Power and energy from continuous to discrete

[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.

 

[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?