The z-transform

A growing signal, $1.25^nu[n]$, divided by a faster growth, $r^n$. Before you press play, decide: which $r$ is the border?

Divide by a faster growth

Some sampled signals grow, such as $x[n]=1.25^nu[n]$, which grows by a quarter each sample. The sum of The DTFT (12.2) has no value for it, because $\sum\lvert x[n]\rvert$ has no limit. The Laplace transform (9.1) met the same problem in continuous time and divided by a faster exponential first.

I do the same here with a sequence, as the picture at the top of this page does: divide by $r^n$, a test sequence that grows when $r$ is above 1. Look at $x[n]r^{-n}=(1.25/r)^n$. This is a geometric sequence with ratio $1.25/r$. It dies out once $r$ is past 1.25, and then the divided samples add to a number.

At $r=1$ nothing is divided. Each sample is 1.25 times the last, the sum has no limit, and this $x$ has no DTFT. At $r=1.25$ dividing by $1.25^n$ exactly cancels the growth: every stem is 1, so each added sample raises the sum by 1, and the sum still has no limit. At $r=1.5$ each divided sample is $1.25/1.5=5/6$ of the last, and the sum settles at 6.00. At $r=2.5$ each is half the last, and the sum is 2.00. Every $r$ above 1.25 gives a number: the border is $x$‘s own growth, 1.25. When the clip ends, drag the marker on the strip to choose $r$ yourself.

The definition

The numbers agree with a geometric sum. For $r>1.25$, the divided samples add to

At $r=1.5$ this is 6, at $r=2$ it is 2.667, and at $r=2.5$ it is 2. For $r\le1.25$ the ratio is 1 or more and the sum has no limit.

Now let the test sequence spin as well. Write a point of the plane as $z=re^{j\Omega}$: its length is $r$ and its angle is $\Omega$ in rad/sample, as in Frequency in discrete time (12.1). Dividing by $z^n$ is multiplying by $z^{-n}$, and the z-transform adds up the result:

Because $z^{-n}=r^{-n}e^{-j\Omega n}$, at a fixed $r$ the sum is the DTFT of $x[n]r^{-n}$. So the new transform is the old one, applied after dividing by $r^n$, the same relation that 9.1 found between its two transforms.

For $a^nu[n]$ the geometric sum, run to infinity, gives

So $1.25^nu[n]\leftrightarrow1/(1-1.25z^{-1})$ for $\lvert z\rvert > 1.25$, which gives 6 at $z=1.5$ and 2 at $z=2.5$, the sums the first picture reads. The set of $z$ where the sum has a value is the region of convergence, written ROC. Where the formula blows up, here at $z=1.25$, is a pole, the word of 9.1, and I draw it as a cross. Only $\lvert z\rvert$ decides whether the sum settles, so the ROC is made of whole circles: a ring.

One note on notation. From here on a delay box in a block diagram is labelled $z^{-1}$. The next lesson, Properties and the inverse z-transform (16.2), shows why.

Walk round the unit circle

The plane of $z$ is the plane of positions of First- and second-order systems (6.3), not the plane of rates. A point $z$ is what each sample is multiplied by, and its axes are “real part” and “imaginary part”. The points with $r=1$ form the unit circle, which is the frequency circle of 12.1: the point at angle $\Omega$ is $e^{j\Omega}$. On that circle $X(z)$ is the DTFT.

I use the signal of 12.2, $x[n]=0.8^nu[n]$, so that $X(z)=1/(1-0.8z^{-1})$ with ROC $\lvert z\rvert > 0.8$. The picture below shades the plane by $\lvert X(z)\rvert$, as 9.1 did, and walks a dot round the unit circle. Before you press play, decide: where on the circle will the profile be tallest?

