To compute the convolution $f * g$ of two functions $f(t) = t^2$ for $t \geq 0$ and $g(t) = u(t - 1)$, we proceed as follows:

1. Identify the convolution formula: The convolution of two functions $f(t)$ and $g(t)$ is defined as: $(f * g)(t) = \int_{0}^{t} f(\tau) g(t - \tau) \, d\tau$

2. Substitute $f(t)$ and $g(t)$: Given $f(t) = t^2$ and $g(t) = u(t - 1)$, where $u(t - 1)$ is the unit step function that "turns on" at $t = 1$. This means that $g(t - \tau) = u(t - \tau - 1)$, which is 1 when $t - \tau \geq 1$ or $\tau \leq t - 1$, and 0 otherwise.

3. Set up the integral: The step function implies that the convolution integral will only have a nonzero contribution when $\tau \leq t - 1$. Thus, $(f * g)(t) = \int_{0}^{t-1} \tau^2 \cdot 1 \, d\tau = \int_{0}^{t-1} \tau^2 \, d\tau$

4. Evaluate the integral: $(f * g)(t) = \int_{0}^{t-1} \tau^2 \, d\tau$ Using the power rule for integration, $\int \tau^2 \, d\tau = \frac{\tau^3}{3}$, we get: $(f * g)(t) = \left[ \frac{\tau^3}{3} \right]_{0}^{t-1} = \frac{(t-1)^3}{3} - 0 = \frac{(t-1)^3}{3}$

Therefore, the convolution $f * g$ is: $(f * g)(t) = \frac{(t - 1)^3}{3} \quad \text{for } t \geq 1$

For $t < 1$, $(f * g)(t) = 0$ because the unit step function $g(t) = u(t - 1)$ does not contribute.

Final Answer

\begin{cases} 0, & t < 1 \\ \frac{(t - 1)^3}{3}, & t \geq 1 \end{cases}$$ Would you like further clarification on the steps, or have any additional questions? --- ### Related Questions 1. How would the convolution change if $$g(t) = u(t - a)$$ for a different constant $$a$$? 2. What would be the result of convolving $$f(t) = t^2$$ with a delta function $$\delta(t - 1)$$? 3. How does the unit step function impact the limits of integration in convolution problems? 4. Can we compute $$f * g$$ when $$f$$ and $$g$$ are defined on $$t < 0$$? 5. What is the physical interpretation of convolution in signal processing? ### Tip When working with convolutions involving step functions, always check where each function is nonzero to simplify the integration limits.