# Power and energy from continuous to discrete

• jashua
In summary: Let's see:x_s(t) = \sum^{N-1}_{n=0} x[n] \delta(t-nT_s)= \sum^{N-1}_{n=0} x[n] e^{j2\pi fnT_s}where f=\frac{1}{T}.Now, the energy spectral density is found to be:\epsilon_x (f) = |X_s(f)|^2 = | \sum^{N-1}_{n=0} x[n] e^{-j2\pi fnT_s} |^2= |\sum^{N-1}_{n=0} x[n]|^2 = |x_s(t)|^2Therefore, we can
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:
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:

## 1. What is the difference between continuous and discrete energy?

Continuous energy refers to energy that can take on any value within a certain range, such as the energy of an object in motion. Discrete energy, on the other hand, can only take on specific values, such as the energy levels of an atom.

## 2. How is power related to energy?

Power is the rate at which energy is transferred or converted. It is the amount of energy per unit of time. The more power an object has, the faster it can transfer or convert energy.

## 3. Why is it important to convert continuous energy to discrete energy?

In certain systems, it may be more efficient or practical to use discrete energy levels. For example, in electronics, discrete energy levels can represent binary information and allow for digital processing.

## 4. How does quantum mechanics relate to the concept of discrete energy?

In quantum mechanics, energy is quantized, meaning it can only exist in discrete levels. This is known as the energy quantization principle and is a fundamental concept in the field.

## 5. Can energy be converted from discrete to continuous?

Yes, energy can be converted from discrete to continuous through processes such as superposition and wave packet spreading. However, this conversion is subject to the laws of quantum mechanics and may not be possible in all cases.

Replies
2
Views
1K
Replies
1
Views
2K
Replies
17
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
4
Views
6K
Replies
3
Views
2K
Replies
4
Views
941
Replies
3
Views
8K
Replies
3
Views
885