1. **State the problem:** We are given the power spectral density (PSD) of a process as $$S_{xx}(\omega) = \frac{\omega^2 + 9}{\omega^4 + 5\omega^2 + 4}$$ and asked to find the mean square value of the process. 2. **Recall the formula:** The mean square value of a stationary process can be found by integrating its PSD over all frequencies: $$\text{Mean square value} = \frac{1}{2\pi} \int_{-\infty}^{\infty} S_{xx}(\omega) \, d\omega$$ 3. **Simplify the denominator:** Factor the denominator: $$\omega^4 + 5\omega^2 + 4 = (\omega^2 + 4)(\omega^2 + 1)$$ 4. **Rewrite the integral:** $$\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{\omega^2 + 9}{(\omega^2 + 4)(\omega^2 + 1)} \, d\omega$$ 5. **Use symmetry:** The integrand is an even function (depends on $\omega^2$), so $$= \frac{1}{\pi} \int_0^{\infty} \frac{\omega^2 + 9}{(\omega^2 + 4)(\omega^2 + 1)} \, d\omega$$ 6. **Partial fraction decomposition:** Let $$\frac{\omega^2 + 9}{(\omega^2 + 4)(\omega^2 + 1)} = \frac{A}{\omega^2 + 4} + \frac{B}{\omega^2 + 1}$$ Multiply both sides by denominator: $$\omega^2 + 9 = A(\omega^2 + 1) + B(\omega^2 + 4) = (A + B)\omega^2 + (A + 4B)$$ Equate coefficients: - For $\omega^2$: $1 = A + B$ - Constant: $9 = A + 4B$ 7. **Solve for A and B:** From $1 = A + B$, we get $A = 1 - B$ Substitute into $9 = A + 4B$: $$9 = (1 - B) + 4B = 1 + 3B \implies 3B = 8 \implies B = \frac{8}{3}$$ Then, $$A = 1 - \frac{8}{3} = -\frac{5}{3}$$ 8. **Rewrite integral:** $$\frac{1}{\pi} \int_0^{\infty} \left( \frac{-\frac{5}{3}}{\omega^2 + 4} + \frac{\frac{8}{3}}{\omega^2 + 1} \right) d\omega = \frac{1}{\pi} \left(-\frac{5}{3} \int_0^{\infty} \frac{d\omega}{\omega^2 + 4} + \frac{8}{3} \int_0^{\infty} \frac{d\omega}{\omega^2 + 1} \right)$$ 9. **Evaluate integrals:** Recall: $$\int_0^{\infty} \frac{d\omega}{\omega^2 + a^2} = \frac{\pi}{2a}$$ So, $$\int_0^{\infty} \frac{d\omega}{\omega^2 + 4} = \frac{\pi}{2 \times 2} = \frac{\pi}{4}$$ $$\int_0^{\infty} \frac{d\omega}{\omega^2 + 1} = \frac{\pi}{2}$$ 10. **Substitute back:** $$\frac{1}{\pi} \left(-\frac{5}{3} \times \frac{\pi}{4} + \frac{8}{3} \times \frac{\pi}{2} \right) = \frac{1}{\pi} \left(-\frac{5\pi}{12} + \frac{8\pi}{6} \right) = \frac{1}{\pi} \left(-\frac{5\pi}{12} + \frac{16\pi}{12} \right) = \frac{1}{\pi} \times \frac{11\pi}{12} = \frac{11}{12}$$ **Final answer:** The mean square value of the process is $$\boxed{\frac{11}{12}}$$.