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<TITLE>Fourier analysis of non-periodic signals</TITLE>
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<H1><A NAME="SECTION001330000000000000000">
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Fourier analysis of non-periodic signals</A>
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</H1>
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<P>
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Most signals aren't periodic, and even a periodic one might have an unknown
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period. So we should be prepared to do Fourier analysis on signals without
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making the comforting assumption that the signal to analyze repeats at a fixed
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period <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img3.png"
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ALT="$N$">. Of course, we can simply take <IMG
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WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img3.png"
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ALT="$N$"> samples of the signal and
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<I>make</I> it periodic; this is essentially what we did in the previous
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section, in which a pure sinusoid gave us the complicated Fourier transform of
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Figure <A HREF="node170.html#fig09.03">9.3</A> (part b).
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<P>
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However, it would be better to get a result in which the response to a pure
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sinusoid were better localized around the corresponding value of <IMG
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WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img58.png"
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ALT="$k$">. We
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can accomplish this using the enveloping technique first introduced in Figure
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<A HREF="node29.html#fig02.07">2.7</A> (Page <A HREF="node29.html#fig02.07"><IMG ALIGN="BOTTOM" BORDER="1" ALT="[*]"
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SRC="crossref.png"></A>). Applying this technique to Fourier
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analysis will not only improve our analyses, but will also shed new light on
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the enveloping looping sampler of Chapter 2.
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<P>
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Given a signal <IMG
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WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img669.png"
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ALT="$X[n]$">, periodic or not, defined on the points from
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<IMG
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WIDTH="11" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img179.png"
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ALT="$0$"> to <IMG
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WIDTH="45" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img171.png"
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ALT="$N-1$">,
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the technique is to envelope the signal before doing the Fourier analysis.
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The envelope shape is known as a
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<A NAME="12499"></A><I>window function</I>.
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Given a window function <IMG
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WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1123.png"
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ALT="$w[n]$">, the
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<A NAME="12501"></A><I>windowed Fourier transform</I>
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is:
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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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{\cal FT} \left \{ w[n] X[n] \right \} (k)
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\end{displaymath}
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-->
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<IMG
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WIDTH="130" HEIGHT="28" BORDER="0"
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SRC="img1124.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ w[n] X[n] \right \} (k)
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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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Much ink has been spilled over the design of suitable window functions for
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particular situations, but here we will consider the simplest one, named the
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<A NAME="12504"></A><A NAME="12505"></A><I>Hann</I>
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window function (the name is sometimes corrupted to "Hanning" in DSP circles).
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The Hann window is:
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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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w[n] = {1\over 2} - {1\over 2} \cos(2\pi n / N)
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\end{displaymath}
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-->
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<IMG
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WIDTH="178" HEIGHT="38" BORDER="0"
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SRC="img1125.png"
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ALT="\begin{displaymath}
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w[n] = {1\over 2} - {1\over 2} \cos(2\pi n / 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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It is easy to analyze the effect of multiplying a signal by the Hann window
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before taking the Fourier transform, because the Hann window can be written
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as a sum of three complex exponentials:
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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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w[n] = {1\over 2} - {1\over 4} {U^n} - {1\over 4} {U^{-n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="177" HEIGHT="38" BORDER="0"
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SRC="img1126.png"
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ALT="\begin{displaymath}
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w[n] = {1\over 2} - {1\over 4} {U^n} - {1\over 4} {U^{-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 as before, <IMG
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WIDTH="16" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img1048.png"
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ALT="$U$"> is the unit-magnitude complex number with argument
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<IMG
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WIDTH="42" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img184.png"
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ALT="$2\pi/N$">.
