1. **Problem Statement:** Find the power spectral density (PSD) of a random process with autocorrelation function $$R(t) = \begin{cases} \lambda^2, & |t| > \varepsilon \\ \lambda^2 + \lambda \left(1 - \frac{|t|}{\varepsilon}\right), & |t| \leq \varepsilon \end{cases}$$ 2. **Recall the relationship:** The power spectral density $S(\omega)$ is the Fourier transform of the autocorrelation function $R(t)$: $$S(\omega) = \int_{-\infty}^{\infty} R(t) e^{-j \omega t} dt$$ 3. **Break the integral into two parts:** Since $R(t)$ is piecewise, $$S(\omega) = \int_{|t| \leq \varepsilon} \left(\lambda^2 + \lambda \left(1 - \frac{|t|}{\varepsilon}\right)\right) e^{-j \omega t} dt + \int_{|t| > \varepsilon} \lambda^2 e^{-j \omega t} dt$$ 4. **Simplify the second integral:** For $|t| > \varepsilon$, $R(t) = \lambda^2$. The integral over $(-\infty, -\varepsilon)$ and $(\varepsilon, \infty)$ of $\lambda^2 e^{-j \omega t}$ is $$\lambda^2 \left( \int_{-\infty}^{-\varepsilon} e^{-j \omega t} dt + \int_{\varepsilon}^{\infty} e^{-j \omega t} dt \right)$$ 5. **Note about the process:** Since $R(t)$ must be absolutely integrable for PSD to exist, and $\lambda^2$ is constant for $|t| > \varepsilon$, this implies the process is not WSS or the autocorrelation is defined only on $[-\varepsilon, \varepsilon]$ with zero elsewhere. Usually, autocorrelation tends to zero at infinity. 6. **Assuming $R(t) = 0$ for $|t| > \varepsilon$ (typical for such problems), then:** $$R(t) = \lambda^2 + \lambda \left(1 - \frac{|t|}{\varepsilon}\right), \quad |t| \leq \varepsilon$$ 7. **Rewrite $R(t)$ for $|t| \leq \varepsilon$:** $$R(t) = \lambda^2 + \lambda - \lambda \frac{|t|}{\varepsilon} = (\lambda^2 + \lambda) - \lambda \frac{|t|}{\varepsilon}$$ 8. **Calculate the PSD:** $$S(\omega) = \int_{-\varepsilon}^{\varepsilon} \left((\lambda^2 + \lambda) - \lambda \frac{|t|}{\varepsilon}\right) e^{-j \omega t} dt$$ 9. **Split the integral:** $$S(\omega) = (\lambda^2 + \lambda) \int_{-\varepsilon}^{\varepsilon} e^{-j \omega t} dt - \frac{\lambda}{\varepsilon} \int_{-\varepsilon}^{\varepsilon} |t| e^{-j \omega t} dt$$ 10. **Evaluate the first integral:** $$\int_{-\varepsilon}^{\varepsilon} e^{-j \omega t} dt = \frac{2 \sin(\omega \varepsilon)}{\omega}$$ 11. **Evaluate the second integral:** Since $|t|$ is even and $e^{-j \omega t} = \cos(\omega t) - j \sin(\omega t)$, only the cosine part contributes (even function), sine part integrates to zero. $$\int_{-\varepsilon}^{\varepsilon} |t| e^{-j \omega t} dt = 2 \int_0^{\varepsilon} t \cos(\omega t) dt$$ 12. **Calculate $\int_0^{\varepsilon} t \cos(\omega t) dt$ using integration by parts:** Let $u = t$, $dv = \cos(\omega t) dt$. Then $du = dt$, $v = \frac{\sin(\omega t)}{\omega}$. $$\int_0^{\varepsilon} t \cos(\omega t) dt = \left. t \frac{\sin(\omega t)}{\omega} \right|_0^{\varepsilon} - \int_0^{\varepsilon} \frac{\sin(\omega t)}{\omega} dt = \frac{\varepsilon \sin(\omega \varepsilon)}{\omega} - \frac{1}{\omega} \int_0^{\varepsilon} \sin(\omega t) dt$$ 13. **Evaluate $\int_0^{\varepsilon} \sin(\omega t) dt$:** $$= \left. -\frac{\cos(\omega t)}{\omega} \right|_0^{\varepsilon} = \frac{1 - \cos(\omega \varepsilon)}{\omega}$$ 14. **Substitute back:** $$\int_0^{\varepsilon} t \cos(\omega t) dt = \frac{\varepsilon \sin(\omega \varepsilon)}{\omega} - \frac{1}{\omega} \cdot \frac{1 - \cos(\omega \varepsilon)}{\omega} = \frac{\varepsilon \sin(\omega \varepsilon)}{\omega} - \frac{1 - \cos(\omega \varepsilon)}{\omega^2}$$ 15. **Therefore:** $$\int_{-\varepsilon}^{\varepsilon} |t| e^{-j \omega t} dt = 2 \left( \frac{\varepsilon \sin(\omega \varepsilon)}{\omega} - \frac{1 - \cos(\omega \varepsilon)}{\omega^2} \right)$$ 16. **Final expression for $S(\omega)$:** $$S(\omega) = (\lambda^2 + \lambda) \frac{2 \sin(\omega \varepsilon)}{\omega} - \frac{\lambda}{\varepsilon} \cdot 2 \left( \frac{\varepsilon \sin(\omega \varepsilon)}{\omega} - \frac{1 - \cos(\omega \varepsilon)}{\omega^2} \right)$$ Simplify: $$S(\omega) = 2 \frac{\sin(\omega \varepsilon)}{\omega} (\lambda^2 + \lambda) - 2 \lambda \left( \frac{\sin(\omega \varepsilon)}{\omega} - \frac{1 - \cos(\omega \varepsilon)}{\varepsilon \omega^2} \right)$$ $$= 2 \frac{\sin(\omega \varepsilon)}{\omega} (\lambda^2 + \lambda - \lambda) + 2 \lambda \frac{1 - \cos(\omega \varepsilon)}{\varepsilon \omega^2}$$ $$= 2 \lambda^2 \frac{\sin(\omega \varepsilon)}{\omega} + 2 \lambda \frac{1 - \cos(\omega \varepsilon)}{\varepsilon \omega^2}$$ 17. **Interpretation:** This is the power spectral density of the given process. **Final answer:** $$\boxed{S(\omega) = 2 \lambda^2 \frac{\sin(\omega \varepsilon)}{\omega} + 2 \lambda \frac{1 - \cos(\omega \varepsilon)}{\varepsilon \omega^2}}$$