# Power and energy from continuous to discrete

by jashua
Tags: continuous, discrete, energy, power
 P: 43 The energy and the power contents of a signal x(t), denoted by $E_x$ and $P_x$, respectively, are defined as $(1) E_x = \int ^{\infty}_{-\infty} |x(t)|^2 dt$ $(2) P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt$ Let us use the discrete time (sampled) signal, with sampling period $T_s$. Then, the first problem is to show that these definitions are equivalent to the following ones: $(3) E_x = T_s \sum ^{\infty}_{n=-\infty} |x[n]|^2$ $(4) P_x = lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2$ Now, if we use the FFT, that is, if the length of the sequence is finite and the sequence is repeated. Then, the next problem is to show that these definitions (#1 and #2) are equivalent to the following ones (i think only one period should be considered for $E_x$): $(5) E_x = T_s \sum ^{N-1}_{n=0} |x[n]|^2$ $(6) P_x = \frac{1}{N} \sum ^{N-1}_{n=0} |x[n]|^2$ My work to solve these questions are as follows: Let us define the sampled waveform $x_s(t)$: $x_s(t) = x(t) \sum ^{\infty}_{n=-\infty}\delta (t-nTs).$ $= \sum ^{\infty}_{n=-\infty}x(nT_s)\delta (t-nTs).$ $= \sum ^{\infty}_{n=-\infty}x[n]\delta (t-nTs).$ Then, the Fourier transform of $x_s(t)$, $X_s(f)$ is $X_s(f) = \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s}.$ which is nothing but the DFT of the discrete time sequence x[n]. On the other hand, we can express $X_s(f)$ in terms of $X(f)$ (using Poissson's sum formula) as follows: $X_s(f) = X(f) \ast \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}\delta (n-\frac{n}{T_s}).$ $= \frac{1}{T_s} \sum ^{\infty}_{n=-\infty}X(f-\frac{n}{T_s}).$ where $\ast$ is the convolution operator. Hence, we have $X_s(f)=\frac{1}{T_s}X(f)$, $|f|  P: 43 Let me give an answer to my question :) As the sequence is repeated we can rewrite the definition # 4 as follows (assuming the period [itex]T=NT_s$): $P_x = lim_{M\rightarrow \infty} \frac{1}{MN} M \sum^{N-1}_{n=0} |x[n]|^2$ $= \frac{1}{N}\sum^{N-1}_{n=0} |x[n]|^2$ Hence, the definitions #2 and #6 are equivalent. Now, since the energy of a periodic sequence is calculated over one period (by its definition) only, we can rewrite the definition # 3 as follows: $E_x = T_s \sum^{N-1}_{n=0} |x[n]|^2$. However, can we obtain these results using the sampled waveform over one period directly?