Properties and the inverse z-transform

Where this page ends up: a ratio in $z^{-1}$ split into one geometric sequence per pole. Watch the two sequences add, sample by sample, to the signal.

Six rules, each one line from the sum

The z-transform (16.1) turned a sequence $x[n]$ into a function $X(z)=\sum_n x[n]z^{-n}$, with a region of convergence, the ROC. This page does two jobs. It lists the rules that let me work on $X(z)$ instead of on the samples, and then it goes back from $X(z)$ to the samples.

Every rule is proved from the sum. I use the pairs of 16.1: $a^nu[n]\leftrightarrow\dfrac1{1-az^{-1}}$ for $\lvert z\rvert > \lvert a\rvert$, and its left-sided partner.

Linearity. The sum is linear, so $ax[n]+by[n]\leftrightarrow aX(z)+bY(z)$. The ROC contains at least the overlap of the two ROCs. It can be larger, if a pole cancels.

Delay. Put $m=n-n_0$ in the sum:

A delay by $n_0$ samples multiplies $X(z)$ by $z^{-n_0}$. The ROC stays the same, except possibly at $z=0$ and at infinity. In particular, one delay box is $z^{-1}$.

Multiplying by $a^n$. Write $a^nz^{-n}=(z/a)^{-n}$. Then

Wherever $X$ had a pole $p$, $X(z/a)$ has a pole at $ap$. The plane is stretched by the factor $\lvert a\rvert$ and turned by the angle of $a$. Take $a=-1$ and $x[n]=0.5^nu[n]$. Then $(-1)^n0.5^nu[n]=(-0.5)^nu[n]$, whose transform is $\dfrac1{1+0.5z^{-1}}$. The pole moved from 0.5 to $-0.5$: with $a=-1$ the plane turns half a turn.

Multiplying by $n$. Differentiate the sum term by term: $\dfrac{dX}{dz}=\sum_n(-n)\,x[n]\,z^{-n-1}$. Multiply by $-z$ and the power goes back to $z^{-n}$:

Try it on $0.5^nu[n]$, with $X=(1-0.5z^{-1})^{-1}$. The derivative is $-(1-0.5z^{-1})^{-2}\cdot0.5z^{-2}$, so

The ratio has the pole 0.5 twice. Running this ratio as a recursion with an impulse in gives the samples 0, 0.5, 0.5, 0.375, 0.25, 0.15625, and $n\,0.5^n$ at $n=0$ to 5 is the same list.

Convolution. For the sequences $x$ and $h$, the convolution sum from Discrete convolution (5.2) is $\sum_k x[k]\,h[n-k]$. Put it into the z-sum and swap the order of the two sums. The delay rule turns the inner sum into $z^{-k}H(z)$, and what is left is

This is the rule of Properties of the DTFT (12.3), now at every $z$ in the ROC, not only on the unit circle. The ROC contains at least the overlap of the two. I will check it on numbers below.

The first sample. Suppose $x$ is causal, which means $x[n]=0$ for $n<0$. Then $X(z)=x[0]+x[1]z^{-1}+x[2]z^{-2}+\cdots$. Let $\lvert z\rvert$ grow without limit. Every term but the first carries a $z^{-n}$ with $n\ge1$ and shrinks to 0, so

This is the initial value theorem. Take $Y(z)=\dfrac{1+2z^{-1}}{1-0.5z^{-1}}$, which appears again below. At $z=10$ it is 1.263, at $z=100$ it is 1.025 and at $z=10^4$ it is 1.00025. It is heading for $y[0]=1$.

Every delay box is z⁻¹

In Difference equations (6.1) the equation $y[n]=x[n]+2x[n-1]+0.5y[n-1]$ was drawn as wires, delay boxes and gains. The delay rule says what each box does to the transforms: it multiplies by $z^{-1}$.

Read the diagram as algebra. The adder collects the input, the delayed and doubled input, and the delayed and halved output:

Collect the $Y$ terms to get $(1-0.5z^{-1})Y=(1+2z^{-1})X$. With an impulse in, $X=1$, so

The rest of the page goes the other way, from a ratio like this back to its samples.

One sequence per pole

Take

right-sided, with ROC $\lvert z\rvert > 0.5$. Multiply out the two brackets and you get the bottom on the left, so the factoring is right. The poles are 0.5 and $-0.25$.

The idea is the one from Properties and the inverse Laplace transform (9.2), done in $z^{-1}$. Split $X$ into one term per pole, with a weight $r_k$ for the pole $p_k$:

To find $r_1$, multiply both sides by $(1-0.5z^{-1})$ and set $z=0.5$, so that $z^{-1}=2$. The second term is multiplied by zero and vanishes. On the left what remains is $r_1=\dfrac1{1+0.25\cdot2}=\dfrac23$. Likewise, setting $z=-0.25$, so that $z^{-1}=-4$, gives $r_2=\dfrac1{1-0.5\cdot(-4)}=\dfrac13$.

Each term is a pair from 16.1, $a^nu[n]$ with its weight, so $x[n]=\tfrac23\,0.5^nu[n]+\tfrac13\,(-0.25)^nu[n]$. Make a guess. The second pole is negative. What will its sequence do from one sample to the next? Then look again at the picture at the top of this page, which plays this split.

Pole 0.5 gives $\tfrac23\,0.5^n$: it starts at 0.667 and halves every sample. Pole $-0.25$ gives $\tfrac13(-0.25)^n$: it starts at 0.333, shrinks to a quarter and flips sign every sample, as you guessed. Added, they give $x[n]=1$, 0.25, 0.1875, 0.0781, …, the same numbers the recursion $y[n]=0.25y[n-1]+0.125y[n-2]+x[n]$ gives for an impulse. When the clip has finished, drag along the samples to read the two terms and their sum at any $n$: at $n=1$ it is $0.333-0.083=0.250$.

