Achievable performance for systems with RHP zeros/poles











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I have often read that RHP zeros and poles set limits on the maximum achievable performance of LTI systems.



However, what does that exactly mean and how can you compute these performance limits?



For example, take these three transfer functions:



$$
begin{align}
G_1(s) &= frac{1}{(s + 1)(s - 2)} quadtext{(RHP pole)} \
G_2(s) &= frac{(s - 3)}{(s + 1)(s + 2)} quadtext{(RHP zero)} \
G_3(s) &= frac{(s - 3)}{(s + 1)(s + 2)(s - 5)} quadtext{(RHP pole and zero)}
end{align}
$$



What would be the limits of performance for these transfer functions? And are there differences between RHP zeros and RHP poles in terms of achievable performance?










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  • How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
    – Kwin van der Veen
    Nov 21 at 14:18










  • @KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
    – SampleTime
    Nov 21 at 14:36










  • I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
    – Kwin van der Veen
    Nov 22 at 3:06















up vote
1
down vote

favorite












I have often read that RHP zeros and poles set limits on the maximum achievable performance of LTI systems.



However, what does that exactly mean and how can you compute these performance limits?



For example, take these three transfer functions:



$$
begin{align}
G_1(s) &= frac{1}{(s + 1)(s - 2)} quadtext{(RHP pole)} \
G_2(s) &= frac{(s - 3)}{(s + 1)(s + 2)} quadtext{(RHP zero)} \
G_3(s) &= frac{(s - 3)}{(s + 1)(s + 2)(s - 5)} quadtext{(RHP pole and zero)}
end{align}
$$



What would be the limits of performance for these transfer functions? And are there differences between RHP zeros and RHP poles in terms of achievable performance?










share|cite|improve this question






















  • How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
    – Kwin van der Veen
    Nov 21 at 14:18










  • @KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
    – SampleTime
    Nov 21 at 14:36










  • I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
    – Kwin van der Veen
    Nov 22 at 3:06













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have often read that RHP zeros and poles set limits on the maximum achievable performance of LTI systems.



However, what does that exactly mean and how can you compute these performance limits?



For example, take these three transfer functions:



$$
begin{align}
G_1(s) &= frac{1}{(s + 1)(s - 2)} quadtext{(RHP pole)} \
G_2(s) &= frac{(s - 3)}{(s + 1)(s + 2)} quadtext{(RHP zero)} \
G_3(s) &= frac{(s - 3)}{(s + 1)(s + 2)(s - 5)} quadtext{(RHP pole and zero)}
end{align}
$$



What would be the limits of performance for these transfer functions? And are there differences between RHP zeros and RHP poles in terms of achievable performance?










share|cite|improve this question













I have often read that RHP zeros and poles set limits on the maximum achievable performance of LTI systems.



However, what does that exactly mean and how can you compute these performance limits?



For example, take these three transfer functions:



$$
begin{align}
G_1(s) &= frac{1}{(s + 1)(s - 2)} quadtext{(RHP pole)} \
G_2(s) &= frac{(s - 3)}{(s + 1)(s + 2)} quadtext{(RHP zero)} \
G_3(s) &= frac{(s - 3)}{(s + 1)(s + 2)(s - 5)} quadtext{(RHP pole and zero)}
end{align}
$$



What would be the limits of performance for these transfer functions? And are there differences between RHP zeros and RHP poles in terms of achievable performance?







dynamical-systems control-theory linear-control






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asked Nov 19 at 19:44









SampleTime

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  • How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
    – Kwin van der Veen
    Nov 21 at 14:18










  • @KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
    – SampleTime
    Nov 21 at 14:36










  • I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
    – Kwin van der Veen
    Nov 22 at 3:06


















  • How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
    – Kwin van der Veen
    Nov 21 at 14:18










  • @KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
    – SampleTime
    Nov 21 at 14:36










  • I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
    – Kwin van der Veen
    Nov 22 at 3:06
















How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
– Kwin van der Veen
Nov 21 at 14:18




How would you define performance? The maximal achievable bandwidth while still having a sensitivity below a certain threshold?
– Kwin van der Veen
Nov 21 at 14:18












@KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
– SampleTime
Nov 21 at 14:36




@KwinvanderVeen Well yes, but that would be part of the question tbh. Because I have often read that statement, but without any specification what exactly is meant with that. Maximumal achievable bandwidth sounds interesting, can you elaborate please?
– SampleTime
Nov 21 at 14:36












