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<H2><A ID="SECTION001322000000000000000"></A>
<A ID="sect9.shift"></A>
<BR>
Shifts and phase changes
</H2>
<P>
Section <A HREF="node107.html#sect7.phase">7.2</A> showed how time-shifting a signal changes the
phases of its sinusoidal components, and Section <A HREF="node154.html#sect8.singlesideband">8.4.3</A>
showed how multiplying a signal by a complex sinusoid shifts its component
frequencies. These two effects have corresponding identities
involving the Fourier transform.
<P>
First we consider a time shift. If <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$">, as usual, is a complex-valued
signal that repeats every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples, let <IMG
WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img717.png"
ALT="$Y[n]$"> be <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> delayed <IMG
WIDTH="11" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img28.png"
ALT="$d$">
samples:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Y[n] = X[n-d]
\end{displaymath}
-->
<IMG
WIDTH="112" HEIGHT="28" BORDER="0"
SRC="img671.png"
ALT="\begin{displaymath}
Y[n] = X[n-d]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
which also repeats every <IMG
WIDTH="18" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img3.png"
ALT="$N$"> samples since <IMG
WIDTH="17" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img670.png"
ALT="$X$"> does. We can reduce the Fourier
transform of <IMG
WIDTH="34" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img717.png"
ALT="$Y[n]$"> this way:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT} \left \{ Y[n] \right \} (k) =
{V ^ {0}} Y[0] +
{V ^ {1}} Y[1] +
\cdots +
{V ^ {N-1}} Y[N-1]
\end{displaymath}
-->
<IMG
WIDTH="400" HEIGHT="28" BORDER="0"
SRC="img1101.png"
ALT="\begin{displaymath}
{\cal FT} \left \{ Y[n] \right \} (k) =
{V ^ {0}} Y[0] +
{V ^ {1}} Y[1] +
\cdots +
{V ^ {N-1}} Y[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
=
{V ^ {0}} X[-d] +
{V ^ {1}} X[-d+1] +
\cdots +
{V ^ {N-1}} X[-d+N-1]
\end{displaymath}
-->
<IMG
WIDTH="393" HEIGHT="28" BORDER="0"
SRC="img1102.png"
ALT="\begin{displaymath}
=
{V ^ {0}} X[-d] +
{V ^ {1}} X[-d+1] +
\cdots +
{V ^ {N-1}} X[-d+N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
=
{V ^ {d}} X[0] +
{V ^ {d+1}} X[1] +
\cdots +
{V ^ {d+N-1}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="333" HEIGHT="28" BORDER="0"
SRC="img1103.png"
ALT="\begin{displaymath}
=
{V ^ {d}} X[0] +
{V ^ {d+1}} X[1] +
\cdots +
{V ^ {d+N-1}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {V^d} \left (
{V ^ {0}} X[0] +
{V ^ {1}} X[1] +
\cdots +
{V ^ {N-1}} X[N-1]
\right )
\end{displaymath}
-->
<IMG
WIDTH="336" HEIGHT="30" BORDER="0"
SRC="img1104.png"
ALT="\begin{displaymath}
= {V^d} \left (
{V ^ {0}} X[0] +
{V ^ {1}} X[1] +
\cdots +
{V ^ {N-1}} X[N-1]
\right )
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {V^d} {\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}
-->
<IMG
WIDTH="136" HEIGHT="28" BORDER="0"
SRC="img1105.png"
ALT="\begin{displaymath}
= {V^d} {\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
(The third line is just the second one with the terms summed in a
different order). We therefore get the Time Shift Formula for Fourier
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Transforms: <A ID="12441"></A>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT} \left \{ X[n-d] \right \} (k) =
\left ( {
\parbox[t][0.1in]{0in}{\mbox{}}
\cos(-dk\omega) + i\sin(-dk\omega)
} \right )
{\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}
-->
<IMG
WIDTH="447" HEIGHT="35" BORDER="0"
SRC="img1106.png"
ALT="\begin{displaymath}
{\cal FT} \left \{ X[n-d] \right \} (k) =
\left ( {
\par...
