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Some general hints for this type of problem If the function is defined piecewise than chances are your calculation and solution will also have to be done piecewise in sections. Determine first where the result is zero, i.e. where $x(\tau)$ and $h(t-\tau)$ don't overlap. This determines the start and end of the sections that you need to consider. Hint: if ...

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This exercise is meant to help the student appreciate the fact that if the response $y_1(t)$ of an LTI system to an input $x_1(t)$ is known, then the response to an input $$x_2(t)=\sum_{k=1}^Ka_kx_1(t-t_k)\tag{1}$$ is given by $$y_2(t)=\sum_{k=1}^Ka_ky_1(t-t_k)\tag{2}$$ which is a direct consequence of linearity and time-invariance. Consequently, if an input ...

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It should be clear that a property of a system, such as causality, cannot be determined by looking at its input signals. For a linear time-invariant system, it is its impulse response $h(t)$ from which properties such as causality or stability can be determined. Only the second definition in the question is correct: a causal LTI system has an impulse ...

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The key is being even, notice that $f_{N+k} = f_{k}$, and when you do a circular wrapping of an infinite function you should have $f_{k} = \sum_{i=-\infty}^\infty f((i \cdot N + k) T_s)$, then naturally $f_{N-k} = f_{k}$, in practice you want to choose $f$ and $N$ so that $f$ vanishes before $N/2$, then you are safe to use only one term of the summation that ...

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Well, you know that $u[n]=0$ for all $n<0$. So from the two terms in your sum, all terms will be zero where either $k+2 < 0$ (i.e., $k < -2$) or $n-2-k<0$ (i.e., $k>n-2$). From this your sum will need to run from $-2$ to $n-2$. Can you take it from here?

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