Block Two


  • Chapter 5. Reduction of multiple subsystems


    block diagrams
    signal flow graphs
    similarity transforms *

    Block diagrams and signal flow graphs

     When developing or describing a control system, diagrams showing signal connections are of great benefit.


    Example 5.2, fig 5.11, p 243 / 241


    \begin{align*} \frac{C(s)}{R(s)} &= \frac{G_1(s) G_3(s) [1+G_2(s)]} {[1+G_2(s)H_2(s)+G_1(s)G_2(s)H_1(s)][1+G_3(s)H_3(s)]} \end{align*}
    (Control Engineering 3 notes: weeks 1-6)


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    Example 5.2, fig 5.11, p 243 / 241



    Example 5.2, fig 5.19, p250 / 248
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    Example 5.8 as a Scilab Xcos model

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    Phase variable block diagram


    Fig 3.12, example 3.5, p137
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    Cascade state space


    Fig 5.24(b), p258
    \begin{align*} \dot x_1 &= -4 x_1 && + x_2 && && \\ \dot x_2 &= && -3 x_2 && + x_3 && \\ \dot x_3 &= && && -2 x_3 && + 24 r \end{align*}
    \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -4 & 1 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & -2 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 0 \\ 0 \\ 24 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 1 & 0 & 0 \end{pmatrix} \underset{\sim}{ x} \end{align*}
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    Controller canonical form
    Controller canonical form: renumber phase variables in reverse order (p261).

    Phase variable form: \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -24 & -26 & -9 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 2 & 7 & 1 \end{pmatrix} \underset{\sim}{x} \end{align*} Controller canonical form: \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -9 & -26 & -24 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 1 & 7 & 2 \end{pmatrix} \underset{\sim}{x} \end{align*}
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    Chapter 5 - The Laplace transform

    Transfer functions
    Decomposing block diagrams into regions with a single input and single output (possibly moving blocks to achieve this), block diagrams can be reduced to transfer functions for analysis in the frequency domain.

    Block diagrams can also be used to develop Simulink models directly, using a series of “Transfer Function” blocks.

    Signal flow diagrams are alternatives to block diagrams that emphasise the model variables (states, outputs, etc.) instead of the component transfer functions.
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    Example 5.8, fig 5.30, p 264 / 261

    Signal flow diagrams can help to understand or visualise the relationship between states in a state space model. \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -3 & 1 \\ -200 & -102 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 0 \\ 100 \end{pmatrix} \underset{\sim}{r} \\ \underset{\sim}{y} &= \begin{pmatrix} 2 & 1 \end{pmatrix} \underset{\sim}{x} \end{align*}
    When drawing or analysing signal flow graphs, it is usually easiest to work right to left.

    Start with the outputs and work backwards, adding states as necessary, an integrator ($1/s$) for each state, and the links between the states and their rates. \begin{align*} C &= 5 X_1 + \dot X_1 \qquad= 5 X_1 + (X_2 -3 X_1) \qquad= 2 X_1 + X_2 \\ \dot X_1 &= -3 X_1 + X_2 \\ \dot X_2 &= -2 X_2 + 100 (R - C) \qquad= -200 X_1 - 102 X_2 + 100 R \end{align*}
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    Alternative state space representations

    Phase variables
    Phase variables are the states that we encountered in Chapter 3, where the states comprise successive derivatives of each other. \begin{align*} \ddot x_1 & &+ a &\dot x_1 & + b &x_1 & &=& c u \\ \downarrow& & &\downarrow& &\downarrow& && \\ \downarrow& & & \chi_2 & & \chi_1 & && \\ \downarrow& & &\downarrow& &\downarrow& && \\ \dot \chi_2& & & \dot \chi_1& & \chi_1 & && \end{align*} \begin{align*} \begin{pmatrix} \dot\chi_1 \\ \dot\chi_2 \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ -b& -a& \end{pmatrix} \begin{pmatrix} \chi_1 \\ \chi_2 \end{pmatrix} + \begin{pmatrix} 0 \\ c \end{pmatrix} u \end{align*}
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    Cascade form signal graph


