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Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even experimented with cooling the receiver front end in a liquid helium bath to reduce the noise temperature to about $5^\circ$K. Be that as is may, the receiver has filters (and amplifiers) so that the noise at the receiver output is definitely band-limited. Because the front-end noise has been amplified greatly, the contribution of the impedances of later receiver stages to the thermal noise is negligible. Indeed, the noise at later stages is a (stationary) band-pass Gaussian process $\left\{N(t)\colon -\infty < t < \infty\right\}$ where $$N(t) = X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\tag{1}$$ with $\left\{X(t)\colon -\infty < t < \infty\right\}$ and $\left\{Y(t)\colon -\infty < t < \infty\right\}$ being independent zero-mean low-pass Gaussian processes with common autocorrelation function $R(\tau)$. And Yes, I do want a $-$ sign in $(1)$.

The autocorrelation function of the $\{N(t)\}$ process is \begin{align}\require{cancel}R_N( \tau) &= E[N(t)N(t+\tau)]\\ &= E\big[\big(X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\big)\\ &= \hspace{0.2in}\cdot \big(X(t+\tau)\cos(2\pi f_c(t+\tau)) - Y(t+\tau)\sin(2\pi f_c(t+\tau))\big)\big]\\ &= R(\tau)\big[\cos(2\pi f_c t)\cos(2\pi f_c(t+\tau))+\sin(2\pi f_c t)\sin(2\pi f_c(t+\tau))\big]\\ &= \hspace{0.2in}~ - \cancelto{0}{R_{X,Y}(t)}\big[\cos(2\pi f_c t)\sin(2\pi f_c(t+\tau))-\sin(2\pi f_c t)\cos(2\pi f_c(t+\tau))\big]\\ \\&= R(\tau)\cos(2\pi f_c \tau) \end{align} which is a band-pass function. Note that we can also express the $\{N(t)\}$ process in terms of its complex baseband equivalent process $\{\mathcal Z(t)\colon -\infty < t < \infty\}$ where $\mathcal Z(t) = X(t) + jY(t)$ as $$N(t) = \mathrm{Re}\left( \mathcal Z(t)e^{j2\pi f_c t}\right)\tag{2}$$ But, if we choose to represent the complex number $\mathcal Z(t)$ in polar form as $B(t)e^{j\Theta(t)}$ where $B(t) \geq 0$ and $\Theta(t) \in [0,2\pi)$ are real numbers, then $B(t) = \sqrt{X^2(t)+Y^2(t)}$ is a Rayleigh random variable and $\Theta(t)$ is uniformly distributed on $[0,2\pi)$. (This idea is the basis of the Box-Muller method of simulating Gaussian random variables. The usual probability transform method does not work well for Gaussian random variables because the Gaussian CDF does not have an inverse that is an elementary function whereas a Rayleigh CDF $1-\exp(-t^2/2\sigma^2)$ is readily invertible. So, the Box-Muller method calls a random number generator twice to simulate $B$ and $\Theta$ and then sets $X = B\cos(\Theta)$ and $Y = B\sin(\Theta)$ to get realizations of two independent Gaussian random variables).

The other place where Rayleigh's name comes in is in the transmission of signals over fading channels. Real-life signals are real-valued, not complex-valued. When a sinusoid (say $\cos(2\pi f_ct), -\infty < t < \infty$) is transmitted over a (slowly fading) channel, the received signal is of form \begin{align}r(t) &= b(t)\cos(2\pi f_c t + \theta(t)) \tag{3}\\ &= b(t)\cos(\theta(t))\cos(2\pi f_c t) - b(t)\sin(\theta(t))\sin(2\pi f_c t) \tag{4}\end{align} where $b(t)$ and $\theta(t)$ are slowly varying signals. For the purposes of digital communications, it is reasonable to treat $b(t)$ and $\theta(t)$ as constants over the duration of transmission of one bit/symbol, so that $(4)$ simplifies to $$b\cos(\theta)\cos(2\pi f_c t) - b\sin(\theta)\sin(2\pi f_c t).\tag{5}$$ Slow fading occurs in ionospheric channels because the transmitted signal is reflected back to earth from the ions in a ion cloud and the received signal is a sum of a very large number of very weak reflections with nearly equal delays. As a consequence, with a lot of handwaving and genuflections in the direction of the Central Limit Theorem, it is possible to regard $b\cos(\theta)$ and $b\sin(theta)$$b\sin(\theta)$ in $(5)$ as realizations of two independent Gaussian random variables which allows us to treat $b$ and $\theta$ as realizations of Rayleigh and uniform random variables respectively. Hence, the complex baseband equivalent of a Rayleigh fading channel is that complex input $\mathrm{Re}(\mathcal Ae^{j2\pi f_c t}$ is received as output $\mathrm{Re}(\mathcal{AB}e^{j2\pi f_c t}$ where $\mathcal B = Be^{j\Theta}$ with $B$ and $\Theta$ being independent Rayleigh and uniform random variables respectively. Fading is sometimes referred to as multiplicative noise since the fading multiplies the input amplitude instead of adding to it as additive noise does.

