Week 13

We know that the frequency response of a system is the fourier transform of the impulse response. When we pass a signal through such a system, we get a product the the frequency responses in the frequency domain. In other words, convolution in the time domain corresponds to multiplication in the frequency domain:

$$ y(n) = (x*h)(n) \ftp Y(\omega) = X(\omega)H(\omega) $$

This tells us the the frequency response of the system can be described as a ratio of the input and output signal frequency responses:

$$ H(\omega) = \frac{Y(\omega)}{X(\omega)} $$


\begin{example}
Find the frequency response of the causal linear constant coefficient difference equation (LCCDE) given by
$$ \sum \limits_{k=0}^{N}a_ky(n-k) = \sum \limits_{\ell=0}^{M}b_\ell x(n-\ell) $$

We can use the following properties of the DTFT:

\begin{align*}
x(n) &\ftp X(\omega) \\
x(n-\ell) &\ftp e^{-i\ell\omega}X(\omega) \\
y(n-k) &\ftp e^{-i k \omega}Y(\omega) \\
\end{align*}

Now we can solve the LCCDE:

\begin{align*}
\sum \limits_{k=0}^{N}a_ky(n-k) &= \sum \limits_{\ell=0}^{M}b_\ell x(n-\ell) \\
\sum \limits_{k=0}^{N}a_ke^{-i k \omega}Y(\omega) &= \sum \limits_{\ell=0}^{M}b_\ell e^{-i \ell \omega}X(\omega) \\
Y(\omega) \lr{\sum \limits_{k=0}^{N}a_ke^{-i k \omega}} &= X(\omega) \lr{\sum \limits_{\ell=0}^{M}b_\ell e^{-i \ell \omega}} \\
\frac{Y(\omega)}{X(\omega)} &= \frac{\sum \limits_{\ell=0}^{M}b_\ell e^{-i \ell \omega}}{\sum \limits_{k=0}^{N}a_ke^{-i k \omega}} \\
H(\omega) &= \frac{\sum \limits_{\ell=0}^{M}b_\ell e^{-i \ell \omega}}{\sum \limits_{k=0}^{N}a_ke^{-i k \omega}} \\
\end{align*}

It turns out that $H(\omega)$ is a ratio of polynomials in $e^{i\omega}$. This is a characteristic of systems that have LCCDEs. Note that if we define $z = e^{i\omega}$, we can write this expression as 

$$ \hat{H}(z) = \frac{\sum \limits_{\ell=0}^{M}b_\ell z^{-\ell}}{\sum \limits_{k=0}^{N}a_k z^{-k}} = \frac{P(z)}{Q(z)}  $$

\end{example}

\begin{example}
What is the LCCDE of the frequency response given by $H(\omega) = \frac{1}{1-\alpha e^{-i\omega}}$?

\begin{align*}
H(\omega) &= \frac{1}{1-\alpha e^{-i\omega}} \\
\frac{Y(\omega)}{X(\omega)} &= \frac{1}{1-\alpha e^{-i\omega}} \\
Y(\omega) - \alpha e^{-i\omega}Y(\omega) &= X(\omega) \\
y(n) - \alpha y(n-1) &= x(n) \quad \mbox{inverse fourier transform of both sides} \\
\end{align*}

\end{example}


Causal LCCDEs

This \emph{Delay-Adder-Gain} (DAG) block diagram represents a system with 3 components required to implement an LCCDE given by $y(n) = \alpha y(n-1) + x(n)$.  The components required are adders, multipliers, and delay blocks. The minimum number of delay blocks is the order of the filter.


%\includegraphics{images/feedback/dag.ps}

\begin{center}
\begin{pspicture}(0,-3.5)(8,2)

\rput(0,0){$x$}
\rput(7.2,0){$y$}
\rput(5.5,-1.25){$z^{-1}$}
\rput(3.5,-2){$\alpha$}
\rput(1.5,-2.5){$H$}

% plus or minus for adder
\rput(2.1,-0.5){$-$}
\rput(1.45,0.35){$+$}

\psframe(1,-3)(6.5,1)

\pspolygon(4,-1.5)(4,-2.5)(2.5,-2)

\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(0.25,0)(1.5,0)
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\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(5.5,-2)(4,-2)
\psline[linewidth=1.25 pt](2.5,-2)(1.75,-2)
\psline[linewidth=1.25 pt, arrowscale=1.5]{->}(2,0)(7,0)
\psline[linewidth=1.25 pt](5.5,-1.7)(5.5,-2)
\psline[linewidth=1.25 pt]{->}(5.5,0)(5.5,-0.75)

\psframe(5, -1.7)(6,-0.75)

\pscircle(1.75,0){0.25}
\psline(1.5,0)(2,0)
\psline(1.75,0.25)(1.75,-0.25)

\end{pspicture}
\end{center}


Notice that the DAG block diagram shows the delay block with $z^{-1}$. This is because $y(n-1) \ftp e^{-i\omega}Y(\omega) = z^{-1}Y(\omega)$. The DAG shows a system that is rational in $z=e^{i\omega}$. Here is an example of a signal whose DTFT is not rational in $z=e^{i\omega}$:

$$ x(n) = \frac{\sin(An)}{\pi n} $$

The fourier transform of this signal is the ideal low-pass filter:


