Why more poles than zeroes?

Question

I read that an "improper system" "has more zeros than poles; it is not causal, cannot be implemented, has a strictly proper inverse and has infinite high-frequency gain."

Does causality fail due to instant signal change? I try to make up a recurrence, where $x_n$ depends on both $x_{n-1}$ as well as $x_{n+1}$ but they all can be interpreted as $x_{n+1}$ dependence on $x_{n}$ and $x_{n-1}$ and, thus, are causal.

Is it related with solving recurrent relations saying that P(x) in generating function $f(x) = f_0 + f_1 x + f_2 x^2 + \cdots = P(x)/R(x) \ \ \ \ \ $ must have degree less than R(x) as Miguel A. Lerma says in Generating function of Linear Homogenous Recurrence. Here, $R(x) = 1 + r_1 x^1 + r_2 x^2 + \cdots$ is a reciprocal of the characteristic polynomial of our recurrence $x_n + c_1 x_{n+1} + \cdots = 0.$ This indeed seems to happen whenever which R(x) I try.

On the other hand, https://ccrma.stanford.edu/~jos/fp/Existence_Z_Transform.html seem to identify causality with pole values (if your pole > 1 then series does not converge, which means anticausality) rather than their number and others say that causality is $x_n = 0$ for all $n < 0.$ I do not see how this is related with poles/zeros ratio and my guess that poles > zeroes = causality may be wrong. I have asked to relate various kinds of causality here.

Answer 1

I think the system is always non-causal if it has more zeros than poles. @Hilmar, your implementation might be possible digitally but if you see a transfer function like this: $H(z) = \frac{(z-1)(z-2)}{(z-3)}$, taking the inverse Z-Transform, $y(n-1)-3y(n-2) = x(n)-3x(n-1)+2x(n-2)$ is clearly non-causal and cannot be implemented real-time(of course you can store the signals and do processing offline, but real time implementation is impossible). So I guess Non-Causality is the answer, apart from the infinite high-frequency gain.

Update by Val Ok, I understand now. In wikipedia, we see that $$a_0 y_n + a_1 y_{n-1} + a_2 y_{n-2} + \cdots = b_0 + b_1 x_{n-1} + b_2 x_{n-2} + \cdots$$ is identical to $$H(z) = {b_0 + b_1 z^{-1} + b_2 z^{-2} + \cdots \over a_0 z^{0} + a_1 z^{-1} +\cdots}.$$ I have finally noted that there are no positive powers. That is, $a_0$ corresponds to the current output, $z^0 = y[n]$, and $\cdots = a_{-2} = a_{-1} = \mathbf{0} = b_{-1} = b_{-2} = \cdots$. If we had more zeroes then we have more power in the nominator and, after normalizing $H(z)$ so that $a_0$ has $z^0$, we still have a positive power of z in the nominator, which means that output depends on future input.

In the simplest case, $Y(z) = zX(z)-x_0$ corresponds to $y[i] = x[i+1]$, that is output depends on future input (we can also drop $x_0$ assuming our sequences are causal, i.e. $0 = y_{-1} = x_0$ aka $z^{-1}Y = X$ or $Y = zX$). It is curious that causal is an example of non-causal. Could anybody clarify it further?

It also worth noting that $(1-az^{-1})(1-bz^{-1})$ is not a zero of order two. It has equal number of zeroes and poles, since it is equal to $(z-a)(z-b)\over z^2.$

Answer 2

In a causal discrete time system, the poles delay the impulse response so that no future samples are required. If you don't have enough poles, you may instead need a time machine (or do processing offline or with a "hidden" delay).

Note that the phase represented by walking CCW around a pole well inside the unit circle is about one sample time delay.

Answer 3

If there are more zeros than poles, then there is added lags(power of Z^(-1)) in the output which makes the output more lagged while the input remain the same. This makes the system non-causal.

