1. **State the problem:** Given the discrete-time signal $$x[n] = \frac{1}{2} n u[n-3]$$, where $$u[n-3]$$ is the unit step function shifted by 3, find the values of $$x[n]$$ for different $$n$$. 2. **Recall the unit step function:** The unit step function $$u[n-k]$$ is defined as: $$ u[n-k] = \begin{cases} 0 & n < k \\ 1 & n \geq k \end{cases}$$ 3. **Apply the definition:** Since $$u[n-3] = 0$$ for $$n < 3$$ and $$1$$ for $$n \geq 3$$, the signal $$x[n]$$ is zero for $$n < 3$$. 4. **For $$n \geq 3$$,** the signal is: $$x[n] = \frac{1}{2} n \times 1 = \frac{n}{2}$$ 5. **Summary:** $$x[n] = \begin{cases} 0 & n < 3 \\ \frac{n}{2} & n \geq 3 \end{cases}$$ This means the signal starts at zero until $$n=3$$, then increases linearly with slope $$\frac{1}{2}$$. **Final answer:** $$x[n] = \frac{n}{2} u[n-3]$$