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1. How to use this iBook
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2. Prologue
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3. Introduction
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4. Characterization of Random Signals
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5. Correlations and Spectra
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6. Filtering of Stochastic Signals
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6. Filtering of Stochastic Signals
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Interpretation of the convolution result
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The mean
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The autocorrelation function
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Problem 6.5
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Problem 6.6
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Laboratory Exercise 6.2
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Laboratory Exercise 6.3
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Laboratory Exercise 6.4
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Laboratory Exercise 6.5
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Laboratory Exercise 6.6
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Laboratory Exercise 6.7
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Laboratory Exercise 6.8
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8. Characterizing Signal-to-Noise Ratios
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The autocorrelation function
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Problem 6.2
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Problem 6.3
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Laboratory Exercise 6.2
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Laboratory Exercise 6.3
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Laboratory Exercise 6.4
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Laboratory Exercise 6.6
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Laboratory Exercise 6.7
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<h1 id="filtering-of-stochastic-signals">Filtering of Stochastic Signals<a class="headerlink" href="#filtering-of-stochastic-signals" title="Permanent link">¶</a></h1>
<p>We now have the tools necessary to describe what happens when a stochastic signal is processed through a linear, time-invariant (LTI) system. These tools consist of measures on the random signals which describe and<em>/</em>or characterize the signals. The two most important of these are the mean value of the signal and the autocorrelation function of the signal. Further, we can characterize the relation between two random signals through the cross-correlation function. In all of our discussions <em>we will assume that our random processes are complex as well as ergodic (and thus stationary)</em>.</p>
<p>We recall that:</p>
<div class="" id="eq:review_acorr">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.1)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{\varphi _{xx}}[k]}&{ = E\left\{ {x[n]\,{x^*}[n + k]} \right\}}\\
{\,\,\,}&{ = \! < x[n]\,{x^*}[n + k] > \;}\\
{\,\,\,}&\begin{array}{l}
= \mathop {\lim }\limits_{N \to \infty } \frac{1}{{2N + 1}}\sum\limits_{n = - N}^{ + N} {x[n]\,{x^*}[n + k]} \\
\,\,\,
\end{array}
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<div class="" id="eq:review_crosscorr">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.2)</td>
<td class="eqTableEq">
<div>$$\begin{array}{l}
\begin{array}{*{20}{l}}
{{\varphi _{xy}}[k]}&{ = E\left\{ {x[n]\,{y^*}[n + k]} \right\}}\\
{\,\,\,}&{ = \! < x[n]\,{y^*}[n + k] > \;}\\
{\,\,\,}&{ = \mathop {\lim }\limits_{N \to \infty } \frac{1}{{2N + 1}}\sum\limits_{n = - N}^{ + N} {x[n]\,{y^*}[n + k]} }
\end{array}\\
\,\,\,
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<div class="" id="eq:review_mean">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.3)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{m_x}}&{ = E\left\{ {x[n]} \right\} = \! < x[n] > }\\
{\,\,\,}&{ = \mathop {\lim }\limits_{N \to \infty } \frac{1}{{2N + 1}}\sum\limits_{n = - N}^{ + N} {x[n]} }
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>To understand what happens when stochastic signals are processed by LTI
systems, let us begin with discrete-time convolution.</p>
<h2 id="interpretation-of-the-convolution-result">Interpretation of the convolution result<a class="headerlink" href="#interpretation-of-the-convolution-result" title="Permanent link">¶</a></h2>
<p>First we can say that the relation between input and output is still
given by the convolution sum, that is, we can write the input as:</p>
<div class="" id="eq:review_conv1">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.4)</td>
<td class="eqTableEq">
<div>$$x[n] = \sum\limits_{k = - \infty }^{ + \infty } {x[k]\,\delta [n - k]}$$</div>
</td>
</tr>
</table>
</div>
<p>The total signal is a weighted sum of unit impulse functions where the
weights <span class="arithmatex">\(\left\{ {x[k]} \right\}\)</span> have random values according to some probability
function. The basic concepts of convolution theory, as argued below,
still hold. As a reminder of what <a href="Chap_6.html#eq:review_conv1">Equation 6.4</a> means, see <a href="Chap_3.html#fig3_1">Figure 3.1</a>. </p>
<p>The deterministic impulse function <span class="arithmatex">\(\delta [n]\)</span> produces the impulse response
<span class="arithmatex">\(h[n]\)</span> from the LTI system. A delayed version <span class="arithmatex">\(\delta [n - k]\)</span> produces a delayed output <span class="arithmatex">\(h[n - k].\)</span> Multiplication of the input by a scale factor produces multiplication of
the output by the same factor <em>even if that factor is a random variable</em>.</p>
<p>Thus <span class="arithmatex">\(x[k]\,\delta [n - k]\)</span> will produce <span class="arithmatex">\(x[k]\,h[n - k]\)</span> as an output.
