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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 2006-03-03