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