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@@ -8,7 +8,7 @@ Tags
: [Finite Element Model]({{< relref "finite_element_model" >}})
Reference
: ([Rankers 1998](#org1e7e43d))
: ([Rankers 1998](#org9a37ad0))
Author(s)
: Rankers, A. M.
@@ -163,13 +163,13 @@ The basic questions that are addressed in this thesis are:
### Basic Control Aspects {#basic-control-aspects}
A block diagram representation of a typical servo-system is shown in Figure [1](#org8d51e97).
A block diagram representation of a typical servo-system is shown in Figure [1](#orgf92c352).
The main task of the system is achieve a desired positional relation between two or more components of the system.
Therefore, a sensor measures the position which is then compared to the desired value, and the resulting error is used to generate correcting forces.
In most systems, the "actual output" (e.g. position of end-effector) cannot be measured directly, and the feedback will therefore be based on a "measured output" (e.g. encoder signal at the motor).
It is important to realize that these two outputs can differ, first due to resilience in the mechanical system, and second because of geometrical imperfections in the mechanical transmission between motor and end-effector.
<a id="org8d51e97"></a>
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{{< figure src="/ox-hugo/rankers98_basic_el_mech_servo.png" caption="Figure 1: Basic elements of mechanical servo system" >}}
@@ -180,10 +180,10 @@ The correction force \\(F\\) is defined by:
F = k\_p \epsilon + k\_d \dot{\epsilon} + k\_i \int \epsilon dt
\end{equation}
It is illustrative to see that basically the proportional and derivative part of such a position control loop is very similar to a mechanical spring and damper that connect two points (Figure [2](#org7b90722)).
It is illustrative to see that basically the proportional and derivative part of such a position control loop is very similar to a mechanical spring and damper that connect two points (Figure [2](#org30b866e)).
If \\(c\\) and \\(d\\) represent the constant mechanical stiffness and damping between points \\(A\\) and \\(B\\), and a reference position profile \\(h(t)\\) is applied at \\(A\\), then an opposing force \\(F\\) is generated as soon as the position \\(x\\) and speed \\(\dot{x}\\) of point \\(B\\) does not correspond to \\(h(t)\\) and \\(\dot{h}(t)\\).
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{{< figure src="/ox-hugo/rankers98_basic_elastic_struct.png" caption="Figure 2: Basic Elastic Structure" >}}
@@ -199,9 +199,9 @@ These properties are very essential since they introduce the issue of **servo st
An important aspect of a feedback controller is the fact that control forces can only result from an error signal.
Thus any desired set-point profile first leads to a position error before the corresponding driving forces are generated.
Most modern servo-systems have not only a feedback section, but also a feedforward section, as indicated in Figure [3](#org4adf108).
Most modern servo-systems have not only a feedback section, but also a feedforward section, as indicated in Figure [3](#org48cf617).
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{{< figure src="/ox-hugo/rankers98_feedforward_example.png" caption="Figure 3: Mechanical servo system with feedback and feedforward control" >}}
@@ -246,9 +246,9 @@ Basically, machine dynamics can have two deterioration effects in mechanical ser
#### Actuator Flexibility {#actuator-flexibility}
The basic characteristics of what is called "actuator flexibility" is the fact that in the frequency range of interest (usually \\(0-10\times \text{bandwidth}\\)) the driven system no longer behaves as one rigid body (Figure [4](#org72d5adf)) due to compliance between the motor and the load.
The basic characteristics of what is called "actuator flexibility" is the fact that in the frequency range of interest (usually \\(0-10\times \text{bandwidth}\\)) the driven system no longer behaves as one rigid body (Figure [4](#org7e67aa4)) due to compliance between the motor and the load.
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{{< figure src="/ox-hugo/rankers98_actuator_flexibility.png" caption="Figure 4: Actuator Flexibility" >}}
@@ -258,9 +258,9 @@ The basic characteristics of what is called "actuator flexibility" is the fact t
The second category of dynamic phenomena results from the limited stiffness of the guiding system in combination with the fact the the device is driven in such a way that it has to rely on the guiding system to suppress motion in an undesired direction (in case of a linear direct drive system this occurs if the driving force is not applied at the center of gravity).
In general, a rigid actuator possesses six degrees of freedom, five of which need to be suppressed by the guiding system in order to leave one mobile degree of freedom.
In the present discussion, a planar actuator with three degrees of freedom will be considered (Figure [5](#orgc8208b1)).
In the present discussion, a planar actuator with three degrees of freedom will be considered (Figure [5](#org8db4207)).
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{{< figure src="/ox-hugo/rankers98_guiding_flexibility_planar.png" caption="Figure 5: Planar actuator with guiding system flexibility" >}}
@@ -280,9 +280,9 @@ The last category of dynamic phenomena results from the limited mass and stiffne
In contrast to many textbooks on mechanics and machine dynamics, it is good practice always to look at the combination of driving force on the moving part, and reaction force on the stationary part, of a positioning device.
When doing so, one has to consider what the effect of the reaction force on the systems performance will be.
In the discussion of the previous two dynamic phenomena, the stationary part of the machine was assumed to be infinitely stiff and heavy, and therefore the effect of the reaction force was negligible.
However, in general the stationary part is neither infinitely heavy, nor is it connected to its environment with infinite stiffness, so the stationary part will exhibit a resonance that is excited by the reaction forces (Figure [6](#org1252ea7)).
However, in general the stationary part is neither infinitely heavy, nor is it connected to its environment with infinite stiffness, so the stationary part will exhibit a resonance that is excited by the reaction forces (Figure [6](#org235784d)).
