Update content - 2026-10-14

content/zettels/electronic_active_filters.md

@@ -16,13 +16,13 @@ Tags
 \end{equation}
 
 \begin{equation}
-  \frac{V\_o}{V\_i}(s) = \frac{1}{\frac{s^2}{\omega\_0^2} + 2 \xi \frac{s}{\omega\_0} + 1}
+  \frac{V\_o}{V\_i}(s) = \frac{1}{\frac{s^2}{\omega\_0^2} + 2 \zeta \frac{s}{\omega\_0} + 1}
 \end{equation}
 
 With:
 
 -   \\(\omega\_0 = \frac{1}{R\sqrt{C\_1 C\_2}}\\)
--   \\(\xi = \frac{C\_2}{C\_1}\\)
+-   \\(\zeta = \frac{C\_2}{C\_1}\\)
 
 The input impedance is \\(V\_i/i\_i\\).
 
@@ -42,7 +42,7 @@ Same as [Figure 1](#figure--fig:analog-act-filt-second-order-lpf) but by exchang
 With:
 
 -   \\(\omega\_0 = \frac{1}{R\sqrt{C\_1 C\_2}}\\)
--   \\(\xi = \frac{C\_2}{C\_1}\\)
+-   \\(\zeta = \frac{C\_2}{C\_1}\\)
 
 
 ## PID controller {#pid-controller}

content/zettels/electronic_passive_filters.md

@@ -76,7 +76,7 @@ Z = (1 + R*C*s)/(C*s); % Filter Impedance
 {{< figure src="/ox-hugo/analog_filt_second_order_lpf.png" caption="<span class='figure-number'>Figure 5: </span>Second Order Low Pass Filter using an RLC circuit" >}}
 
 \begin{equation}
-\boxed{\frac{V\_o}{V\_i}(s) = \frac{1}{1 + 2 \xi \frac{s}{\omega\_0} + \frac{s^2}{\omega\_0^2}}, \quad \omega\_0 = \frac{1}{\sqrt{LC}},\quad \xi = \frac{1}{2R\sqrt{LC}}}
+\boxed{\frac{V\_o}{V\_i}(s) = \frac{1}{1 + 2 \zeta \frac{s}{\omega\_0} + \frac{s^2}{\omega\_0^2}}, \quad \omega\_0 = \frac{1}{\sqrt{LC}},\quad \zeta = \frac{1}{2R\sqrt{LC}}}
 \end{equation}
 
 \begin{equation}

content/zettels/feedforward_control.md

@@ -111,22 +111,22 @@ In order to implement a fourth order trajectory, look at [this](https://www.math
 See (<a href="#citeproc_bib_item_2">Schmidt, Schitter, and Rankers 2020</a>) (Section 4.2.1).
 
 Suppose we have a second order plant (could typically be a piezoelectric stage):
-\\[ G(s) = \frac{C\_f \omega\_0^2}{s^2 + 2\xi \omega\_0 s + \omega\_0^2} \\]
+\\[ G(s) = \frac{C\_f \omega\_0^2}{s^2 + 2\zeta \omega\_0 s + \omega\_0^2} \\]
 
 <a id="figure--fig:feedforward-second-order-plant"></a>
 
 {{< figure src="/ox-hugo/feedforward_second_order_plant.png" caption="<span class='figure-number'>Figure 4: </span>Bode plot of a second order system with fitted model" >}}
 
 The idea is to design a feedforward controller that corresponds to the plant inverse:
-\\[ C\_{ff}(s) = \frac{s^2 + 2\xi \omega\_0 s + \omega\_0^2}{C\_f \omega\_0^2} \\]
+\\[ C\_{ff}(s) = \frac{s^2 + 2\zeta \omega\_0 s + \omega\_0^2}{C\_f \omega\_0^2} \\]
 
 This controller has a pair of zeros, corresponding to an anti-resonance at the eigenfrequency of the first eigenmode of the system, with equal damping.
 The controller needs to be modified in such a way that it becomes realisable.
 In this case it is decided to create a resulting overall transfer function of the controller and the plant that acts like a well damped mass-spring system with the same natural frequency as the plant and an additional reduction of the excitation of higher frequency eigenmodes.
 In order to realise this controller first two poles have to be added, placed at the same frequency as the resonance but with a higher damping ratio.
-Typically a damping ratio between aperiodic and critical (\\(0.7 < \xi < 1\\)) is applied to avoid oscillations.
-For \\(\xi = 1\\) this results in the following transfer function:
-\\[ C\_{ff}(s) = \frac{s^2 + 2\xi \omega\_0 s + \omega\_0^2}{s^2 + 2 \cdot 1 \cdot \omega\_0 s + \omega\_0^2}\\]
+Typically a damping ratio between aperiodic and critical (\\(0.7 < \zeta < 1\\)) is applied to avoid oscillations.
+For \\(\zeta = 1\\) this results in the following transfer function:
+\\[ C\_{ff}(s) = \frac{s^2 + 2\zeta \omega\_0 s + \omega\_0^2}{s^2 + 2 \cdot 1 \cdot \omega\_0 s + \omega\_0^2}\\]
 
