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Energy signals, i.e., signals with finite energy $$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty$$ have zero power and, consequently, a power spectrum that is equal to zero. They do have an energy density spectrum, which is the squared magnitude of their Fourier transform. You will only get a non-zero power spectrum, according to the definition in your ...


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A more general expression states that for $ M \geq N$: $$ \sum_{n= N}^{n = M} c = (M-N+1) \cdot c $$ where the derivation simply relies on fact that the epxression has (M-N+1) terms : $$ \sum_{n= N}^{n = M} c = \{ c + c + ... + c\} = (M-N+1) \cdot c $$ And when applied for your particular case (with $N = -M$) it becomes: $$ \sum_{n= -M}^{n = M} c = (M-(-...


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For instance, from $-3$ to $3$, you have $-3,\,-2\,-1,\,0\,1,\,2,\,3$, hence $2\times 3+1$ terms. More generally, the sum from $-M$ to $M$ is composed of $2M+1$ terms: indices with $m$ strictly negative (a total of $M$), those which $m$ strictly positive (a total of $M$), plus one at zero ($1$). If all terms are the same constant $c$, the total is $(2M+...


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You are missing the effect of the integration interval on the integrand function... If you choose a period; T = 2 , and a periodic signal whose base period is $s_0(t) = t$ in the interval [0,2]... Then the integrand in the shifted interval [-1,1] will be different as given by $$ s_1(t) = \begin{cases}{ t + 2 ~~~, -1 < t < 0 \\ t ~~~~~~~~~~ , ~ ...


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