The profile is tallest at $\Omega=0$, where $z=1$ is the point of the circle nearest the pole: $\lvert X\rvert=5.00$, the sum $1+0.8+0.64+\cdots$, as in 12.2. At $\Omega=0.5\pi$, $z=j$ is farther from the pole and $\lvert X\rvert=0.78$. At $\Omega=\pi$, $z=-1$ is the far side and $\lvert X\rvert=0.56$. This is 12.2’s size curve, read off the plane. Back at $z=1$ after one turn, the curve repeats every $2\pi$. When the clip ends, drag the dot round the circle to read $\lvert X\rvert$ at any $\Omega$.

The off-circle values read from the same formula. Away from the circle at $z=0.9$ it is 9, at $z=1.2$ it is 3, at $z=-1.5$ it is 0.652, and at $z=1.5j$ it is 0.882.

On the unit circle $r=1$, so $X(e^{j\Omega})$ is $X(z)$ at $z=e^{j\Omega}$. The DTFT exists exactly when the unit circle lies in the ROC, which is 12.2’s condition that $\sum\lvert x[n]\rvert$ is finite. For a real signal the lower half of the walk mirrors the upper half. The signal $1.25^nu[n]$ has ROC $\lvert z\rvert > 1.25$, which misses the circle: it has no DTFT, as the first picture found at $r=1$.

One formula, two signals

Is the formula enough to name the signal? Take the left-sided signal $-0.8^nu[-n-1]$, which is nonzero only for $n\le-1$. Its samples at $n=-1,-2,-3,-4$ are $-1.25$, $-1.5625$, $-1.953$ and $-2.441$, so it grows going back in time. Its sum is $-\sum_{m\ge1}(z/0.8)^m$, which has a value only for $\lvert z\rvert < 0.8$, and there it equals $1/(1-0.8z^{-1})$ again. At $z=0.5$ the sum is $-1.667$, and the formula gives $1/(1-0.8/0.5)=-1.667$ too.

The picture below puts the two on one page: both have $X(z)=1/(1-0.8z^{-1})$, and only the region of convergence tells them apart. Before you tap, decide: will the unit circle be in the second region?

The right-sided $0.8^nu[n]$ dies away, and its sum has a value outside the circle $\lvert z\rvert=0.8$, a region that holds the unit circle. The left-sided $-0.8^nu[-n-1]$ grows going back, and its sum has a value only inside $\lvert z\rvert=0.8$, yet the formula is the same. The formula alone does not say which signal you have; the region does. The unit circle lies outside the second region, so $-0.8^nu[-n-1]$ has no DTFT.

What a region looks like

The same few rules hold for every signal in this lesson. I state them for sums of exponentials like these, and the next lessons use them again.

1. The ROC is a ring centred on 0, since only $\lvert z\rvert$ matters.
2. It contains no pole.
3. A right-sided signal, one that is zero before some sample, has an ROC outside its outermost pole. If it is also causal (zero before $n=0$, as in 5.4), the ROC runs out to infinity.
4. A left-sided signal has an ROC inside its innermost pole.
5. A two-sided signal has a ring between poles, or no ROC at all.
6. A signal of finite length has the whole plane as its ROC, except perhaps $z=0$ or infinity.

A two-sided example is $0.5^{\lvert n\rvert}$. Its right half $0.5^nu[n]$ needs $\lvert z\rvert > 0.5$, and its left half needs $\lvert z\rvert < 2$, so

The ring contains the unit circle, so there is a DTFT. Its size is 3 at $\Omega=0$, 0.6 at $0.5\pi$ and 0.333 at $\pi$.

The table of pairs is in the reference card. Three rows need more than the geometric sum. The step $u[n]$ is the case $a=1$. The ramp times an exponential, $n\,a^nu[n]$, is derived in 16.2. The damped cosine and sine come from writing each as two spinning exponentials with $a=re^{\pm j\theta}$. I checked those two rows against a direct sum at $z=1.3e^{j0.4}$, with $r=0.9$ and $\theta=\pi/4$, and the two sides agree to about $10^{-15}$. At $z=2$ the step gives 2 and $n\,0.5^nu[n]$ gives 0.444.

