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I have the diagram above. I found the transfer function below from it;

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The question asks me to find out if the system is causal and stable, but didn't it have to indicate whether it was left-sided or right-sided?

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It looks like a Homework problem so I will only provide small hints:

  1. Check from the block diagram implementation of the system whether the current output sample $y[n]$ depends on only current input sample $x[n]$ and past input & output samples $x[n-n_0]$ and $y[n-m_0]$. If so, the system is causal, else it will be non-causal.

  2. You can use the information from the previous point (causal or non-causal) along with the pole location of $H(z)$ to check if the system is BIBO stable or not.

Any Causal System with poles inside unit-circle are BIBO stable.

You can refer to this Pole-Zero Analysis for better understanding.

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  • $\begingroup$ Yes you are right it is a homework problem, thank you very much for your hints really helped me. $\endgroup$ – cem ekkazan Jun 23 at 14:56
  • $\begingroup$ Note that given only the difference equation you can't say anything about causality. $\endgroup$ – Matt L. Jun 23 at 16:05
  • $\begingroup$ But the question wants it first? The question is not in English but it gives the system diagram and says "Determine the casuality of the system.". I have only the diagram about system. $\endgroup$ – cem ekkazan Jun 23 at 16:11
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    $\begingroup$ It's indeed in the diagram, the difference equation doesn't help here. The difference equation is helpful to obtain the transfer function, but causality can only be determined from the implementation shown in the diagram. A difference equation by itself has the same information as a transfer function without ROC. $\endgroup$ – Matt L. Jun 23 at 18:06
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    $\begingroup$ @MattL. Ok. We can rearrange the terms in the Difference Equation to make it seem like anti-causal system or causal system. So, there is an underlying ambiguity if only difference equation is given. Understood. And, the block diagram showing implementation or ROC basically clears this ambiguity. Thanks. I will change my answer to reflect this information. $\endgroup$ – DSP Rookie Jun 23 at 19:32

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