z 变换
引入| English | 释义 | 拼音 |
|---|---|---|
| z-transform | z 变换 | z biàn huàn |
Choosing the coefficients
- A digital equaliser ships as a short list of coefficients. Change one number and the bass, the treble or the stability changes.
- The z-transform is how those numbers are chosen. It does for discrete time what the Laplace transform does for circuits.
The z-transform
- The z-transform z 变换 is $X(z) = \displaystyle\sum_{n} x[n]\,z^{-n}$. A delay of one sample multiplies by $z^{-1}$.
- The key pair: $a^{n}u[n] \leftrightarrow \dfrac{1}{1 - az^{-1}}$, a geometric series that converges for $|z| > |a|$.
In the z-domain, delaying a signal by one sample multiplies its transform by z^(−1).
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Each term $x[n-1]z^{-n}$ re-indexes to $z^{-1}\,x[m]z^{-m}$. That is why difference equations become algebra in $z$.
Poles inside the unit circle
- The series converges outside a circle through the pole. A causal system is stable when that region includes the unit circle, so every pole must satisfy $|p| < 1$.
- The unit circle $z = e^{j\omega}$ is where the frequency response lives: angle is frequency, from DC at $z = 1$ to half the sampling rate at $z = -1$.
A causal system has poles at z = 0.9 and z = −1.2. It is
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Every pole must lie inside the unit circle. The pole at −1.2 gives a mode $(-1.2)^{n}$ that grows while flipping sign.
Match each point on the unit circle to the frequency it represents.
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On $z = e^{j\omega}$ the angle is the frequency: 0 at DC, $\dfrac{\pi}{2}$ at $\dfrac{f_{s}}{4}$, $\pi$ at $\dfrac{f_{s}}{2}$.
Worked example: the step response
- $H(z) = \dfrac{1}{1 - 0.5z^{-1}}$ has its pole at 0.5: stable. With a step input, $Y(z) = \dfrac{z^{2}}{(z - 0.5)(z - 1)}$.
- Partial fractions give $y[n] = 2(1 - 0.5^{n+1})$. Check: $y[0] = 1$, as the iteration gave, and $y[\infty] = 2$, the DC gain.
Using y[n] = 2(1 − 0.5^(n+1)), what is y[2]?
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$2(1 - 0.125) = 1.75$, matching the iteration $1, 1.5, 1.75$.
From the s-plane to the z-plane
- Sampling every $T$ seconds maps $s$ to $z = e^{sT}$. The $j\omega$ axis wraps onto the unit circle, and a decay rate $\sigma$ becomes a radius $e^{\sigma T}$.
- The left half-plane maps inside the unit circle, which is why both stability rules agree.
Under z = e^(sT), the left half of the s-plane maps to
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$\text{Re}(s) < 0$ gives $|z| = e^{\sigma T} < 1$. So the two stability rules are one rule.
Worked example: a pole's time constant
- With $T = 1$ ms, the pole at $z = 0.5$ corresponds to $\sigma = \dfrac{\ln 0.5}{0.001} = -693$ per second: a time constant of 1.44 ms.
- Check: the gap halves every sample, and $0.5 = e^{-0.693}$ per 1 ms sample is the same decay.
A pole at z = 0.9 with T = 1 ms. What is the time constant, in ms?
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$\sigma = \dfrac{\ln 0.9}{0.001} = -105.4$ per second, so $\tau = \dfrac{1}{105.4} = 9.49$ ms. A pole nearer the circle decays more slowly.
Three common slips
- $H(z)$ without its ROC. The same expression can describe a causal or an anti-causal system.
- Reading frequency off the real axis. The frequency response is on $|z| = 1$.
- Mixing $z$ and $z^{-1}$. With $z^{-1}$ as the delay, causal transforms converge outside the poles.
H(z) = 1/(1 − 0.5z^(−1)) with ROC |z| < 0.5 describes
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An ROC inside the pole gives $h[n] = -0.5^{n}u[-n-1]$, nonzero only for negative $n$. The expression alone does not fix the system; the ROC does.
Recap
- $a^{n}u[n] \leftrightarrow \dfrac{1}{1 - az^{-1}}$ for $|z| > |a|$; $z^{-1}$ is one sample of delay.
- Causal and stable: all poles inside the unit circle.
- $z = e^{sT}$: frequency becomes angle, decay becomes radius.