521 lines
14 KiB
HTML
521 lines
14 KiB
HTML
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<H1><A ID="SECTION001410000000000000000">
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Symmetries and Fourier series</A>
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</H1>
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<P>
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Before making a quantitative analysis of the Fourier series of the classical
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waveforms, we pause to make two useful observations about symmetries in
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waveforms and the corresponding symmetries in the Fourier series. First, a
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Fourier series might consist only of even or odd-numbered harmonics; this is
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reflected in symmetries comparing a waveform to its displacement by half
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a cycle. Second, the Fourier series may contain only real-valued or pure
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imaginary-valued coefficients (corresponding to the cosine or sine functions).
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This is reflected in symmetries comparing the waveform to its reversal in time.
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<P>
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In this section we will assume that our waveform has an integer 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="$N$">, and
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furthermore, for simplicity, that <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="$N$"> is even (if it isn't we can just
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up-sample by a factor of two). We know from Chapter 9 that any (real or
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complex valued) waveform <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="$X[n]$"> can be written as a Fourier series (whose
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coefficients we'll denote by <IMG
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WIDTH="33" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1251.png"
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ALT="$A[k]$">):
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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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X[n] = A[0] + A[1]{U^n} + \cdots + A[N-1]{U^{(N-1)n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="331" HEIGHT="28" BORDER="0"
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SRC="img1252.png"
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ALT="\begin{displaymath}
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X[n] = A[0] + A[1]{U^n} + \cdots + A[N-1]{U^{(N-1)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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or, equivalently,
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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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X[n] =
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A[0] + A[1](\cos(\omega n) + i \sin(\omega n)) + \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="313" HEIGHT="28" BORDER="0"
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SRC="img1253.png"
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ALT="\begin{displaymath}
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X[n] =
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A[0] + A[1](\cos(\omega n) + i \sin(\omega n)) + \cdots
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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[N-1](\cos(\omega (N-1) n) + i \sin(\omega (N-1) n))
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\end{displaymath}
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-->
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<IMG
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WIDTH="327" HEIGHT="28" BORDER="0"
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SRC="img1254.png"
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ALT="\begin{displaymath}
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+ A[N-1](\cos(\omega (N-1) n) + i \sin(\omega (N-1) 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 <!-- MATH
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$\omega = 2\pi/N$
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-->
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<IMG
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WIDTH="74" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1047.png"
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ALT="$\omega=2\pi/N$"> is the fundamental frequency of the waveform,
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and
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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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U = \cos(\omega) + i \sin(\omega)
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\end{displaymath}
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-->
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<IMG
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WIDTH="146" HEIGHT="28" BORDER="0"
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SRC="img1049.png"
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ALT="\begin{displaymath}
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U = \cos(\omega) + i \sin(\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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is the unit-magnitude complex number whose argument is <IMG
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WIDTH="14" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
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SRC="img27.png"
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ALT="$\omega $">.
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<P>
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To analyze the first symmetry we delay the signal <IMG
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WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img669.png"
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ALT="$X[n]$"> by a half-cycle. Since
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<!-- MATH
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${U^{N/2}} = -1$
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-->
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<IMG
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WIDTH="81" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
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SRC="img1255.png"
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ALT="${U^{N/2}} = -1$"> we get:
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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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X[n+N/2] = A[0] - A[1]{U^n} + A[2]{U^{2n}} \pm \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="313" HEIGHT="28" BORDER="0"
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SRC="img1256.png"
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ALT="\begin{displaymath}
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X[n+N/2] = A[0] - A[1]{U^n} + A[2]{U^{2n}} \pm \cdots
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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[N-2]{U^{(N-2)n}} - A[N-1]{U^{(N-1)n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="274" HEIGHT="28" BORDER="0"
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SRC="img1257.png"
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ALT="\begin{displaymath}
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+ A[N-2]{U^{(N-2)n}} - A[N-1]{U^{(N-1)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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In effect, a half-cycle delay changes the sign of every other term in the
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Fourier series.
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We combine this with the original series in two different ways. Letting
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<IMG
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WIDTH="22" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1258.png"
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ALT="$X'$"> denote half the sum of the two:
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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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X'[n] = {{X[n] + X[n+N/2]}\over 2} =
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A[0] + A[2]{U^{2n}} + \cdots + A[N-2]{U^{(N-2)n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="500" HEIGHT="40" BORDER="0"
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SRC="img1259.png"
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ALT="\begin{displaymath}
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X'[n] = {{X[n] + X[n+N/2]}\over 2} =
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A[0] + A[2]{U^{2n}} + \cdots + A[N-2]{U^{(N-2)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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and <IMG
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WIDTH="26" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1260.png"
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ALT="$X''$"> half the difference:
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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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X''[n] = {{X[n] - X[n+N/2]}\over 2} =
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A[1]{U^n} + A[3]{U^{3n}} + \cdots + A[N-1]{U^{(N-1)n}}
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\end{displaymath}
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-->
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<IMG
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WIDTH="525" HEIGHT="40" BORDER="0"
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SRC="img1261.png"
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ALT="\begin{displaymath}
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X''[n] = {{X[n] - X[n+N/2]}\over 2} =
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A[1]{U^n} + A[3]{U^{3n}} + \cdots + A[N-1]{U^{(N-1)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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we see that <IMG
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WIDTH="22" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1258.png"
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ALT="$X'$"> consists only of even-numbered harmonics (including DC) and
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<IMG
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WIDTH="26" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1260.png"
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ALT="$X''$"> only of odd ones.
