491 lines
11 KiB
HTML
491 lines
11 KiB
HTML
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original version by: Nikos Drakos, CBLU, University of Leeds
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* revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan
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* with significant contributions from:
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Jens Lippmann, Marek Rouchal, Martin Wilck and others -->
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<B> Next:</B> <A NAME="tex2html3105"
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HREF="node170.html">Fourier transform of a</A>
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HREF="node167.html">Properties of Fourier transforms</A>
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HREF="node168.html">Fourier transform of DC</A>
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HREF="node201.html">Index</A></B>
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<H2><A NAME="SECTION001322000000000000000"></A>
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<A NAME="sect9.shift"></A>
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<BR>
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Shifts and phase changes
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</H2>
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<P>
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Section <A HREF="node107.html#sect7.phase">7.2</A> showed how time-shifting a signal changes the
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phases of its sinusoidal components, and Section <A HREF="node154.html#sect8.singlesideband">8.4.3</A>
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showed how multiplying a signal by a complex sinusoid shifts its component
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frequencies. These two effects have corresponding identities
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involving the Fourier transform.
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<P>
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First we consider a time shift. 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="$X[n]$">, as usual, is a complex-valued
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signal 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, let <IMG
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WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img717.png"
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ALT="$Y[n]$"> be <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="$X[n]$"> delayed <IMG
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WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
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SRC="img28.png"
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ALT="$d$">
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samples:
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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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Y[n] = X[n-d]
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\end{displaymath}
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-->
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<IMG
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WIDTH="112" HEIGHT="28" BORDER="0"
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SRC="img671.png"
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ALT="\begin{displaymath}
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Y[n] = X[n-d]
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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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which also 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 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="$X$"> does. We can reduce the Fourier
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transform of <IMG
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WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img717.png"
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ALT="$Y[n]$"> this way:
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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 \{ Y[n] \right \} (k) =
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{V ^ {0}} Y[0] +
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{V ^ {1}} Y[1] +
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\cdots +
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{V ^ {N-1}} Y[N-1]
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\end{displaymath}
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-->
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<IMG
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WIDTH="400" HEIGHT="28" BORDER="0"
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SRC="img1101.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ Y[n] \right \} (k) =
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{V ^ {0}} Y[0] +
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{V ^ {1}} Y[1] +
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\cdots +
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{V ^ {N-1}} Y[N-1]
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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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=
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{V ^ {0}} X[-d] +
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{V ^ {1}} X[-d+1] +
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\cdots +
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{V ^ {N-1}} X[-d+N-1]
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\end{displaymath}
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-->
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<IMG
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WIDTH="393" HEIGHT="28" BORDER="0"
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SRC="img1102.png"
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ALT="\begin{displaymath}
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=
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{V ^ {0}} X[-d] +
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{V ^ {1}} X[-d+1] +
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\cdots +
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{V ^ {N-1}} X[-d+N-1]
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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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=
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{V ^ {d}} X[0] +
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{V ^ {d+1}} X[1] +
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\cdots +
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{V ^ {d+N-1}} X[N-1]
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\end{displaymath}
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-->
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<IMG
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WIDTH="333" HEIGHT="28" BORDER="0"
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SRC="img1103.png"
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ALT="\begin{displaymath}
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=
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{V ^ {d}} X[0] +
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{V ^ {d+1}} X[1] +
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\cdots +
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{V ^ {d+N-1}} X[N-1]
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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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= {V^d} \left (
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{V ^ {0}} X[0] +
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{V ^ {1}} X[1] +
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\cdots +
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{V ^ {N-1}} X[N-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="336" HEIGHT="30" BORDER="0"
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SRC="img1104.png"
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ALT="\begin{displaymath}
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= {V^d} \left (
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{V ^ {0}} X[0] +
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{V ^ {1}} X[1] +
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\cdots +
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{V ^ {N-1}} X[N-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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<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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= {V^d} {\cal FT} \left \{ X[n] \right \} (k)
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\end{displaymath}
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-->
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<IMG
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WIDTH="136" HEIGHT="28" BORDER="0"
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SRC="img1105.png"
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ALT="\begin{displaymath}
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= {V^d} {\cal FT} \left \{ 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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(The third line is just the second one with the terms summed in a
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different order). We therefore get the Time Shift Formula for Fourier
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Transforms: <A NAME="12441"></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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{\cal FT} \left \{ X[n-d] \right \} (k) =
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\left ( {
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\parbox[t][0.1in]{0in}{\mbox{}}
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\cos(-dk\omega) + i\sin(-dk\omega)
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} \right )
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{\cal FT} \left \{ X[n] \right \} (k)
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\end{displaymath}
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-->
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<IMG
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WIDTH="447" HEIGHT="35" BORDER="0"
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SRC="img1106.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ X[n-d] \right \} (k) =
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\left ( {
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\par...