I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
– Kwin van der Veen
Nov 22 at 3:06




I do not have any proof for this, but I believe you can use a rule of thumb that: a system with unstable pole has a lower bound for the bandwidth roughly equal to the absolute value of the unstable pole; a system with unstable zero has an upper bound for the bandwidth roughly equal to the absolute value of the unstable zero. You can have bandwidths which violate these bounds, but at the cost of a large peak in your sensitivity. This would mean that for $G_3$ you might not be able to get the sensitivity below a desired threshold.
– Kwin van der Veen
Nov 22 at 3:06










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Although this is not a rigorous proof, it should at least demonstrate the limitation that RHP poles and zeros put on the bandwidth in combination with the peak of the sensitivity.



If you have a system with only one RHP pole or only one RHP zero then, although bad practice, you can always cancel the remaining poles, zeros and gain in the controller. In order ensure that the controller has a proper transfer function one can always add a high bandwidth low-pass filter of sufficiently high order. This low-pass filter shouldn't effect the closed loop much, since this is mainly dominated around the frequency range where the magnitude of the open loop (system times controller) crosses the 0 dB line. Furthermore by using time scaling the RHP pole or zero can always be normalized to $-1$.





In the case of only one RHP pole the considered system can be generalized to



$$
G(s) = frac{1}{s - 1}.
$$



By using a controller of the form



$$
C(s) = frac{a,s + b}{s}
$$



then the sensitivity transfer function would look like



$$
S(s) = frac{s(s - 1)}{s^2 + (a - 1)s + b}.
$$



Using $b = omega^2$ and $a=2,zeta,omega+1$ gives a more standard form



$$
S(s) = frac{s(s - 1)}{s^2 + 2,zeta,omega,s + omega^2},
$$



where $omega$ can be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB.



When $omega < 1$ then the slope of the assymptote of $S(s)$ after a frequency of $omega$ will decrease by two to $-1$ and eventually increase to the final assymptote with slope zero after a frequency of $1$. So, before the assymptote goes to the 0 dB line, the slope is negative which means that the magnitude of $S(s)$ during that interval is above 0 dB. The further $omega$ lies below one, the higher the magnitude of $S(s)$ will go above 0 dB.



When $omega > 1$ then the slope of the assymptote of $S(s)$ after a frequency of $1$ will increase by one to $+2$ and eventually decrease to the final assymptote with slope zero after a frequency of $omega$. So, before the assymptote goes to the 0 dB line, the slope is always positive which means that there shouldn't be a magnitude of $S(s)$ which goes significantly above 0 dB.



The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



enter image description here





In the case of only one RHP zero the considered system can be generalized to



$$
G(s) = frac{s - 1}{s + p}.
$$



The pole at $p>0$ is just added to make the system proper. Now by using a controller of the form



$$
C(s) = frac{a(s + p)}{s^2 + b,s}
$$



then the sensitivity transfer function would look like



$$
S(s) = frac{s (s + b)}{s^2 + (a + b)s - a}.
$$



Using $a = -omega^2$ and $b=omega(2,zeta + omega)$ again gives a more standard form



$$
S(s) = frac{s (s + omega(2,zeta + omega))}{s^2 + 2,zeta,omega,s + omega^2},
$$



where $omega$ can again be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB. The transition for the magnitude of the peak of the sensitivity now does not lie near $omega=1$ but roughly at $omega=2,zeta$.



When $omega < 2,zeta$ then the zero of $S(s)$ is of the same magnitude as the bandwidth (assuming a normal value for $zeta$). This means that the slope of the assymptote of $S(s)$ a little after a frequency of $omega$ will eventually decrease by one to zero (decrease by two and increase by one). The damping coefficient can influence this a little, namely the zero might lie a little ahead or behind $omega$, but for realistic values for $zeta$ the assymptote does not change much.



When $omega > 2,zeta$ then the zero of $S(s)$ scales with the square of the bandwidth. So between the frequencies $omega$ and roughly $omega^2$ the assymptote of $S(s)$ will be $-1$ and thus its magnitude of $S(s)$ will have a significant portion above 0 dB.



