Let's solve the problem step-by-step: ### **Given:** \[ X(j\omega) = \frac{1}{(j\omega)^2 + 7j\omega + 10} \] We are to find the **inverse Fourier Transform** using **partial fraction expansion**. --- ## **Step 1: Factor the Denominator** \[ (j\omega)^2 + 7j\omega + 10 = (j\omega + 5)(j\omega + 2) \] So, \[ X(j\omega) = \frac{1}{(j\omega + 5)(j\omega + 2)} \] --- ## **Step 2: Partial Fraction Expansion** Let: \[ \frac{1}{(j\omega + 5)(j\omega + 2)} = \frac{A}{j\omega + 5} + \frac{B}{j\omega + 2} \] Multiply both sides by the denominator: \[ 1 = A(j\omega + 2) + B(j\omega + 5) \] --- ## **Step 3: Solve for A and B** Let's solve for \(A\) and \(B\): - Set \(j\omega = -5\): \[ 1 = A(-5 + 2) + B(0) \implies 1 = A(-3) \implies A = -\frac{1}{3} \] - Set \(j\omega = -2\): \[ 1 = A(0) + B(-2 + 5) \implies 1 = B(3) \implies B = \frac{1}{3} \] So, \[ X(j\omega) = \frac{-1/3}{j\omega + 5} + \frac{1/3}{j\omega + 2} \] --- ## **Step 4: Use the Inverse Fourier Transform Table** Recall: \[ \mathcal{F}^{-1}\left[\frac{1}{j\omega + a}\right] = e^{-a t} u(t) \] where \(u(t)\) is the unit step function. Apply this to both terms: - For \(\frac{1}{j\omega + 5}\): \(a = 5\) - For \(\frac{1}{j\omega + 2}\): \(a = 2\) --- ## **Step 5: Write the Final Answer** \[ x(t) = -\frac{1}{3} e^{-5t} u(t) + \frac{1}{3} e^{-2t} u(t) \] Or, more simply, \[ \boxed{ x(t) = \frac{1}{3} \left[ e^{-2t} - e^{-5t} \right] u(t) } \] --- ### **Summary** - **Partial fractions:** \(\frac{-1/3}{j\omega+5} + \frac{1/3}{j\omega+2}\) - **Inverse FT:** \(x(t) = \frac{1}{3} (e^{-2t} - e^{-5t}) u(t)\) If you have any further questions or want the derivation of the FT table used, let me know!