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{{< figure src="/ox-hugo/rankers98_limited_m_k_stationary_machine_part.png" caption="Figure 6: Limited Mass and Stiffness of Stationary Machine Part" >}}
@@ -393,7 +393,7 @@ q\_1 \\ q\_2 \\ \vdots \\ q\_n
For the i-th modal coordinate \\(q\_i\\) the equation of motion is:
\begin{equation}
\begin{equation} \label{eq:eoq\_modal\_i}
m\_i \ddot{q\_i}(t) + k\_i q\_i(t) = \phi\_i^T f(t)
\end{equation}
@@ -401,6 +401,35 @@ which is a simple second order differential equation similar to that of a single
Using basic formulae that are derived for a simple mass spring system, one is now able to analyse the time and frequency response of all individual modes.
Having done that, the total motion of the system can simply be obtained by summing the contributions of all modes.
Characterisation of the dynamics of a mechanical system in terms of frequency response behavior plays a major role in the stability analysis of the control loop of a mechatronic device.
In such an analysis one is typically interested in the transfer function between a measured displacement \\(x\_l\\) and a force \\(f\_k\\), which acts at the physical DoF \\(x\_k\\).
Applying the principle of modal decomposition, any transfer function can be derived by first calculating the behavior of the individual modes, and then summing all modal contributions.
The contribution of one single mode \\(i\\) to the transfer function \\(x\_l/f\_k\\) can be derived by first considering the response of the modal DoF \\(q\_i\\) to a force vector \\(f\\) with only one non-zero component \\(f\_k\\).
In that case, equation \eqref{eq:eoq_modal_i} is reduced to:
\begin{equation}
m\_i \ddot{q}\_i(t) + k\_i q\_i(t) = \phi\_{ik} f\_k(t)
\end{equation}
After a Laplace transformation and some rearrangement:
\begin{equation}
q\_i(s) = f\_k(s) \frac{\phi\_{ik}}{m\_i s^2 + k\_i}
\end{equation}
Once the modal response \\(q\_i\\) is known, the response of the physical DoF \\(x\_l\\) is found by a simple premultiplication with \\(\phi\_{il}\\), which finally leads to the following expression for the contribution of mode \\(i\\) to the transfer function:
\begin{equation}
\left( \frac{x\_l}{f\_k} \right)\_i = \frac{\phi\_{ik}\phi\_{il}}{m\_i s^2 + k\_i}
\end{equation}
The overall transfer function can be found by summation of the individual modal contributions, which all have the same structure:
\begin{equation}
\left( \frac{x\_l}{f\_k} \right) = \sum\_{i = 1}^n \left( \frac{x\_l}{f\_k} \right)\_i = \sum\_{i = 1}^n \frac{\phi\_{ik} \phi\_{il}}{m\_i s^2 + k\_i}
\end{equation}
### Graphical Representation {#graphical-representation}
@@ -422,6 +451,46 @@ Having done that, the total motion of the system can simply be obtained by summi
### Basic Characteristics of Mechanical FRF {#basic-characteristics-of-mechanical-frf}
Consider the position control loop of Figure [7](#orgd7fce0d).
<a id="orgd7fce0d"></a>
{{< figure src="/ox-hugo/rankers98_mechanical_servo_system.png" caption="Figure 7: Mechanical position servo-system" >}}
In the ideal situation the mechanical system behaves as one rigid body with mass \\(m\\), so the mechanical transfer function can be written as:
\begin{equation}
\frac{x\_{\text{servo}}}{F\_{\text{servo}}} = \frac{1}{m s^2}
\end{equation}
<a id="org6bb431c"></a>
{{< figure src="/ox-hugo/rankers98_ideal_bode_nyquist.png" caption="Figure 8: FRF of an ideal system with no resonances" >}}
In the case of one extra modal contribution, the equation for the mechanical transfer function needs to be extended with one extra term:
\begin{equation} \label{eq:effect\_one\_mode}
\frac{x\_{\text{servo}}}{F\_{\text{servo}}} = \frac{1}{m s^2} + \frac{\phi\_{i,\text{servo}} \phi\_{i,\text{force}}}{m\_i s^2 + k\_i} = \frac{1}{m s^2} + \frac{\phi\_{i,\text{servo}} \phi\_{i,\text{force}}}{m\_i s^2 + m\_i \omega\_i^2}
\end{equation}
The final transfer function and the exact interaction between the two parts depends on the values of the various parameters.
Let's introduce a variable \\(\alpha\\), which relates the high-frequency contribution of the mode to that of the rigid-body motion:
\begin{equation} \label{eq:alpha}
\alpha = \frac{\frac{\phi\_{i,\text{servo}} \phi\_{i,\text{force}}}{m\_i}}{\frac{1}{m}}
\end{equation}
which simplifies equation \eqref{eq:effect_one_mode} to:
\begin{equation}
\frac{x\_{\text{servo}}}{F\_{\text{servo}}} = \frac{1}{ms^2} + \frac{\alpha}{m s^2 + m \omega\_i^2}
\end{equation}
<a id="org1d0aa47"></a>
{{< figure src="/ox-hugo/rankers98_frf_effect_alpha.png" caption="Figure 9: Contribution of rigid-body motion and modal dynamics to the amplitude and phase of FRF for various values of \\(\alpha\\)" >}}
### Destabilising Effect of Modes {#destabilising-effect-of-modes}
@@ -468,4 +537,4 @@ Through the enormous performance drive in mechatronics systems, much has been le
## Bibliography {#bibliography}
<a id="org1e7e43d"></a>Rankers, Adrian Mathias. 1998. “Machine Dynamics in Mechatronic Systems: An Engineering Approach.” University of Twente.
<a id="org9a37ad0"></a>Rankers, Adrian Mathias. 1998. “Machine Dynamics in Mechatronic Systems: An Engineering Approach.” University of Twente.