Week 13

Given $x(n) \ftp X(\omega)$, what is the transform of a time-shifted version $x(n-n_0)$?


We can use the DTFT directly:

\begin{align*}
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty}\hat{x}(n)e^{-i\omega n} \\
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n-n_0)e^{-i\omega n} \\
\hat{X}(\omega) &= \sum \limits_{k=-\infty}^{\infty}x(k)e^{-i\omega (k+n_0)} \quad \mbox{let }k=n-n_0 \\
\hat{X}(\omega) &= \sum \limits_{k=-\infty}^{\infty}x(k)e^{-i\omega k}e^{-i\omega n_0} \\
\hat{X}(\omega) &= X(\omega)e^{-i\omega n_0} \\
\end{align*}

Therefore, a time-shifted version of the signal in the time domain by $n_0$ corresponds to multiplication by $e^{-i\omega n_0}$ in the frequency domain:

\begin{nicebox}
$$ x(n-n_0) \ftp X(\omega)e^{-i\omega n_0} $$
\end{nicebox}



Translation - Time-Shifting

Given $x(n) \ftp X(\omega)$, what is the inverse transform of $X(\omega-\omega_0)$?


We can determine this using the synthesis equation:

\begin{align*}
\hat{x}(n) &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle}\hat{X}(\omega)e^{i\omega n}d\omega \\
\hat{x}(n) &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle}X(\omega-\omega_0)e^{i\omega n}d\omega \\
\hat{x}(n) &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle}X(\Omega)e^{i(\Omega + \omega_0) n}d\Omega \quad \mbox{let } \Omega = \omega-\omega_0 \\
\hat{x}(n) &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle}X(\Omega)e^{i\Omega n}e^{i\omega_0 n}d\Omega \\
\hat{x}(n) &= \lr{\frac{1}{2\pi} \int_{\langle 2 \pi \rangle}X(\Omega)e^{i\Omega n}d\Omega}e^{i\omega_0 n} \\
\hat{x}(n) &= x(n)e^{i\omega_0 n} \\
\end{align*}

Therefore frequency-shifting in the frequency domain by $\omega_0$ corresponds to multiplication in the time domain by $e^{i\omega_0 n}$

\begin{nicebox}
$$ X(\omega - \omega_0) \iftp x(n)e^{i\omega_0 n} $$ 
\end{nicebox}


Frequency-Shifting

Given $x(n) \ftp X(\omega)$, what is the transform of a time-reversed version $x(-n)$?


We can use the DTFT directly:
\begin{align*}
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty} \hat{x}(n) e^{-i\omega n} \\
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty} x(-n) e^{-i\omega n} \\
\hat{X}(\omega) &= \sum \limits_{m=-\infty}^{\infty} x(m) e^{i\omega m} \quad \mbox{let }m=-n\\
\hat{X}(\omega) &= \sum \limits_{m=-\infty}^{\infty} x(m) e^{-i(-\omega) m} \\
\hat{X}(\omega) &= X(-\omega)\\
\end{align*}

Therefore time-reversal in the time-domain corresponds to frequency reversal in the frequency domain:

\begin{nicebox}
$$ x(-n) \ftp X(-\omega) $$
\end{nicebox}

Time-Reversal

Given $x(n) \ftp X(\omega)$, what is the transform of the conjugate signal $x^*(n)$?



We can use the DTFT directly:
\begin{align*}
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty}\hat{x}(n)e^{-i\omega n} \\
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty}x^*(n)e^{-i\omega n} \\
\hat{X}^*(\omega) &= \lr{\sum \limits_{n=-\infty}^{\infty}x^*(n)e^{-i\omega n}}^* \\
\hat{X}^*(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n)e^{i\omega n} \\
\hat{X}^*(\omega) &= \sum \limits_{n=-\infty}^{\infty}x(n)e^{-i(-\omega) n} \\
\hat{X}^*(\omega) &= X(-\omega) \\
\hat{X}(\omega) &= X^*(-\omega) \\
\end{align*}

Therefore conjugation in the time domain corresponds to conjugate frequency reversal in the frequency domain:

\begin{nicebox}
$$ x^*(n) \ftp X^*(-\omega) $$ 
\end{nicebox}

Note that you can also factor-out the conjugation:

\begin{align*}
\hat{X}(\omega) &= \sum \limits_{n=-\infty}^{\infty}x^*(n)e^{-i\omega n} \\
\hat{X}(\omega) &= \lr{\sum \limits_{n=-\infty}^{\infty}x(n)e^{i\omega n}}^* \\
\hat{X}(\omega) &= \lr{X(-\omega)}^* \\
\hat{X}(\omega) &= X^*(-\omega) \\
\end{align*}

Conjugation

Given $x(n) \ftp X(\omega)$, what is the transform of $x(n)=e^{i\omega_0 n}$?