    Fig 5.23, p257


    Fig 5.24(b), p258
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    Parallel (decoupled) states


    Fig 5.25, p259
    \begin{align*} \frac{C(s)}{R(s)} &= \frac{24}{(s+2)(s+3)(s+4)} = \frac{12}{s+2}-\frac{24}{s+3}+\frac{12}{s+4} \end{align*} \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -2 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -4 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 12 \\ -24 \\ 12 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 1 & 1 & 1 \end{pmatrix} \underset{\sim}{x} \end{align*}
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    Observer canonical form
    Observer canonical form: transpose of controller canonical form (p263).

    Controller canonical: \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -9 & -26 & -24 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 1 & 7 & 2 \end{pmatrix} \underset{\sim}{x} \end{align*} Observer canonical: \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -9 & 1 & 0 \\ -26 & 0 & 1 \\ -24 & 0 & 0 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} 1 \\ 7 \\ 2 \end{pmatrix} r \\ \underset{\sim}{y} &= \begin{pmatrix} 1 & 0 & 0 \end{pmatrix} \underset{\sim}{x} \end{align*}
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    Uncoupling systems

    Matrix diagonalisation can be used to diagonalise the state matrix.

    A diagonalised state matrix implies a decoupled system, in which each state is affected only by itself and the inputs; not the other states.

    Decoupled systems are generally easier to analyse and develop control systems for.
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    Diagonalised matrix

    If matrix $\mathrm{P}$ comprises the eigenvectors of matrix $\mathrm{A}$, then the matrix product $\mathrm{P^{-1}}\mathrm{A}\mathrm{P}$ will be a diagonal matrix.

    Hence this transformation can be used to decouple a state space model.
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    Diagonalisation with Matlab


    A = [ 0, 1; -2, -3]
    B = [ 0; 1 ]
    C = [ 1, 0 ]
    D = [ 0 ]
    [ P, Ad ] = eig(A)
    assert((inv(P)*A*P - Ad) = zeros(size(A))
    Bd = inv(P)*B
    Cd = C*P
    G1 = ss( A, B, C, D)
    figure(1)
    step(G1)
    G2 = ss( Ad, Bd, Cd, D)
    figure(2)
    step(G2)
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    Similarity transforms

    Similarity transforms

    ``Oh, what a tangled web we weave. When first we desire to'' ... model engineering systems.
    (apologies to Walter Scott)


    Fig 5.31, p265

    \begin{align*} \underset{\sim}{\dot x} &= \begin{pmatrix} -4 & 0 \\ 0 & -6 \end{pmatrix} \underset{\sim}{x} + \begin{pmatrix} -\frac{1}{2} \\ \frac{3}{2} \end{pmatrix} \underset{\sim}{r} & \underset{\sim}{y} &= \begin{pmatrix} 1 & 1 \end{pmatrix} \underset{\sim}{x} \end{align*}
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    Similarity Transformation ($\S$5.8, p266)

    Coupled state space system: \begin{align*} \underset{\sim}{\dot x} &= \mathrm{A} \underset{\sim}{x} + \mathrm{B} \underset{\sim}{u} \\ \underset{\sim}{y} &= \mathrm{C} \underset{\sim}{x} + \mathrm{D} \underset{\sim}{u} \end{align*}
    Let $\underset{\sim}{z} = \mathrm{P^{-1}} \underset{\sim}{x} \;$ (i.e. $\underset{\sim}{x} \rightarrow \mathrm{P}\underset{\sim}{z}$, $\underset{\sim}{\dot x} \rightarrow \mathrm{P}\underset{\sim}{ \dot z}$).

    \begin{align*} \mathrm{P} \underset{\sim}{\dot z} &= \mathrm{A} \mathrm{P} \underset{\sim}{z} + \mathrm{B} \underset{\sim}{u} \\ \underset{\sim}{y} &= \mathrm{C} \mathrm{P} \underset{\sim}{z} + \mathrm{D} \underset{\sim}{u} \end{align*}
    \begin{align*} \underset{\sim}{\dot z} &= \mathrm{P^{-1}} \mathrm{A} \mathrm{P} \underset{\sim}{z} + \mathrm{P^{-1}} \mathrm{B} \underset{\sim}{u} \\ \underset{\sim}{y} &= \mathrm{C} \mathrm{P} \underset{\sim}{z} + \mathrm{D} \underset{\sim}{u} \end{align*}
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    Diagonalisation example

