text: # Impulse response of non-causal system
  
  Suppose I have the following equation, where $x[n]$ is the input and $y[n]$ is the output: $$x[n]-x[n-1]=-10y[n-1]+2y[n]+2y[n+1].$$
  
  The question asks me to compute the impulse response from this equation.
  
  Attempts: $x[n]=1$ when $n=0$; $0$ otherwise. I need to find $h[n]=y[n]$, but, from the equation, it always depends on the future value $y[n+1]$, which is unknown.
  
  One way to solve such a difference equation is to use the $\mathcal{Z}$-transform. Transforming the given difference equation yields
  
  $$X(z)\left(1-z^{-1}\right)=Y(z)\left(-10z^{-1}+2+2z\right)\tag{1}$$
  
  which results in the following transfer function:
  
  $$H(z)=\frac{Y(z)}{X(z)}=\frac{1-z^{-1}}{2z+2-10z^{-1}}=\frac12\frac{z^{-1}-z^{-2}}{1+z^{-1}-5z^{-2}}$$
  
  The poles of $H(z)$, i.e. the zeros of the denominator polynomial in $(1)$, are both real-valued:
  
  $$1+z^{-1}-5z^{-2}=(1-az^{-1})(1-bz^{-1})\tag{2}$$
  
  with $a=\frac{-1+\sqrt{21}}{2}\approx 1.79$ and $b=\frac{-1-\sqrt{21}}{2}\approx -2.79$. Note that $1<|a|<|b|$ is satisfied, so both real-valued poles are outside the unit circle of the complex plane.
  
  With $(2)$, $H(z)$ can be rewritten as
  
  $$H(z)=\frac12\frac{z^{-1}-z^{-2}}{(1-az^{-1})(1-bz^{-1})}=\frac{1}{2(b-a)}\left[\frac{(1-a)z^{-1}}{1-az^{-1}}-\frac{(1-b)z^{-1}}{1-bz^{-1}}\right]\tag{3}$$
  
  This transfer function can have three possible regions of convergence (ROCs), each of them corresponding to a different impulse response:
  
  \begin{align}(1)\quad &|z|>|b|>|a|\\ (2)\quad & |b|>|z|>|a|\\ (3)\quad & |b|>|a|>|z|\end{align}
  
  ROC $(1)$ corresponds to a right-sided impulse response, ROC $(2)$ corresponds to a two-sided impulse response, and ROC $(3)$ corresponds to a left-sided sequence. Only ROC $(3)$ includes the unit circle $|z|=1$, so only the impulse response corresponding to this ROC results in a stable, yet anti-causal, system. The three impulse responses corresponding to the three ROCs given above are
  
  \begin{align}(1)\quad & h_1[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}-(1-b)b^{n-1}\right)u[n-1]\\ (2)\quad & h_2[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}u[n-1]+(1-b)b^{n-1}u[-n]\right)\\ (3)\quad & h_3[n]=\frac{1}{2(b-a)}\left(-(1-a)a^{n-1}+(1-b)b^{n-1}\right)u[-n]\end{align}
  
  These impulse responses describe three different systems, all of them characterized by the same transfer function given by Eq. $(3)$. $h_1[n]$ corresponds to a causal unstable system, $h_2[n]$ to a non-causal unstable system, and $h_3[n]$ to an anti-causal stable system.
  
  • Actually the question does not state that this system is non-causal, it is my assumption based on the equation. The equation is derived from a 2nd order ODE in continous time, then mapping to DT using Euler approximation. I think your answer might apply in this case – TonyJ Nov 11 '15 at 11:58
  • @TonyJ: But the causal impulse response is not stable, so it's not useful in practice. – Matt L. Nov 11 '15 at 12:45
explanation: The document is a detailed explanation of how to compute the impulse response of a non-causal system from a given difference equation. It includes the problem statement, the approach to solving it using the Z-transform, and the derivation of the transfer function and its poles. The document also discusses the different regions of convergence (ROCs) and the corresponding impulse responses, including their stability and causality properties. The document contains a question and an answer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Impulse response of non-causal system
  
  Suppose I have the following equation, where $x[n]$ is the input and $y[n]$ is the output: $$x[n]-x[n-1]=-10y[n-1]+2y[n]+2y[n+1].$$
  
  The question asks me to compute the impulse response from this equation.
  
