Add test / doc / list of sections
This commit is contained in:
@@ -6,224 +6,9 @@ s = zpk('s');
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addpath('./mat/');
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% Steps
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% Data Loading
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% The measured data is loaded and the first 25 seconds of data corresponding to transient data are removed.
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load('force_sensor_steps.mat', 't', 'encoder', 'u', 'v');
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figure;
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tiledlayout(2, 1, 'TileSpacing', 'None', 'Padding', 'None');
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nexttile;
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plot(t, v);
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xlabel('Time [s]'); ylabel('Measured voltage [V]');
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nexttile;
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plot(t, u);
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xlabel('Time [s]'); ylabel('Actuator Voltage [V]');
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% #+name: fig:force_sen_steps_time_domain
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% #+caption: Time domain signal during the 3 actuator voltage steps
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% #+RESULTS:
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% [[file:figs/force_sen_steps_time_domain.png]]
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% Three steps are performed at the following time intervals:
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t_s = [ 2.5, 23;
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23.8, 35;
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35.8, 50];
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% Fit function:
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f = @(b,x) b(1).*exp(b(2).*x) + b(3);
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% We are interested by the =b(2)= term, which is the time constant of the exponential.
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tau = zeros(size(t_s, 1),1);
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V0 = zeros(size(t_s, 1),1);
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for t_i = 1:size(t_s, 1)
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t_cur = t(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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t_cur = t_cur - t_cur(1);
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y_cur = v(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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nrmrsd = @(b) norm(y_cur - f(b,t_cur)); % Residual Norm Cost Function
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B0 = [0.5, -0.15, 2.2]; % Choose Appropriate Initial Estimates
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[B,rnrm] = fminsearch(nrmrsd, B0); % Estimate Parameters ‘B’
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tau(t_i) = 1/B(2);
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V0(t_i) = B(3);
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end
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% #+RESULTS:
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% | $tau$ [s] | $V_0$ [V] |
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% |-----------+-----------|
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% | 6.47 | 2.26 |
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% | 6.76 | 2.26 |
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% | 6.49 | 2.25 |
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% With the capacitance being $C = 4.4 \mu F$, the internal impedance of the Speedgoat ADC can be computed as follows:
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Cp = 4.4e-6; % [F]
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Rin = abs(mean(tau))/Cp;
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ans = Rin
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% #+RESULTS:
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% : 1494100.0
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% The input impedance of the Speedgoat's ADC should then be close to $1.5\,M\Omega$ (specified at $1\,M\Omega$).
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% #+begin_important
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% How can we explain the voltage offset?
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% #+end_important
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% As shown in Figure [[fig:force_sensor_model_electronics]] (taken from cite:reza06_piezoel_trans_vibrat_contr_dampin), an input voltage offset is due to the input bias current $i_n$.
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% #+name: fig:force_sensor_model_electronics
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% #+caption: Model of a piezoelectric transducer (left) and instrumentation amplifier (right)
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% [[file:figs/force_sensor_model_electronics.png]]
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% The estimated input bias current is then:
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in = mean(V0)/Rin;
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ans = in
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% #+RESULTS:
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% : 1.5119e-06
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% An additional resistor in parallel with $R_{in}$ would have two effects:
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% - reduce the input voltage offset
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% \[ V_{off} = \frac{R_a R_{in}}{R_a + R_{in}} i_n \]
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% - increase the high pass corner frequency $f_c$
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% \[ C_p \frac{R_{in}R_a}{R_{in} + R_a} = \tau_c = \frac{1}{f_c} \]
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% \[ R_a = \frac{R_i}{f_c C_p R_i - 1} \]
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% If we allow the high pass corner frequency to be equals to 3Hz:
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fc = 3;
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Ra = Rin/(fc*Cp*Rin - 1);
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ans = Ra
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% #+RESULTS:
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% : 79804
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% With this parallel resistance value, the voltage offset would be:
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V_offset = Ra*Rin/(Ra + Rin) * in;
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ans = V_offset
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% Add Parallel Resistor
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% A resistor $R_p \approx 100\,k\Omega$ is added in parallel with the force sensor as shown in Figure [[fig:force_sensor_model_electronics_without_R]].