    State space model:
    \begin{align*} \mathrm{A}&=\begin{pmatrix}0&1\\-2&-3\end{pmatrix}& \mathrm{B}&=\begin{pmatrix}0\\1\end{pmatrix}\\ \mathrm{C}&=\begin{pmatrix}1&0\end{pmatrix}& \mathrm{D}&=\begin{pmatrix}0\end{pmatrix} \end{align*} $eig \; (\mathrm{A}): |\lambda I - A|=0$ \begin{align*} \left|\begin{matrix} \lambda & -1 \\ 2 & \lambda+3 \end{matrix}\right| & = 0 & \lambda^2 + 3\lambda + 2 &= 0 \\ &&(\lambda + 1)(\lambda +2) &= 0 \end{align*}
    \begin{align*} \left|\begin{matrix} \lambda & -1 \\ 2 & \lambda+3 \end{matrix}\right| & = 0 & \lambda = -1 \; \text{or} \; -2 \end{align*}
    \begin{align*} \begin{pmatrix} -1 & -1 \\ 2 & -1+3 \end{pmatrix} \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \begin{pmatrix}0\\0\end{pmatrix} \; \implies \underset{\sim}{x} &= \begin{pmatrix} -1 \\ 1 \end{pmatrix} \\ \begin{pmatrix} -2 & -1 \\ 2 & -2+3 \end{pmatrix} \begin{pmatrix}x_1\\x_2\end{pmatrix} &= \begin{pmatrix}0\\0\end{pmatrix} \; \implies \underset{\sim}{x} &= \begin{pmatrix} -1 \\ 2 \end{pmatrix} \\ \mathrm{P} &= \begin{pmatrix} -1 & -1 \\ 1 & 2 \end{pmatrix} \\ \mathrm{P}^{-1}\mathrm{A}\mathrm{P} &= \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix} \end{align*}
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    Routh Hurwitz criteria

    Design tool

    The Matlab command pzmap can show the pole locations of a given system, to allow stability and transient performance analysis, but is less helpful when controller parameters are unknown.

    The Routh-Hurwitz stability criteria allow constraints on unknown parameters to be calculated, hence assisting in the design process.

    These constraint equations can be used to guide the design of physical system parameters or controller gains.
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    Routh array construction

    The Routh array has as many rows as the order of the characteristic polynomial, i.e. the highest power of $s$.

    The coefficients of $s$ (in descending order) are placed alternately in the first two rows of the array.

    Elements in subsequent rows are calculated using the two rows above, making use of the left-most column and the column to the right.

    Routh array (2nd order)
    \begin{align*} G(s) &= \frac{\mathrm{blah}}{s^2+a_1 s+a_0} \end{align*} \begin{array}{l|ll} s^2: & 1 & a_0 \\ s^1: & a_1 & 0 \\ s^0: & b_1 \end{array} \begin{align*} b_1 &= -\frac{\left| \begin{matrix} 1 & a_0 \\ a_1 & 0 \end{matrix} \right|} {a_1} \end{align*}
    Routh array (3nd order)

    \begin{align*} G(s) &= \frac{\mathrm{blah}}{s^3+a_2 s^2+a_1 s+a_0} \end{align*} \begin{array}{l|ll} s^3: & 1 & a_1 \\ s^2: & a_2 & a_0 \\ s^1: & b_1 & 0 \\ s^0: & c_1 \end{array} \begin{align*} b_1 &= -\frac{\left| \begin{matrix} 1 & a_1 \\ a_2 & a_0 \end{matrix} \right|} {a_2} & c_1 &= -\frac{\left| \begin{matrix} a_2 & a_0 \\ b_1 & 0 \end{matrix} \right|} {b_1} \end{align*}
    Routh array (4nd order)

