Scilab Function

rtitr - discrete time response (transfer matrix)

Calling Sequence

[y]=rtitr(Num,Den,u [,up,yp])

Parameters

Description

y=rtitr(Num,Den,u [,up,yp]) returns the time response of the discrete time linear system with transfer matrix Den^-1 Num for the input u, i.e y and u are such that Den y = Num u at t=0,1,...

If d1=maxi(degree(Den)), and d2=maxi(degree(Num)) the polynomial matrices Den(z) and Num(z) may be written respectively as:

  D(z)= D_0  + D_1  z + ... + D_d1   z^d1
  N(z)= N_0  + N_1  z + ... + N_d2   z^d2
   

and Den y = Num u is interpreted as the recursion:

  D(0)y(t)+D(1)y(t+1)+...+ D(d1)y(t+d1)= N(0) u(t) +....+ N(d2) u(t+d2)
   

It is assumed that D(d1) is non singular.

The columns of u are the inputs of the system at t=0,1,...,T:

  u=[u(0) , u(1),...,u(T)]
   

The outputs at t=0,1,...,T+d1-d2 are the columns of the matrix y:

  y=[y(0), y(1),  .... y(T+d1-d2)]
   

up and yp define the initial conditions for t < 0 i.e

  up=[u(-d1), ..., u(-1)  ]
  yp=[y(-d1), ...  y(-1)  ]
   

Depending on the relative values of d1 and d2, some of the leftmost components of up, yp are ignored. The default values of up and yp are zero: up = 0*ones(m,d1), yp=0*ones(n,d1)

Examples

See Also