Step & Impulse Time Responses

(Uses Matlab): step, impulse & State-Space representation
For (Sample 1),


x' = A x + B u
y = C x + D u

where in MATLAB the entry is:

EDU>> A = [-1,0,-4000; 0,-2,2000; 0.5,-0.5,0]

EDU>> B = [1,0; 0,2000; 0,0]

EDU>> C = [1,0,0; 0,1,0; 0,0,1]

EDU>> D = [0,0; 0,0; 0,0]

The Unit Step Response is,

EDU>> step(A,B,C,D)

Signals: Y1, Y2 & Y3 represent the state variables: v1, v2 & i
Signals: U1 & U2 represent the inputs: vi & is


and the Unit Impulse Response is,

EDU>> impulse(A,B,C,D)



Notice the responses due to input vi is unclear, this is due in part they share the same units with those responses due to input ii. To solve this problem separate the responses between the two inputs.

For step responses:

Due to input vi we use,

EDU>> step(A,B,C,D,1)

Due to input is use,

EDU>> step(A,B,C,D,2)

For impulse responses:

Due to input vi we use,

EDU>> impulse(A,B,C,D,1)

Due to input is use,

EDU>> impulse(A,B,C,D,2)



For (Sample 2):

A = [ 0,1,0; -(k1+k2)/m, 0, k1/m; k1/b,0,-k1/b]
B = [0;1;0]
C = [1,0,0]
D = [0]

We can model a large bump at high speed as an impulse. Lets execute the m-script,

% ap3_4.m
b=2;
A=[0,1,0; -3,0,2; 2/b,0,-2/b];
B=[0;1;0];
C=[1,0,0];
D=0;
impulse(A,B,C,D)

The response is,


(Try it out): The impulse bump produces response below the ground. Fix the problem, assuming the bike bounces each time it hit the ground surface. For simplicity, make the bouce a perfect mirror image of the negative response.

For small bump at low speed we use a step. The response is,