, sia a tempo continuo che a tempo

Calcolare le risposte (soluzioni) libere e forzate nello stato e nell’uscita per i sistemi dinamici caratterizzati dalle seguenti matrici, e per le seguenti scelte dell’ingresso u(t) e dello stato iniziale x

⁰

, sia a tempo continuo che a tempo

discreto

1. A =

 − 2 0

− 3 −1



; B =  1 0



; C =  0 1  ; D = 0;

x

⁰

=

 − 1 0



;  u(t) = exp(−2t), t ∈ R u(t) = −2(−1)

^t

, t ∈ Z

2. A =





− 2 3 0 0 − 1 0 2 3 − 1



 ; B =



 1 0 1



 ; C =  0 1 1  ; D = 0;

x

⁰

=



 2 0 0



 ;  u(t) = exp(2t), t ∈ R u(t) = 4(2)

^t

, t ∈ Z

3. A =





− 2 0 0 4 − 2 0 2 0 − 4



 ; B =



 4 0 0



 ; C =  1 0 1  ; D = 1;

x

⁰

=





− 2 0 0



 ;  u(t) = exp(−4t), t ∈ R u(t) = 2(−4)

^t

, t ∈ Z

Soluzioni:

1.

t ∈ R



 

 

 

 



 

 

 

 



x

`

(t) =

"

− e

^−2t

− 3 e

^−2t

+ 3 e

^−t

#

x

f

(t) =

"

te

^−2t

− 3 e

^−t

− 3 (−t − 1) e

^−2t

#

y

`

(t) = −3 e

^−2t

+ 3 e

^−t

y

f

(t) = 3 e

^−2t

+ 3 te

^−2t

− 3 e

^−t

t ∈ Z



 

 

 

 



 

 

 

 



x

`

(t) =

"

− (−2)

^t

− 3 (−2)

^t

+ 3 (−1)

^t

#

x

f

(t) =

"

2 (−2)

^t

− 2 (−1)

^t

− 6 (−1)

^t

− 6 (−1)

^t

t + 6 (−2)

^t

#

y

`

(t) = −3 (−2)

^t

+ 3 (−1)

^t

y

f

(t) = −6 (−1)

^t

− 6 (−1)

^t

t + 6 (−2)

^t

1

2.

t ∈ R



 

 

 

 

 

 

 



 

 

 

 

 

 

 



x

`

(t) =









2 e

^−2t

0

− 4 e

^−2t

+ 4 e

^−t









x

f

(t) =









1/4 e

^2t

− 1/4 e

^−2t

0

1/2 e

^2t

− e

^−t

+ 1/2 e

^−2t







 y

`

(t) = −4 e

^−2t

+ 4 e

^−t

y

f

(t) = 1/2 e

^2t

− e

^−t

+ 1/2 e

^−2t

t ∈ Z



 

 

 

 

 

 

 



 

 

 

 

 

 

 



x

`

(t) =









2 (−2)

^t

0

− 4 (−2)

^t

+ 4 (−1)

^t









x

f

(t) =









2

^t

− (−2)

^t

0

2 2

^t

+ 2 (−2)

^t

− 4 (−1)

^t







 y

`

(t) = −4 (−2)

^t

+ 4 (−1)

^t

y

f

(t) = 2 2

^t

+ 2 (−2)

^t

− 4 (−1)

^t

3.

t ∈ R



 

 

 

 

 

 

 



 

 

 

 

 

 

 



x

`

(t) =









− 2 e

^−2t

− 8 te

^−2t

2 e

^−4t

− 2 e

^−2t









x

f

(t) =









− e

^−4t

+ 2 e

^−2t

5 e

^−4t

+ 8 te

^−2t

− 4 e

^−2t

− 4 te

^−4t

− e

^−4t

+ 2 e

^−2t







 y

`

(t) = 2 e

^−4t

− 4 e

^−2t

y

f

(t) = −3 e

^−4t

− 4 te

^−4t

+ 4 e

^−2t

t ∈ Z



 

 

 

 

 

 

 



 

 

 

 

 

 

 



x

`

(t) =









− 2 (−2)

^t

4 (−2)

^t

t 2 (−4)

^t

− 2 (−2)

^t









x

f

(t) =









− 2 (−4)

^t

+ 4 (−2)

^t

10 (−4)

^t

− 8 (−2)

^t

− 8 (−2)

^t

t

− 2 (−4)

^t

+ 2 (−4)

^t

t + 4 (−2)

^t







 y

`

(t) = 2 (−4)

^t

− 4 (−2)

^t

y

f

(t) = −6 (−4)

^t

+ 2 (−4)

^t

t + 8 (−2)

^t

2