## Why does $h_1(t)$ work when sampling with Zero-Order Hold

$x_0(t) = h_1(t) * (x(t)p(t))$

$x_0(t) = h_1(t) * \sum_{n = -\infty}^{\infty}x(t) \delta(t-nT)$

$x_0(t) = h_1(t) * \sum_{n = -\infty}^{\infty}x(nt) \delta(t-nT)$

$x_0(t) = \sum_{n = -\infty}^{\infty}x(nt) h_1(t-nT)$

$h_1(t)$ works because $p(t)$ is a shifted delta function, and when you convolve a specific function with a shifted delta function, you get back the specific function but shifted in time.