Final expression for $S(f)$:** $$ S(f) = \lambda^2 \delta(f) + \lambda \frac{1 - \cos(6 \pi f)}{6 \pi^2 f^2} $$ 12. **Interpretation:** - The PSD has a delta spike at zero frequency due to the constant term $\lambda^2$. - The second term is a continuous function describing the spectral content from the linear part of $R(\tau)$. **Answer:** $$ \boxed{S(f) = \lambda^2 \delta(f) + \lambda \frac{1 - \cos(6 \pi f)}{6 \pi^2 f^2}} $$