    \begin{align*} G(s) &= \frac{\mathrm{blah}}{s^4+a_3 s^3+a_2 s^2+a_1 s+a_0} \end{align*} \begin{array}{l|lll} s^4: & 1 & a_2 & a_0\\ s^3: & a_3 & a_1 & 0\\ s^2: & b_1 & b_2 & 0\\ s^1: & c_1 & 0 \\ s^0: & d_1 & 0 \end{array} \begin{align*} b_1 &= -\frac{\left| \begin{matrix} 1 & a_2 \\ a_3 & a_1 \end{matrix} \right|} {a_3} & b_2 &= -\frac{\left| \begin{matrix} 1 & a_0 \\ a_3 & 0 \end{matrix} \right|} {a_3} \\ c_1 &= -\frac{\left| \begin{matrix} a_3 & a_1 \\ b_1 & b_2 \end{matrix} \right|} {b_1} & d_1 &= -\frac{\left| \begin{matrix} b_1 & b_2 \\ c_1 & 0 \end{matrix} \right|} {c_1} \end{align*}
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    State space stability

    State space stability

    Recall the state transition matrix: \begin{align*} \phi(s) &= (s\mathrm{I}-\mathrm{A})^{-1} \\ &= \frac{Adj(s\mathrm{I}-\mathrm{A})}{|s\mathrm{I}-\mathrm{A}|} \end{align*} The system poles are located at the eigenvalues of matrix $\mathrm{A}$.

    For stability we require that the eigenvalues of the state matrix $\mathrm{A}$ should be located in the left half plane (LHP) of the $s$-domain, i.e. negative real part.

    Unlike mathematicians, as Engineers you should be able to take systems with complex eigenvalues and eigenvectors in your stride. You have been studying such systems for years without realising it!
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    Chapter 6. Stability

    Routh Hurwitz criteria
    Stability in state space

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    Routh-Hurwitz criteria

    For stability, three criteria must be satisfied. These criteria are based on the denominator of the closed loop transfer function, i.e. the characteristic polynomial.

    1. All coefficients of $s$ must be non-zero
    2. No sign changes in the coefficients of $s$
    3. No sign changes in the first column of the Routh array

    Criterion 1 arises because systems with missing coefficients of $s$ imply poles on the imaginary ($j\omega$) axis, implying marginal stability. For design purposes, we can consider marginally stable systems to be unstable.

    Criterion 2 arises because the number of sign changes equals the number of poles in the right half plane (RHP); we require none!
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    Special cases

    Zeros in the first column of the Routh array would lead to division by zero. These should be replaced by $\large \epsilon$, representing a very small number.

    Rows of zeros imply that the CLTF has an even polynomial as a factor. These can be dealt with by generating an even polynomial from the row above and differentiating, as shown in Example 6.8, p316.

    These special cases will not appear in the exam.


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    Steady state errors

    Context

    Along with analysis of stability and transient performance, steady state error often forms an integral part of the specification for a system's behaviour.

    Evaluation of steady state error only makes sense in the context of a stable system. Analysing steady state error in an unstable system would lead to perverse results.

    In the $s$-domain, steady state errors can be analysed using the Final Value Theorem.

    It can be instructive to consider the steady state errors arising from systems subjected to various standard inputs, such as step, ramp and parabolic inputs.
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    Ramp input: SSE


    Fig 7.2b, p342 / 336: ramp input
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    Steady state error for unit step input

    \begin{align*} e_{step}(\infty) &= \lim_{s\rightarrow 0} \frac{s R(s)}{1+G(s)} & R(s) &= \frac{1}{s} \\ \ \\ e_{step}(\infty) &= \lim_{s\rightarrow 0} \frac{\require{cancel}\cancel{s} \times 1 / \cancel{s}}{1+G(s)} \\ &= \frac{1}{1+\lim_{s \rightarrow 0} G(s)} \end{align*}
    The term $\lim_{s\rightarrow 0}G(s)$ is known as the DC gain of the forward path in electrical engineering, or its steady state gain more generally.
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    Integral action