"But, but, but," you say, "In a Rayleigh channel model, the fading parameters $B_{n+1}$ and $\Theta_{n+1}$ for the next bit/symbol $B_{n+1}$ and $\Theta_{n+1}$ isinterval are independent of the fadingparameters $B_n$ and $\Theta_n$ for the current bitinterval. How does that fit into your model of $b(t)$ and $\theta(t)$ being slowlyslowly varying functions? If $b(t)$ and $\theta(t)$ really vary so slowly that you are treatingcan treat them as constant over the bit interval, then $B_{n+1}$ and $B_{n}$ should be very highly correlated, as should with $\Theta_{n+1}$ and $\Theta_n$." Well, the dirty little secret that no one will ever tell you about is that communication systems intended for use over Rayleigh fading channels are very heavily interleaved so that what appearsappears to be the next symbol in the mathematical model is actually a symbol that occurs $N\gg 1$ bit durations later where $N$ is chosen to be large enough that heavily-faded consecutive symbols appear in different frames rather than wiping out four bars of Bethoven's Fifth Symphony or a whole stripe from an image of Old Glory if interleaving were not used.

Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even experimented with cooling the receiver front end in a liquid helium bath to reduce the noise temperature to about $5^\circ$K. Be that as is may, the receiver has filters (and amplifiers) so that the noise at the receiver output is definitely band-limited. Because the front-end noise has been amplified greatly, the contribution of the impedances of later stages to the thermal noise is negligible. Indeed, the noise at later stages is a (stationary) band-pass Gaussian process $\left\{N(t)\colon -\infty < t < \infty\right\}$ where $$N(t) = X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\tag{1}$$ with $\left\{X(t)\colon -\infty < t < \infty\right\}$ and $\left\{Y(t)\colon -\infty < t < \infty\right\}$ being independent zero-mean low-pass Gaussian processes with common autocorrelation function $R(\tau)$. And Yes, I do want a $-$ sign in $(1)$.

The autocorrelation function of the $\{N(t)\}$ process is \begin{align}\require{cancel}R_N( \tau) &= E[N(t)N(t+\tau)]\\ &= E\big[\big(X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\big)\\ &= \hspace{0.2in}\cdot \big(X(t+\tau)\cos(2\pi f_c(t+\tau)) - Y(t+\tau)\sin(2\pi f_c(t+\tau))\big)\big]\\ &= R(\tau)\big[\cos(2\pi f_c t)\cos(2\pi f_c(t+\tau))+\sin(2\pi f_c t)\sin(2\pi f_c(t+\tau))\big]\\ &= \hspace{0.2in}~ - \cancelto{0}{R_{X,Y}(t)}\big[\cos(2\pi f_c t)\sin(2\pi f_c(t+\tau))-\sin(2\pi f_c t)\cos(2\pi f_c(t+\tau))\big]\\ \\&= R(\tau)\cos(2\pi f_c \tau) \end{align} which is a band-pass function. Note that we can also express the $\{N(t)\}$ process in terms of its complex baseband equivalent process $\{\mathcal Z(t)\colon -\infty < t < \infty\}$ where $\mathcal Z(t) = X(t) + jY(t)$ as $$N(t) = \mathrm{Re}\left( \mathcal Z(t)e^{j2\pi f_c t}\right)\tag{2}$$ But, if we choose to represent the complex number $\mathcal Z(t)$ in polar form as $B(t)e^{j\Theta(t)}$ where $B(t) \geq 0$ and $\Theta(t) \in [0,2\pi)$ are real numbers, then $B(t) = \sqrt{X^2(t)+Y^2(t)}$ is a Rayleigh random variable and $\Theta(t)$ is uniformly distributed on $[0,2\pi)$. (This idea is the basis of the Box-Muller method of simulating Gaussian random variables. The usual probability transform method does not work well for Gaussian random variables because the CDF does not have an inverse that is an elementary function whereas a Rayleigh CDF $1-\exp(-t^2/2\sigma^2)$ is readily invertible. So, the Box-Muller method calls a random number generator twice to simulate $B$ and $\Theta$ and then sets $X = B\cos(\Theta)$ and $Y = B\sin(\Theta)$ to get realizations of two independent Gaussian random variables).

