ss2tf

A one-dimensional discrete-time oscillating system consists of a unit mass, $$m$$, attached to a wall by a spring of unit elastic constant. A sensor samples the acceleration, $$a$$, of the mass at $$F_s=5$$ Hz.

Generate 50 time samples. Define the sampling interval $$\Delta t=1/F_s$$.

Fs = 5;
dt = 1/Fs;
N = 50;
t = dt*(0:N-1);

The oscillator can be described by the state-space equations

where $$x={\left(\begin{array}{cc}r& v\end{array}\right)}^T$$ is the state vector, $$r$$ and $$v$$ are respectively the position and velocity of the mass, and the matrices

A = [cos(dt) sin(dt);-sin(dt) cos(dt)];
B = [1-cos(dt);sin(dt)];
C = [-1 0];
D = 1;

The system is excited with a unit impulse in the positive direction. Use the state-space model to compute the time evolution of the system starting from an all-zero initial state.

u = [1 zeros(1,N-1)];

x = [0;0];
for k = 1:N
    y(k) = C*x + D*u(k);
    x = A*x + B*u(k);
end

Plot the acceleration of the mass as a function of time.

stem(t,y,'filled')
xlabel('t')

Compute the time-dependent acceleration using the transfer function H(z) to filter the input. Plot the result.

[b,a] = ss2tf(A,B,C,D);
yt = filter(b,a,u);

stem(t,yt,'filled')
xlabel('t')

The transfer function of the system has an analytic expression:

Use the expression to filter the input. Plot the response.

bf = [1 -(1+cos(dt)) cos(dt)];
af = [1 -2*cos(dt) 1];
yf = filter(bf,af,u);

stem(t,yf,'filled')
xlabel('t')

The result is the same in all three cases.

An ideal one-dimensional oscillating system consists of two unit masses, $$m_1$$ and $$m_2$$, confined between two walls. Each mass is attached to the nearest wall by a spring of unit elastic constant. Another such spring connects the two masses. Sensors sample $$a_1$$ and $$a_2$$, the accelerations of the masses, at $$F_s=16$$ Hz.

Specify a total measurement time of 16 s. Define the sampling interval $$\Delta t=1/F_s$$.

Fs = 16;
dt = 1/Fs;
N = 257;
t = dt*(0:N-1);

The system can be described by the state-space model

where $$x={\left(\begin{array}{cccc}{r}_1& v_1& r_2& v_2\end{array}\right)}^T$$ is the state vector and $$r_i$$ and $$v_i$$ are respectively the location and the velocity of the $$i$$th mass. The input vector $$u={\left(\begin{array}{cc}{u}_1& u_2\end{array}\right)}^T$$ and the output vector $$y={\left(\begin{array}{cc}{a}_1& a_2\end{array}\right)}^T$$. The state-space matrices are

the continuous-time state-space matrices are

and $$I$$ denotes an identity matrix of the appropriate size.

Ac = [0 1 0 0; -2 0 1 0; 0 0 0 1; 1 0 -2 0];
A = expm(Ac*dt);
Bc = [0 0; 1 0; 0 0; 0 1];
B = Ac\(A-eye(4))*Bc;
C = [-2 0 1 0; 1 0 -2 0];
D = eye(2);

The first mass, $$m_1$$, receives a unit impulse in the positive direction.

ux = [1 zeros(1,N-1)];
u0 = zeros(1,N);
u = [ux;u0];

Use the model to compute the time evolution of the system starting from an all-zero initial state.

x = [0 0 0 0]';
y = zeros(2,N);

for k = 1:N
    y(:,k) = C*x + D*u(:,k);
    x = A*x + B*u(:,k);
end

Plot the accelerations of the two masses as functions of time.

stem(t,y','.')
xlabel('t')
legend('a_1','a_2')
title('Mass 1 Excited')
grid

Convert the system to its transfer function representation. Find the response of the system to a positive unit impulse excitation on the first mass.

[b1,a1] = ss2tf(A,B,C,D,1);
y1u1 = filter(b1(1,:),a1,ux);
y1u2 = filter(b1(2,:),a1,ux);

Plot the result. The transfer function gives the same response as the state-space model.

stem(t,[y1u1;y1u2]','.')
xlabel('t')
legend('a_1','a_2')
title('Mass 1 Excited')
grid

The system is reset to its initial configuration. Now the other mass, $$m_2$$, receives a unit impulse in the positive direction. Compute the time evolution of the system.

u = [u0;ux];

x = [0;0;0;0];
for k = 1:N
    y(:,k) = C*x + D*u(:,k);
    x = A*x + B*u(:,k);
end

Plot the accelerations. The responses of the individual masses are switched.

stem(t,y','.')
xlabel('t')
legend('a_1','a_2')
title('Mass 2 Excited')
grid

Find the response of the system to a positive unit impulse excitation on the second mass.

[b2,a2] = ss2tf(A,B,C,D,2);
y2u1 = filter(b2(1,:),a2,ux);
y2u2 = filter(b2(2,:),a2,ux);

Plot the result. The transfer function gives the same response as the state-space model.

stem(t,[y2u1;y2u2]','.')
xlabel('t')
legend('a_1','a_2')
title('Mass 2 Excited')
grid