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We can now calculate the windowed Fourier transform of a
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sinusoid <IMG
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WIDTH="24" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img653.png"
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ALT="$Z^n$"> with angular frequency <IMG
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WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img7.png"
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ALT="$\alpha $"> as before. The phases
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come out messy and we'll replace them with simplified approximations:
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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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{\cal FT} \left \{ w[n] {Z^n} \right \} (k)
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\end{displaymath}
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-->
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<IMG
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WIDTH="118" HEIGHT="28" BORDER="0"
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SRC="img1127.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ w[n] {Z^n} \right \} (k)
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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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= {\cal FT} \left \{ {1\over 2} {Z^n} - {1\over 4} {(UZ)^n}
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- {1\over 4} {({U^{-1}}Z)^n}\right \} (k)
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\end{displaymath}
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-->
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<IMG
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WIDTH="294" HEIGHT="45" BORDER="0"
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SRC="img1128.png"
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ALT="\begin{displaymath}
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= {\cal FT} \left \{ {1\over 2} {Z^n} - {1\over 4} {(UZ)^n}
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- {1\over 4} {({U^{-1}}Z)^n}\right \} (k)
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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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\approx \left [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
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M(k - {{\alpha } \over {\omega}})
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\end{displaymath}
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-->
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<IMG
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WIDTH="257" HEIGHT="35" BORDER="0"
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SRC="img1129.png"
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ALT="\begin{displaymath}
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\approx \left [ \cos(\Phi(k)) + i \sin(\Phi(k))\right ]
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M(k - {{\alpha } \over {\omega}})
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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 the (approximate) phase term is:
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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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\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}})
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\end{displaymath}
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-->
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<IMG
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WIDTH="140" HEIGHT="35" BORDER="0"
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SRC="img1130.png"
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ALT="\begin{displaymath}
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\Phi(k) = - \pi \cdot (k - {{\alpha } \over {\omega}})
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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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and the magnitude function is:
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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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M(k) =
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{\left [ {
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{1\over 2}{D_N}(k)
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+ {1\over 4}{D_N}(k + 1)
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+ {1\over 4}{D_N}(k - 1)
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} \right ] }
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\end{displaymath}
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-->
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<IMG
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WIDTH="340" HEIGHT="45" BORDER="0"
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SRC="img1131.png"
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ALT="\begin{displaymath}
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M(k) =
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{\left [ {
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{1\over 2}{D_N}(k)
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+ {1\over 4}{D_N}(k + 1)
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+ {1\over 4}{D_N}(k - 1)
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} \right ] }
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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 magnitude function <IMG
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WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1132.png"
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ALT="$M(k)$"> is graphed in Figure <A HREF="#fig09.05">9.5</A>. The three
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Dirichlet kernel components are also shown separately.
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<P>
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<DIV ALIGN="CENTER"><A NAME="fig09.05"></A><A NAME="12536"></A>
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<TABLE>
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<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.5:</STRONG>
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The magnitude M(k) of the Fourier transform of the Hann
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window function. It is the sum of three (shifted and magnified) copies of the Dirichlet
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kernel <IMG
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WIDTH="28" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img62.png"
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ALT="$D_N$">, with <IMG
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WIDTH="63" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img63.png"
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ALT="$N=100$">. </CAPTION>
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<TR><TD><IMG
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WIDTH="409" HEIGHT="207" BORDER="0"
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SRC="img1133.png"
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ALT="\begin{figure}\psfig{file=figs/fig09.05.ps}\end{figure}"></TD></TR>
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</TABLE>
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</DIV>
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<P>
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The main lobe of <IMG
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WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1132.png"
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ALT="$M(k)$"> is four harmonics wide, twice the width of the
|
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main lobe of the Dirichlet kernel. The sidelobes, on the other hand, have
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much smaller magnitude. Each sidelobe of <IMG
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WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1132.png"
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ALT="$M(k)$"> is a sum of three sidelobes
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of <IMG
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WIDTH="46" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1134.png"
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ALT="${D_n}(k)$">, one attenuated by <IMG
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WIDTH="27" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img98.png"
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ALT="$1/2$"> and the others, opposite in sign,
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attenuated by <IMG
|
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WIDTH="27" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1135.png"
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ALT="$1/4$">. They do not cancel out perfectly but they do cancel out
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fairly well.