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% #+name: fig:force_sensor_model_electronics_without_R
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% #+caption: Model of a piezoelectric transducer (left) and instrumentation amplifier (right) with added resistor $R_p$
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% [[file:figs/force_sensor_model_electronics_without_R.png]]
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load('force_sensor_steps_R_82k7.mat', 't', 'encoder', 'u', 'v');
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figure;
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tiledlayout(2, 1, 'TileSpacing', 'None', 'Padding', 'None');
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nexttile;
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plot(t, v);
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xlabel('Time [s]'); ylabel('Measured voltage [V]');
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nexttile;
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plot(t, u);
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xlabel('Time [s]'); ylabel('Actuator Voltage [V]');
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% #+name: fig:force_sen_steps_time_domain_par_R
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% #+caption: Time domain signal during the actuator voltage steps
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% #+RESULTS:
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% [[file:figs/force_sen_steps_time_domain_par_R.png]]
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% Three steps are performed at the following time intervals:
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t_s = [1.9, 6;
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8.5, 13;
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15.5, 21;
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22.6, 26;
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30.0, 36;
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37.5, 41;
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46.2, 49.5]
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% Fit function:
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f = @(b,x) b(1).*exp(b(2).*x) + b(3);
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% We are interested by the =b(2)= term, which is the time constant of the exponential.
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tau = zeros(size(t_s, 1),1);
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V0 = zeros(size(t_s, 1),1);
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for t_i = 1:size(t_s, 1)
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t_cur = t(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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t_cur = t_cur - t_cur(1);
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y_cur = v(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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nrmrsd = @(b) norm(y_cur - f(b,t_cur)); % Residual Norm Cost Function
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B0 = [0.5, -0.2, 0.2]; % Choose Appropriate Initial Estimates
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[B,rnrm] = fminsearch(nrmrsd, B0); % Estimate Parameters ‘B’
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tau(t_i) = 1/B(2);
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V0(t_i) = B(3);
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end
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% #+RESULTS:
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% | $tau$ [s] | $V_0$ [V] |
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% |-----------+-----------|
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% | 0.43 | 0.15 |
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% | 0.45 | 0.16 |
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% | 0.43 | 0.15 |
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% | 0.43 | 0.15 |
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% | 0.45 | 0.15 |
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% | 0.46 | 0.16 |
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% | 0.48 | 0.16 |
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% Knowing the capacitance value, we can estimate the value of the added resistor (neglecting the input impedance of $\approx 1\,M\Omega$):
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Cp = 4.4e-6; % [F]
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Rin = abs(mean(tau))/Cp;
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ans = Rin
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% #+RESULTS:
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% : 101200.0
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% And we can verify that the bias current estimation stays the same:
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in = mean(V0)/Rin;
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ans = in
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% Sinus
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load('force_sensor_sin.mat', 't', 'encoder', 'u', 'v');
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@@ -232,8 +17,7 @@ v = v(t>25);
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encoder = encoder(t>25) - mean(encoder(t>25));
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t = t(t>25);
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% Excitation signal and corresponding displacement
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% The driving voltage is a sinus at 0.5Hz centered on 3V and with an amplitude of 3V (Figure [[fig:force_sensor_sin_u]]).
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@@ -248,7 +32,9 @@ xlabel('Time [s]'); ylabel('Control Voltage [V]');
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% #+RESULTS:
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% [[file:figs/force_sensor_sin_u.png]]
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% The full stroke as measured by the encoder is:
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% The corresponding displacement as measured by the encoder is shown in Figure [[fig:force_sensor_sin_encoder]].
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% The full stroke is:
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max(encoder)-min(encoder)
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@@ -257,54 +43,41 @@ max(encoder)-min(encoder)
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% #+RESULTS:
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% : 5.005e-05
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% Its signal is shown in Figure [[fig:force_sensor_sin_encoder]].
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figure;
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plot(t, encoder)
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xlabel('Time [s]'); ylabel('Encoder [m]');
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% #+name: fig:force_sensor_sin_encoder
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% #+caption: Encoder measurement
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% #+RESULTS:
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% [[file:figs/force_sensor_sin_encoder.png]]
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% The generated voltage by the stack is shown in Figure
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% Generated Voltage
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% The generated voltage by the stack is shown in Figure [[fig:force_sensor_sin_stack]].
