8 replaced http://math.stackexchange.com/ with https://math.stackexchange.com/ edited Apr 13 '17 at 12:19 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 causalcausal, 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 $$(z-a)(z-b)\over z^2.$$ 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 $$(z-a)(z-b)\over z^2.$$ 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 $$(z-a)(z-b)\over z^2.$$ 7 drop x_0, causal non-causal edit approved Jul 31 '14 at 13:53 Val 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 $$(z-a)(z-b)\over z^2.$$ 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. 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 $$(z-a)(z-b)\over z^2.$$ 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 $$(z-a)(z-b)\over z^2.$$ 6 single-zero example edited Jul 31 '14 at 4:29 jojek♦ 8,99366 gold badges2525 silver badges6060 bronze badges 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]$$y[i] = x[i+1]$$, that is output depends on future input. 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 $$(z-a)(z-b)\over z^2.$$ 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. 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 $$(z-a)(z-b)\over z^2.$$ 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. 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 $$(z-a)(z-b)\over z^2.$$ 5 single-zero example edit approved Jul 31 '14 at 4:29 Val 1 4 explained that more power in nominator means positive power of z after reduction edit approved Aug 7 '13 at 14:26 Val 1 3 generalized the answer edit approved Aug 6 '13 at 19:12 Val 1 2 added 2 characters in body edited Aug 5 '13 at 20:56 Sudarsan 40922 silver badges99 bronze badges 1 answered Aug 5 '13 at 20:37 Sudarsan 40922 silver badges99 bronze badges