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<P>
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The sidelobes reach their maximum amplitudes near their midpoints, and we
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|
can estimate their amplitudes there, using the approximation:
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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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|
{D_N}(k) \approx {
|
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|
{N \sin(\pi k) } \over {\pi k}
|
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|
}
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|
\end{displaymath}
|
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|
-->
|
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|
<IMG
|
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|
WIDTH="138" HEIGHT="40" BORDER="0"
|
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|
|
SRC="img1136.png"
|
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|
|
ALT="\begin{displaymath}
|
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|
|
{D_N}(k) \approx {
|
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|
|
{N \sin(\pi k) } \over {\pi k}
|
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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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|
Setting <!-- MATH
|
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|
$k = 3/2, 5/2, \ldots$
|
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|
-->
|
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|
<IMG
|
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|
|
WIDTH="113" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
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|
|
SRC="img1137.png"
|
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|
ALT="$k = 3/2, 5/2, \ldots$"> gives sidelobe amplitudes, relative to the
|
|
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|
|
peak height <IMG
|
|
|
|
|
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img3.png"
|
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|
|
|
ALT="$N$">, of:
|
|
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|
|
<BR><P></P>
|
|
|
|
|
<DIV ALIGN="CENTER">
|
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|
|
<!-- MATH
|
|
|
|
|
\begin{displaymath}
|
|
|
|
|
{2 \over {3 \pi}} \approx -13 \mathrm{dB} , \;
|
|
|
|
|
{2 \over {5 \pi}} \approx -18 \mathrm{dB} , \;
|
|
|
|
|
{2 \over {7 \pi}} \approx -21 \mathrm{dB} , \;
|
|
|
|
|
{2 \over {9 \pi}} \approx -23 \mathrm{dB} ,
|
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|
|
|
\ldots
|
|
|
|
|
\end{displaymath}
|
|
|
|
|
-->
|
|
|
|
|
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="420" HEIGHT="38" BORDER="0"
|
|
|
|
|
SRC="img1138.png"
|
|
|
|
|
ALT="\begin{displaymath}
|
|
|
|
|
{2 \over {3 \pi}} \approx -13 \mathrm{dB} , \;
|
|
|
|
|
{2 \over {5...
|
|
|
|
|
...{dB} , \;
|
|
|
|
|
{2 \over {9 \pi}} \approx -23 \mathrm{dB} ,
|
|
|
|
|
\ldots
|
|
|
|
|
\end{displaymath}">
|
|
|
|
|
</DIV>
|
|
|
|
|
<BR CLEAR="ALL">
|
|
|
|
|
<P></P>
|
|
|
|
|
The sidelobes drop off progressively more slowly so that the tenth one is only
|
|
|
|
|
attenuated about 30 dB and the 32nd one about -40 dB. On the
|
|
|
|
|
other hand, the Hann window sidelobes
|
|
|
|
|
are attenuated by:
|
|
|
|
|
<BR><P></P>
|
|
|
|
|
<DIV ALIGN="CENTER">
|
|
|
|
|
<!-- MATH
|
|
|
|
|
\begin{displaymath}
|
|
|
|
|
{2 \over {5 \pi}} - {1\over 2} [ {2 \over {3 \pi}} + {2 \over {7 \pi}} ]
|
|
|
|
|
\approx -32.30 \mathrm{dB}
|
|
|
|
|
\end{displaymath}
|
|
|
|
|
-->
|
|
|
|
|
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="210" HEIGHT="38" BORDER="0"
|
|
|
|
|
SRC="img1139.png"
|
|
|
|
|
ALT="\begin{displaymath}
|
|
|
|
|
{2 \over {5 \pi}} - {1\over 2} [ {2 \over {3 \pi}} + {2 \over {7 \pi}} ]
|
|
|
|
|
\approx -32.30 \mathrm{dB}
|
|
|
|
|
\end{displaymath}">
|
|
|
|
|
</DIV>
|
|
|
|
|
<BR CLEAR="ALL">
|
|
|
|
|
<P></P>
|
|
|
|
|
and <IMG
|
|
|
|
|
WIDTH="31" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1140.png"
|
|
|
|
|
ALT="$-42$">, <IMG
|
|
|
|
|
WIDTH="31" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1141.png"
|
|
|
|
|
ALT="$-49$">, <IMG
|
|
|
|
|
WIDTH="31" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1142.png"
|
|
|
|
|
ALT="$-54$">, and <IMG
|
|
|
|
|
WIDTH="31" HEIGHT="29" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1143.png"
|
|
|
|
|
ALT="$-59$"> dB for the next four sidelobes.