Check the answer another way. The two brackets are the transforms of $0.5^nu[n]$ and $(-0.25)^nu[n]$, and a product of transforms is a convolution. Convolve those two sequences and the first samples are again 1, 0.25, 0.1875, 0.078125, the same as the sum of the two terms. That is the convolution rule on real numbers.

SciPy does the splitting. residuez(b, a) takes the coefficients of the top and bottom in powers of $z^{-1}$ and returns the weights r, the poles p, and a direct term k. The direct term is non-empty when the top is as long as the bottom or longer. For this $X$, residuez([1], [1, -0.25, -0.125]) returns the weights 0.3333 at $-0.25$ and 0.6667 at 0.5, and no direct term.

Three variations. A mirror pair of poles has mirror weights, as in 9.2, and the two terms add to a damped cosine. A repeated pole gives an $n\,a^nu[n]$ term, the multiplying-by-$n$ row. For $n\,0.5^nu[n]$ above, residuez returns the pole 0.5 twice, with the weights $-1$ and 1 for the terms in $\dfrac1{1-0.5z^{-1}}$ and $\dfrac1{(1-0.5z^{-1})^2}$. And for another ROC the fractions are the same, but a pole outside the ROC gives a left-sided term $-a^nu[-n-1]$, as in 16.1.

Divide, and the samples come out one by one

There is a method with no poles in it. Take $Y(z)=\dfrac{1+2z^{-1}}{1-0.5z^{-1}}$ from the diagram, where the input is an impulse. Divide the top by the bottom as with numbers, but with powers of $z^{-1}$, highest power first, which means $z^0$ first.

It works like dividing 1 by 7 on paper: each step gives one digit of 0.142857… Here each step gives one sample. Before you watch, make a guess. $1-0.5z^{-1}$ goes into $1+2z^{-1}$ once. What is left over?

Then watch each remainder: after the first step it is half the one before.

What is left after the first step is $2.5z^{-1}$. The quotient is 1, 2.5, 1.25, 0.625, 0.3125, …: 6.1’s impulse response, one sample per step of the division. Why does this give the same numbers as running the equation? After the first step, each remainder is $0.5$ times the sample just found, pushed one place later. That is what the delay box holds: the previous output, halved. Each step of the division is one pass of the loop.

For a recursion the division never ends. It gives the power series of a right-sided signal, one more sample for each extra step, and I stop when I have enough. For a left-sided signal I would divide in powers of $z$ instead.

The same $Y$ by partial fractions has a direct term, because the top is as long as the bottom. residuez([1, 2], [1, -0.5]) returns the weight 5 at the pole 0.5 and the direct term $-4$. So $y[n]=-4\,\delta[n]+5\cdot0.5^nu[n]$, which gives 1, 2.5, 1.25, 0.625, … again: at $n=0$, $-4+5=1$, and for $n\ge1$ it is $5\cdot0.5^n$. The initial value theorem agrees: $Y(z)\to1$ as $z\to\infty$. A third method is to read the table by eye, when the ratio is simple enough.

Worked example

1. Partial fractions. $X=\dfrac1{1-0.25z^{-1}-0.125z^{-2}}$. The cover-up gives $r=\tfrac23$ at $p=0.5$ and $r=\tfrac13$ at $p=-0.25$, so $x[n]=\tfrac23\,0.5^n+\tfrac13(-0.25)^n$ for $n\ge0$. The first seven samples are 1, 0.25, 0.1875, 0.07813, 0.04297, 0.02051 and 0.0105. Running the recursion and convolving $0.5^nu[n]$ with $(-0.25)^nu[n]$ give the same list.
2. Long division. $Y=\dfrac{1+2z^{-1}}{1-0.5z^{-1}}$ gives the quotient 1, 2.5, 1.25, 0.625, 0.3125, 0.15625 and the remainders $2.5z^{-1}$, $1.25z^{-2}$, $0.625z^{-3}$, $0.3125z^{-4}$, $0.15625z^{-5}$, $0.078125z^{-6}$. With residuez, $y[n]=-4\delta[n]+5\cdot0.5^n$ gives the same samples.
3. Initial value. $Y(z)$ at $z=10$, 100 and $10^4$ is 1.263, 1.025 and 1.00025, tending to $y[0]=1$. For the first example’s $X$ the values are 1.027, 1.0025 and 1.000025.
4. Multiplying by $n$. $n\,0.5^nu[n]\leftrightarrow\dfrac{0.5z^{-1}}{(1-0.5z^{-1})^2}$. Its impulse response is 0, 0.5, 0.5, 0.375, 0.25, 0.15625, which is $n\,0.5^n$.
5. Multiplying by $a^n$. $(-1)^n0.5^nu[n]=(-0.5)^nu[n]\leftrightarrow\dfrac1{1+0.5z^{-1}}$. The pole moved from 0.5 to $-0.5$.

Where you’ll meet this

Digital filters are described by ratios of polynomials in $z^{-1}$. Partial fractions split such a ratio into one first-order piece per pole, and each piece is one geometric sequence. A filter built from first-order sections in parallel is exactly that sum.

Long division is the checking tool: a few steps give the first samples of the impulse response without running any code, and they must agree with lfilter. The next pages use the same ratios to read behaviour from where the poles sit. See Transfer functions, poles and zeros (16.3), Stability and causality (16.4), where the left-sided terms matter, and The unilateral z-transform (16.5), where starting values enter through the delay rule.

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