# Power and energy from continuous to discrete

1. Jun 8, 2011

### jashua

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

$X_s(f)=\frac{1}{T_s}X(f)$, $|f|<W$.

Then, the energy spectral density is found to be $\epsilon_x (f) = |X(f)|^2 = {T_s}^2 |X_s(f)|^2$. Now we can find $E_x$ using the following formula:

$E_x = \int ^{\infty}_{-\infty} \epsilon_x (f) df$

$= {T_s}^2 \int ^{W}_{-W} |X_s(f)|^2 df$

$= {T_s}^2 \int ^{W}_{-W} \sum ^{\infty}_{n=-\infty}x[n]e^{-j2\pi f n T_s} \sum ^{\infty}_{m=-\infty}x[m]^\ast e^{j2\pi f m T_s}df$

$= {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast \int ^{W}_{-W} e^{j2\pi f (m-n) T_s}df$

$= {T_s}^2 \sum ^{\infty}_{n=-\infty}x[n] \sum ^{\infty}_{m=-\infty}x[m]^\ast 2W sinc(2WT_s(m-n))$

$= {T_s}^2 2W \sum ^{\infty}_{n=-\infty} |x[n]|^2$

where $x[m]^\ast$ is the conjugate of $x[m]$. If $T_s=\frac{1}{2W}$, then we have shown the definitions #1 and #3 are equivalent.

In order to show that the definitions #2 and #4 are equivalent, let $T=(2N+1)T_s$ such that

$x_{s_T}(t)= \sum ^{N}_{n=-N}x[n]\delta (t-nTs)$

Then, we have,

$lim_{T\rightarrow \infty}X_{s_T}(f) = lim_{N\rightarrow \infty} \frac{1}{T_s} \sum ^{N}_{n=-N}X(f-\frac{n}{T_s})$

And hence,

$lim_{T\rightarrow \infty} X_{s_T}(f)=\frac{1}{T_s}X(f)$, $|f|<W$.

Substituting $X(f)=T_s X_{s_T}(f)$ into the definition # 1 (after applying Parseval relation), we get

$P_x = lim_{T\rightarrow \infty} \frac{1}{T} \int ^{T/2}_{-T/2} |x(t)|^2 dt$

$= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|x(t)|^2 dt$

$= lim_{T\rightarrow \infty} \frac{1}{T} \int ^{\infty}_{-\infty}|X(f)|^2 df$

$= lim_{T\rightarrow \infty} \frac{1}{T} {T_s}^2 \int ^{W}_{-W}|X_{s_T}(f)|^2 df$

Now substituting $X_{s_T}(f)=\sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s}$, and $T=(2N+1)T_s$, we get

$P_x = lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 \int ^{W}_{-W} \sum ^{N}_{n=-N}x[n]e^{-j2\pi f n T_s} \sum ^{N}_{m=-N}x[m]^\ast e^{j2\pi f m T_s} df$

$= lim_{N\rightarrow \infty} \frac{1}{(2N+1)T_s} {T_s}^2 2W \sum ^{N}_{n=-N} |x[n]|^2$

If $T_s=\frac{1}{2W}$, then,

$P_x= lim_{N\rightarrow \infty} \frac{1}{2N+1} \sum ^{N}_{n=-N} |x[n]|^2$

Hence we have shown the definitions #2 and #4 are equivalent.

Am I correct up to this point?

I'm still trying to get some satisfactory solutions to show the relation between the definitions #1 and # 5 and the definitions #2 and # 6 and waiting for some help.

Last edited: Jun 8, 2011
2. Jun 9, 2011

### jashua

Let me give an answer to my question :)

As the sequence is repeated we can rewrite the definition # 4 as follows (assuming the period $T=NT_s$):

$P_x = lim_{M\rightarrow \infty} \frac{1}{MN} M \sum^{N-1}_{n=0} |x[n]|^2$

$= \frac{1}{N}\sum^{N-1}_{n=0} |x[n]|^2$

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?

Last edited: Jun 9, 2011