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<P>
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Furthermore, if <IMG
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WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img670.png"
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ALT="$X$"> happens to be equal to itself shifted a half cycle, that
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is, if <!-- MATH
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$X[n] = X[n+N/2]$
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-->
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<IMG
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WIDTH="138" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1262.png"
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ALT="$X[n] = X[n+N/2]$">, then (looking at the definitions of <IMG
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WIDTH="22" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1258.png"
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ALT="$X'$"> and <IMG
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WIDTH="26" HEIGHT="15" ALIGN="BOTTOM" BORDER="0"
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SRC="img1260.png"
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ALT="$X''$">)
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we get <IMG
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WIDTH="94" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1263.png"
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ALT="$X'[n] = X[n]$"> and <IMG
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WIDTH="73" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1264.png"
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ALT="$X''[n] = 0$">. This implies that, in this case,
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<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="$X[n]$"> has only even numbered harmonics. Indeed, this should be no surprise,
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since in this case <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="$X[n]$"> would have to repeat every <IMG
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WIDTH="32" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1265.png"
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ALT="$N/2$"> samples, so its
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fundamental frequency is twice as high as normal for 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="$N$">.
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<P>
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In the same way, if <!-- MATH
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$X[n] = -X[n+N/2$
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-->
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<IMG
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WIDTH="146" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1266.png"
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ALT="$X[n] = -X[n+N/2$">], then <IMG
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WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img670.png"
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ALT="$X$"> can have only odd-numbered
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harmonics. This allows us easily to split any desired waveform into its
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even- and odd-numbered harmonics. (This is equivalent to using a comb filter
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to extract even or odd harmonics; see Chapter <A HREF="node104.html#chapter-delay">7</A>.)
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<P>
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To derive the second symmetry relation we compare <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="$X[n]$"> with its time
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reversal, <IMG
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WIDTH="48" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1267.png"
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ALT="$X[-n]$"> (or, equivalently, since <IMG
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WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img670.png"
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ALT="$X$"> repeats every <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, with
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<IMG
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WIDTH="70" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1268.png"
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ALT="$X[N-n]$">). The Fourier series 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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X[-n] =
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A[0] + A[1](\cos(\omega n) - i \sin(\omega n)) + \cdots
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\end{displaymath}
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-->
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<IMG
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WIDTH="326" HEIGHT="28" BORDER="0"
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SRC="img1269.png"
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ALT="\begin{displaymath}
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X[-n] =
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A[0] + A[1](\cos(\omega n) - i \sin(\omega n)) + \cdots
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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[N-1](\cos(\omega (N-1) n) - i \sin(\omega (N-1) n))
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\end{displaymath}
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-->
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<IMG
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WIDTH="327" HEIGHT="28" BORDER="0"
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SRC="img1270.png"
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ALT="\begin{displaymath}
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+ A[N-1](\cos(\omega (N-1) n) - i \sin(\omega (N-1) 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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(since the cosine function is even and the sine function is odd). In the same
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way as before we can extract the cosines by forming <IMG
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WIDTH="40" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1271.png"
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ALT="$X'[n]$"> as half the sum:
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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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X'[n] = {{X[n] + X[-n]}\over 2} =
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A[0] + A[1]\cos(\omega n)+ \cdots + A[N-1]\cos(\omega (N-1) n)
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\end{displaymath}
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-->
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<IMG
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WIDTH="544" HEIGHT="40" BORDER="0"
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SRC="img1272.png"
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ALT="\begin{displaymath}
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X'[n] = {{X[n] + X[-n]}\over 2} =
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A[0] + A[1]\cos(\omega n)+ \cdots + A[N-1]\cos(\omega (N-1) 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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and <IMG
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WIDTH="44" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1273.png"
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ALT="$X''[n]$"> as half the difference divided by <IMG
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WIDTH="9" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img646.png"
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ALT="$i$">:
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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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X''[n] = {{X[n] - X[-n]}\over {2i}} =
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A[1]\sin(\omega n)+ \cdots + A[N-1]\sin(\omega (N-1) n)
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\end{displaymath}
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-->
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<IMG
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WIDTH="496" HEIGHT="40" BORDER="0"
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SRC="img1274.png"
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ALT="\begin{displaymath}
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X''[n] = {{X[n] - X[-n]}\over {2i}} =
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A[1]\sin(\omega n)+ \cdots + A[N-1]\sin(\omega (N-1) 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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<P>
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So if <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="$X[n]$"> satisfies <IMG
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WIDTH="102" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1275.png"
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ALT="$X[-n] = X[n]$"> the Fourier series consists of cosine
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terms only; if <IMG
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WIDTH="115" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1276.png"
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ALT="$X[-n] = -X[n]$"> it consists of sine terms only; and as
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before we can decompose any <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="$X[n]$"> (that repeats every <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) as a sum
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of the two.
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<P>
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<BR><HR>
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<!--Table of Child-Links-->
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<A ID="CHILD_LINKS"><STRONG>Subsections</STRONG></A>
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<LI><A ID="tex2html3343"
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HREF="node186.html">Sawtooth waves and symmetry</A>
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<BR>
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<B> Next:</B> <A ID="tex2html3342"
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HREF="node186.html">Sawtooth waves and symmetry</A>
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<B> Up:</B> <A ID="tex2html3336"
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HREF="node184.html">Classical waveforms</A>
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<B> Previous:</B> <A ID="tex2html3330"
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HREF="node184.html">Classical waveforms</A>
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<B> <A ID="tex2html3338"
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HREF="node4.html">Contents</A></B>
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<B> <A ID="tex2html3340"
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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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