%\includegraphics{images/idealpass/idealpass.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(1,-1)(3,1.3)

%axis
\psline(1,0)(3,0)
% 0 
\psline(2,0.1)(2,-0.1)

% pi and -pi
\psline(1,0.1)(1,-0.1)
\rput(1,-0.25){$-\pi$}
\psline(3,0.1)(3,-0.1)
\rput(3,-0.25){$\pi$}

\rput(3.3,-0.25){$\omega$}
\rput(3,1){$H(\omega)$}
\rput(1.35,1){$(1)$}

\psline(1.5,0)(1.5,1)
\psline(1.5,1)(2.5,1)
\psline(2.5,1)(2.5,0)

\rput(2,-0.25){0}
\rput(1.5,-0.25){-A}
\rput(2.5,-0.25){A}



\end{pspicture}
\end{center}

This system is an "ideal low-pass" filter because we cannot implement it. The cuttoff/rolloff can never be vertical like the filter shows in the frequency domain. If the filter has an LCCDE, it is implementable, but otherwise we cannot.

Delay-Adder-Gain Block Diagrams

We know $x(n) = \alpha u(n) \ftp \frac{1}{1-\alpha e^{-i\omega}}$ where $\abs{\alpha} \lt 1$. What about a translated version of $x(n)$? We know that $x(n - n_0) \ftp e^{-i\omega n_0}X(\omega)$, so we can determine that 

$$ x(n-n_0) = \alpha^{n-n_0}u(n-n_0) \ftp \frac{e^{-i\omega n_0}}{1-\alpha e^{-i\omega}} $$

Any shifted version of this signal will produce a signal that is rational in $z=e^{i\omega}$.

\begin{example}

Given an impulse response $g(n) = -2^nu(-n-1)$, find the impulse response $G(\omega)$.



First note that $u(-n-1) = \begin{cases} 1 & -n-1\geq0 \\ 0 & \mbox{otherwise} \end{cases}$, which translates to $u(-n-1) = \begin{cases} 1 & n\leq-1 \\ 0 & \mbox{otherwise} \end{cases}$. Then we can start to solve for the frequency response:

\begin{align*}
G(\omega) &= \sum \limits_{n=-\infty}^{\infty}-2^n u(-n-1) e^{-i\omega n} \\
G(\omega) &= \sum \limits_{n=-\infty}^{-1}-2^n e^{-i\omega n} \\
G(\omega) &= -\sum \limits_{n=-\infty}^{-1}\lr{2 e^{-i\omega }}^n \\
G(\omega) &= -\sum \limits_{m=1}^{\infty}\lr{2 e^{-i\omega }}^{-m} \quad \mbox{let }m=-n\\
G(\omega) &= -\sum \limits_{m=1}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^m \\
G(\omega) &= -\sum \limits_{m=0}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^{m+1} \quad \mbox{shift limits}\\
G(\omega) &= -\frac{1}{2}e^{i\omega}\sum \limits_{m=0}^{\infty}\lr{\frac{1}{2} e^{i\omega }}^{m} \\
G(\omega) &= -\frac{1}{2}e^{i\omega} \frac{1}{1-\frac{1}{2}e^{i\omega}} \\
G(\omega) &= \frac{1}{1-2e^{-i\omega}} \\
\end{align*}

Note that we could have solved for $g(-n) = -\frac{1}{2}^nu(n-1)$, and then used the DTFT property that $g(-n) \ftp G(-\omega)$ to solve the problem.

\end{example}

Rational DTFTs

What happens when we differentiate in the frequency domain?

\begin{align*}
\frac{d H(\omega)}{d\omega} &= \frac{d}{d\omega}\sum \limits_{n=-\infty}^{\infty}h(n)e^{-i\omega n} \\
\frac{d H(\omega)}{d\omega} &= \sum \limits_{n=-\infty}^{\infty}h(n) (-in) e^{-i\omega n} \\
\frac{d H(\omega)}{d\omega} &= -i \cdot \sum \limits_{n=-\infty}^{\infty}n \cdot h(n) e^{-i\omega n} \\
i \cdot \frac{d H(\omega)}{d\omega} &= \sum \limits_{n=-\infty}^{\infty}n \cdot h(n) e^{-i\omega n} \\
\end{align*}

This gives us the DTFT differentiation property:

\begin{nicebox}
$$ nh(n) \ftp  i\frac{d H(\omega)}{d\omega} $$
\end{nicebox}

\begin{example}

$f(n) = n \alpha^n u(n)$ for $\abs{\alpha} \lt 1$. Find the frequency response.

\begin{align*}
F(\omega) &= \ftrans{nh(n)} \\
F(\omega) &= i\frac{d H(\omega)}{d\omega} \\
F(\omega) &= i\frac{d}{d\omega} \lr{\frac{1}{1-\alpha e^{-i\omega}}} \\
F(\omega) &= i \cdot \frac{0 - i\alpha e^{-i\omega}}{\lr{1-\alpha e^{-i\omega}}^2} \\
F(\omega) &= \frac{\alpha e^{-i\omega}}{\lr{1-\alpha e^{-i\omega}}^2} \\
\end{align*}

\end{example}

13.2 DT-LTI Systems