The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



enter image description here





For a RHP pole the maximal magnitude of the sensitivity quickly goes up when the bandwidth is chosen below the break frequency of the pole. The opposite is true for a RHP zero, so when the bandwidth is chosen above the break frequency of the zero. So it is possible to place the bandwidth anywhere you want if you have a single RHP pole or zero, but you will have poor performance. Namely a large maximal magnitude of the sensitivity transfer function means large amplification of disturbances that act of the system. So often one would want to keep the magnitude of the sensitivity below roughly 6 dB.






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    up vote
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    down vote



    accepted










    Although this is not a rigorous proof, it should at least demonstrate the limitation that RHP poles and zeros put on the bandwidth in combination with the peak of the sensitivity.



    If you have a system with only one RHP pole or only one RHP zero then, although bad practice, you can always cancel the remaining poles, zeros and gain in the controller. In order ensure that the controller has a proper transfer function one can always add a high bandwidth low-pass filter of sufficiently high order. This low-pass filter shouldn't effect the closed loop much, since this is mainly dominated around the frequency range where the magnitude of the open loop (system times controller) crosses the 0 dB line. Furthermore by using time scaling the RHP pole or zero can always be normalized to $-1$.





    In the case of only one RHP pole the considered system can be generalized to



    $$
    G(s) = frac{1}{s - 1}.
    $$



    By using a controller of the form



    $$
    C(s) = frac{a,s + b}{s}
    $$



    then the sensitivity transfer function would look like



    $$
    S(s) = frac{s(s - 1)}{s^2 + (a - 1)s + b}.
    $$



    Using $b = omega^2$ and $a=2,zeta,omega+1$ gives a more standard form



    $$
    S(s) = frac{s(s - 1)}{s^2 + 2,zeta,omega,s + omega^2},
    $$



    where $omega$ can be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB.



    When $omega < 1$ then the slope of the assymptote of $S(s)$ after a frequency of $omega$ will decrease by two to $-1$ and eventually increase to the final assymptote with slope zero after a frequency of $1$. So, before the assymptote goes to the 0 dB line, the slope is negative which means that the magnitude of $S(s)$ during that interval is above 0 dB. The further $omega$ lies below one, the higher the magnitude of $S(s)$ will go above 0 dB.



    When $omega > 1$ then the slope of the assymptote of $S(s)$ after a frequency of $1$ will increase by one to $+2$ and eventually decrease to the final assymptote with slope zero after a frequency of $omega$. So, before the assymptote goes to the 0 dB line, the slope is always positive which means that there shouldn't be a magnitude of $S(s)$ which goes significantly above 0 dB.



    The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



    enter image description here





    In the case of only one RHP zero the considered system can be generalized to



    $$
    G(s) = frac{s - 1}{s + p}.
    $$



    The pole at $p>0$ is just added to make the system proper. Now by using a controller of the form



    $$
    C(s) = frac{a(s + p)}{s^2 + b,s}
    $$



    then the sensitivity transfer function would look like



    $$
    S(s) = frac{s (s + b)}{s^2 + (a + b)s - a}.
    $$



    Using $a = -omega^2$ and $b=omega(2,zeta + omega)$ again gives a more standard form



    $$
    S(s) = frac{s (s + omega(2,zeta + omega))}{s^2 + 2,zeta,omega,s + omega^2},
    $$



    where $omega$ can again be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB. The transition for the magnitude of the peak of the sensitivity now does not lie near $omega=1$ but roughly at $omega=2,zeta$.



    When $omega < 2,zeta$ then the zero of $S(s)$ is of the same magnitude as the bandwidth (assuming a normal value for $zeta$). This means that the slope of the assymptote of $S(s)$ a little after a frequency of $omega$ will eventually decrease by one to zero (decrease by two and increase by one). The damping coefficient can influence this a little, namely the zero might lie a little ahead or behind $omega$, but for realistic values for $zeta$ the assymptote does not change much.



    When $omega > 2,zeta$ then the zero of $S(s)$ scales with the square of the bandwidth. So between the frequencies $omega$ and roughly $omega^2$ the assymptote of $S(s)$ will be $-1$ and thus its magnitude of $S(s)$ will have a significant portion above 0 dB.



    The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



    enter image description here





    For a RHP pole the maximal magnitude of the sensitivity quickly goes up when the bandwidth is chosen below the break frequency of the pole. The opposite is true for a RHP zero, so when the bandwidth is chosen above the break frequency of the zero. So it is possible to place the bandwidth anywhere you want if you have a single RHP pole or zero, but you will have poor performance. Namely a large maximal magnitude of the sensitivity transfer function means large amplification of disturbances that act of the system. So often one would want to keep the magnitude of the sensitivity below roughly 6 dB.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Although this is not a rigorous proof, it should at least demonstrate the limitation that RHP poles and zeros put on the bandwidth in combination with the peak of the sensitivity.