 <a id="figure--fig:feedforward-compensated-system"></a>
 
@@ -211,10 +211,10 @@ Therefore, it is very important to match the delay of the plant:
 {{< figure src="/ox-hugo/feedforward_4th_order.png" >}}
 
 Dynamics from \\(F\\) to \\(x\_2\\) is:
-\\[ G(s) = \frac{x\_2}{F} = \frac{cs + k}{m\_1m\_2 s^2(s^2 + 2\xi\omega\_0 s + \omega\_0^2)} \\]
+\\[ G(s) = \frac{x\_2}{F} = \frac{cs + k}{m\_1m\_2 s^2(s^2 + 2\zeta\omega\_0 s + \omega\_0^2)} \\]
 We take the inverse for the feedforward controller:
-\\[ K\_{ff}(s) = G^{-1}(s) = \frac{m\_1m\_2 s^2(s^2 + 2\xi\omega\_0 s + \omega\_0^2)}{cs + k} \\]
-If we neglect damped: \\(\xi = 0\\), and we get:
+\\[ K\_{ff}(s) = G^{-1}(s) = \frac{m\_1m\_2 s^2(s^2 + 2\zeta\omega\_0 s + \omega\_0^2)}{cs + k} \\]
+If we neglect damped: \\(\zeta = 0\\), and we get:
 \\[ K\_{ff}(s) = \underbrace{\frac{m\_1m\_2}{k} s^4}\_{\text{snap FF}} + \underbrace{(m\_1 + m\_2) s^2}\_{\text{acc FF}} \\]
 This can be solved by using **snap feedforward**
 

content/zettels/mass_spring_damper_systems.md

@@ -31,24 +31,24 @@ Let's consider [Figure 1](#figure--fig:mass-spring-damper-system) where:
 Transmissibility:
 
 \begin{equation}
-  \frac{x}{w}(s) = \frac{c s + k}{m s^2 + c s + k} = \frac{2 \xi \frac{s}{\omega\_0} + 1}{\frac{s^2}{\omega\_0^2} + 2 \xi \frac{s}{\omega\_0} + 1}
+  \frac{x}{w}(s) = \frac{c s + k}{m s^2 + c s + k} = \frac{2 \zeta \frac{s}{\omega\_0} + 1}{\frac{s^2}{\omega\_0^2} + 2 \zeta \frac{s}{\omega\_0} + 1}
 \end{equation}
 
 Compliance:
 
 \begin{equation}
-  \frac{x}{F}(s) = \frac{x}{F\_d}(s) = \frac{1}{m s^2 + c s + k} = \frac{1/k}{\frac{s^2}{\omega\_0^2} + 2 \xi \frac{s}{\omega\_0} + 1}
+  \frac{x}{F}(s) = \frac{x}{F\_d}(s) = \frac{1}{m s^2 + c s + k} = \frac{1/k}{\frac{s^2}{\omega\_0^2} + 2 \zeta \frac{s}{\omega\_0} + 1}
 \end{equation}
 
 with:
 
 -   \\(\omega\_0 = \sqrt{k/m}\\) the natural frequency in [rad/s]
--   \\(\xi = \frac{1}{2} \frac{c}{\sqrt{km}}\\) the damping ratio [unit-less]
+-   \\(\zeta = \frac{1}{2} \frac{c}{\sqrt{km}}\\) the damping ratio [unit-less]
 
 A quality factor \\(Q\\) can also be defined:
 
 \begin{equation}
-Q = \frac{1}{2\xi}
+Q = \frac{1}{2\zeta}
 \end{equation}
 
 This corresponds to the amplification at the natural frequency \\(\omega\_0\\).

content/zettels/tuned_mass_damper.md

@@ -51,7 +51,7 @@ The optimal parameters of the tuned mass damper can be roughly estimated as foll
 -   This allows to compute the stiffness of the TMD:
     \\[ k\_2 = \nu^2 k\_1 \mu = k\_1 \frac{\mu}{(1 + \mu)^2} \\]
 -   Finally, the optimal damping of the TMD is:
-    \\[ \xi\_2 = \sqrt{\frac{3 \mu}{8 (1 + \mu)}} \Longrightarrow c\_2 = 2 \xi\_2 \sqrt{k\_2 m\_2} \\]
+    \\[ \zeta\_2 = \sqrt{\frac{3 \mu}{8 (1 + \mu)}} \Longrightarrow c\_2 = 2 \zeta\_2 \sqrt{k\_2 m\_2} \\]
 
 
 ## Simple TMD model {#simple-tmd-model}