The z-transform
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| unit circle/ˈjuːnɪt ˈsɜːkl/ | A circle with radius one, usually centred at the origin. |
The unit circle is the frequency axis
- $X(z) = \sum x[n]z^{-n}$; the pair $a^{n}u[n] \leftrightarrow 1/(1-az^{-1})$ with ROC $|z| > |a|$; poles inside the unit circle mean stability.
$H(z) = 1/(1-0.5z^{-1})$: step $y[n] = 2(1-0.5^{n+1})$ — y[0] = 1 ✓ matches the iteration, limit 2 ✓.
{exercise:uni-ee0671-pole}
H(z) = 1/(1 − 0.5z^{-1}) is stable because —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
The geometric ROC |z| > 0.5 contains the unit circle (|z| = 1), so the frequency response exists and the mode 0.5^n decays.
Quoting H(z) without its ROC —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
1/(1 − 0.5z^{-1}) with |z| > 0.5 is causal-decaying; with |z| < 0.5 it is anti-causal-growing. The rational function alone does not choose.
One river, two banks
- $z = e^{sT}$ maps decay to radius and frequency to angle: the pole at 0.5 with T = 1 ms is $\ln(0.5)/0.001 = -693\ \mathrm{s}^{-1}$ — a 1.44 ms time constant, the same decay the iteration showed.
{exercise:uni-ee0672-map}
A pole at z = 0.5 with T = 1 ms corresponds to a continuous decay rate of about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
ln(0.5)/0.001 = −693 s^{-1}; the gap halves per sample, and 0.693/T per second is the same rate the iteration showed.