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Shifts and phase changes

Section 7.2 showed how time-shifting a signal changes the phases of its sinusoidal components, and section 8.4.3 showed how multiplying a signal by a complex sinusoid shifts its component frequencies. These two effects each correspond to an identity involving the Fourier transform.

First we consider a time shift. If $X[n]$, as usual, is a complex-valued signal that repeats every $N$ samples, let $Y[n]$ be $X[n]$ delayed $d$ samples:

\begin{displaymath}
Y[n] = X[n-d]
\end{displaymath}

which also repeats every $N$ samples since $X$ does. We can reduce the Fourier transform of $Y[n]$ this way:

\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}


\begin{displaymath}
=
{V ^ {0}} X[-d] +
{V ^ {1}} X[-d+1] +
\cdots +
{V ^ {N-1}} X[-d+N-1]
\end{displaymath}


\begin{displaymath}
=
{V ^ {d}} X[0] +
{V ^ {d+1}} X[1] +
\cdots +
{V ^ {d+N-1}} X[N-1]
\end{displaymath}


\begin{displaymath}
= {V^d} \left (
{V ^ {d}} X[0] +
{V ^ {d+1}} X[1] +
\cdots +
{V ^ {d+N-1}} X[N-1]
\right )
\end{displaymath}


\begin{displaymath}
= {V^d} {\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}

We therefore get the TIME SHIFT FORMULA FOR FOURIER TRANSFORMS:

\begin{displaymath}
{\cal FT} \left \{ Y[n] \right \} (k) =
\left (\cos(-dk\o...
...sin(-dk\omega) \right )
{\cal FT} \left \{ X[n] \right \} (k)
\end{displaymath}

So the Fourier transform of $Y[n]$ is a phase term times the Fourier transform of $X[n]$. The phase is changed by $-dk\omega$, a multiple of the frequency $k$.

Now suppose instead that we change our starting signal $X[n]$ by multiplying it by a complex exponential $Z^n$ with angular frequency $\alpha $:

\begin{displaymath}
Y[n] = {Z^n} X[n]
\end{displaymath}


\begin{displaymath}
Z = \cos(\alpha) + i \sin(\alpha)
\end{displaymath}

The Fourier transform is:

\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}


\begin{displaymath}
=
{V ^ {0}} X[0] +
{V ^ {1}} Z X[1] +
\cdots +
{V ^ {N-1}} {Z^{N-1}} X[N-1]
\end{displaymath}


\begin{displaymath}
=
{{(VZ)} ^ {0}} X[0] +
{{(VZ)} ^ {1}} X[1] +
\cdots +
{{(VZ)} ^ {N-1}} X[N-1]
\end{displaymath}


\begin{displaymath}
= {\cal FT} \left \{ X[n] \right \} (k + {{\alpha } \over {\omega}})
\end{displaymath}

We therefore get the PHASE SHIFT FORMULA FOR FOURIER TRANSFORMS:

\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}


next up previous contents index
Next: Fourier transform of a Up: Properties of Fourier transforms Previous: Fourier transform of DC   Contents   Index
Miller Puckette 2005-04-01