      If you have a system with only one RHP pole or only one RHP zero then, although bad practice, you can always cancel the remaining poles, zeros and gain in the controller. In order ensure that the controller has a proper transfer function one can always add a high bandwidth low-pass filter of sufficiently high order. This low-pass filter shouldn't effect the closed loop much, since this is mainly dominated around the frequency range where the magnitude of the open loop (system times controller) crosses the 0 dB line. Furthermore by using time scaling the RHP pole or zero can always be normalized to $-1$.





      In the case of only one RHP pole the considered system can be generalized to



      $$
      G(s) = frac{1}{s - 1}.
      $$



      By using a controller of the form



      $$
      C(s) = frac{a,s + b}{s}
      $$



      then the sensitivity transfer function would look like



      $$
      S(s) = frac{s(s - 1)}{s^2 + (a - 1)s + b}.
      $$



      Using $b = omega^2$ and $a=2,zeta,omega+1$ gives a more standard form



      $$
      S(s) = frac{s(s - 1)}{s^2 + 2,zeta,omega,s + omega^2},
      $$



      where $omega$ can be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB.



      When $omega < 1$ then the slope of the assymptote of $S(s)$ after a frequency of $omega$ will decrease by two to $-1$ and eventually increase to the final assymptote with slope zero after a frequency of $1$. So, before the assymptote goes to the 0 dB line, the slope is negative which means that the magnitude of $S(s)$ during that interval is above 0 dB. The further $omega$ lies below one, the higher the magnitude of $S(s)$ will go above 0 dB.



      When $omega > 1$ then the slope of the assymptote of $S(s)$ after a frequency of $1$ will increase by one to $+2$ and eventually decrease to the final assymptote with slope zero after a frequency of $omega$. So, before the assymptote goes to the 0 dB line, the slope is always positive which means that there shouldn't be a magnitude of $S(s)$ which goes significantly above 0 dB.



      The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



      enter image description here





      In the case of only one RHP zero the considered system can be generalized to



      $$
      G(s) = frac{s - 1}{s + p}.
      $$



      The pole at $p>0$ is just added to make the system proper. Now by using a controller of the form



      $$
      C(s) = frac{a(s + p)}{s^2 + b,s}
      $$



      then the sensitivity transfer function would look like



      $$
      S(s) = frac{s (s + b)}{s^2 + (a + b)s - a}.
      $$



      Using $a = -omega^2$ and $b=omega(2,zeta + omega)$ again gives a more standard form



      $$
      S(s) = frac{s (s + omega(2,zeta + omega))}{s^2 + 2,zeta,omega,s + omega^2},
      $$



      where $omega$ can again be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB. The transition for the magnitude of the peak of the sensitivity now does not lie near $omega=1$ but roughly at $omega=2,zeta$.



      When $omega < 2,zeta$ then the zero of $S(s)$ is of the same magnitude as the bandwidth (assuming a normal value for $zeta$). This means that the slope of the assymptote of $S(s)$ a little after a frequency of $omega$ will eventually decrease by one to zero (decrease by two and increase by one). The damping coefficient can influence this a little, namely the zero might lie a little ahead or behind $omega$, but for realistic values for $zeta$ the assymptote does not change much.



      When $omega > 2,zeta$ then the zero of $S(s)$ scales with the square of the bandwidth. So between the frequencies $omega$ and roughly $omega^2$ the assymptote of $S(s)$ will be $-1$ and thus its magnitude of $S(s)$ will have a significant portion above 0 dB.



      The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



      enter image description here





      For a RHP pole the maximal magnitude of the sensitivity quickly goes up when the bandwidth is chosen below the break frequency of the pole. The opposite is true for a RHP zero, so when the bandwidth is chosen above the break frequency of the zero. So it is possible to place the bandwidth anywhere you want if you have a single RHP pole or zero, but you will have poor performance. Namely a large maximal magnitude of the sensitivity transfer function means large amplification of disturbances that act of the system. So often one would want to keep the magnitude of the sensitivity below roughly 6 dB.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Although this is not a rigorous proof, it should at least demonstrate the limitation that RHP poles and zeros put on the bandwidth in combination with the peak of the sensitivity.



