S10.

Solution:

Let’s use the relation between the weighting function and transfer characteristic at first:
$$ w(t)= \mathfrak{F}^{-1} \left \{ w(j\omega) \right \} = \mathfrak{F}^{-1} \left \{ \frac{2(1+j\omega)}{4+3j\omega} \right \}= \frac{2}{3} \mathfrak{F}^{-1} \left \{ \frac{j\omega+1}{j\omega+\frac{4}{3}} \right \}= \frac{2}{3} \mathfrak{F}^{-1} \left \{ \frac{j\omega+\frac{4}{3}-\frac{1}{3}}{j\omega+\frac{4}{3}} \right \}$

$$ w(t)= \frac{2}{3} \mathfrak{F}^{-1} \left \{ 1-\frac{1}{3}\frac{1}{j\omega+\frac{4}{3}} \right \} = \frac{2}{3} \left [ \delta(t)-\frac{1}{3}e^{-\frac{4}{3}t}1(t)} \right ]$
Let’s use, $ w(t) is general step function.

Step response function can be derived by the weighting function as follow:
$$ h(t)= \int_0^tw(\tau)d\tau=\int_0^t\frac{2}{3} \left [ \delta(\tau)-\frac{1}{3}e^{-\frac{4}{3}\tau}1(\tau) \right ]d\tau=\int_0^t\frac{2}{3} \delta(\tau)d\tau+ \int_{0}^{t}\left[ \left ( -\frac{1}{3} \right ) \frac{2}{3} e^{-\frac{4}{3}\tau}1(\tau)\right] d\tau$

$$ h(t)=1(t) \frac{2}{3}+\left ( -\frac{1}{3} \right ) \left [ \frac{2}{3} \frac{-3}{4} e^{-\frac{4}{3}t} \right ]_0^t 1(t)$

$$ h(t)= \left [ \frac{2}{3}+ \frac{1}{6} e^{-\frac{4}{3}t} \right ] 1(t)$

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