    If the open loop transfer function $G(s)$ has one or more uncancelled poles at the origin (i.e. a pure integrator), there will be no steady state error in response to a step input.
    Proof: \begin{align*} \text{Let } G(s) &= \frac{(s+z_1)(s+z_2)\ldots}{s^n(s+p_1)(s+p_2)\ldots}\\ \ \\ \lim_{s\rightarrow 0} G(s) &= \frac{(z_1)(z_2)\ldots}{0^n(p_1)(p_2)\ldots} \rightarrow \infty \\ \ \\ e_{step}(\infty) &= \frac{1}{1+\lim_{s\rightarrow 0}G(s)} \rightarrow \frac{1}{1+\infty} \rightarrow 0 \end{align*}
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    Steady state error for parabola input

    \begin{align*} e_{parabola}(\infty) &= \lim_{s\rightarrow 0} \frac{s R(s)}{1+G(s)} & R(s) &= \frac{1}{s^3} \\ e_{parabola}(\infty) &= \lim_{s\rightarrow 0} \frac{\cancel{s} \times 1 / s^{\cancelto{2}{3}}}{1+G(s)} \\ &= \lim_{s\rightarrow 0} \frac{1}{s^2+s^2 G(s)} \\ &= \frac{1}{\lim_{s \rightarrow 0} s^2 G(s)} \end{align*} To eliminate steady state error for a parabolic input, there must be three or more uncancelled poles at the origin of the open loop (forward path) transfer function.
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    Chapter 7 - Steady state errors

    Steady state error specifications
    Steady state errors for systems in state space

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    Step input: SSE


    Fig 7.2a, p342 / 336: step input
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    Final value theorem

    Recall the Final value theorem: \begin{align*} f(\infty) &= \lim_{s\rightarrow 0} s F(s) \\ \end{align*}

    Fig 7.3b, p342 / 336: open loop TF G(s) with unity feedback

    \begin{align*} E(s) &= \frac{R(s)}{1+G(s)} & e(\infty) &= \lim_{s\rightarrow 0} \frac{s R(s)}{1+G(s)} \end{align*}
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    Using $T(s)$ instead

    Note, if we calculate the closed loop transfer function $T$ (or $G_c(s)$) instead, we can derive a similar result: \begin{align*} T(s) &= \frac{G(s)}{1+G(s)} \\ \\ y_{step}(\infty) &= \lim_{s\rightarrow 0} s T(s) R(s) \\ \\ e_{step}(\infty) &= \lim_{s\rightarrow 0} \cancel{s} \left(1 - T(s) \right) \cancel{R(s)} & R(s) &= \frac{1}{s} \\ \\ e_{step}(\infty) &= 1 - \lim_{s\rightarrow 0} T(s) \end{align*}
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    Steady state error for unit ramp input
    \begin{align*} e_{ramp}(\infty) &= \lim_{s\rightarrow 0} \frac{s R(s)}{1+G(s)} & R(s) &= \frac{1}{s^2} \\ e_{ramp}(\infty) &= \lim_{s\rightarrow 0} \frac{\cancel{s} \times 1 / s^{\cancel{2}}}{1+G(s)} \\ &= \lim_{s\rightarrow 0} \frac{1}{s+s G(s)} \\ &= \frac{1}{\lim_{s \rightarrow 0} s G(s)} \end{align*}
    To eliminate steady state error for a ramp input, there must be two or more uncancelled poles at the origin of the open loop (forward path) transfer function.



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    Relationships between input, SSE and system type


    Table 7.2, p353 / 346
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    Sensitivity

    Sensitivity analysis allows determination of the dependence of a function on a change of parameter.