Because the input <span class="arithmatex">\(x[n]\)</span> can be expressed as the sum of inputs of this
form, we can apply the linearity conditions, <a href="Chap_4.html#eq:additive">Equation 4.13</a> and <a href="Chap_4.html#eq:homogeneous">Equation 4.14</a>, to write the output as:</p>
<div class="" id="eq:review_conv2">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.5)</td>
<td class="eqTableEq">
<div>$$y[n] = \sum\limits_{k = - \infty }^{ + \infty } {x[k]\,h[n - k]} = x[n] \otimes h[n]$$</div>
</td>
</tr>
</table>
</div>
<p>The model for this situation is shown in <a href="#fig:filt_noise">Figure 6.1</a>.</p>
<figure class="figaltcap fullsize" id="fig:filt_noise"><img src="images/Fig_6_1.gif" /><figcaption><strong>Figure 6.1:</strong> A stochastic signal <span class="arithmatex">\(x\lbrack n\rbrack\)</span> filtered by a deterministic LTI filter <span class="arithmatex">\(h\lbrack n\rbrack\)</span> to produce a stochastic output signal <span class="arithmatex">\(y\lbrack n\rbrack\)</span>.</figcaption>
</figure>
<p>It is easy to see from <a href="Chap_6.html#eq:review_conv2">Equation 6.5</a> that <span class="arithmatex">\(y[n]\)</span> will be a stochastic signal
if <span class="arithmatex">\(x[n]\)</span> is a stochastic signal. The questions now arise:</p>
<ol>
<li>If the input mean is <span class="arithmatex">\({m_x},\)</span> what is the output mean <span class="arithmatex">\({m_y}?\)</span></li>
<li>If the input autocorrelation function is <span class="arithmatex">\({\varphi _{xx}}[k],\)</span> what form does the output autocorrelation function <span class="arithmatex">\({\varphi _{yy}}[k]\)</span> have?</li>
<li>Assuming that the input is stationary, is the output also stationary?</li>
</ol>
<h2 id="the-mean">The mean<a class="headerlink" href="#the-mean" title="Permanent link">¶</a></h2>
<p>Using the standard definition we have:</p>
<div class="" id="eq:filt_mean1">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.6)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{m_y}}&{ = E\left\{ {y[n]} \right\} = E\left\{ {\sum\limits_{k = - \infty }^{ + \infty } {x[k]\,h[n - k]} } \right\}}\\
{\,\,\,}&{ = \sum\limits_{k = - \infty }^{ + \infty } {E\left\{ {x[k]\,h[n - k]} \right\}} }
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>This last statement is true because, as we have observed earlier in <a href="Chap_4.html#eq:additive">Equation 4.13</a>, the averaging operator (or expectation operator) distributes over addition.</p>
<p>Only the weights <span class="arithmatex">\(\left\{ {x[k]} \right\}\)</span> are random variables. The impulse response,
<span class="arithmatex">\(h[n],\)</span> of the LTI system is <em>not</em> random and the term <span class="arithmatex">\(h[n - k]\)</span> is a
constant with respect to the averaging process over the random variable
<span class="arithmatex">\(x[k].\)</span> We can, therefore, rewrite this equation as:</p>
<div class="" id="eq:filt_mean2">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.7)</td>
<td class="eqTableEq">
<div>$${m_y} = \sum\limits_{k = - \infty }^{ + \infty } {E\left\{ {x[k]\,h[n - k]} \right\}} = \sum\limits_{k = - \infty }^{ + \infty } {E\left\{ {x[k]} \right\}\,} h[n - k]$$</div>
