If phi(t)=begin{cases} 1, & text{for } 0 leq | vedclass

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(B) We are given $\phi(t) = 1$ for $0 \leq t < 1$ and $0$ otherwise. This means $\phi(t-k) = 1$ for $k \leq t < k+1$ and $0$ otherwise. The integral i

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(B) We are given $\phi(t) = 1$ for $0 \leq t The integral is $I = \int_{-3000}^{3000} \sum_{r'=2014}^{2016} \phi(t-r') \phi(t-2016) dt$. Expanding the summation: $I = \int_{-3000}^{3000} [\phi(t-2014)\phi(t-2016) + \phi(t-2015)\phi(t-2016) + \phi(t-2016)\phi(t-2016)] dt$. Note that $\phi(t-2014)\phi(t-2016) = 0$ because the intervals $[2014, 2015)$ and $[2016, 2017)$ are disjoint. Similarly, $\phi(t-2015)\phi(t-2016) = 0$ because the intervals $[2015, 2016)$ and $[2016, 2017)$ are disjoint. Thus, the expression simplifies to $\int_{-3000}^{3000} \phi(t-2016)^2 dt$. Since $\phi(t-2016) = 1$ for $2016 \leq t Therefore, $I = \int_{2016}^{2017} 1 dt = [t]_{2016}^{2017} = 2017 - 2016 = 1$.

If $\phi(t)=\begin{cases} 1, & \text{for } 0 \leq t < 1 \\ 0, & \text{otherwise} \end{cases}$, then $\int_{-3000}^{3000} \left( \sum_{r'=2014}^{2016} \phi(t-r') \phi(t-2016) \right) dt$ is

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