...-dk\omega)
} \right )
{\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
The Fourier transform of <IMG
WIDTH="64" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img1107.png"
ALT="$X[n-d]$"> is a phase term times the Fourier transform
of <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$">. The phase is changed by <IMG
WIDTH="43" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img1108.png"
ALT="$-dk\omega$">, a
linear function of the frequency <IMG
WIDTH="12" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img58.png"
ALT="$k$">.
<P>
Now suppose instead that we change our starting signal <IMG
WIDTH="36" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img669.png"
ALT="$X[n]$"> by multiplying
it by a complex exponential <IMG
WIDTH="24" HEIGHT="14" ALIGN="BOTTOM" BORDER="0"
SRC="img653.png"
ALT="$Z^n$"> with angular frequency <IMG
WIDTH="13" HEIGHT="13" ALIGN="BOTTOM" BORDER="0"
SRC="img7.png"
ALT="$\alpha $">:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Y[n] = {Z^n} X[n]
\end{displaymath}
-->
<IMG
WIDTH="104" HEIGHT="28" BORDER="0"
SRC="img1109.png"
ALT="\begin{displaymath}
Y[n] = {Z^n} X[n]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
Z = \cos(\alpha) + i \sin(\alpha)
\end{displaymath}
-->
<IMG
WIDTH="145" HEIGHT="28" BORDER="0"
SRC="img1110.png"
ALT="\begin{displaymath}
Z = \cos(\alpha) + i \sin(\alpha)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
The Fourier transform is:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT} \left \{ Y[n] \right \} (k) =
{V ^ {0}} Y[0] +
{V ^ {1}} Y[1] +
\cdots +
{V ^ {N-1}} Y[N-1]
\end{displaymath}
-->
<IMG
WIDTH="400" HEIGHT="28" BORDER="0"
SRC="img1101.png"
ALT="\begin{displaymath}
{\cal FT} \left \{ Y[n] \right \} (k) =
{V ^ {0}} Y[0] +
{V ^ {1}} Y[1] +
\cdots +
{V ^ {N-1}} Y[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
=
{V ^ {0}} X[0] +
{V ^ {1}} Z X[1] +
\cdots +
{V ^ {N-1}} {Z^{N-1}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="353" HEIGHT="28" BORDER="0"
SRC="img1111.png"
ALT="\begin{displaymath}
=
{V ^ {0}} X[0] +
{V ^ {1}} Z X[1] +
\cdots +
{V ^ {N-1}} {Z^{N-1}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
=
{{(VZ)} ^ {0}} X[0] +
{{(VZ)} ^ {1}} X[1] +
\cdots +
{{(VZ)} ^ {N-1}} X[N-1]
\end{displaymath}
-->
<IMG
WIDTH="373" HEIGHT="28" BORDER="0"
SRC="img1112.png"
ALT="\begin{displaymath}
=
{{(VZ)} ^ {0}} X[0] +
{{(VZ)} ^ {1}} X[1] +
\cdots +
{{(VZ)} ^ {N-1}} X[N-1]
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
= {\cal FT} \left \{ X[n] \right \} (k - {{\alpha } \over {\omega}})
\end{displaymath}
-->
<IMG
WIDTH="149" HEIGHT="35" BORDER="0"
SRC="img1113.png"
ALT="\begin{displaymath}
= {\cal FT} \left \{ X[n] \right \} (k - {{\alpha } \over {\omega}})
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
We therefore get the Phase Shift Formula for Fourier Transforms:
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<A ID="12464"></A>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
{\cal FT} \left \{ (\cos(\alpha) + i \sin(\alpha)) X[n] \right \} (k) =
{\cal FT} \left \{ X[n] \right \} (k - {{\alpha N} \over {2 \pi}})
\end{displaymath}
-->
<IMG
WIDTH="396" HEIGHT="39" BORDER="0"
SRC="img1114.png"
ALT="\begin{displaymath}
{\cal FT} \left \{ (\cos(\alpha) + i \sin(\alpha)) X[n] \ri...
...l FT} \left \{ X[n] \right \} (k - {{\alpha N} \over {2 \pi}})
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
<P>
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2006-12-30
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