ss2tf

Convert state-space representation to transfer function

Syntax

[b,a] =
ss2tf(A,B,C,D)

[b,a]
= ss2tf(A,B,C,D,ni)

Description

example

[b,a] = ss2tf(A,B,C,D) converts a state-space representation of a system into an equivalent transfer function. ss2tf returns the Laplace-transform transfer function for continuous-time systems and the Z-transform transfer function for discrete-time systems.

example

[b,a] = ss2tf(A,B,C,D,ni) returns the transfer function that results when the nith input of a system with multiple inputs is excited by a unit impulse.

Examples

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Mass-Spring System

Open Live Script

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 $Δ t = 1 / F_{s}$ .

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

The oscillator can be described by the state-space equations

$\begin{array}{c} x (k + 1) = A x (k) + B u (k), \\ y (k) = C x (k) + D u (k), \end{array}$

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

$A = (\begin{array}{c} \cos Δ t & \sin Δ t \\ - \sin Δ t & \cos Δ t \end{array}), B = (\begin{array}{c} 1 - \cos Δ t \\ \sin Δ t \end{array}), C = (\begin{array}{c} - 1 & 0 \end{array}), D = (\begin{array}{c} 1 \end{array}) .$

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:

$H (z) = \frac{1 - z^{- 1} (1 + \cos Δ t) + z^{- 2} \cos Δ t}{1 - 2 z^{- 1} \cos Δ t + z^{- 2}} .$

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.

Two-Body Oscillator

Open Live Script

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 $Δ t = 1 / F_{s}$ .

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

The system can be described by the state-space model

$\begin{array}{c} x (n + 1) = A x (n) + B u (n), \\ y (n) = C x (n) + D u (n), \end{array}$

where $x = {(\begin{array}{c} r_{1} & v_{1} & r_{2} & v_{2} \end{array})}^{T}$ 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 $u = {(\begin{array}{c} u_{1} & u_{2} \end{array})}^{T}$ and the output vector $y = {(\begin{array}{c} a_{1} & a_{2} \end{array})}^{T}$ . The state-space matrices are

$A = \exp (A_{c} Δ t), B = A_{c}^{- 1} (A - I) B_{c}, C = (\begin{array}{c} - 2 & 0 & 1 & 0 \\ 1 & 0 & - 2 & 0 \end{array}), D = I,$

the continuous-time state-space matrices are

$A_{c} = (\begin{array}{c} 0 & 1 & 0 & 0 \\ - 2 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & - 2 & 0 \end{array}), B_{c} = (\begin{array}{c} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{array}),$

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];
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

Input Arguments

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`A` — State matrix
matrix

State matrix, specified as a matrix. If the system has p inputs and q outputs and is described by n state variables, then A is n-by-n.