        If you have a system with only one RHP pole or only one RHP zero then, although bad practice, you can always cancel the remaining poles, zeros and gain in the controller. In order ensure that the controller has a proper transfer function one can always add a high bandwidth low-pass filter of sufficiently high order. This low-pass filter shouldn't effect the closed loop much, since this is mainly dominated around the frequency range where the magnitude of the open loop (system times controller) crosses the 0 dB line. Furthermore by using time scaling the RHP pole or zero can always be normalized to $-1$.





        In the case of only one RHP pole the considered system can be generalized to



        $$
        G(s) = frac{1}{s - 1}.
        $$



        By using a controller of the form



        $$
        C(s) = frac{a,s + b}{s}
        $$



        then the sensitivity transfer function would look like



        $$
        S(s) = frac{s(s - 1)}{s^2 + (a - 1)s + b}.
        $$



        Using $b = omega^2$ and $a=2,zeta,omega+1$ gives a more standard form



        $$
        S(s) = frac{s(s - 1)}{s^2 + 2,zeta,omega,s + omega^2},
        $$



        where $omega$ can be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB.



        When $omega < 1$ then the slope of the assymptote of $S(s)$ after a frequency of $omega$ will decrease by two to $-1$ and eventually increase to the final assymptote with slope zero after a frequency of $1$. So, before the assymptote goes to the 0 dB line, the slope is negative which means that the magnitude of $S(s)$ during that interval is above 0 dB. The further $omega$ lies below one, the higher the magnitude of $S(s)$ will go above 0 dB.



        When $omega > 1$ then the slope of the assymptote of $S(s)$ after a frequency of $1$ will increase by one to $+2$ and eventually decrease to the final assymptote with slope zero after a frequency of $omega$. So, before the assymptote goes to the 0 dB line, the slope is always positive which means that there shouldn't be a magnitude of $S(s)$ which goes significantly above 0 dB.



        The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



        enter image description here





        In the case of only one RHP zero the considered system can be generalized to



        $$
        G(s) = frac{s - 1}{s + p}.
        $$



        The pole at $p>0$ is just added to make the system proper. Now by using a controller of the form



        $$
        C(s) = frac{a(s + p)}{s^2 + b,s}
        $$



        then the sensitivity transfer function would look like



        $$
        S(s) = frac{s (s + b)}{s^2 + (a + b)s - a}.
        $$



        Using $a = -omega^2$ and $b=omega(2,zeta + omega)$ again gives a more standard form



        $$
        S(s) = frac{s (s + omega(2,zeta + omega))}{s^2 + 2,zeta,omega,s + omega^2},
        $$



        where $omega$ can again be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB. The transition for the magnitude of the peak of the sensitivity now does not lie near $omega=1$ but roughly at $omega=2,zeta$.



        When $omega < 2,zeta$ then the zero of $S(s)$ is of the same magnitude as the bandwidth (assuming a normal value for $zeta$). This means that the slope of the assymptote of $S(s)$ a little after a frequency of $omega$ will eventually decrease by one to zero (decrease by two and increase by one). The damping coefficient can influence this a little, namely the zero might lie a little ahead or behind $omega$, but for realistic values for $zeta$ the assymptote does not change much.



        When $omega > 2,zeta$ then the zero of $S(s)$ scales with the square of the bandwidth. So between the frequencies $omega$ and roughly $omega^2$ the assymptote of $S(s)$ will be $-1$ and thus its magnitude of $S(s)$ will have a significant portion above 0 dB.



        The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



        enter image description here





        For a RHP pole the maximal magnitude of the sensitivity quickly goes up when the bandwidth is chosen below the break frequency of the pole. The opposite is true for a RHP zero, so when the bandwidth is chosen above the break frequency of the zero. So it is possible to place the bandwidth anywhere you want if you have a single RHP pole or zero, but you will have poor performance. Namely a large maximal magnitude of the sensitivity transfer function means large amplification of disturbances that act of the system. So often one would want to keep the magnitude of the sensitivity below roughly 6 dB.






        share|cite|improve this answer














        Although this is not a rigorous proof, it should at least demonstrate the limitation that RHP poles and zeros put on the bandwidth in combination with the peak of the sensitivity.