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...-dk\omega)
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} \right )
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{\cal FT} \left \{ 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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The Fourier transform of <IMG
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WIDTH="64" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
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SRC="img1107.png"
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ALT="$X[n-d]$"> is a phase term times the Fourier transform
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of <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="$X[n]$">. The phase is changed by <IMG
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WIDTH="43" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
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SRC="img1108.png"
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ALT="$-dk\omega$">, a
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linear function of the frequency <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="$k$">.
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<P>
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Now suppose instead that we change our starting 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 multiplying
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it by a complex exponential <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 $">:
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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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Y[n] = {Z^n} X[n]
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\end{displaymath}
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-->
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<IMG
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WIDTH="104" HEIGHT="28" BORDER="0"
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SRC="img1109.png"
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ALT="\begin{displaymath}
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Y[n] = {Z^n} X[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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Z = \cos(\alpha) + i \sin(\alpha)
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\end{displaymath}
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-->
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<IMG
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WIDTH="145" HEIGHT="28" BORDER="0"
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SRC="img1110.png"
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ALT="\begin{displaymath}
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Z = \cos(\alpha) + i \sin(\alpha)
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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 Fourier transform 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 \{ Y[n] \right \} (k) =
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{V ^ {0}} Y[0] +
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{V ^ {1}} Y[1] +
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\cdots +
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{V ^ {N-1}} Y[N-1]
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\end{displaymath}
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-->
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<IMG
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WIDTH="400" HEIGHT="28" BORDER="0"
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SRC="img1101.png"
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ALT="\begin{displaymath}
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{\cal FT} \left \{ Y[n] \right \} (k) =
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{V ^ {0}} Y[0] +
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{V ^ {1}} Y[1] +
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<!-- MATH
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\begin{displaymath}
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=
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{{(VZ)} ^ {0}} X[0] +
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=
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{{(VZ)} ^ {0}} X[0] +
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\cdots +
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<!-- MATH
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\begin{displaymath}
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= {\cal FT} \left \{ X[n] \right \} (k - {{\alpha } \over {\omega}})
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<IMG
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= {\cal FT} \left \{ X[n] \right \} (k - {{\alpha } \over {\omega}})
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<P></P>
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We therefore get the Phase Shift Formula for Fourier Transforms:
|
||
|
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<A NAME="12464"></A>
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<BR><P></P>
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<DIV ALIGN="CENTER">
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<!-- MATH
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||
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\begin{displaymath}
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{\cal FT} \left \{ (\cos(\alpha) + i \sin(\alpha)) X[n] \right \} (k) =
|
||
|
|
{\cal FT} \left \{ X[n] \right \} (k - {{\alpha N} \over {2 \pi}})
|
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\end{displaymath}
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-->
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<IMG
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WIDTH="396" HEIGHT="39" BORDER="0"
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ALT="\begin{displaymath}
|
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{\cal FT} \left \{ (\cos(\alpha) + i \sin(\alpha)) X[n] \ri...
|
||
|
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...l FT} \left \{ X[n] \right \} (k - {{\alpha N} \over {2 \pi}})
|
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\end{displaymath}">
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</DIV>
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<P></P>
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<P>
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<HR>
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<BR>
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<B> Next:</B> <A NAME="tex2html3105"
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HREF="node170.html">Fourier transform of a</A>
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<B> Up:</B> <A NAME="tex2html3099"
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HREF="node167.html">Properties of Fourier transforms</A>
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<B> Previous:</B> <A NAME="tex2html3093"
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HREF="node168.html">Fourier transform of DC</A>
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<B> <A NAME="tex2html3101"
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HREF="node4.html">Contents</A></B>
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|
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<B> <A NAME="tex2html3103"
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HREF="node201.html">Index</A></B>
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<ADDRESS>
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|
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Miller Puckette
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|
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2006-12-30
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</ADDRESS>
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</HTML>
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