nass-simscape/org/uncertainty_payload.org

Effect of Uncertainty on the payload's dynamics on the isolation platform dynamics

• Introduction
• Simple Introductory Example
  • Introduction
  • Equations of motion
  • Initialization of the payload dynamics
  • Initialization of the isolation platform
  • Comparison
  • Conclusion
• Generalization to arbitrary dynamics
  • Introduction
  • Equations of motion
  • Impedance $G^\prime(s)$ of a mass-spring-damper payload
  • First Analytical analysis
  • Impedance of the Payload and Dynamical Uncertainty
  • Equivalent Inverse Multiplicative Uncertainty
  • Effect of the Isolation platform Stiffness
  • Reduce the Uncertainty on the plant
    • Introduction
    • Effect of the platform's stiffness $k$
    • Effect of the platform's damping $c$
    • Effect of the platform's mass $m$
  • Conclusion

Introduction   ignore

In this document we will consider an isolation platform (e.g. the nano-hexapod) with a payload on top (e.g. the the sample to be positioned).

The goal is to study:

• how does the dynamics of the payload influence the dynamics of the isolation platform
• similarly: how does the uncertainty on the payload's dynamics will be transferred to uncertainty on the plant
• what design choice should be made in order to minimize the resulting uncertainty on the plant

Two models are made to study these effects:

• In section sec:introductory_example, simple mass-spring-damper systems are chosen to model both the isolation platform and the payload
• In section sec:arbitrary_dynamics, we consider arbitrary payload dynamics with multiplicative input uncertainty to study the unmodelled dynamics of the payload

Simple Introductory Example

<<sec:introductory_example>>

Introduction   ignore

Let's consider the system shown in Figure fig:2dof_system_stiffness_uncertainty_payload consisting of:

• An isolation platform represented by a mass $m$, a stiffness $k$ and a dashpot $c$ and an actuator $F$
• A payload represented by a mass $m^\prime$, a stiffness $k^\prime$ and a dashpot $c^\prime$

The goal is to stabilize $x$ using $F$ in spite of uncertainty on the payload mechanical properties.

  \begin{tikzpicture}
    % ====================
    % Parameters
    % ====================
    \def\massw{2.2}  % Width of the masses
    \def\massh{0.8}  % Height of the masses
    \def\spaceh{1.2} % Height of the springs/dampers
    \def\dispw{0.3}  % Width of the dashed line for the displacement
    \def\disph{0.5}  % Height of the arrow for the displacements
    \def\bracs{0.05} % Brace spacing vertically
    \def\brach{-10pt} % Brace shift horizontaly
    % ====================


    % ====================
    % Ground
    % ====================
    \draw (-0.5*\massw, 0) -- (0.5*\massw, 0);
    \draw[dashed] (0.5*\massw, 0) -- ++(\dispw, 0) coordinate(dlow);
    % \draw[->] (0.5*\massw+0.5*\dispw, 0) -- ++(0, \disph) node[right]{$x_{w}$};

    % ====================
    % Micro Station
    % ====================
    \begin{scope}[shift={(0, 0)}]
      % Mass
      \draw[fill=white] (-0.5*\massw, \spaceh) rectangle (0.5*\massw, \spaceh+\massh) node[pos=0.5]{$m$};

      % Spring, Damper, and Actuator
      \draw[spring] (-0.4*\massw, 0) -- (-0.4*\massw, \spaceh) node[midway, left=0.1]{$k$};
      \draw[damper] (0, 0)           -- ( 0, \spaceh)          node[midway, left=0.2]{$c$};
      \draw[actuator] ( 0.4*\massw, 0) -- (0.4*\massw, \spaceh) node[midway, left=0.1](F){$F$};

      % Displacements
      \draw[dashed] (0.5*\massw, \spaceh) -- ++(\dispw, 0);
      \draw[->] (0.5*\massw+0.5*\dispw, \spaceh) -- ++(0, \disph) node[right]{$x$};

      % Legend
      \draw[decorate, decoration={brace, amplitude=8pt}, xshift=\brach] %
        (-0.5*\massw, \bracs) -- (-0.5*\massw, \spaceh+\massh-\bracs) %
        node[midway,rotate=90,anchor=south,yshift=10pt,align=center]{Isolation\\Platform};
    \end{scope}

