3 added 196 characters in body edited Nov 8 '18 at 13:52 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In the general case you need additional information, and that information is given by the ROC. Information about the ROC can also be given in the form of a causality constraint, which basically tells you that the ROC is a right half-plane, or - in discrete time - the region outside the pole with the largest radius. As explained above, the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. Also take a look at this answer for a concrete example of three impulse responses corresponding to the same transfer function with different ROCs. There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In the general case you need additional information, and that information is given by the ROC. Information about the ROC can also be given in the form of a causality constraint, which basically tells you that the ROC is a right half-plane, or - in discrete time - the region outside the pole with the largest radius. As explained above, the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In the general case you need additional information, and that information is given by the ROC. Information about the ROC can also be given in the form of a causality constraint, which basically tells you that the ROC is a right half-plane, or - in discrete time - the region outside the pole with the largest radius. As explained above, the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. Also take a look at this answer for a concrete example of three impulse responses corresponding to the same transfer function with different ROCs. 2 added 318 characters in body edited Nov 8 '18 at 11:39 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In the general case you need additional information, and that information is given by the ROC. Information about the ROC can also be given in the form of a causality constraint, which basically tells you that the ROC is a right half-plane, or - in discrete time - the region outside the pole with the largest radius. In all other casesAs explained above, the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In all other cases the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In the general case you need additional information, and that information is given by the ROC. Information about the ROC can also be given in the form of a causality constraint, which basically tells you that the ROC is a right half-plane, or - in discrete time - the region outside the pole with the largest radius. As explained above, the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle. 1 answered Nov 8 '18 at 9:53 Matt L. 52.4k22 gold badges3939 silver badges9898 bronze badges There is no principal difference between continuous-time and discrete-time systems when judging stability. The imaginary axis of the $$s$$-plane corresponds to the unit circle in the $$z$$-plane, and the region inside the unit circle in the $$z$$-plane corresponds to the left half-plane of the the $$s$$-plane. In both cases the region of convergence (ROC) is essential. Given only the pole locations, you generally cannot say whether a system is stable or not, unless the poles are on the imaginary axis (the unit circle), in which case the system is unstable. In all other cases the ROC is essential for deciding if the system is stable or not. As you've mentioned in your question, if the ROC contains the imaginary axis (the unit circle) then the system is stable. The ROC of causal systems is to the right of the right-most pole (or outside the pole with the largest radius in discrete time). That's why all poles of a causal and stable system must lie in the left half-plane (inside the unit circle). For an anti-causal (left-sided) system, the opposite is true. For (non-causal) systems with two-sided impulse responses, the ROC is a strip (annulus) in the complex plane, limited by two or more poles. It is important to understand that pole locations do not uniquely define a system. That's why they are not sufficient to decide if the system is stable or not. Take as an example a discrete-time system with two poles, one outside the unit circle and one inside. There are three systems (impulse responses) with these same poles. One causal unstable system, one anti-causal unstable system, and one non-causal (two-sided) stable system. Only the two-sided system is stable because its ROC is between the two poles and it includes the unit circle. The ROCs of the other two systems do not include the unit circle.