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figure;
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plot(t, v)
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xlabel('Time [s]'); ylabel('Force Sensor Output [V]');
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% #+name: fig:force_sensor_sin_stack
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% #+caption: Voltage measured on the stack used as a sensor
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% #+RESULTS:
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% [[file:figs/force_sensor_sin_stack.png]]
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% Generated Charge
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% The capacitance of the stack is
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Cp = 4.4e-6; % [F]
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% The voltage and charge across a capacitor are related through the following equation:
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% \begin{equation}
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% U_C = \frac{Q}{C}
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% \end{equation}
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% where $U_C$ is the voltage in Volts, $Q$ the charge in Coulombs and $C$ the capacitance in Farads.
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% The corresponding generated charge is then shown in Figure [[fig:force_sensor_sin_charge]].
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figure;
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plot(t, 1e6*Cp*(v-mean(v)))
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xlabel('Time [s]'); ylabel('Generated Charge [$\mu C$]');
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% #+name: fig:force_sensor_sin_charge
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% #+caption: Generated Charge
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% #+RESULTS:
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% [[file:figs/force_sensor_sin_charge.png]]
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% Generated Voltage/Charge as a function of the displacement
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% The relation between the generated voltage and the measured displacement is almost linear as shown in Figure [[fig:force_sensor_linear_relation]].
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206
matlab/parallel_resistor.m
Normal file
206
matlab/parallel_resistor.m
Normal file
@@ -0,0 +1,206 @@
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%% Clear Workspace and Close figures
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clear; close all; clc;
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%% Intialize Laplace variable
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s = zpk('s');
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addpath('./mat/');
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% Excitation steps and measured generated voltage
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% The measured data is loaded.
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load('force_sensor_steps.mat', 't', 'encoder', 'u', 'v');
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% The excitation signal (steps) and measured voltage across the sensor stack are shown in Figure [[fig:force_sen_steps_time_domain]].
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figure;
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tiledlayout(2, 1, 'TileSpacing', 'None', 'Padding', 'None');
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nexttile;
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plot(t, v);
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xlabel('Time [s]'); ylabel('Measured voltage [V]');
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nexttile;
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plot(t, u);
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xlabel('Time [s]'); ylabel('Actuator Voltage [V]');
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% Estimation of the voltage offset and discharge time constant
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% The measured voltage shows an exponential decay which indicates that the charge across the capacitor formed by the stack is discharging into a resistor.
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% This corresponds to an RC circuit with a time constant $\tau = RC$.
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% In order to estimate the time domain, we fit the data with an exponential.
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% The fit function is:
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f = @(b,x) b(1).*exp(b(2).*x) + b(3);
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% Three steps are performed at the following time intervals:
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t_s = [ 2.5, 23;
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23.8, 35;
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35.8, 50];
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% We are interested by the =b(2)= term, which is the time constant of the exponential.
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tau = zeros(size(t_s, 1),1);
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V0 = zeros(size(t_s, 1),1);
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for t_i = 1:size(t_s, 1)
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t_cur = t(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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t_cur = t_cur - t_cur(1);
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y_cur = v(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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nrmrsd = @(b) norm(y_cur - f(b,t_cur)); % Residual Norm Cost Function
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B0 = [0.5, -0.15, 2.2]; % Choose Appropriate Initial Estimates
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[B,rnrm] = fminsearch(nrmrsd, B0); % Estimate Parameters ‘B’
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tau(t_i) = 1/B(2);
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V0(t_i) = B(3);
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end
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% Estimation of the ADC input impedance
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% With the capacitance being $C = 4.4 \mu F$, the internal impedance of the Speedgoat ADC can be computed as follows:
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Cp = 4.4e-6; % [F]
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Rin = abs(mean(tau))/Cp;
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ans = Rin
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% Explanation of the Voltage offset
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% As shown in Figure [[fig:force_sen_steps_time_domain]], the voltage across the Piezoelectric sensor stack shows a constant voltage offset.
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% We can explain this offset by looking at the electrical model shown in Figure [[fig:force_sensor_model_electronics_without_R]] (taken from cite:reza06_piezoel_trans_vibrat_contr_dampin).
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% The differential amplifier in the Speedgoat has some input bias current $i_n$ that produces a voltage offset across its own internal resistance.
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% Note that the impedance of the piezoelectric stack is much larger that that at DC.