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
This shows that applying a Hann window before taking the Fourier transform
|
|
|
|
|
will better allow us to isolate sinusoidal
|
|
|
|
|
components. If a signal has many sinusoidal components, the sidelobes
|
|
|
|
|
engendered by each one will interfere with the main lobe of all the others.
|
|
|
|
|
Reducing the amplitude of the sidelobes reduces this interference.
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
|
|
|
|
|
<DIV ALIGN="CENTER"><A NAME="fig09.06"></A><A NAME="12558"></A>
|
|
|
|
|
<TABLE>
|
|
|
|
|
<CAPTION ALIGN="BOTTOM"><STRONG>Figure 9.6:</STRONG>
|
|
|
|
|
The Hann-windowed Fourier transform of a signal with two
|
|
|
|
|
sinusoidal components, at frequencies 5.3 and 10.6 times the fundamental,
|
|
|
|
|
and with different complex amplitudes.</CAPTION>
|
|
|
|
|
<TR><TD><IMG
|
|
|
|
|
WIDTH="482" HEIGHT="245" BORDER="0"
|
|
|
|
|
SRC="img1144.png"
|
|
|
|
|
ALT="\begin{figure}\psfig{file=figs/fig09.06.ps}\end{figure}"></TD></TR>
|
|
|
|
|
</TABLE>
|
|
|
|
|
</DIV>
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
Figure <A HREF="#fig09.06">9.6</A> shows a Hann-windowed Fourier analysis of a signal with
|
|
|
|
|
two sinusoidal components. The two are separated by about 5 times the
|
|
|
|
|
fundamental frequency <IMG
|
|
|
|
|
WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img27.png"
|
|
|
|
|
ALT="$\omega $">, and for each we see clearly the shape of the
|
|
|
|
|
Hann window's Fourier transform. Four points of the Fourier analysis lie
|
|
|
|
|
within the main lobe of <IMG
|
|
|
|
|
WIDTH="41" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1132.png"
|
|
|
|
|
ALT="$M(k)$"> corresponding to each sinusoid. The amplitude
|
|
|
|
|
and phase of the individual sinusoids are reflected in those of the
|
|
|
|
|
(four-point-wide) peaks. The four points within a peak which happen to fall at
|
|
|
|
|
integer values <IMG
|
|
|
|
|
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img58.png"
|
|
|
|
|
ALT="$k$"> are successively about one half cycle out of phase.
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
To fully resolve the partials of a signal, we should choose an analysis size
|
|
|
|
|
<IMG
|
|
|
|
|
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img3.png"
|
|
|
|
|
ALT="$N$"> large enough so that <IMG
|
|
|
|
|
WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
|
|
|
|
|
SRC="img1047.png"
|
|
|
|
|
ALT="$\omega=2\pi/N$"> is no more than a quarter of the
|
|
|
|
|
frequency separation between neighboring partials. For a periodic signal, for
|
|
|
|
|
example, the partials are separated by the fundamental frequency. For the
|
|
|
|
|
analysis to fully resolve the partials, the analysis period <IMG
|
|
|
|
|
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
|
|
|
|
|
SRC="img3.png"
|
|
|
|
|
ALT="$N$"> must be
|
|
|
|
|
at least four periods of the signal.
|
|
|
|
|
|
|
|
|
|
<P>
|
|
|
|
|
In some applications it works to allow the peaks to overlap as long as the
|
|
|
|
|
center of each peak is isolated from all the other peaks; in this case the
|
|
|
|
|
four-period rule may be relaxed to three or even slightly less.
|
|
|
|
|
|
|
|
|
|
<P>
|
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|
|
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2006-12-30
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