From the s-plane to the z-plane

Sampling $e^{st}$ every $T_s$ seconds gives $e^{snT_s}=(e^{sT_s})^n$. So a rate $s$ becomes the multiplier $z=e^{sT_s}$. The $a=e^{-T_s/RC}$ of Differential equations and analog systems (6.2) is one case of it, with $s=-1/RC$.

The figure uses $T_s=0.25$ s, so $f_s=4$ Hz. The first panel is the plane of rates of Poles, zeros and the s-plane (9.3), and the second is the plane of $z$. Five marks are matched. The rate $s=0$ becomes $z=1$ (a filled dot). The rate $s=-1$ becomes $z=0.779$ (a filled square). The pair $s=-1\pm2j$ becomes $0.779$ at angles $\pm28.6°$ (crosses). The rate $s=j2\pi$, which is 1 Hz, becomes $z=j$ (a filled diamond).

The left half of the s-plane lands inside the unit circle, because $\lvert e^{sT_s}\rvert=e^{\sigma T_s}$ is below 1 when $\sigma$ is negative. The spin-rate axis wraps round the circle. A spin rate of $j4\pi$, 2 Hz and so $f_s/2$, lands at $z=-1$. A positive rate such as $s=0.5$ lands outside the circle, at $1.133$. Spin rates $2\pi/T_s=25.13$ rad/s apart land on the same $z$, which is aliasing seen from the plane.

Worked example

1. Growing. $1.25^nu[n]\leftrightarrow1/(1-1.25z^{-1})$, ROC $\lvert z\rvert > 1.25$. The value is 6 at $z=1.5$, 2.667 at $z=2$, 2 at $z=2.5$ and 1.714 at $z=3$. The first 16 divided samples at $r=1.5$ add to 5.675, and the rest add 0.325. At $r=1$ the 16th stem is $1.25^{15}=28.4$.
2. On the circle. For $0.8^nu[n]$, $\lvert X\rvert$ at $\Omega=0$, $0.25\pi$, $0.5\pi$, $0.75\pi$ and $\pi$ is 5, 1.402, 0.781, 0.601 and 0.556. The angles are $0°$, $-52.48°$, $-38.66°$, $-19.86°$ and $0°$, as in 12.2. Off the circle, $\lvert X(0.9)\rvert=9$, $\lvert X(1.2)\rvert=3$, $\lvert X(-1.5)\rvert=0.652$ and $\lvert X(1.5j)\rvert=0.882$.
3. Left-sided. $-0.8^nu[-n-1]$ at $n=-1,-2,-3,-4$ is $-1.25$, $-1.5625$, $-1.953$ and $-2.441$. At $z=0.5$ its sum is $-\sum_{m\ge1}0.625^m=-1.667$, which is $1/(1-0.8/0.5)$.
4. Two-sided. $0.5^{\lvert n\rvert}\leftrightarrow\dfrac{-1.5z^{-1}}{1-2.5z^{-1}+z^{-2}}$ with ROC $0.5 < \lvert z\rvert < 2$. On the circle it is 3 at $\Omega=0$, 0.6 at $0.5\pi$ and 0.333 at $\pi$.
5. s to z with $T_s=0.25$ s. $s=-1$ gives $0.7788$. The pair $s=-1\pm2j$ gives $0.7788$ at $\pm28.65°$. A rate of 1 Hz gives $j$, and 2 Hz gives $-1$. The rate $s=0.5$ gives $1.133$, outside the circle.

Where you’ll meet this

Every transfer function of a digital filter is a ratio of polynomials in $z^{-1}$, and its poles and zeros are points of this plane. Transfer functions, poles and zeros (16.3) reads a filter’s behaviour from their positions, just as 9.3 did for analog systems. Stability and causality (16.4) turns this lesson’s rules about the ROC into a test: a causal system is stable when its ROC contains the unit circle.

Starting values come from the one-sided sum of The unilateral z-transform (16.5), and the map $z=e^{sT_s}$ returns in filter design, where it turns analog designs into digital ones.

Reference card