We know that this represents a dirac delta in the frequency domain centered at $\omega_0$. This gives us a frequency response of $A\delta(\omega -\omega_0)$ for some constant $A$.

%\includegraphics{images/pulsetrains/freqpulse.ps}

\begin{center}
\begin{pspicture}(-2,-1)(6,1.5)

\psline{->}(-1,0)(6,0)
\rput(6,-0.25){$\omega$}
\rput(6,1){$X(\omega)$}
\rput(6,0.5){$\cdots$}
\rput(-1,0.5){$\cdots$}
\rput(2.8,1){$(A)$}

\psline{->}(0,0)(0,1)
\rput(0,-0.25){$\omega_0-2\pi$}

\psline{->}(2.5,0)(2.5,1)
\rput(2.5,-0.25){$\omega_0$}

\psline{->}(5,0)(5,1)
\rput(5,-0.25){$\omega_0+2\pi$}

\end{pspicture}
\end{center}


So we can use the synthesis equation over an interval of $2\pi$:

\begin{align*}
x(n) &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle } X(\omega) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{1}{2\pi} \int_{\langle 2 \pi \rangle } A \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} \int_{-\pi}^{\pi} \delta(\omega-\omega_0) e^{i\omega n}d\omega \\
e^{i\omega_0 n} &= \frac{A}{2\pi} e^{i\omega_0 n} \\
A &= 2\pi \\
\end{align*}

This tells us that $X(\omega) = 2\pi\delta(\omega - \omega_0)$, but we know that $X(\omega + 2\pi k) = X(\omega) $ $ \forall k \in \Z$, so we must write the frequency response in the form:

$$ X(\omega) = \sum \limits_{k=-\infty}^{\infty} 2\pi \delta(\omega-\omega_0 +2\pi k) $$

If we restricted the domain, for example, $-\pi \leq \omega \lt \pi$, then $X(\omega) = 2\pi\delta(\omega-\omega_0)$. 


Complex Exponential

Given $f(n) \ftp F(\omega)$ and $g(n) \ftp G(\omega)$, what is the transform of a convolved signal $(f*g)(n)$?


We can use a combination of the convolution sum and the DTFT:

\begin{align*}
H(\omega) &= \sum \limits_{n=-\infty}^{\infty} h(n) e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n=-\infty}^{\infty} (f*g)(n) e^{-i\omega n} \\
H(\omega) &= \sum \limits_{n=-\infty}^{\infty} \lr{\sum \limits_{\ell = -\infty}^{\infty}f(\ell)g(n-\ell)}e^{-i\omega n} \\
H(\omega) &= \sum \limits_{\ell=-\infty}^{\infty} \sum \limits_{n = -\infty}^{\infty}f(\ell)g(n-\ell)e^{-i\omega n} \quad \mbox{swap the summations}\\
H(\omega) &= \sum \limits_{\ell=-\infty}^{\infty} f(\ell)\sum \limits_{m = -\infty}^{\infty}g(m)e^{-i\omega (m+\ell)} \quad \mbox{let } m = n -\ell\\
H(\omega) &= \sum \limits_{\ell=-\infty}^{\infty} f(\ell)\sum \limits_{m = -\infty}^{\infty}g(m)e^{-i\omega m} e^{-i\omega\ell} \\
H(\omega) &= \sum \limits_{\ell=-\infty}^{\infty} f(\ell)e^{-i\omega\ell} \sum \limits_{m = -\infty}^{\infty}g(m)e^{-i\omega m} \\
H(\omega) &= F(\omega)G(\omega)
\end{align*}

Therefore we can conclude that convolution in the time-domain corresponds to multiplication in the frequency domain:

\begin{nicebox}
$$ (f*g)(n) \ftp F(\omega)G(\omega) $$
\end{nicebox}


13.1 Properties of the DTFT