Demo on the interpretation of the norm of linear systems:

1. SISO (Single Input Single Output) system:

Let us consider the second order system:

Numerical application:

w0=2;
xi=0.1;
G=tf(w0*w0,[1 2*xi*w0 w0^2]);

The

norm of

using Matlab function norm:

norm(G,2)
ans = 
   2.2361e+00

using the analytical solution (see exercise on slide 45):

sqrt(w0/4/xi)
ans = 
   2.2361e+00

1.1 Time-domain interpretation:

The impulse response of

figure

impulse(G)

The square root of the numerical intregration of the impulse response squared:

[Y,T]=impulse(G);
Z=trapz(T,Y.^2);
sqrt(Z)
ans = 
   2.2360e+00

except the numerical integration error, we recognize the time-domain interpretation of the

norm :

1.2 Frequency-domain interpretation:

The Bode response:

figure

bode(G)

The square root of the numerical integration of the frequency-domain response squared and divided by

[MAG,PHASE,W]=bode(G);
sqrt(trapz(W,MAG.^2)/pi)
ans = 
   2.2390e+00

except the numerical integration error, we recognize the frequency-domain interpretation of the

norm :

since the frequency-domain response magnitude is an even function of

1.3 Stochastic interpretation:

It can be also shown that the

norm is also the standard deviation of the random signal corresponding to the steady-state of the output signal

when the input signal

is a normalized centered white noise (the PSD (Powed spectral density) of

is : 1).

To simulate such a response one have to keep in mind that is not possible to simulate a white noise on a digital computer because the variance of a white noise is infinite. We use a discrete-time approximation: the signal

is sampled with a sampling period

and hold over

seconds on the value of a centered normal random variable . Each sample is stochatically independant from the other. Let

be the variance of the this normal random variable, then the PSD of such a signal is:

So to approximate a unitary withe noise, one have to choose

The sampling period

must be chosen such the half sampling frequency

is very big with respect to the bandwidth of the system

dt=0.01;

T=[0:dt:1000]; % the stop-time is large to simulate the steady state

U=randn(size(T))/sqrt(dt);

Y=lsim(G,U,T);

figure

plot(T,U)

hold on

plot(T,Y); legend('u(t)','y(t)')

The standard deviation of the output in steady state is computed statistically with the Matlab function std:

std(Y(10000:end))  % it is assumed that the steady steady is reached after 100s
ans = 
   2.1542e+00

except the numerical error:

Now we can consider multi-input multi-output (MIMO) systems.

2. MIMO systems

Let us consider a stable system

with

states

outputs and

inputs randomly generated by Matlab function rss

n=5;p=3;m=2;
G=rss(n,p,m);

The direct-feedtrough matrix

must be null otherwise its

norm is infinite. It is obvious from the:

frequency-domain interpretation : the magnitude of the frequency-domain response nevers goes back to 0 in high frequency when . So the integral over frequency of the magnitude squared is always infinite,
stochastic interpretation: when the white noise on the input signal is transmitted to the output and the variance of a white noise is always infinite.

G.d=0*G.d;

The

norm of

using Matlab function norm:

norm(G,2)
ans = 
   2.1244e+00

using the analytical solution (see slide 43):

From the time-domain definition of the

norm of a MIMO system defined by the state-space matrices

Let

(the observability grammian) then

is solution of the Lyapunov equation:

and

Qo=lyap(G.a',G.c'*G.c)
Qo = 5×5
   3.0918e+00  -4.5208e-02  -1.0684e+00  -9.5959e-02   8.4245e-02
  -4.5208e-02   8.3381e-01   2.9562e-01  -7.5605e-02   3.2785e-01
  -1.0684e+00   2.9562e-01   1.1538e+00   6.6253e-02   2.1174e-01
  -9.5959e-02  -7.5605e-02   6.6253e-02   2.4290e-01  -5.8832e-03
   8.4245e-02   3.2785e-01   2.1174e-01  -5.8832e-03   5.1811e-01
sqrt(trace(G.b'*Qo*G.b))
ans = 
   2.1244e+00

2.1 Time-domain interpretation:

The impulse response of

figure

impulse(G)

Computation of

[Y,T]=impulse(G);
size(Y)
ans = 1×3
   288     3     2

is the

matrix of the impulse responses:

Y(i,j,k) is the value at the time T(i) of the response of the j-th output to an single impulse on the k-th input.

Z=zeros(length(Y),1);
for j=1:p,for k=1:m,Z=Z+Y(:,j,k).^2;end;end;

The square root of the numerical intregration of

sqrt(trapz(T,Z))
ans = 
   2.1256e+00

except the numerical integration integration errors, we found

2.2 Frequency-domain interpretation:

The Bode response:

figure

bode(G)

[MAG,PHASE,W]=bode(G);

size(MAG)

ans = 1×3

     3     2    83

is the

matrix of the frequency-domaine responses:

MAG(i,j,k) is the value at the time T(k) of the magnitude of transfer from the j-th input to the i-th output

Z=zeros(length(MAG),1);
for i=1:p,for j=1:m,Z=Z+squeeze(MAG(i,j,:).^2);end;end;
sqrt(trapz(W,Z)/pi)
ans = 
   2.1144e+00

except the numerical integration integration errors, we found again

2.3 Stochastic interpretation:

It is the direct extension of the approach presented for SISO systems. One should keep in mind that the m normalized centered white noises on the m inputs of the system must be stochastically independent from each other. Then the results is:

In the following Matlab sequence, since

is generated randomly:

the sampling period is computed such that the sampling frequency is 100 times the highest eigenvalue magnitude (in ),
the steaty state is assumed to start at Tst equal to 10 times the higest time-constant of ,
the time range to compute the variance in steady state considers 10000 samples after Tst.

dt=2*pi/(100*max(abs(eig(G))))
dt = 
   2.1218e-02
Tst=10/(min(abs(eig(G))))
Tst = 
   1.0888e+01
Tfin=Tst+10000*dt;
T=[0:dt:Tfin];
U=[];
for i=1:m, U=[U;randn(size(T))/sqrt(dt)];end;
Y=lsim(G,U,T);
sig2=0;
for i=1:p,sig2=sig2+var(Y(end-10000:end,i));end;
sqrt(sig2)
ans = 
   2.1699e+00