    % ====================
    % Nano Station
    % ====================
    \begin{scope}[shift={(0, \spaceh+\massh)}]
      % Mass
      \draw[fill=white] (-0.5*\massw, \spaceh) rectangle (0.5*\massw, \spaceh+\massh) node[pos=0.5]{$m^\prime$};

      % Spring, Damper, and Actuator
      \draw[spring] (-0.4*\massw, 0) -- (-0.4*\massw, \spaceh) node[midway, left=0.1]{$k^\prime$};
      \draw[damper] (0, 0)           -- ( 0, \spaceh)          node[midway, left=0.2]{$c^\prime$};

      % Displacements
      \draw[dashed] (0.5*\massw, \spaceh) -- ++(\dispw, 0) coordinate(dhigh);
      \draw[->] (0.5*\massw+0.5*\dispw, \spaceh) -- ++(0, \disph) node[right]{$x^\prime$};

      % Legend
      \draw[decorate, decoration={brace, amplitude=8pt}, xshift=\brach] %
        (-0.5*\massw, \bracs) -- (-0.5*\massw, \spaceh+\massh-\bracs) %
        node[midway,rotate=90,anchor=south,yshift=10pt,align=center]{Payload};
    \end{scope}
  \end{tikzpicture}

Two degrees-of-freedom system

Equations of motion

If we write the equation of motion of the system in Figure fig:2dof_system_stiffness_uncertainty_payload, we obtain:

\begin{align} ms^2 x &= F - (cs + k) x + (c^\prime s + k^\prime) (x^\prime - x) \\ m^\prime s^2 x^\prime &= - (c^\prime s + k^\prime) (x^\prime - x) \end{align}

After eliminating $x^\prime$, we obtain:

\begin{equation} \frac{x}{F} = \frac{m^\prime s^2 + c^\prime s + k^\prime}{(ms^2 + cs + k)(m^\prime s^2 + c^\prime s + k^\prime) + m^\prime s^2(c^\prime s + k^\prime)} \end{equation}

Initialization of the payload dynamics

Let the payload have:

• a nominal mass of $m^\prime = 50\ [kg]$
• a nominal stiffness of $k^\prime = 5 \cdot 10^6\ [N/m]$
• a nominal damping of $c^\prime = 3 \cdot 10^3\ [N/(m/s)]$

  mpi = 50;
  kpi = 5e6;
  cpi = 3e3;

  kpi = (2*pi*50)^2*mpi;
  cpi = 0.2*sqrt(kpi*mpi);

Let's also consider some uncertainty in those parameters:

  mp = ureal('m', mpi, 'Range', [1, 100]);
  cp = ureal('c', cpi, 'Percentage', 30);
  kp = ureal('k', kpi, 'Percentage', 30);

The compliance of the payload without the isolation platform is $\frac{1}{m^\prime s^2 + c^\prime s + k^\prime}$ and its bode plot is shown in Figure fig:nominal_payload_compliance_dynamics.

One can see that the payload has a resonance frequency of $\omega_0^\prime = 250\ Hz$.

<<plt-matlab>>

Nominal compliance of the payload (png, pdf)

Initialization of the isolation platform

Let's first fix the mass of the isolation platform:

  m = 10;

And we generate three isolation platforms:

• A soft one with $\omega_0 = 0.1 \omega_0^\prime = 5\ Hz$
• A medium stiff one with $\omega_0 = \omega_0^\prime = 50\ Hz$
• A stiff one with $\omega_0 = 10 \omega_0^\prime = 500\ Hz$

Comparison

The obtained dynamics from $F$ to $x$ for the three isolation platform are shown in Figure fig:plant_dynamics_uncertainty_payload_variability.

<<plt-matlab>>

Obtained plant for the three isolation platforms considered (png, pdf)

Conclusion

The stiff platform dynamics does not seems to depend on the dynamics of the payload.

Generalization to arbitrary dynamics

<<sec:arbitrary_dynamics>>

Introduction

Let's now consider a general payload described by its impedance $G^\prime(s) = \frac{F^\prime}{x}$ as shown in Figure fig:general_payload_impedance.