The other place where Rayleigh's name comes in is in the transmission of signals over fading channels. Real-life signals are real-valued, not complex-valued. When a sinusoid (say $\cos(2\pi f_ct), -\infty < t < \infty$) is transmitted over a (slowly fading) channel, the received signal is of form \begin{align}r(t) &= b(t)\cos(2\pi f_c t + \theta(t)) \tag{3}\\ &= b(t)\cos(\theta(t))\cos(2\pi f_c t) - b(t)\sin(\theta(t))\sin(2\pi f_c t) \tag{4}\end{align} where $b(t)$ and $\theta(t)$ are slowly varying signals. For the purposes of digital communications, it is reasonable to treat $b(t)$ and $\theta(t)$ as constants over the duration of transmission of one bit/symbol, so that $(4)$ simplifies to $$b\cos(\theta)\cos(2\pi f_c t) - b\sin(\theta)\sin(2\pi f_c t).\tag{5}$$ Slow fading occurs in ionospheric channels because the transmitted signal is reflected back to earth from the ions in a ion cloud and the received signal is a sum of a very large number of very weak reflections with nearly equal delays. As a consequence, with a lot of handwaving and genuflections in the direction of the Central Limit Theorem, it is possible to regard $b\cos(\theta)$ and $b\sin(theta)$ in $(5)$ as realizations of two independent Gaussian random variables which allows us to treat $b$ and $\theta$ as realizations of Rayleigh and uniform random variables respectively. Hence, the complex baseband equivalent of a Rayleigh fading channel is that complex input $\mathrm{Re}(\mathcal Ae^{j2\pi f_c t}$ is received as output $\mathrm{Re}(\mathcal{AB}e^{j2\pi f_c t}$ where $\mathcal B = Be^{j\Theta}$ with $B$ and $\Theta$ being independent Rayleigh and uniform random variables respectively. Fading is sometimes referred to as multiplicative noise since the fading multiplies the input amplitude instead of adding to it as additive noise does.

"But, but, but," you say, "In a Rayleigh channel model, the fading for the next bit/symbol $B_{n+1}$ and $\Theta_{n+1}$ is independent of the fading $B_n$ and $\Theta_n$ for the current bit. How does that fit into your model of $b(t)$ and $\theta(t)$ being slowly varying functions? If $b(t)$ and $\theta(t)$ really vary so slowly that you are treating them as constant over the bit interval, then $B_{n+1}$ and $B_{n}$ should be very highly correlated, as should with $\Theta_{n+1}$ and $\Theta_n$." Well, the dirty little secret that no one will ever tell you about is that communication systems intended for use over Rayleigh fading channels are very heavily interleaved so that what appears to be the next symbol in the mathematical model is actually a symbol that occurs $N\gg 1$ bit durations later where $N$ is chosen to be large enough that heavily-faded consecutive symbols appear in different frames rather than wiping out four bars of Bethoven's Fifth Symphony or a whole stripe from an image of Old Glory if interleaving were not used.

Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even experimented with cooling the receiver front end in a liquid helium bath to reduce the noise temperature to about $5^\circ$K. Be that as is may, the receiver has filters (and amplifiers) so that the noise at the receiver output is definitely band-limited. Because the front-end noise has been amplified greatly, the contribution of the impedances of later receiver stages to the thermal noise is negligible. Indeed, the noise at later stages is a (stationary) band-pass Gaussian process $\left\{N(t)\colon -\infty < t < \infty\right\}$ where $$N(t) = X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\tag{1}$$ with $\left\{X(t)\colon -\infty < t < \infty\right\}$ and $\left\{Y(t)\colon -\infty < t < \infty\right\}$ being independent zero-mean low-pass Gaussian processes with common autocorrelation function $R(\tau)$. And Yes, I do want a $-$ sign in $(1)$.

The autocorrelation function of the $\{N(t)\}$ process is \begin{align}\require{cancel}R_N( \tau) &= E[N(t)N(t+\tau)]\\ &= E\big[\big(X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\big)\\ &= \hspace{0.2in}\cdot \big(X(t+\tau)\cos(2\pi f_c(t+\tau)) - Y(t+\tau)\sin(2\pi f_c(t+\tau))\big)\big]\\ &= R(\tau)\big[\cos(2\pi f_c t)\cos(2\pi f_c(t+\tau))+\sin(2\pi f_c t)\sin(2\pi f_c(t+\tau))\big]\\ &= \hspace{0.2in}~ - \cancelto{0}{R_{X,Y}(t)}\big[\cos(2\pi f_c t)\sin(2\pi f_c(t+\tau))-\sin(2\pi f_c t)\cos(2\pi f_c(t+\tau))\big]\\ \\&= R(\tau)\cos(2\pi f_c \tau) \end{align} which is a band-pass function. Note that we can also express the $\{N(t)\}$ process in terms of its complex baseband equivalent process $\{\mathcal Z(t)\colon -\infty < t < \infty\}$ where $\mathcal Z(t) = X(t) + jY(t)$ as $$N(t) = \mathrm{Re}\left( \mathcal Z(t)e^{j2\pi f_c t}\right)\tag{2}$$ But, if we choose to represent the complex number $\mathcal Z(t)$ in polar form as $B(t)e^{j\Theta(t)}$ where $B(t) \geq 0$ and $\Theta(t) \in [0,2\pi)$ are real numbers, then $B(t) = \sqrt{X^2(t)+Y^2(t)}$ is a Rayleigh random variable and $\Theta(t)$ is uniformly distributed on $[0,2\pi)$. (This idea is the basis of the Box-Muller method of simulating Gaussian random variables. The usual probability transform method does not work well for Gaussian random variables because the Gaussian CDF does not have an inverse that is an elementary function whereas a Rayleigh CDF $1-\exp(-t^2/2\sigma^2)$ is readily invertible. So, the Box-Muller method calls a random number generator twice to simulate $B$ and $\Theta$ and then sets $X = B\cos(\Theta)$ and $Y = B\sin(\Theta)$ to get realizations of two independent Gaussian random variables).

The other place where Rayleigh's name comes in is in the transmission of signals over fading channels. Real-life signals are real-valued, not complex-valued. When a sinusoid (say $\cos(2\pi f_ct), -\infty < t < \infty$) is transmitted over a (slowly fading) channel, the received signal is of form \begin{align}r(t) &= b(t)\cos(2\pi f_c t + \theta(t)) \tag{3}\\ &= b(t)\cos(\theta(t))\cos(2\pi f_c t) - b(t)\sin(\theta(t))\sin(2\pi f_c t) \tag{4}\end{align} where $b(t)$ and $\theta(t)$ are slowly varying signals. For the purposes of digital communications, it is reasonable to treat $b(t)$ and $\theta(t)$ as constants over the duration of transmission of one bit/symbol, so that $(4)$ simplifies to $$b\cos(\theta)\cos(2\pi f_c t) - b\sin(\theta)\sin(2\pi f_c t).\tag{5}$$ Slow fading occurs in ionospheric channels because the transmitted signal is reflected back to earth from the ions in a ion cloud and the received signal is a sum of a very large number of very weak reflections with nearly equal delays. As a consequence, with a lot of handwaving and genuflections in the direction of the Central Limit Theorem, it is possible to regard $b\cos(\theta)$ and $b\sin(\theta)$ in $(5)$ as realizations of two independent Gaussian random variables which allows us to treat $b$ and $\theta$ as realizations of Rayleigh and uniform random variables respectively. Hence, the complex baseband equivalent of a Rayleigh fading channel is that complex input $\mathrm{Re}(\mathcal Ae^{j2\pi f_c t}$ is received as output $\mathrm{Re}(\mathcal{AB}e^{j2\pi f_c t}$ where $\mathcal B = Be^{j\Theta}$ with $B$ and $\Theta$ being independent Rayleigh and uniform random variables respectively. Fading is sometimes referred to as multiplicative noise since the fading multiplies the input amplitude instead of adding to it as additive noise does.