Answer 4

There are answers which considers inputs X in the nominator and characteristic polynomial in the denominator. Let me explain in terms of simple signal x[n]. It's z-transform, $$Z\{x[n]\} = x_0 + x_1 z^{-1} + x_2 z^{-2} + \cdots,$$ by definition. For instance,$$Z\{a_0, a_1, a_2\} = a_0 + a_1 z^{-1} + a_2 z^{-2} = {a_0 z^2 + a_1 z^1 + a_2 \over z^2 }.$$ Note that signal is causal by definion when it has all negative time values equal 0. That is, you cannot have $a_{-3}$ in the series. This means that you cannot have z with positive powers in the series, which means that in the fractional form, nominator powers cannot be greater than denominator.

Now, System is z-transform of the impulse response, which can be considered as a signal in time domain or a rational z-function $H(z)=Y(z)/X(z)$, which naturally comes out from system equation $\sum a_n y_n = \sum b_n x_n.$ System is causal if it's impulse response is causal.

Answer 5

You must put the transfer function in $$H(z)$$ form, not $$H(z^{-1})$$ form, when counting the number of poles and zeros. There are a lot of replies in this thread that are confused for this reason. If you make sure everything is in terms of positive powers of $z$, all this confusion will go away. That is when the statement "the number of zeros must be less than or equal to the number of poles in $H(z)$ to be causal" applies.

Answer 6

EDIT: based on the comments, the answer below only applies to discrete time systems and poles and zeroes in the Z-plane. NOT to time continuous system and poles and zeros in the s-plane.

A time discrete system with all zeroes and no poles is simply an FIR filter which is always stable and causal. A system with more zeros than poles can be easily split into a cascaded FIR and IIR filter where the excess zeros are end up in the FIR filter.

I don't see any problem with that. It's easily possible to create a system with more zeros then poles that' stable and causal. Here is a simple MATLAB example

z = [.1 .2 .3 .5 .6 .7]; p = [.15 .25 .85]; % 6 zeros & 3 poles more or less random
% plot the impulse response
x = zeros(128,1); x(1) = 1;
[b,a] = zp2tf(z',p',1);
y = filter(b,a,x);
plot(y);

Answer 7

Answer 2: I'll try one more time and then I give up:

Please consider the the difference equation:

$$y[n]-0.5\cdot y[n-1] = x[n] - 3\cdot x[n-1] + 2 \cdot x[n-2] $$

This is a slight modification of Sudarsan's example. I just put the pole at 0.5 so that the whole thing is stable. So we can all hopefully agree on the following

1. This is a valid difference equation
2. It describes a linear time invariant system
3. It is stable
4. It is causal
5. It can be implemented in real time

Let's derive the Z-transform $$H(z) = \frac{1-3\cdot z^{-1}+2\cdot z^{-2}}{1-0.5\cdot z^{-1}}$$

This can be rewritten in terms of poles and zeros as follows $$H(z) = \frac{(1-1\cdot z^{-1})\cdot (1-2\cdot z^{-1}) }{1-0.5\cdot z^{-1}}$$ or $$H(z) = \frac{(1-\frac{1}{z})\cdot (1-\frac{2}{z}) }{1-\frac{0.5}{z}}$$

See for example https://ccrma.stanford.edu/~jos/filters/Z_Transform_Difference_Equations.html and https://ccrma.stanford.edu/~jos/filters/Factored_Form.html on how this is done.

In this representation we can clearly see that there are two zeros (at z=1 and z=2) and one pole (z=0.5). This is a specific example of a time discrete filter that is causal, stable and real time implementable and has MORE zeros than poles.

Re the real time implementation of FIR filters. Let's consider the simple FIR filter with an simple impulse response h[n] = [1 -3 2], which is basically the FIR portion of our filter above. Again, it's perfectly causal and stable. You would implement this as follows (in C)_

static float x1 = 0.0F;
static float x2 = 0.0F;

float firFilterExample(float x)
{
  float y;
  // calculate output
  y = x - 3*x1 + 2*x2;
  // update state variables
  x2 = x1;
  x1 = x;
  return(y);
}

There is one immediate output per input sample. There is no inherent delay.