Note here that we use the term impedance, however, the mechanical impedance is usually defined as the ratio of the force over the velocity $F^\prime/\dot{x}$. We should refer to resistance instead of impedance.

  \begin{tikzpicture}
    \def\massw{2.2}  % Width of the masses
    \def\massh{0.8}  % Height of the masses
    \def\spaceh{1.2} % Height of the springs/dampers
    \def\dispw{0.3}  % Width of the dashed line for the displacement
    \def\disph{0.5}  % Height of the arrow for the displacements
    \def\bracs{0.05} % Brace spacing vertically
    \def\brach{-10pt} % Brace shift horizontaly

    \node[piezo={2.2}{3}{10}] (piezo) at (0, 0){};
    \draw[] ($(piezo.north)+(-1.2, 0)$) -- ++(2.4, 0);
    \draw[] ($(piezo.south)+(-1.2, 0)$) -- ++(2.4, 0);
    \draw[dashed] (piezo.south east) -- ++(\dispw, 0) coordinate(dhigh);
    \draw[->] ($(piezo.south east)+(0.5*\dispw, 0)$) -- ++(0, \disph) node[right]{$x$};
    \draw[->] (piezo.south) node[branch]{} -- ++(0, -1) node[above right]{$F^\prime$};

    \draw[decorate, decoration={brace, amplitude=8pt}, xshift=\brach] %
      ($(piezo.south west) + (-10pt, 0)$) -- ($(piezo.north west) + (-10pt, 0)$) %
      node[midway,rotate=90,anchor=south,yshift=10pt,align=center]{General Payload};
  \end{tikzpicture}

General support

Now let's consider the system consisting of a mass-spring-system (the isolation platform) supporting the general payload as shown in Figure fig:general_payload_with_isolator.

  \begin{tikzpicture}
    \def\massw{2.2}  % Width of the masses
    \def\massh{0.8}  % Height of the masses
    \def\spaceh{1.2} % Height of the springs/dampers
    \def\dispw{0.3}  % Width of the dashed line for the displacement
    \def\disph{0.5}  % Height of the arrow for the displacements
    \def\bracs{0.05} % Brace spacing vertically
    \def\brach{-10pt} % Brace shift horizontaly

    % Mass
    \draw[fill=white] (-0.5*\massw, \spaceh) rectangle node[left=6pt]{$m$} (0.5*\massw, \spaceh+\massh);

    % Spring, Damper, and Actuator
    \draw[spring] (-0.4*\massw, 0) -- (-0.4*\massw, \spaceh) node[midway, left=0.1]{$k$};
    \draw[damper] (0, 0)           -- ( 0, \spaceh)          node[midway, left=0.2]{$c$};
    \draw[actuator] ( 0.4*\massw, 0) -- (0.4*\massw, \spaceh) node[midway, left=0.1](F){$F$};

    % Ground
    \draw (-0.5*\massw, 0) -- (0.5*\massw, 0);

    % Displacements
    \draw[dashed] (0.5*\massw, \spaceh+\massh) -- ++(\dispw, 0) coordinate(dhigh);
    \draw[->] (0.5*\massw+0.5*\dispw, \spaceh+\massh) -- ++(0, \disph) node[right]{$x$};

    % Legend
    \draw[decorate, decoration={brace, amplitude=8pt}, xshift=\brach] %
      (-0.5*\massw, \bracs) -- (-0.5*\massw, \spaceh+\massh-\bracs) %
      node[midway,rotate=90,anchor=south,yshift=10pt,align=center]{Isolation\\Platform};

    \begin{scope}[shift={(0, \spaceh+\massh)}]
      \node[piezo={2.2}{1.5}{6}, anchor=south] (piezo) at (0, 0){};
      \draw[->] (0,0)node[branch]{} -- ++(0, -0.6)node[above right]{$F^\prime$}

      \draw[decorate, decoration={brace, amplitude=8pt}, xshift=\brach] %
        ($(piezo.south west) + (-10pt, 0)$) -- ($(piezo.north west) + (-10pt, 0)$) %
        node[midway,rotate=90,anchor=south,yshift=10pt,align=center]{Payload};
    \end{scope}
  \end{tikzpicture}

Mass-Spring-Damper (isolation platform) supporting a general payload

Equations of motion

We have to following equations of motion:

\begin{align} ms^2 x &= F - (cs + k) x - F^\prime \\ F^\prime &= G^\prime(s) x \end{align}

And by eliminating $F^\prime$, we find the plant dynamics $G(s) = \frac{x}{F}$.

\begin{equation} \frac{x}{F} = \frac{1}{ms^2 + cs + k + G^\prime(s)} \end{equation}

