Update Content - 2022-11-02
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		| @@ -90,14 +90,17 @@ Let's fix `m` (payload mass) and `Df` (wanted motion induced by the voice coil). | |||||||
| Then, let's vary `f0` and compute the corresponding `Dg`, `Fmax` and `k`. | Then, let's vary `f0` and compute the corresponding `Dg`, `Fmax` and `k`. | ||||||
|  |  | ||||||
| ```matlab | ```matlab | ||||||
| g = 9.8; | %% Fixed Parameters | ||||||
|  | g = 9.8; % [m/s^2] | ||||||
| m = 5; % [kg] | m = 5; % [kg] | ||||||
| Df = 5e-3; % [m] | Df = 5e-3; % [m] | ||||||
|  |  | ||||||
|  | %% Suspension resonance is varied | ||||||
| f0 = 0.1:0.1:20; % [Hz] | f0 = 0.1:0.1:20; % [Hz] | ||||||
| ``` | ``` | ||||||
|  |  | ||||||
| ```matlab | ```matlab | ||||||
|  | %% Other parameters are computed | ||||||
| k = m * (2*pi*f0).^2; % [N/m] | k = m * (2*pi*f0).^2; % [N/m] | ||||||
| Dg = m * g ./ k; % [m] | Dg = m * g ./ k; % [m] | ||||||
| Fmax = k * Df; % [N] | Fmax = k * Df; % [N] | ||||||
| @@ -114,14 +117,17 @@ Let's fix `m` (payload mass) and `Fmax` (maximum force applied by the Voice coil | |||||||
| Then, let's vary `Df` and compute the corresponding `Dg`, `f0` and `k`. | Then, let's vary `Df` and compute the corresponding `Dg`, `f0` and `k`. | ||||||
|  |  | ||||||
| ```matlab | ```matlab | ||||||
| g = 9.8; | %% Fixed Parameters | ||||||
|  | g = 9.8; % [m/s^2] | ||||||
| m = 5; % [kg] | m = 5; % [kg] | ||||||
| Fmax = 50; % [N] | Fmax = 50; % [N] | ||||||
|  |  | ||||||
|  | %% Wanted stroke is varied | ||||||
| Df = 1e-3:1e-4:10e-3; % [m] | Df = 1e-3:1e-4:10e-3; % [m] | ||||||
| ``` | ``` | ||||||
|  |  | ||||||
| ```matlab | ```matlab | ||||||
|  | %% Other parameters are computed | ||||||
| k = Fmax./Df; % [N/m] | k = Fmax./Df; % [N/m] | ||||||
| f0 = sqrt(k/m)/2/pi; % [Hz] | f0 = sqrt(k/m)/2/pi; % [Hz] | ||||||
| Dg = m * g ./ k; % [m] | Dg = m * g ./ k; % [m] | ||||||
|   | |||||||
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