Signal Power and Energy

# Signal Power and Energy View the book with " 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 energy spectral density ## 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 power spectral density - 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 WSS 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 the power spectral density $$ S_{x}(f):=\mathbb{E}\left[\lim_{T\to\infty}\frac{1}{2T}\left\lvert X_{T}(u,f)\right\rvert^{2}\right]=\int^{\infty}_{-\infty}R_{x}(\tau)e^{-i2\pi f\tau}d\tau $$ provided that the RHS exists. ::: :::warning #### Definition Let $R_{x}(\tau)$ be the correlation function of a second order WSS random process $X(u, t)$. 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 two second order jointly WSS random processes $X(u, t)$ 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 second order WSS random process $X(u, t)$; 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 white process 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 does not exist as the signal has $\infty$ power. ###### tags: `Digital Communication` `Book`