</td>
</tr>
</table>
</div>
<p>Because the random process <span class="arithmatex">\(\left\{ {x[k]} \right\}\)</span> is stationary (independent of <span class="arithmatex">\(n\)</span>)
we have:</p>
<div class="" id="eq:filt_mean3">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.8)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{m_y}}&{ = \sum\limits_{k = - \infty }^{ + \infty } {E\left\{ {x[k]} \right\}h[n - k]} = \sum\limits_{k = - \infty }^{ + \infty } {{m_x}h[n - k]} }\\
{\,\,\,}&{ = {m_x}\sum\limits_{k = - \infty }^{ + \infty } {h[n - k]} }
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>Using the Fourier transform of the impulse response <span class="arithmatex">\(H(\Omega ) = {\mathscr{F}}\left\{ {h[n]} \right\}\)</span> (<a href="Chap_3.html#eq:fourtranspair">Equation 3.3</a>), this expression simplifies to:</p>
<div class="mainresult" id="eq:filt_mean4">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.9)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{m_y}}&{ = {m_x}\sum\limits_{k = - \infty }^{ + \infty } {h[n - k]} = {m_x}\sum\limits_{n = - \infty }^{ + \infty } {h[n]} }\\
{\,\,\,}&{ = {m_x}H(\Omega = 0)}
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p><em>What does this mean?</em> From our knowledge of LTI filter theory, the expression <span class="arithmatex">\({m_y} = {m_x}H(\Omega = 0)\)</span> makes sense. The expression says the average value of the input random signal is multiplied by the constant gain factor—the gain at (<span class="arithmatex">\(\Omega = 0\)</span>)—of the linear filter. Because the average value is, indeed, just a constant (DC) value, the non-fluctuating component, this is a realistic result. Note that <span class="arithmatex">\(H(\Omega ),\)</span> <span class="arithmatex">\({m_x}\)</span> and <span class="arithmatex">\({m_y}\)</span> are all complex in this derivation.</p>
<p>We also see through this derivation that, because the input average is
stationary, the output average is also stationary.</p>
<h2 id="the-autocorrelation-function">The autocorrelation function<a class="headerlink" href="#the-autocorrelation-function" title="Permanent link">¶</a></h2>
<p>The development of the output correlation proceeds along similar lines:</p>
<div class="" id="eq:filt_acorr1">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.10)</td>
<td class="eqTableEq">
<div>$$\begin{array}{l}
{\varphi _{yy}}[n,n + k] = E\left\{ {y[n]{y^*}[n + k]} \right\}\\
\,\,\,\,\,\,\,\, = E\left\{ {\left( {\sum\limits_{m = - \infty }^{ + \infty } {h[m ]x[n - m ]} } \right)\left( {\sum\limits_{r = - \infty }^{ + \infty } {{h^*}[r]{x^*}[n + k - r]} } \right)} \right\}\\
\,\,\,\,\,\,\,\, = E\left\{ {\sum\limits_{m = - \infty }^{ + \infty } {\sum\limits_{r = - \infty }^{ + \infty } {h[m ]{h^*}[r]x[n - m ]{x^*}[n + k - r]} } } \right\}
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>Using the distributive property once again:</p>
<div class="" id="eq:filt_acorr2">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.11)</td>
<td class="eqTableEq">
