317 lines
8.6 KiB
HTML
317 lines
8.6 KiB
HTML
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<H1><A ID="SECTION00560000000000000000"></A>
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<A ID="sect1.combine"></A>
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<BR>
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Superposing Signals
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</H1>
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<P>
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If a signal <IMG
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WIDTH="31" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img80.png"
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ALT="$x[n]$"> has a peak or RMS amplitude <IMG
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WIDTH="15" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img23.png"
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ALT="$A$"> (in some fixed window), then
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the scaled signal <IMG
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WIDTH="51" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img122.png"
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ALT="$k \cdot x[n]$"> (where <IMG
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WIDTH="41" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img123.png"
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ALT="$k \ge 0$">) has amplitude <IMG
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WIDTH="24" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img124.png"
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ALT="$kA$">. The
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mean power of the scaled signal changes by a factor of <IMG
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WIDTH="19" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
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SRC="img125.png"
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ALT="$k^2$">. The situation gets
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more complicated when two different signals are added together; just knowing
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the amplitudes of the two does not suffice to know the amplitude of the sum.
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The two amplitude measures do at least obey triangle inequalities; for any
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two signals <IMG
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WIDTH="31" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img80.png"
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ALT="$x[n]$"> and <IMG
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WIDTH="30" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img2.png"
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ALT="$y[n]$">,
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{A_{\mathrm{peak}}} \{x[n]\} +
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{A_{\mathrm{peak}}} \{y[n]\} \ge
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{A_{\mathrm{peak}}} \{x[n]+y[n]\}
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\end{displaymath}
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-->
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<IMG
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WIDTH="332" HEIGHT="29" BORDER="0"
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SRC="img126.png"
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ALT="\begin{displaymath}
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{A_{\mathrm{peak}}} \{x[n]\} +
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{A_{\mathrm{peak}}} \{y[n]\} \ge
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{A_{\mathrm{peak}}} \{x[n]+y[n]\}
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{A_{\mathrm{RMS}}} \{x[n]\} +
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{A_{\mathrm{RMS}}} \{y[n]\} \ge
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{A_{\mathrm{RMS}}} \{x[n]+y[n]\}
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\end{displaymath}
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-->
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<IMG
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WIDTH="337" HEIGHT="28" BORDER="0"
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SRC="img127.png"
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ALT="\begin{displaymath}
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{A_{\mathrm{RMS}}} \{x[n]\} +
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{A_{\mathrm{RMS}}} \{y[n]\} \ge
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{A_{\mathrm{RMS}}} \{x[n]+y[n]\}
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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If we fix a window from <IMG
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WIDTH="20" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img86.png"
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ALT="$M$"> to <IMG
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WIDTH="82" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img128.png"
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ALT="$N+M-1$"> as usual, we can write out the
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mean power of the sum of two signals:
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<A ID="eq-meanpowersum"></A>
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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P \{x[n] + y[n]\} = P \{x[n]\} + P \{y[n]\}
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+ 2 \cdot {\mathrm{COV}} \{ x[n] , y[n] \}
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\end{displaymath}
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-->
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<IMG
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WIDTH="405" HEIGHT="28" BORDER="0"
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SRC="img129.png"
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ALT="\begin{displaymath}
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P \{x[n] + y[n]\} = P \{x[n]\} + P \{y[n]\}
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+ 2 \cdot {\mathrm{COV}} \{ x[n] , y[n] \}
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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where we have introduced the
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<A ID="1161"></A><I>covariance</I> of two signals:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{\mathrm{COV}} \{ x[n] , y[n] \} =
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{
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{x[M]y[M] + \cdots + x[M+N-1]y[M+N-1]}
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\over
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N
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}
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\end{displaymath}
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-->
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<IMG
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WIDTH="454" HEIGHT="40" BORDER="0"
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SRC="img130.png"
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ALT="\begin{displaymath}
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{\mathrm{COV}} \{ x[n] , y[n] \} =
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{
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{x[M]y[M] + \cdots + x[M+N-1]y[M+N-1]}
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\over
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N
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}
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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The covariance may be positive, zero, or negative. Over a sufficiently large
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window, the covariance of two sinusoids with different frequencies is
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negligible compared to the mean power. Two signals which have no covariance
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are called <I>uncorrelated</I> (the correlation is the covariance normalized
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to lie between -1 and 1).
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In general, for two uncorrelated
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signals, the power of the
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sum is the sum of the powers:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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P \{x[n] + y[n]\} = P \{x[n]\} + P \{y[n]\} , \hspace{0.1in}
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\mathrm{whenever}
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\ {\mathrm{COV}} \{ x[n] , y[n] \} = 0
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\end{displaymath}
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-->
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<IMG
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WIDTH="483" HEIGHT="28" BORDER="0"
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SRC="img131.png"
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ALT="\begin{displaymath}
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P \{x[n] + y[n]\} = P \{x[n]\} + P \{y[n]\} , \hspace{0.1in}
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\mathrm{whenever}
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\ {\mathrm{COV}} \{ x[n] , y[n] \} = 0
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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Put in terms of amplitude, this becomes:
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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\begin{displaymath}
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{{\left ( {A_{\mathrm{RMS}}} \{x[n]+y[n]\} \right ) } ^ 2} =
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{{\left ( {A_{\mathrm{RMS}}} \{x[n]\} \right ) } ^ 2} +
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{{\left ( {A_{\mathrm{RMS}}} \{y[n]\} \right ) } ^ 2} .
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\end{displaymath}
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-->
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<IMG
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WIDTH="398" HEIGHT="28" BORDER="0"
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SRC="img132.png"
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ALT="\begin{displaymath}
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{{\left ( {A_{\mathrm{RMS}}} \{x[n]+y[n]\} \right ) } ^ 2} ...
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...2} +
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{{\left ( {A_{\mathrm{RMS}}} \{y[n]\} \right ) } ^ 2} .
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\end{displaymath}">
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</DIV>
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<BR CLEAR="ALL">
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<P></P>
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This is the familiar Pythagorean relation. So uncorrelated signals can be
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thought of as vectors at right angles to each other; positively correlated ones
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as having an acute angle between them, and negatively correlated as having an
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obtuse angle between them.
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<P>
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For example, if two uncorrelated signals both have RMS amplitude <IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img4.png"
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ALT="$a$">,
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the sum will have RMS amplitude <IMG
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WIDTH="33" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
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SRC="img133.png"
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ALT="${\sqrt 2} a$">. On the other hand if the two
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signals happen to be equal--the most correlated possible--the sum will have
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amplitude <IMG
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WIDTH="19" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img134.png"
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ALT="$2a$">, which is the maximum allowed by the triangle inequality.
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<P>
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<B> Next:</B> <A ID="tex2html736"
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HREF="node14.html">Periodic Signals</A>
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<B> Up:</B> <A ID="tex2html730"
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HREF="node7.html">Sinusoids, amplitude and frequency</A>
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<B> Previous:</B> <A ID="tex2html724"
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HREF="node12.html">Synthesizing a sinusoid</A>
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<B> <A ID="tex2html732"
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HREF="node4.html">Contents</A></B>
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<B> <A ID="tex2html734"
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HREF="node201.html">Index</A></B>
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<ADDRESS>
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Miller Puckette
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2006-12-30
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