        If you have a system with only one RHP pole or only one RHP zero then, although bad practice, you can always cancel the remaining poles, zeros and gain in the controller. In order ensure that the controller has a proper transfer function one can always add a high bandwidth low-pass filter of sufficiently high order. This low-pass filter shouldn't effect the closed loop much, since this is mainly dominated around the frequency range where the magnitude of the open loop (system times controller) crosses the 0 dB line. Furthermore by using time scaling the RHP pole or zero can always be normalized to $-1$.





        In the case of only one RHP pole the considered system can be generalized to



        $$
        G(s) = frac{1}{s - 1}.
        $$



        By using a controller of the form



        $$
        C(s) = frac{a,s + b}{s}
        $$



        then the sensitivity transfer function would look like



        $$
        S(s) = frac{s(s - 1)}{s^2 + (a - 1)s + b}.
        $$



        Using $b = omega^2$ and $a=2,zeta,omega+1$ gives a more standard form



        $$
        S(s) = frac{s(s - 1)}{s^2 + 2,zeta,omega,s + omega^2},
        $$



        where $omega$ can be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB.



        When $omega < 1$ then the slope of the assymptote of $S(s)$ after a frequency of $omega$ will decrease by two to $-1$ and eventually increase to the final assymptote with slope zero after a frequency of $1$. So, before the assymptote goes to the 0 dB line, the slope is negative which means that the magnitude of $S(s)$ during that interval is above 0 dB. The further $omega$ lies below one, the higher the magnitude of $S(s)$ will go above 0 dB.



        When $omega > 1$ then the slope of the assymptote of $S(s)$ after a frequency of $1$ will increase by one to $+2$ and eventually decrease to the final assymptote with slope zero after a frequency of $omega$. So, before the assymptote goes to the 0 dB line, the slope is always positive which means that there shouldn't be a magnitude of $S(s)$ which goes significantly above 0 dB.



        The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



        enter image description here





        In the case of only one RHP zero the considered system can be generalized to



        $$
        G(s) = frac{s - 1}{s + p}.
        $$



        The pole at $p>0$ is just added to make the system proper. Now by using a controller of the form



        $$
        C(s) = frac{a(s + p)}{s^2 + b,s}
        $$



        then the sensitivity transfer function would look like



        $$
        S(s) = frac{s (s + b)}{s^2 + (a + b)s - a}.
        $$



        Using $a = -omega^2$ and $b=omega(2,zeta + omega)$ again gives a more standard form



        $$
        S(s) = frac{s (s + omega(2,zeta + omega))}{s^2 + 2,zeta,omega,s + omega^2},
        $$



        where $omega$ can again be used as a measure of the bandwidth. As expected for a sensitivity transfer function at really low frequencies the assymptote of $S(s)$ has a positive slope, namely $+1$, and at really high frequencies the assymptote of $S(s)$ has a slope of zero and a magnitude of 0 dB. The transition for the magnitude of the peak of the sensitivity now does not lie near $omega=1$ but roughly at $omega=2,zeta$.



        When $omega < 2,zeta$ then the zero of $S(s)$ is of the same magnitude as the bandwidth (assuming a normal value for $zeta$). This means that the slope of the assymptote of $S(s)$ a little after a frequency of $omega$ will eventually decrease by one to zero (decrease by two and increase by one). The damping coefficient can influence this a little, namely the zero might lie a little ahead or behind $omega$, but for realistic values for $zeta$ the assymptote does not change much.



        When $omega > 2,zeta$ then the zero of $S(s)$ scales with the square of the bandwidth. So between the frequencies $omega$ and roughly $omega^2$ the assymptote of $S(s)$ will be $-1$ and thus its magnitude of $S(s)$ will have a significant portion above 0 dB.



        The two cases above (and the case when $omega=1$) are also illustrated in the figure below which uses $zeta = tfrac{1}{2}sqrt{2}$:



        enter image description here





        For a RHP pole the maximal magnitude of the sensitivity quickly goes up when the bandwidth is chosen below the break frequency of the pole. The opposite is true for a RHP zero, so when the bandwidth is chosen above the break frequency of the zero. So it is possible to place the bandwidth anywhere you want if you have a single RHP pole or zero, but you will have poor performance. Namely a large maximal magnitude of the sensitivity transfer function means large amplification of disturbances that act of the system. So often one would want to keep the magnitude of the sensitivity below roughly 6 dB.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 7:40

























        answered Nov 22 at 7:25









        Kwin van der Veen

        5,2602826




        5,2602826






























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