"But, but, but," you say, "In a Rayleigh channel model, the fading parameters $B_{n+1}$ and $\Theta_{n+1}$ for the next bit/symbol interval are independent of the parameters $B_n$ and $\Theta_n$ for the current interval. How does that fit into your model of $b(t)$ and $\theta(t)$ being slowly varying functions? If $b(t)$ and $\theta(t)$ really vary so slowly that you can treat them as constant over the interval, then $B_{n+1}$ and $B_{n}$ should be very highly correlated, as should with $\Theta_{n+1}$ and $\Theta_n$." Well, the dirty little secret that no one will ever tell you about is that communication systems intended for use over Rayleigh fading channels are very heavily interleaved so that what appears to be the next symbol in the mathematical model is actually a symbol that occurs $N\gg 1$ bit durations later where $N$ is chosen to be large enough that heavily-faded consecutive symbols appear in different frames rather than wiping out four bars of Bethoven's Fifth Symphony or a whole stripe from an image of Old Glory if interleaving were not used.

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There are too many things named after Rayleigh floating around in this problem.

Consider first additive white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$. This is generally taken as the model for channel noise, though in fact the source of the noise actually is thermal noise in the front-end of the receiver. NASA once even experimented with cooling the receiver front end in a liquid helium bath to reduce the noise temperature to about $5^\circ$K. Be that as is may, the receiver has filters (and amplifiers) so that the noise at the receiver output is definitely band-limited. Because the front-end noise has been amplified greatly, the contribution of the impedances of later stages to the thermal noise is negligible. Indeed, the noise at later stages is a (stationary) band-pass Gaussian process $\left\{N(t)\colon -\infty < t < \infty\right\}$ where $$N(t) = X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\tag{1}$$ with $\left\{X(t)\colon -\infty < t < \infty\right\}$ and $\left\{Y(t)\colon -\infty < t < \infty\right\}$ being independent zero-mean low-pass Gaussian processes with common autocorrelation function $R(\tau)$. And Yes, I do want a $-$ sign in $(1)$.

The autocorrelation function of the $\{N(t)\}$ process is \begin{align}\require{cancel}R_N( \tau) &= E[N(t)N(t+\tau)]\\ &= E\big[\big(X(t)\cos(2\pi f_ct) - Y(t)\sin(2\pi f_ct)\big)\\ &= \hspace{0.2in}\cdot \big(X(t+\tau)\cos(2\pi f_c(t+\tau)) - Y(t+\tau)\sin(2\pi f_c(t+\tau))\big)\big]\\ &= R(\tau)\big[\cos(2\pi f_c t)\cos(2\pi f_c(t+\tau))+\sin(2\pi f_c t)\sin(2\pi f_c(t+\tau))\big]\\ &= \hspace{0.2in}~ - \cancelto{0}{R_{X,Y}(t)}\big[\cos(2\pi f_c t)\sin(2\pi f_c(t+\tau))-\sin(2\pi f_c t)\cos(2\pi f_c(t+\tau))\big]\\ \\&= R(\tau)\cos(2\pi f_c \tau) \end{align} which is a band-pass function. Note that we can also express the $\{N(t)\}$ process in terms of its complex baseband equivalent process $\{\mathcal Z(t)\colon -\infty < t < \infty\}$ where $\mathcal Z(t) = X(t) + jY(t)$ as $$N(t) = \mathrm{Re}\left( \mathcal Z(t)e^{j2\pi f_c t}\right)\tag{2}$$ But, if we choose to represent the complex number $\mathcal Z(t)$ in polar form as $B(t)e^{j\Theta(t)}$ where $B(t) \geq 0$ and $\Theta(t) \in [0,2\pi)$ are real numbers, then $B(t) = \sqrt{X^2(t)+Y^2(t)}$ is a Rayleigh random variable and $\Theta(t)$ is uniformly distributed on $[0,2\pi)$. (This idea is the basis of the Box-Muller method of simulating Gaussian random variables. The usual probability transform method does not work well for Gaussian random variables because the CDF does not have an inverse that is an elementary function whereas a Rayleigh CDF $1-\exp(-t^2/2\sigma^2)$ is readily invertible. So, the Box-Muller method calls a random number generator twice to simulate $B$ and $\Theta$ and then sets $X = B\cos(\Theta)$ and $Y = B\sin(\Theta)$ to get realizations of two independent Gaussian random variables).