We can learn few things about the obtained transfer function:

• the zeros of $x/F$ will be the poles of $G^\prime(s)$.
• if the impedance of the payload is small $|G^\prime(s)| \ll |ms^2 + cs + k|$, then the payload will have small influence on the obtained dynamics

Impedance $G^\prime(s)$ of a mass-spring-damper payload

In order to verify that the formula is correct, let's take the same mass-spring-damper system used in the system shown in Figure fig:2dof_system_stiffness_uncertainty_payload:

\begin{align*} m^\prime s^2 x^\prime &= (x - x^\prime) (c^\prime s + k^\prime) \\ F^\prime &= (x - x^\prime) (c^\prime s + k^\prime) \end{align*}

By eliminating $x^\prime$ of the equations, we obtain:

\begin{equation} G^\prime(s) = \frac{F^\prime}{x} = \frac{m^\prime s^2 (c^\prime s + k)}{m^\prime s^2 + c^\prime s + k^\prime} \label{eq:impedance_mass_spring_damper} \end{equation}

The impedance of a 1dof mass-spring-damper system is described by Eq. eqref:eq:impedance_mass_spring_damper.

And we obtain

\begin{align*} \frac{x}{F} &= \frac{1}{ms^2 + cs + k + G^\prime(s)} \\ &= \frac{1}{ms^2 + cs + k + \frac{m^\prime s^2 (c^\prime s + k)}{m^\prime s^2 + c^\prime s + k^\prime}} \\ &= \frac{m^\prime s^2 + c^\prime s + k^\prime}{(ms^2 + cs + k) (m^\prime s^2 + c^\prime s + k^\prime) + m^\prime s^2 (c^\prime s + k)} \end{align*}

Which is the same transfer function that was obtained in section sec:introductory_example (Eq. eq:plant_simple_system).

The impedance of the mass-spring-damper system is shown in Figure fig:example_impedance_mass_spring_damper.

• Before the resonance frequency $\omega_0^\prime$, the impedance follows the mass line
• After the resonance, the impedance will follow the stiffness line (depending on the relative values of the stiffness and damping)
• At high frequency, it will follow the damping line

<<plt-matlab>>

Example of the impedance of a mass-spring-damper system (png, pdf)

First Analytical analysis

To summarize, we consider:

• an Isolation platform represented by a mass $m$, a damper $c$ and a stiffness $k$. This system resonate at $\omega_0 = \sqrt{\frac{k}{m}}$
• A payload represented by a mass $m^\prime$, a damper $c^\prime$ and a stiffness $k^\prime$. The payload resonate at $\omega_0^\prime = \sqrt{\frac{k^\prime}{m^\prime}}$

The "impedance" of the payload is represented by: \[ G^\prime(s) = \frac{m^\prime s^2 (c^\prime s + k^\prime)}{m^\prime s^2 + c^\prime s + k^\prime} \]

And the plant is: \[ G(s) = \frac{x}{F} = \frac{1}{ms^2 + cs + k + G^\prime(s)} \]

Let's write the asymptotic behavior of $|G^\prime(j\omega)|$:

• $\lim_{\omega \to 0} |G^\prime(j\omega)| = m^\prime s^2$
• $|G^\prime(j\omega_0)| = \frac{k^\prime \sqrt{1 + (2\xi^\prime)^2}}{2 \xi^\prime}$
• $\lim_{\omega \to \infty} |G^\prime(j\omega)| = c^\prime s + k$

Let's find some conditions in order to have that the dynamics of the payload does not influence to much the dynamics of the plant: \[ |G^\prime(s)| \ll |ms^2 + cs + k| \]

Let's take the case of a stiff payload ($\omega_0^\prime \gg \omega_0$).

Below $\omega_0$, the condition becomes: \[ |G^\prime(s)| \ll k \Leftrightarrow m^\prime \omega_0^2 \ll k \Leftrightarrow m^\prime \ll m \] The payload mass should be small with respect to the isolation platform mass.

Above $\omega_0$: \[ |G^\prime(j\omega)| \ll m \omega^2 \]

Until $\omega_0^\prime$, we have $m^\prime \ll m$ which is the same condition as before. Above $\omega_0^\prime$, we obtain $|jc^\prime \omega + k| \ll m \omega^2$.