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% #+name: fig:force_sensor_model_electronics_without_R
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% #+caption: Model of a piezoelectric transducer (left) and instrumentation amplifier (right)
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% [[file:figs/force_sensor_model_electronics_without_R.png]]
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% The estimated input bias current is then:
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in = mean(V0)/Rin;
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ans = in
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% Effect of an additional Parallel Resistor
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% Be looking at Figure [[fig:force_sensor_model_electronics_without_R]], we can see that an additional resistor in parallel with $R_{in}$ would have two effects:
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% - reduce the input voltage offset
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% \[ V_{off} = \frac{R_a R_{in}}{R_a + R_{in}} i_n \]
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% - increase the high pass corner frequency $f_c$
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% \[ C_p \frac{R_{in}R_a}{R_{in} + R_a} = \tau_c = \frac{1}{f_c} \]
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% \[ R_a = \frac{R_i}{f_c C_p R_i - 1} \]
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% If we allow the high pass corner frequency to be equals to 3Hz:
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fc = 3;
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Ra = Rin/(fc*Cp*Rin - 1);
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ans = Ra
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% #+RESULTS:
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% : 79804
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% With this parallel resistance value, the voltage offset would be:
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V_offset = Ra*Rin/(Ra + Rin) * in;
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ans = V_offset
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% Obtained voltage offset and time constant with the added resistor
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% After the resistor is added, the same steps response is performed.
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load('force_sensor_steps_R_82k7.mat', 't', 'encoder', 'u', 'v');
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% The results are shown in Figure [[fig:force_sen_steps_time_domain_par_R]].
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figure;
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tiledlayout(2, 1, 'TileSpacing', 'None', 'Padding', 'None');
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nexttile;
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plot(t, v);
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xlabel('Time [s]'); ylabel('Measured voltage [V]');
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nexttile;
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plot(t, u);
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xlabel('Time [s]'); ylabel('Actuator Voltage [V]');
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% #+name: fig:force_sen_steps_time_domain_par_R
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% #+caption: Time domain signal during the actuator voltage steps
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% #+RESULTS:
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% [[file:figs/force_sen_steps_time_domain_par_R.png]]
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% Three steps are performed at the following time intervals:
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|
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t_s = [1.9, 6;
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8.5, 13;
|
||||
15.5, 21;
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22.6, 26;
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30.0, 36;
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||||
37.5, 41;
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46.2, 49.5]
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% The time constant and voltage offset are again estimated using a fit function.
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f = @(b,x) b(1).*exp(b(2).*x) + b(3);
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tau = zeros(size(t_s, 1),1);
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V0 = zeros(size(t_s, 1),1);
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for t_i = 1:size(t_s, 1)
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t_cur = t(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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t_cur = t_cur - t_cur(1);
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y_cur = v(t_s(t_i, 1) < t & t < t_s(t_i, 2));
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|
||||
nrmrsd = @(b) norm(y_cur - f(b,t_cur)); % Residual Norm Cost Function
|
||||
B0 = [0.5, -0.2, 0.2]; % Choose Appropriate Initial Estimates
|
||||
[B,rnrm] = fminsearch(nrmrsd, B0); % Estimate Parameters ‘B’
|
||||
|
||||
tau(t_i) = 1/B(2);
|
||||
V0(t_i) = B(3);
|
||||
end
|
||||
|
||||
|
||||
|
||||
% #+RESULTS:
|
||||
% | $tau$ [s] | $V_0$ [V] |
|
||||
% |-----------+-----------|
|
||||
% | 0.43 | 0.15 |
|
||||
% | 0.45 | 0.16 |
|
||||
% | 0.43 | 0.15 |
|
||||
% | 0.43 | 0.15 |
|
||||
% | 0.45 | 0.15 |
|
||||
% | 0.46 | 0.16 |
|
||||
% | 0.48 | 0.16 |
|
||||
|
||||
% Knowing the capacitance value, we can estimate the value of the added resistor (neglecting the input impedance of $\approx 1\,M\Omega$):
|
||||
|
||||
|
||||
Cp = 4.4e-6; % [F]
|
||||
Rin = abs(mean(tau))/Cp;
|
||||
|
||||
ans = Rin
|
||||
|
||||
|
||||
|
||||
% #+RESULTS:
|
||||
% : 101200.0
|
||||
|
||||
% And we can verify that the bias current estimation stays the same:
|
||||
|
||||
in = mean(V0)/Rin;
|
||||
|
||||
ans = in
|
||||
Reference in New Issue
Block a user