The other place where Rayleigh's name comes in is in the transmission of signals over fading channels. Real-life signals are real-valued, not complex-valued. When a sinusoid (say $\cos(2\pi f_ct), -\infty < t < \infty$) is transmitted over a (slowly fading) channel, the received signal is of form \begin{align}r(t) &= b(t)\cos(2\pi f_c t + \theta(t)) \tag{3}\\ &= b(t)\cos(\theta(t))\cos(2\pi f_c t) - b(t)\sin(\theta(t))\sin(2\pi f_c t) \tag{4}\end{align} where $b(t)$ and $\theta(t)$ are slowly varying signals. For the purposes of digital communications, it is reasonable to treat $b(t)$ and $\theta(t)$ as constants over the duration of transmission of one bit/symbol, so that $(4)$ simplifies to $$b\cos(\theta)\cos(2\pi f_c t) - b\sin(\theta)\sin(2\pi f_c t).\tag{5}$$ Slow fading occurs in ionospheric channels because the transmitted signal is reflected back to earth from the ions in a ion cloud and the received signal is a sum of a very large number of very weak reflections with nearly equal delays. As a consequence, with a lot of handwaving and genuflections in the direction of the Central Limit Theorem, it is possible to regard $b\cos(\theta)$ and $b\sin(theta)$ in $(5)$ as realizations of two independent Gaussian random variables which allows us to treat $b$ and $\theta$ as realizations of Rayleigh and uniform random variables respectively. Hence, the complex baseband equivalent of a Rayleigh fading channel is that complex input $\mathrm{Re}(\mathcal Ae^{j2\pi f_c t}$ is received as output $\mathrm{Re}(\mathcal{AB}e^{j2\pi f_c t}$ where $\mathcal B = Be^{j\Theta}$ with $B$ and $\Theta$ being independent Rayleigh and uniform random variables respectively. Fading is sometimes referred to as multiplicative noise since the fading multiplies the input amplitude instead of adding to it as additive noise does.

"But, but, but," you say, "In a Rayleigh channel model, the fading for the next bit/symbol $B_{n+1}$ and $\Theta_{n+1}$ is independent of the fading $B_n$ and $\Theta_n$ for the current bit. How does that fit into your model of $b(t)$ and $\theta(t)$ being slowly varying functions? If $b(t)$ and $\theta(t)$ really vary so slowly that you are treating them as constant over the bit interval, then $B_{n+1}$ and $B_{n}$ should be very highly correlated, as should with $\Theta_{n+1}$ and $\Theta_n$." Well, the dirty little secret that no one will ever tell you about is that communication systems intended for use over Rayleigh fading channels are very heavily interleaved so that what appears to be the next symbol in the mathematical model is actually a symbol that occurs $N\gg 1$ bit durations later where $N$ is chosen to be large enough that heavily-faded consecutive symbols appear in different frames rather than wiping out four bars of Bethoven's Fifth Symphony or a whole stripe from an image of Old Glory if interleaving were not used.