---
# System prepended metadata

title: Signal Power and Energy
tags: [Book, Digital Communication]

---

# Signal Power and Energy

View the book with "<i class="fa fa-book fa-fw"></i> Book Mode".

## Energy 

The energy $\mathcal{E}_{x}$ of a signal $s(t)$ is 
$$
\mathcal{E}_{x}:=\int^{\infty}_{-\infty}\left\lvert x(t) \right\rvert^{2} dt 
$$
provided that the above improper integral converges.

- Energy of bandpass signal
$$
\mathcal{E}_{x}:=\int^{\infty}_{-\infty}\left\lvert x(t) \right\rvert^{2} dt 
$$
- Energy of pre-envelope signal
$$
\mathcal{E}_{x_{+}}:=\int^{\infty}_{-\infty}\left\lvert x_{+}(t) \right\rvert^{2} dt 
$$
- Energy of baseband signal
$$
\mathcal{E}_{x_{l}}:=\int^{\infty}_{-\infty}\left\lvert x_{l}(t) \right\rvert^{2} dt 
$$

We are interested in the relations among $\mathcal{E}_{x}$, $\mathcal{E}_{x_{+}},$ and $\mathcal{E}_{x_{l}}$

:::success
#### Theorem 2 (Energy Relations)
$$
\mathcal{E}_{x_{l}}=2\mathcal{E}_{x}=4\mathcal{E}_{x_{+}}
$$
:::

:::info
#### Proof:

From Parseval’s Theorem we see
$$
\mathcal{E}_{x}=\int^{\infty}_{-\infty}\left\lvert x(t)\right\rvert^{2} dt=\int^{\infty}_{-\infty}\left\lvert X(f)\right\rvert^{2} df
$$
Secondly we know
$$
X(f)=\underbrace{\frac{1}{2}X_{l}(f-f_{c})}_{=X_{+}(f)}+\underbrace{\frac{1}{2}X_{l}^{*}(-f-f_{c})}_{=X_{+}^{*}(-f)}
$$
Thirdly, $f_{c}\gg W$ and 
$$
X_{l}(f-f_{c})X_{l}^{*}(-f-f_{c})=0\textrm{ for all }f
$$
It follows that
$$
\begin{aligned}
\mathcal{E}_{x}&=\int^{\infty}_{-\infty}\left\lvert \frac{1}{2}X_{l}(f-f_{c})
+\frac{1}{2}X_{l}^{*}(-f-f_{c}) \right\rvert^{2} df \\
&=\frac{1}{4}\int^{\infty}_{-\infty}\left\lvert X_{l}(f-f_{c})
\right\rvert^{2} df + \frac{1}{4}\int^{\infty}_{-\infty}\left\lvert X_{l}(-f-f_{c}) \right\rvert^{2} df\\
&+\underbrace{\frac{1}{4}\int^{\infty}_{-\infty} X_{l}(f-f_{c})X^{*}_{l}(-f-f_{c}) df}_{=0} + \underbrace{\frac{1}{4}\int^{\infty}_{-\infty} X_{l}(-f-f_{c})X^{*}_{l}(f-f_{c}) df}_{=0}\\
&= \frac{1}{4}\mathcal{E}_{x_{l}}+\frac{1}{4}\mathcal{E}_{x_{l}}=\frac{1}{2}\mathcal{E}_{x}
\end{aligned}
$$
and 
$$
\mathcal{E}_{x}=\int^{\infty}_{-\infty}\left\lvert X_{+}(f)
+X_{+}(-f) \right\rvert^{2} df = 2\mathcal{E}_{x_{+}}
$$
:::

## Energy Spectral Density
Let $x(t)$ be a Deterministic Energy Signal, i.e.
$$
\mathcal{E}_{x}:=\int^{\infty}_{-\infty}\left\lvert x(t) \right\rvert^{2} dt <\infty
$$
By Parseval Theorem
$$
\mathcal{E}_{x}=\int^{\infty}_{-\infty}\left\lvert x(t)\right\rvert^{2} dt=\int^{\infty}_{-\infty}\underbrace{\left\lvert X(f)\right\rvert^{2}}_{\mathcal{E}_{x}(f)} df
$$
The function 
$$
\mathcal{E}_{x}(f)=\left\lvert X(f) \right\rvert^{2}
$$
plays the role of <font color=#FF0000>energy spectral density</font>

