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4.3: Steady-State Error Improvement

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    24404
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    Steady-State Tracking Error

    A tracking control system is designed to have a low steady-state error in response to a constant (i.e., unit-step) or linearly varying (i.e., unit-ramp) input. The desired error tolerance may be specified as percentage of the reference input. Additionally the system response to a constant disturbance input should stays small.

    We consider the feedback control system (Figure 4.3.1) with input \(r(s)\) and output \(y(s)\). 

    clipboard_e6356cc9b518d7071d9bf638e71c14036.png
    Figure \(\PageIndex{1}\): Copy and Paste Caption here. (Copyright; author via source)

    Let \(T\left(s\right)\) denote the closed-loop transfer function; then \(y\left(s\right)=T\left(s\right)r(s)\). The tracking error, \(e(s)\), in response to a reference signal, \(r(s)\), is defined as:

    \[e\left(s\right)=\left(1-T(s)\right)r\left(s\right) \nonumber \]

    To characterize the tracking error, we consider a unity-gain feedback system (\(H(s)=1\)). Then, the tracking error, \(e(s)\), is given as:

    \[e(s)=\frac{1}{1+KG(s)} r(s) \nonumber \]

    By using the final-value theorem (FVT), the steady-state tracking error is computed as: \(e(\infty )={\mathop{\lim }\limits_{s\to 0}} \:se(s) \)

    The steady-state error to a step reference input, \(\left(r(s)=\frac{1}{s} \right)\), is given as: \(e(\infty )=\frac{1}{1+KG(0)} .\)

    The steady-state error to a ramp reference input, \(\left(r(s)=\frac{1}{s^{2} } \right)\), is given as: \(e(\infty )=\frac{1}{\left. sKG(s)\right|_{s=0} } .\)

    System Error Constants

    The steady-state tracking error in the case of a unity-gain feedback control system is characterized in terms of system position and velocity error constants, which are defined as:

    \[K_ p ={\mathop{\lim }\limits_{s\to 0}} KG(s) \nonumber \]

    \[K_{ v} ={\mathop{\lim }\limits_{s\to 0}} sKG(s) \nonumber \]

    In terms of the error constants, the steady-state tracking error to a step or a ramp input is evaluated as:

    \[\left. e(\infty )\right|_{ step} =\frac{1}{1+K_ p } \nonumber \]

    \[\left. e(\infty )\right|_{ ramp} =\frac{1}{K_v } \nonumber \]

    Example \(\PageIndex{1}\)

    Let \(KG(s)=\frac{K}{s(s+2)}\); then we have: \(K_ p =\infty ,\; \; K_{ v} =\frac{K}{2}\).

    Hence \(\left. e(\infty )\right|_{ step} =0, \;\left. e(\infty )\right|_{ramp} \; =\; \frac{2}{K} .\)

    We note that the presence of an integrator in the loop forces \(K_p\to \infty\), and the steady-state error to a step input to go to zero.

    Example \(\PageIndex{2}\)

    The approximate model of a small DC motor is given as:

    \[KG\left(s\right)=\frac{500K}{\left(s+10\right)\left(s+100\right)} \nonumber \]

    Then we have: \(K_p=0.5K\), and \(K_v=0\). 

    Hence \(\left. e(\infty )\right|_{ step} =\frac{2}{K}, \;\left. e(\infty )\right|_{ramp} \; =\; \infty .\)

    Integral control, \(K\left(s\right)=\frac{k_i}{s}\), is often employed to reduce the steady-state tracking error. The presence of an integrator in the loop forces \(K_p\to \infty\), and the steady-state error to a step input to go to zero.

    Steady-State Error to Ramp Input

    Assuming that the feedback loop contains an integrator, so that the steady-state error to a step input is zero, the steady-state error to a ramp input is expressed as:

    \[e(s)=\left[1-T(s)\right]r(s);\; \, \, r(s)=\frac{1}{s^{2} } . \nonumber \]

    The error is evaluated by the application of FVT: \(\left. e(\infty )\right|_{ramp} ={\mathop{\lim }\limits_{s\to 0}} \frac{1-T(s)}{s} .\)

    By using the L’ Hospital’s rule, we obtain: \(\left. e(\infty )\right|_{ramp} ={\mathop{\lim }\limits_{s\to 0}} \, \left(-\frac{ dT(s)}{ ds} \right)\).

    To evaluate the RHS, we use the natural logarithm to write: \(\frac d{ ds} \ln T(s)=\frac{1}{T(s)} \frac{ dT(s)}{ ds} .\)

    Since \({\mathop{\lim }\limits_{s\to 0}}\; T(s)=1\) by the integrator assumption, we have: \({\mathop{\lim }\limits_{s\to 0}}\;\frac d{ ds} \ln T(s)\; ={\mathop{\lim }\limits_{s\to 0}} \;\frac{ dT(s)}{ ds}\).

    Next, assume that \(T(s)\) is expressed as: \(T(s)=\frac{K(s-z_{1} )\ldots (s-z_{m} )}{(s-p_{1} )\ldots (s-p_{n} )} .\)

    Then, we have: \(\ln T(s)=\ln K+\sum _{{ l}n} (s-z_{i} )-\sum _{{ l}n} (s-p_{i} ),\) and

    \[\frac d{ ds} \ln T(s)=\ln K+\sum \frac{1}{s-z_{i} } -\sum \frac{1}{s-p_{i} } \nonumber \] It follows that: \[\left. e(\infty )\right|_{ ramp} =\sum \frac{1}{p_{i} } -\sum \frac{1}{z_{i} } \nonumber \]

    where \(z_{i}\) and \(p_{i}\) denote the zeros and poles of the closed-loop system.

    Example \(\PageIndex{3}\)

    Let \(KG(s)=\frac{K}{s(s+2)}\); then, the closed-loop characteristic polynomial is: \(\mathit{\Delta}\left(s\right)=s^2+2s+K\).

    Assuming a value of \(K=1\), the closed-loop poles are located at: \(s_{1,2}=-1,\ -1\).

    The resulting steady-state error to a ramp input is given as: \(\left. e(\infty )\right|_{ramp} =\sum \frac{1}{p_{i} } =2\).

    For verification, the closed-loop system response is plotted in Figure 4.3.1.

    clipboard_e9647d2abea90445cb888a23d4c9a9375.png
    Figure \(\PageIndex{1}\): Steady-state error to a ramp input (Example 4.3.3).

     


    This page titled 4.3: Steady-State Error Improvement is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kamran Iqbal.

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