I am having trouble to define in matlab the transfer function of the following block diagram.

The systems Information for Modelling:

G1(s) = Front left motor.

G2(s) = Front right motor.

G3(s) = Rear left motor.

G4(s) = Rear right motor.

For a quadcopter all motors are identical which means, G1(s) = G2(s) = G3(s) = G4(s)

PID(s) = PID controller transfer function

do = Output disturbance

F(s) = Sensor feedback transfer function, defined as:

F(s)= 1100/(s+1100)

G1= 774 / (s^2 + 37.2 * s + 8649)

Kp =1

D= 0.5

Answer 1

[Sam Chak]

Hi @Jake

In order to obtain the transfer function of the plant, you can likely perform the algebraic manipulations in the manner as suggested. However, I must mention that I excluded the consideration of the output disturbance and sensor in the controller design.

syms s u K y

%% Transfer function of the motor
G    = 774/(s^2 + 37.2*s + 8649)

G = 

u1   =  K*y + 1*u - 1*u;    % don't understand why add and cancel out
u2   = -u;
u3   =  u;
u4   = -u;
eqn  = y - (1.1*G*u1 - 1.1*G*u2 + 1.1*G*u3 - 1.1*G*u4) == 0;
ySol = isolate(eqn, y)

ySol = 

ySol = subs(ySol, K, 23064/2365)    % K = 23064/2365 (free parameter to be designed)

ySol = 

%% Transfer function of the Plant
Gp  = tf(12771, [5 186 8649/5])

Gp =
 
         12771
  --------------------
  5 s^2 + 186 s + 1730
 
Continuous-time transfer function.

%% Transfer function of the PIDF controller
Gc  = pidtune(Gp, 'PIDF')

Gc =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = 0.275, Ki = 3.58, Kd = 0.00475, Tf = 0.000441
 
Continuous-time PIDF controller in parallel form.

%% Transfer function of the Closed-loop system
Gcl = feedback(Gc*Gp, 1);

%% Step-response of the Closed-loop system
step(Gcl, 0.6),  grid on