## Power Spectral Density
If $x(t)$ is a deterministic Power Signal, i.e.
$$
P:=\lim_{T\to \infty}\frac{1}{2T}\int^{T}_{-T}\left\lvert x(t) \right\rvert^{2} dt <\infty
$$
Define the truncated version of $x(t)$
$$
x_{T}=x(t)\left[u(t+T)-u(t-T)\right]
$$
By Parseval Theorem 
$$
P=\lim_{T\to \infty}\frac{1}{2T}\int^{T}_{-T}\left\lvert x_{T}(t)\right\rvert^{2}dt = \lim_{T\to \infty}\frac{1}{2T}\int^{T}_{-T}\left\lvert X_{T}(f)\right\rvert^{2}df
$$
The function 
$$
S_{x}(f)=\lim_{T\to\infty}\frac{1}{2T}\left\lvert X_{T}(f)\right\rvert^{2}
$$
plays the role of <font color=#FF0000>power spectral density</font>

- If $x(u,t)$ is a WSS random process,
$$
\mathcal{E}=\mathbb{E}\left[\int^{\infty}_{-\infty}\left\lvert x(u,t)\right\rvert^{2}dt\right]=\int^{\infty}_{-\infty}R_{x}(0)dt \to \infty
$$
Thus $x(u, t)$ cannot be an energy signal.
- $\,$ 
$$
P=\mathbb{E}\left[\lim_{T\to\infty}\frac{1}{2T}\int^{T}_{-T}\left\lvert x(u,t)\right\rvert^{2}dt\right]=R_{x}(0) < \infty
$$
So, $x(u,t)$ is a power signal with power spectral density
$$
S_{X}(f) = \mathbb{E}\left[\lim_{T\to\infty}\frac{1}{2T}\int^{T}_{-T}\left\lvert X(u,f)\right\rvert^{2}\right]
$$

:::success
#### Wiener-Khintchine Theorem
Let $x(u,t)$ be a <font color=#FF0000>WSS</font> random process with $\mathcal{T}=\mathbb{R}$. Define 
$$
x_{T}(u,t)=\begin{cases}
x(u,t)    &\textrm{if } t \in \left[-T,T \right]\\
0    &\textrm{otherwise}
\end{cases}
$$
and set 
$$
X_{T}(u,f)=\int^{\infty}_{-\infty}x_{T}(u,t)e^{-i2\pi tf}dt = \int^{T}_{-T}x(u,t)e^{-i2\pi tf}dt
$$
Then <font color=#FF0000>the power spectral density
$$
S_{x}(f):=\mathbb{E}\left[\lim_{T\to\infty}\frac{1}{2T}\left\lvert X_{T}(u,f)\right\rvert^{2}\right]=\int^{\infty}_{-\infty}R_{x}(\tau)e^{-i2\pi f\tau}d\tau
$$
</font>
provided that the RHS exists.
:::

:::warning
#### Definition
Let $R_{x}(\tau)$ be the correlation function of a <font color=#FF0000>second order WSS random process</font> $X(u, t)$. The Power Spectral Density function of X(u, t) is “defined” as
$$
S_{X}(f)=\int^{\infty}_{-\infty}R_{x}(\tau)e^{-i2\pi f\tau}d \tau
$$
Let $R_{xy}(\tau)$ be the cross correlation function of <font color=#FF0000>two second order jointly WSS random processes</font> $X(u, t)$ and $Y(u,t)$; then the Cross Power Spectral Density function is
$$
S_{XY}(f)=\int^{\infty}_{-\infty}R_{xy}(\tau)e^{-i2\pi f\tau}d \tau
$$
:::

:::success
#### Theorem
Let $S(f)$ be the power spectral density of a <font color=#FF0000>second order WSS random process</font> $X(u, t)$; then
- $S_{X}(f)$ is real-valued $\left(\because R_{x}(\tau)=R_{x}^{*}(-\tau)\right)$
- If $X(u,t)$ is a real random process, then $S_{X}(f)$ is an even function
- $S_{X}(f)\geq 0$ for all $f$
:::

## White Process
A process is called a <font color=#FF0000>white process</font> if
$$
S_{X}(f)=\frac{N_{0}}{2}
$$
meaning
$$
R_{x}(\tau)=\frac{N_{0}}{2}\delta(\tau)
$$
Remark: such signal is only of theoretical interest and <font color=#FF0000>does not exist</font> as the signal has $\infty$ power.
###### tags: `Digital Communication` `Book`