<div>$${\varphi _{yy}}[n,n + k] = \sum\limits_{m = - \infty }^{ + \infty } {\sum\limits_{r = - \infty }^{ + \infty } {h[m ]{h^*}[r]E\left\{ {x[n - m ]{x^*}[n + k - r]} \right\}} }$$</div>
</td>
</tr>
</table>
</div>
<p>The term within the expectation braces <span class="arithmatex">\(E\left\{ \bullet \right\}\)</span> yields <span class="arithmatex">\({\varphi _{xx}}[k + m - r]\)</span> producing:</p>
<div class="" id="eq:filt_acorr3">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.12)</td>
<td class="eqTableEq">
<div>$${\varphi _{yy}}[n,n + k] = \sum\limits_{m = - \infty }^{ + \infty } {\sum\limits_{r = - \infty }^{ + \infty } {h[m ]{h^*}[r]{\varphi _{xx}}[k + m - r]} }$$</div>
</td>
</tr>
</table>
</div>
<p>Because <span class="arithmatex">\(m\)</span> and <span class="arithmatex">\(r\)</span> in <a href="Chap_6.html#eq:filt_acorr3">Equation 6.12</a> are simply dummy variables that
disappear when the two sums are performed, the result of the double sum,
<span class="arithmatex">\({\varphi _{yy}},\)</span> will only be a function of <span class="arithmatex">\(k\)</span> and, of course, the precise forms of
the impulse response <span class="arithmatex">\(h[n]\)</span> and the autocorrelation function <span class="arithmatex">\({\varphi _{xx}}.\)</span></p>
<p>If the right side of <a href="Chap_6.html#eq:filt_acorr3">Equation 6.12</a> is only a function of <span class="arithmatex">\(k\)</span> then the left
side is only a function of <span class="arithmatex">\(k,\)</span> as well, and we have:</p>
<div class="" id="eq:filt_acorr4">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.13)</td>
<td class="eqTableEq">
<div>$${\varphi _{yy}}[n,n + k] = {\varphi _{yy}}[k] = \sum\limits_{m = - \infty }^{ + \infty } {\sum\limits_{r = - \infty }^{ + \infty } {h[m ]{h^*}[r]{\varphi _{xx}}[k + m - r]} }$$</div>
</td>
</tr>
</table>
</div>
<p>From this we can conclude that if <span class="arithmatex">\({\varphi _{xx}}[ \bullet ]\)</span> is stationary then <span class="arithmatex">\({\varphi _{yy}}[ \bullet ]\)</span> is stationary. We can go further. If we set <span class="arithmatex">\(n = r - m,\)</span> then:</p>
<div class="" id="eq:filt_acorr5">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.14)</td>
<td class="eqTableEq">
<div>$$\begin{array}{*{20}{l}}
{{\varphi _{yy}}[k]}&{ = \sum\limits_{m = - \infty }^{ + \infty } {\sum\limits_{n = - \infty }^{ + \infty } {h[m]{h^*}[m + n]{\varphi _{xx}}[k - n]} } }\\
{\,\,\,}&{ = \sum\limits_{n = - \infty }^{ + \infty } {{\varphi _{xx}}[k - n]} \left( {\sum\limits_{m = - \infty }^{ + \infty } {h[m]{h^*}[m + n]} } \right)}
\end{array}$$</div>
</td>
</tr>
</table>
</div>
<p>We recognize from ordinary (deterministic) signal processing theory that
the term within parentheses is the autocorrelation of the impulse
response. If this autocorrelation is <span class="arithmatex">\({\varphi _{hh}}[m]\)</span> then:</p>
<div class="mainresult" id="eq:filt_acorr6">
<table class="eqTable">
<tr>
<td class="eqTableTag">(6.15)</td>
<td class="eqTableEq">
<div>$${\varphi _{yy}}[k] = \sum\limits_{m = - \infty }^{ + \infty } {{\varphi _{xx}}[k - m]} {\varphi _{hh}}[m] = {\varphi _{xx}}[k] \otimes {\varphi _{hh}}[k]$$</div>
</td>