  Attempts: $x[n]=1$ when $n=0$; $0$ otherwise. I need to find $h[n]=y[n]$, but, from the equation, it always depends on the future value $y[n+1]$, which is unknown.
  
  One way to solve such a difference equation is to use the $\mathcal{Z}$-transform. Transforming the given difference equation yields
  
  $$X(z)\left(1-z^{-1}\right)=Y(z)\left(-10z^{-1}+2+2z\right)\tag{1}$$
  
  which results in the following transfer function:
  
  $$H(z)=\frac{Y(z)}{X(z)}=\frac{1-z^{-1}}{2z+2-10z^{-1}}=\frac12\frac{z^{-1}-z^{-2}}{1+z^{-1}-5z^{-2}}$$
  
  The poles of $H(z)$, i.e. the zeros of the denominator polynomial in $(1)$, are both real-valued:
  
  $$1+z^{-1}-5z^{-2}=(1-az^{-1})(1-bz^{-1})\tag{2}$$
  
  with $a=\frac{-1+\sqrt{21}}{2}\approx 1.79$ and $b=\frac{-1-\sqrt{21}}{2}\approx -2.79$. Note that $1<|a|<|b|$ is satisfied, so both real-valued poles are outside the unit circle of the complex plane.
  
  With $(2)$, $H(z)$ can be rewritten as
  
  $$H(z)=\frac12\frac{z^{-1}-z^{-2}}{(1-az^{-1})(1-bz^{-1})}=\frac{1}{2(b-a)}\left[\frac{(1-a)z^{-1}}{1-az^{-1}}-\frac{(1-b)z^{-1}}{1-bz^{-1}}\right]\tag{3}$$
  
  This transfer function can have three possible regions of convergence (ROCs), each of them corresponding to a different impulse response:
  
  \begin{align}(1)\quad &|z|>|b|>|a|\\ (2)\quad & |b|>|z|>|a|\\ (3)\quad & |b|>|a|>|z|\end{align}
  
  ROC $(1)$ corresponds to a right-sided impulse response, ROC $(2)$ corresponds to a two-sided impulse response, and ROC $(3)$ corresponds to a left-sided sequence. Only ROC $(3)$ includes the unit circle $|z|=1$, so only the impulse response corresponding to this ROC results in a stable, yet anti-causal, system. The three impulse responses corresponding to the three ROCs given above are
  
  \begin{align}(1)\quad & h_1[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}-(1-b)b^{n-1}\right)u[n-1]\\ (2)\quad & h_2[n]=\frac{1}{2(b-a)}\left((1-a)a^{n-1}u[n-1]+(1-b)b^{n-1}u[-n]\right)\\ (3)\quad & h_3[n]=\frac{1}{2(b-a)}\left(-(1-a)a^{n-1}+(1-b)b^{n-1}\right)u[-n]\end{align}
  
  These impulse responses describe three different systems, all of them characterized by the same transfer function given by Eq. $(3)$. $h_1[n]$ corresponds to a causal unstable system, $h_2[n]$ to a non-causal unstable system, and $h_3[n]$ to an anti-causal stable system.
  
  • Actually the question does not state that this system is non-causal, it is my assumption based on the equation. The equation is derived from a 2nd order ODE in continous time, then mapping to DT using Euler approximation. I think your answer might apply in this case – TonyJ Nov 11 '15 at 11:58
  • @TonyJ: But the causal impulse response is not stable, so it's not useful in practice. – Matt L. Nov 11 '15 at 12:45
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed explanation of how to compute the impulse response of a non-causal system from a given difference equation. It includes the problem statement, the approach to solving it using the Z-transform, and the derivation of the transfer function and its poles. The document also discusses the different regions of convergence (ROCs) and the corresponding impulse responses, including their stability and causality properties. The document contains a question and an answer.
  
  Label: question-and-answer