ve216_l17.md

VE216 Lecture 17

Discrete Time Frequency Representations

Complex Geometric Sequences

For the DT LTI $h[n]$ with input $x[n] = z^n$, we get $\begin{align}y[n] = (h*x)[n] = \sum^\infty_{k=-\infty} h[k]z^{n-k} = z^n\sum^\infty_{k=-\infty}h[k]z^{-k} = z^nH(z)\end{align}$

So $\begin{align}H(z) = \sum^\infty_{k=-\infty}h[k]z^{-k}\end{align}$, this is for DT Transform.

Remember CT Transform $\begin{align}H(s) = \int^\infty_{-\infty} h(t)e^{-st}dt\end{align}$

Rational System Functions

We can derive $\sum z^{-k} Y = \sum z^{-p} Z$, then we can derive the $H(z) X = Y$ here.

DT Vector Diagram

$\begin{align} H(z_0) = K\frac{\sum^{k_q}{k=0} (z_0 - q_k)}{\sum^{r_p}{r=0}(z_0 - p_r)} \end{align}$, very similar to CT Vector diagrams.

$\begin{align} |H(z_0)| = |K| \frac{\sum^{k_q}{k=0} |z_0 - q_k|}{\sum^{r_p}{r=0}|z_0 - p_r|} \end{align}$ and $\angle H(z_0) = \angle K + \sum^{k_q}{k=0} \angle (z_0-q_k) - \sum^{r_p}{r=0} \angle(z_0 - p_r)$.

DT Frequency Response

$\begin{align}x[n] = \frac 1 2 (e^{j\Omega_0 n} + e^{-j\Omega_0 n}) \leftrightarrow y [n] &= \frac 1 2(H(e^{j\Omega_0}) e^{j\Omega_0 n} + H(e^{-j\Omega_0}) e^{-j\Omega_0 n} )\&=\text{Re}{H(e^{j\Omega_0}) e^{j\Omega_0 n} }\&=\text{Re}{|H(e^{j\Omega_0}) |e^{j\angle H(e^{j\Omega_0})}e^{j\Omega_0 n} }\&=|H(e^{j\Omega_0})|\text{cos}(\Omega_0 n + \angle H(e^{j\Omega_0}))\end{align}$

CT DT Frequency Responses Difference

• CT Frequency Response: $H(s)$ on imaginary axis, $s = j\omega$
• DT Frequency Response: $H(z)$ on unit circle, $z = e^{j\Omega}$

DT Periodicity

Since $e^{j\Omega_2} = e^{j(\Omega_1 + 2\pi k)} = e^{j\Omega_1}$, then the "highest" DT frequency $\Omega = \pi$

DT Fourier Series

$\begin{align} x[n] = x[n + N] = \sum^{N-1}_{k=0} a_k e^{kj\Omega_0n} \end{align}$ with $\Omega_0 = \frac {2\pi}N$

Orthogonality

$\begin{align} \sum^{N-1}{n=0} e^{j\Omega_0 kn}e^{-j\Omega_0ln} = \sum^{N-1}{n=0} e^{j\Omega_0 (k-l)n} = \begin{cases} N &k=l\0 &k\neq l \end{cases} = N \delta[k-l] \end{align}$

So we get $\begin{align} a_k = \frac 1 N \sum^{N-1}_{n=0}x[n]e^{-jk\Omega_0 n} \end{align}$

• Analysis equation: $\begin{align} a_k = a_{k+N} = \frac 1 N \sum_{n=}x[n]e^{-jk\Omega_0 n} \end{align}$ $\Omega_0 = \frac {2\pi}N$
• Synthesis equation: $\begin{align}x[n] = x[n+N] =\sum_{n=}a_k e^{jk\Omega_0 n} \end{align}$