When using a soft isolation platform and a stiff payload such that the payload resonate above the first resonance of the isolation platform, the mass of the payload should be small compared to the isolation platform mass in order to not disturb the dynamics of the isolation platform.

Impedance of the Payload and Dynamical Uncertainty

We model the payload by a mass-spring-damper model with some uncertainty.

Let the payload have:

• a nominal mass of $m^\prime = 50\ [kg]$
• a nominal stiffness of $k^\prime = 5 \cdot 10^6\ [N/m]$
• a nominal damping of $c^\prime = 3 \cdot 10^3\ [N/(m/s)]$

The main resonance of the payload is then $\omega^\prime = \sqrt{\frac{m^\prime}{k^\prime}} \approx 50\ Hz$.

  m0 = 10;
  c0 = 3e2;
  k0 = 5e5;

  Gp0 = (m0*s^2 * (c0*s + k0))/(m0*s^2 + c0*s + k0);

Let's represent the uncertainty on the impedance of the payload by a multiplicative uncertainty (Figure fig:input_uncertainty_set): \[ G^\prime(s) = G_0^\prime(s)(1 + w_I^\prime(s)\Delta_I(s)) \quad |\Delta_I(j\omega)| < 1\ \forall \omega \]

This could represent unmodelled dynamics or unknown parameters of the payload.

Input Multiplicative Uncertainty

We choose a simple uncertainty weight: \[ w_I(s) = \frac{\tau s + r_0}{(\tau/r_\infty) s + 1} \] where $r_0$ is the relative uncertainty at steady-state, $1/\tau$ is the frequency at which the relative uncertainty reaches $100\ \%$, and $r_\infty$ is the magnitude of the weight at high frequency.

The parameters are defined below.

  r0 = 0.5;
  tau = 1/(50*2*pi);
  rinf = 10;

  wI = (tau*s + r0)/((tau/rinf)*s + 1);

We then generate a complex $\Delta$.

  DeltaI = ucomplex('A',0);

We generate the uncertain plant $G^\prime(s)$.

  Gp = Gp0*(1+wI*DeltaI);

A set of uncertainty payload's impedance transfer functions is shown in Figure fig:payload_impedance_uncertainty.

<<plt-matlab>>

Uncertainty of the payload's impedance (png, pdf)

Equivalent Inverse Multiplicative Uncertainty

Let's express the uncertainty of the plant $x/F$ as a function of the parameters as well as of the uncertainty on the platform's compliance:

\begin{align*} \frac{x}{F} &= \frac{1}{ms^2 + cs + k + G_0^\prime(s)(1 + w_I(s)\Delta(s))}\\ &= \frac{1}{ms^2 + cs + k + G_0^\prime(s)} \cdot \frac{1}{1 + \frac{G_0^\prime(s) w_I(s)}{ms^2 + cs + k + G_0^\prime(s)} \Delta(s)}\\ \end{align*}

We can the plant dynamics that as an inverse multiplicative uncertainty (Figure fig:inverse_uncertainty_set):

\begin{equation} \frac{x}{F} = G_0(s) (1 + w_{iI}(s) \Delta(s))^{-1} \end{equation}

with:

• $G_0(s) = \frac{1}{ms^2 + cs + k + G_0^\prime(s)}$
• $w_{iI}(s) = \frac{G_0^\prime(s) w_I(s)}{ms^2 + cs + k + G_0^\prime(s)} = G_0(s) G_0^\prime(s) w_I(s)$

Inverse Multiplicative Uncertainty

Effect of the Isolation platform Stiffness

Let's first fix the mass of the isolation platform:

  m = 20;

And we generate three isolation platforms:

• A soft one with $\omega_0 = 5\ Hz$
• A medium stiff one with $\omega_0 = 50\ Hz$
• A stiff one with $\omega_0 = 500\ Hz$

Soft Isolation Platform:

  k_soft = m*(2*pi*5)^2;
  c_soft = 0.1*sqrt(m*k_soft);

  G_soft = 1/(m*s^2 + c_soft*s + k_soft + Gp);
  G0_soft = 1/(m*s^2 + c_soft*s + k_soft + Gp0);
  wiI_soft = Gp0*G0_soft*wI;

Mid Isolation Platform

  k_mid = m*(2*pi*50)^2;
  c_mid = 0.1*sqrt(m*k_mid);

