Let's solve the problem step by step: --- ### **Given:** \[ x(t) = 3e^{2t}u(t) + 4e^{3t}u(t) \] where \( u(t) \) is the unit step function. --- ## **(a) Does the Fourier Transform (FT) of this function converge?** **Fourier Transform definition:** \[ X(j\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \] For convergence, **\( x(t) \)** must be absolutely integrable: \[ \int_{-\infty}^{\infty} |x(t)| dt < \infty \] For \( t < 0 \), \( u(t) = 0 \), so \( x(t) = 0 \). For \( t \geq 0 \): \[ x(t) = 3e^{2t} + 4e^{3t} \] which **grows exponentially as \( t \to \infty \)**. \[ \int_0^{\infty} (3e^{2t} + 4e^{3t}) dt = \infty \] **Conclusion:** \[ \boxed{\text{No, the Fourier Transform of this function does not converge.}} \] --- ## **(b) For which values of \( \alpha \) does \( e^{-\alpha t} x(t) \) converge?** Consider: \[ \int_{0}^{\infty} e^{-\alpha t} x(t) dt = \int_{0}^{\infty} e^{-\alpha t}(3e^{2t} + 4e^{3t}) dt \] \[ = \int_{0}^{\infty} 3e^{(2-\alpha)t} dt + \int_{0}^{\infty} 4e^{(3-\alpha)t} dt \] Each integral converges only if the exponent is negative: - For \( 3e^{(2-\alpha)t} \): **Converges when \( 2-\alpha < 0 \rightarrow \alpha > 2 \)** - For \( 4e^{(3-\alpha)t} \): **Converges when \( 3-\alpha < 0 \rightarrow \alpha > 3 \)** **Both terms converge when \( \alpha > 3 \)**. **For the given values:** - \( \alpha = 1 \): **Does not converge** - \( \alpha = 2.5 \): **Does not converge** - \( \alpha = 3.5 \): **Converges** **Conclusion:** \[ \boxed{ \begin{align*} \alpha = 1 \quad &\text{No convergence} \\ \alpha = 2.5 \quad &\text{No convergence} \\ \alpha = 3.5 \quad &\text{Converges} \end{align*} } \] --- ## **(c) By definition, find the Laplace Transform of \( x(t) \), with ROC and sketch the poles** **Laplace Transform:** \[ X(s) = \int_{0}^{\infty} x(t) e^{-st}dt \] \[ x(t) = 3e^{2t}u(t) + 4e^{3t}u(t) \] \[ X(s) = \int_{0}^{\infty} 3e^{2t}e^{-st} dt + \int_{0}^{\infty} 4e^{3t}e^{-st} dt \] \[ = 3\int_{0}^{\infty} e^{(2-s)t} dt + 4\int_{0}^{\infty} e^{(3-s)t} dt \] Each integral is \( \int_{0}^{\infty} e^{at}dt = \frac{1}{-a} \) for \( \text{Re}(a) < 0 \). - First term: converges if \( \text{Re}(s) > 2 \) - Second term: converges if \( \text{Re}(s) > 3 \) So, **overall ROC is \( \text{Re}(s) > 3 \)**. \[ X(s) = 3 \left[\frac{1}{s - 2}\right] + 4 \left[\frac{1}{s - 3}\right] \] --- ### **Poles:** - At \( s = 2 \) (from \( 3/(s-2) \)) - At \( s = 3 \) (from \( 4/(s-3) \)) ### **Sketch:** - Poles at \( s = 2 \) and \( s = 3 \) on the real axis. - ROC is to the right of \( s = 3 \) (i.e., \( \text{Re}(s) > 3 \)). --- ## **Final Answers (Summary):** ### a) **No, the Fourier Transform does not converge.** ### b) \[ \begin{align*} \alpha = 1 &: \text{Does not converge} \\ \alpha = 2.5 &: \text{Does not converge} \\ \alpha = 3.5 &: \text{Converges} \end{align*} \] ### c) **Laplace Transform:** \[ X(s) = \frac{3}{s-2} + \frac{4}{s-3} \] **ROC:** \( \text{Re}(s) > 3 \) **Poles:** at \( s = 2 \) and \( s = 3 \) **Pole-Zero Sketch:** (draw two crosses at \( s = 2 \) and \( s = 3 \) on the real axis, shade the region to the right of \( s = 3 \) for the ROC).