    Sensitivity $S_{F:P}$ gives the change in value of function $F$ due to a change in parameter $P$ and is defined: \begin{align*} S_{F:P} &= \lim_{\Delta P\rightarrow 0} \frac{\text{Fractional change in} F}{\text{Fractional change in} P} \\ \ \\ &= \lim_{\Delta P\rightarrow 0} \frac{P\Delta F}{F\Delta P} \\ \ \\ &= \frac{P}{F} \frac{\delta F}{\delta P} \end{align*}
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    Steady state error in state space

    Final value theorem in state space

    \begin{align*} e(\infty) &= \lim_{s\rightarrow 0} s E(s) \\ \ \\ E(s) &= R(s) - C(s) = \left(1 - (\mathrm{C}\mathrm{\Phi}(s)\mathrm{B}+\mathrm{D})\right) R(s) \\ \mathrm{\Phi}(s) &= (s\mathrm{I}-\mathrm{A})^{-1} \\ \ \\ e(\infty) &= \lim_{s\rightarrow 0} s (1-\mathrm{C}\mathrm{\Phi}(s)\mathrm{B}-\mathrm{D}) R(s) \end{align*}
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    Static error constants

    The three limit terms obtained in the steady state error calculations for the step, ramp and parabolic input are known collectively as static error constants.
    \begin{align*} K_p &= \lim_{s\rightarrow 0} G(s) & \quad \text{Position constant} \\ \ \\ K_v &= \lim_{s\rightarrow 0} s G(s) & \quad \text{Velocity constant} \\ \ \\ K_a &= \lim_{s\rightarrow 0} s^2 G(s) & \quad \text{Acceleration constant} \\ \end{align*}
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    Disturbances


    Figs 7.11, 7.12, p356–7 / 349–350: Disturbances
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    Sensitivity example 7.10 (1/3)

    Given a system with uncertain parameter $a$, determine the sensitivity of the closed loop transfer function $T(s)$ to a change in $a$


    Fig 7.19, p362 / 355: Example 7.10
    \begin{align*} T(s) &= \frac{K}{s^2+a s+K} & S_{T:a} &= \frac{a}{T} \frac{\delta T}{\delta a} \\ \end{align*}

    Sensitivity example 7.10 (2/3)
    \begin{align*} T(s) &= \frac{K}{s^2+a s+K} & S_{T:a} &= \frac{a}{T} \frac{\delta T}{\delta a} \\ \end{align*}
    \begin{align*} \frac{\delta T}{\delta a} &= \frac{\delta}{\delta a} \left(K(s^2+a s+K)^{-1}\right) \\ &= (-1)\times K(s^2+a s +K)^{(-1-1)}\times \frac{\delta}{\delta a}\left(s^2+a s+K)\right)\\ &= \frac{- K}{(s^2+a s+K)^2} \times s \\ &= \frac{- K s}{(s^2+a s + K)^2} \end{align*}
    Sensitivity example 7.10 (3/3)
    \begin{align*} T(s) &= \frac{K}{s^2+a s+K} & S_{T:a} &= \frac{a}{T} \frac{\delta T}{\delta a} \\ \end{align*}
    \begin{align*} S_{T:a} &= \frac{a}{T} \times \frac{-K s}{(s^2+a s + K)^2} \\ &= \frac{a}{\frac{\require{cancel}\cancel{K}}{\cancel{s^2+as+K}}} \times \frac{-\cancel{K} s}{(s^2+a s + K)^{\cancel{2}}} \\ &= \frac{-as}{s^2+as+K} \end{align*} Note that increasing $K$ decreases the sensitivity of
    $T(s)$ to $a$.
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    Steady state error for a unit step input
    For a unit step $R(s) = 1/s$
    \begin{align*} e_{step}(\infty) &= \lim_{s\rightarrow 0} (1-\mathrm{C}\mathrm{\Phi}(s)\mathrm{B}-\mathrm{D}) \end{align*} Quick and dirty (works for a step input!): \begin{align*} \lim_{s\rightarrow 0} \mathrm{\Phi}(s) &= \lim_{s\rightarrow 0}(\cancelto{0}{s}\mathrm{I}-\mathrm{A})^{-1} \\ &= (-\mathrm{A})^{-1} \\ \ \\ e_{step}(\infty) &= 1+\mathrm{C}\mathrm{A}^{-1}\mathrm{B}-\mathrm{D} \end{align*}