  G_mid = 1/(m*s^2 + c_mid*s + k_mid + Gp);
  G0_mid = 1/(m*s^2 + c_mid*s + k_mid + Gp0);
  wiI_mid = Gp0*G0_mid*wI;

Stiff Isolation Platform

  k_stiff = m*(2*pi*500)^2;
  c_stiff = 0.1*sqrt(m*k_stiff);

  G_stiff = 1/(m*s^2 + c_stiff*s + k_stiff + Gp);
  G0_stiff = 1/(m*s^2 + c_stiff*s + k_stiff + Gp0);
  wiI_stiff = Gp0*G0_stiff*wI;

The obtained transfer functions $x/F$ for each of the three platforms are shown in Figure fig:plant_uncertainty_impedance_payload.

<<plt-matlab>>

Obtained plant for the three isolators (png, pdf)

Reduce the Uncertainty on the plant

Introduction   ignore

Now that we know the expression of the uncertainty on the plant, we can wonder what parameters of the isolation platform would lower the plant uncertainty, or at least bring the uncertainty to reasonable level.

The uncertainty of the plant is described by an inverse multiplicative uncertainty with the following weight: \[ w_{iI}(s) = \frac{G_0^\prime(s) w_I(s)}{ms^2 + cs + k + G_0^\prime(s)} \]

Let's study separately the effect of the platform's mass, damping and stiffness.

Effect of the platform's stiffness $k$

Let's fix $\xi = \frac{c}{2\sqrt{km}} = 0.1$, $m = 100\ [kg]$ and see the evolution of $|w_{iI}(j\omega)|$ with $k$.

This is first shown for few values of the stiffness $k$ in figure fig:inverse_multiplicative_uncertainty_payload_few_k

<<plt-matlab>>

Norm of the inverse multiplicative uncertainty weight for various values of the the isolation platform's stiffness (png, pdf)

The norm of the uncertainty weight $|w_iI(j\omega)|$ is displayed as a function of $\omega$ and $k$ in Figure fig:inverse_multiplicative_payload_uncertainty_norm_k.

<<plt-matlab>>

Evolution of the norm of the uncertainty weight $|w_{iI}(j\omega)|$ as a function of the platform's stiffness $k$ (png, pdf)

Instead of plotting as a function of the platform's stiffness, we can plot as a function of $\omega_0/\omega_0^\prime$ where:

• $\omega_0$ is the resonance of the platform alone
• $\omega_0^\prime$ is the resonance of the support alone

The obtain plot is shown in Figure fig:inverse_multiplicative_payload_uncertainty_k_normalized_frequency. In that case, we can see that with a platform's resonance frequency 10 times higher than the resonance of the payload, we get less than $1\%$ uncertainty until some fairly high frequency.

<<plt-matlab>>

Evolution of the norm of the uncertainty weight $|w_{iI}(j\omega)|$ as a function of the frequency ratio $\omega_0/\omega_0^\prime$ (png, pdf)

Effect of the platform's damping $c$

Let's fix $k = 10^7\ [N/m]$, $m = 100\ [kg]$ and see the evolution of $|w_{iI}(j\omega)|$ with the isolation platform damping $c$ (Figure fig:inverse_multiplicative_payload_uncertainty_c).

<<plt-matlab>>

Evolution of the norm of the uncertainty weight $|w_{iI}(j\omega)|$ as a function of the platform's damping ratio $\xi$ (png, pdf)

Effect of the platform's mass $m$

Let's fix $k = 10^7\ [N/m]$, $\xi = \frac{c}{2\sqrt{km}} = 0.1$ and see the evolution of $|w_{iI}(j\omega)|$ with the payload mass $m$ (Figure fig:inverse_multiplicative_payload_uncertainty_m).

<<plt-matlab>>

Evolution of the norm of the uncertainty weight $|w_{iI}(j\omega)|$ as a function of the payload mass $m$ (png, pdf)

Conclusion

As was expected from Eq. eq:plant_dynamics_general_payload, it is usually a good idea to maximize the mass, damping and stiffness of the isolation platform in order to be less sensible to the payload dynamics.

The best thing to do is to have a stiff isolation platform.

If a soft isolation platform is to be used, it is first a good idea to damp the isolation platform as shown in Figure fig:inverse_multiplicative_payload_uncertainty_c. This can make the uncertainty quite low until the first resonance of the payload. In that case, maximizing the stiffness of the payload is a good idea.