Theory:

Linear time invariant system may be represented in state space form by the following equations: State equation: $$ \dot{x}(t)=A x(t)+B u(t) \tag{1a} $$ Output equation: $$ y(t)= C x(t) \tag{1b} $$ Deadbeat Control design:
A deadbeat controller is one where the system's output reaches its desired value in the smallest number of steps (the so-called "deadbeat time" or "deadbeat response"). This is achieved by designing the control law such that the poles of the closed-loop system lie at the origin.
The discrete-time state-space model with feedback can be written as:
State equation: $$ {x}[k+1]=F x[k]+g u[k] \tag{2a} $$ Output equation: $$ y[k] = C x[k] \tag{2b} $$ where, x[k] is state vector, y[k] is output vector, u is input or control vector, F is system matrix, g is input matrix, C is output matrix.

where, K is the state feedback gain matrix, r[k] is the reference signal.
The control input is typically given by:
$$ u[k] = - K x[k] + r[k] \tag{3} $$
State feedback design for deadbeat response:
The goal of deadbeat control is to place the poles of the closed-loop system at the origin of the z-plane, i.e., all the eigenvalues of (F - gK) should be at zero. This ensures the system reaches the equilibrium state (desired state) as quickly as possible. For a discrete-time system, the characteristic equation of the closed-loop system is derived from the state-space model with the feedback gain K. The poles of the system are determined by the eigenvalues of (F - gK).

Controllability: The system must be controllable for a deadbeat design to be possible. Controllability ensures that it's possible to drive the system from any initial state to any desired state using the available inputs.

Pole Placement: To achieve a deadbeat response, place the poles of the closed-loop system at the origin. This requires solving for the feedback gain matrix K that ensures that all eigenvalues of the matrix (F - gK) are zero.
Solve for K such that the eigenvalues of (F - gK) are at the origin.
• Determinant of (F - gK) = 0.
• Trace of (F - gK) = 0.

State feedback design:
A necessary and sufficient condition for arbitrary pole placement is that the pair (F, g) must be controllable.
Control input: $$ u[k]= - K x[k] \tag{4} $$ where, K is the state feedback gain vector,i.e., $$ K = [k_1 \ k_2 \ ... \ k_n] $$ With this control input, the closed loop system is as follows: $$ x[k+1]=(F-gK) x[k] $$ The characteristic equation of the closed loop system is, $$ |zI-(F-gK)|=0 $$ For nth order system, the characteristic equation is, $$ z^n + k_n z^{n-1} + k_{n-1} z^{n-2} + ... + k_1=0 \tag{5} $$ The desired closed loop poles (in s domain) are P₁, P₂, P₃, ... , P_n.
The desired closed loop poles (in z domain) are P_z1, P_z2, P_z3, ... , P_zn.

Then the desired characteristic equation is: $$ (z-{P_z}_1)(z-{P_z}_2)(z-{P_z}_3)...(z-{P_z}_n)=0 \tag{6} $$ The required state feedback gain (K) vector elements are obtained by comparing the matching coefficients of (5) and (6).

State Space Model of Mechanical System:
Consider the mechanical system shown in Fig. 1. Assume that the system is linear. The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output. The displacement y(t) is measured from the equilibrium position in the absence of the external force. This system is a single-input, single-output system.

where, m is mass, b is damping friction, k is the spring constant, y is the displacement (output) and u is the external force.

The system equation is:
$$ m\ddot{y} + b \dot{y} + k y = 0 \tag{7} $$ State Space form of the Mechanical system:
The state-space form:

Continuous State Space form:
$$ \begin{bmatrix} \dot{x}_1(t) \\ \dot{x}_2(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{b}{m} \end{bmatrix} \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} + \begin{bmatrix} 0 \\ \frac{1}{m} \end{bmatrix} u(t) $$ $$ y(t) = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} \quad \tag{8} $$ where, x₁ is the displacement, x₂ is the velocity, u(t) is the external force, y(t) is the output.


Discrete State Space form:
$$ \begin{bmatrix} x_1 [k+1] \\ x_2 [k+1] \end{bmatrix} = \begin{bmatrix} 1 & T_s \\ -\frac{k T_s}{m} & 1-\frac{b T_s}{m} \end{bmatrix} \begin{bmatrix} x_1 [k] \\ x_2 [k] \end{bmatrix} + \begin{bmatrix} \frac{{T_s}^2}{2m} \\ \frac{T_s}{m} (1-\frac{b T_s}{2m}) \end{bmatrix} u[k] $$ $$ y [k] = \begin{bmatrix} 0 & 1 \end{bmatrix} \begin{bmatrix} x_1[k] \\ x_2[k] \end{bmatrix} \quad \tag{9} $$