Property of interconnected feedback systems
$begingroup$
In the figure you can see the statespace form of a feedback interconnection system.
Very quick question: is there a reason they have taken $D_1=0$ and $D_2=0$?
It makes workings a lot easier but I see no reason as to why this would be true generally.
dynamical-systems control-theory linear-control systems-theory
edited Jan 5 at 0:20
Arash
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
$endgroup$
add a comment |
$begingroup$
In the figure you can see the statespace form of a feedback interconnection system.
Very quick question: is there a reason they have taken $D_1=0$ and $D_2=0$?
It makes workings a lot easier but I see no reason as to why this would be true generally.
dynamical-systems control-theory linear-control systems-theory
edited Jan 5 at 0:20
Arash
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
$endgroup$
$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08
add a comment |
$begingroup$
In the figure you can see the statespace form of a feedback interconnection system.
Very quick question: is there a reason they have taken $D_1=0$ and $D_2=0$?
It makes workings a lot easier but I see no reason as to why this would be true generally.
dynamical-systems control-theory linear-control systems-theory
edited Jan 5 at 0:20
Arash
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
$endgroup$
In the figure you can see the statespace form of a feedback interconnection system.
Very quick question: is there a reason they have taken $D_1=0$ and $D_2=0$?
It makes workings a lot easier but I see no reason as to why this would be true generally.
dynamical-systems control-theory linear-control systems-theory
dynamical-systems control-theory linear-control systems-theory
edited Jan 5 at 0:20
Arash
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
edited Jan 5 at 0:20
Arash
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
edited Jan 5 at 0:20
Arash
885210
edited Jan 5 at 0:20
Arash
885210
edited Jan 5 at 0:20
Arash
885210
885210
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
asked Jan 2 at 23:56
smoking_huge_doinkssmoking_huge_doinks
132
132
$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08
add a comment |
$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08
$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08
$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In most of practical systems $boldsymbol D=boldsymbol 0$.
If you push the gas pedal in your car, does your car jump?
If there is a jump in the temperature, does your thermometer mercury jump immediately?
In the real-world, it is rare to find plants and sensors with immediate output. But, sometimes there is a system that approximately gives you an immediate response. What to do with them?
In such cases, for continuous time systems, you can assume that there is a very trivial low-pass filter before or after the block. For discrete systems, you can assume there is a small unit delay before or after the block. The low-pass filter and the delay should be design with minimal impact on the model accuracy and the system needs to be robust enough to tolerate that.
answered Jan 5 at 0:16
ArashArash
885210
$endgroup$
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In most of practical systems $boldsymbol D=boldsymbol 0$.
If you push the gas pedal in your car, does your car jump?
If there is a jump in the temperature, does your thermometer mercury jump immediately?
In the real-world, it is rare to find plants and sensors with immediate output. But, sometimes there is a system that approximately gives you an immediate response. What to do with them?
In such cases, for continuous time systems, you can assume that there is a very trivial low-pass filter before or after the block. For discrete systems, you can assume there is a small unit delay before or after the block. The low-pass filter and the delay should be design with minimal impact on the model accuracy and the system needs to be robust enough to tolerate that.
answered Jan 5 at 0:16
ArashArash
885210
$endgroup$
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
add a comment |
$begingroup$
In most of practical systems $boldsymbol D=boldsymbol 0$.
If you push the gas pedal in your car, does your car jump?
If there is a jump in the temperature, does your thermometer mercury jump immediately?
In the real-world, it is rare to find plants and sensors with immediate output. But, sometimes there is a system that approximately gives you an immediate response. What to do with them?
In such cases, for continuous time systems, you can assume that there is a very trivial low-pass filter before or after the block. For discrete systems, you can assume there is a small unit delay before or after the block. The low-pass filter and the delay should be design with minimal impact on the model accuracy and the system needs to be robust enough to tolerate that.
answered Jan 5 at 0:16
ArashArash
885210
$endgroup$
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
add a comment |
$begingroup$
In most of practical systems $boldsymbol D=boldsymbol 0$.
If you push the gas pedal in your car, does your car jump?
If there is a jump in the temperature, does your thermometer mercury jump immediately?
In the real-world, it is rare to find plants and sensors with immediate output. But, sometimes there is a system that approximately gives you an immediate response. What to do with them?
In such cases, for continuous time systems, you can assume that there is a very trivial low-pass filter before or after the block. For discrete systems, you can assume there is a small unit delay before or after the block. The low-pass filter and the delay should be design with minimal impact on the model accuracy and the system needs to be robust enough to tolerate that.
answered Jan 5 at 0:16
ArashArash
885210
$endgroup$
In most of practical systems $boldsymbol D=boldsymbol 0$.
If you push the gas pedal in your car, does your car jump?
If there is a jump in the temperature, does your thermometer mercury jump immediately?
In the real-world, it is rare to find plants and sensors with immediate output. But, sometimes there is a system that approximately gives you an immediate response. What to do with them?
In such cases, for continuous time systems, you can assume that there is a very trivial low-pass filter before or after the block. For discrete systems, you can assume there is a small unit delay before or after the block. The low-pass filter and the delay should be design with minimal impact on the model accuracy and the system needs to be robust enough to tolerate that.
answered Jan 5 at 0:16
ArashArash
885210
answered Jan 5 at 0:16
ArashArash
885210
answered Jan 5 at 0:16
ArashArash
885210
answered Jan 5 at 0:16
ArashArash
885210
885210
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
add a comment |
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
Thank you, this is a fantastic explanation! So I guess if this state-space representation were to be derived completely theoretically and not purely for real world application then it would appropriate to not take D to be the 0 matrix?
$endgroup$
– smoking_huge_doinks
Jan 5 at 20:06
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks That also depends on the context. Maybe the authors used that representation for a purely theoretical proof of some property of the closed loop system, which however is just valid for the special case that $D = 0$. That could be a valid reason as well, so it doesn't necessarily depend on their investigation being completely theoretically or practically. Both is possible.
$endgroup$
– SampleTime
Jan 5 at 20:46
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
$begingroup$
@smoking_huge_doinks, in the most of real systems $boldsymbol{D}=0$. It is hard to find real control systems with $boldsymbol{D}ne0$ when the sample time $T_s$ is small enough. Even with using the most accurate control system, there are a lot of noise and disturbances adding uncertainty to your results. Let's say by adding a lot of complexities you increased the accuracy by 0.1%. How much is the industry interested?
$endgroup$
– Arash
Jan 14 at 11:27
add a comment |
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$begingroup$
Who is "they"? Without context it is hard to say, because there can be different reasons. However, as @Arash said, $D = 0$ is the case in many real world applications, so maybe the authors where just considering systems that represent their actual control problem.
$endgroup$
– SampleTime
Jan 5 at 19:08