dis09B_sol

.pdf

School

University of California, Berkeley *

*We aren’t endorsed by this school

Course

16B

Subject

Electrical Engineering

Date

Oct 30, 2023

Type

pdf

Pages

Uploaded by DeanKangaroo5110

EECS 16B Designing Information Systems and Devices II UC Berkeley Fall 2023 Discussion 9B 1. Uncontrollability Recall that, for a n -dimensional, discrete-time linear system to be controllable, we require that the controllability matrix C = h A n − 1 B A n − 2 B . . . AB B i to be rank n . Consider the following discrete-time system with the given initial state: ⃗ x [ i + 1 ] =    2 0 0 − 3 0 1 0 1 0    ⃗ x [ i ] +    0 0 2    u [ i ] (1) ⃗ x [ 0 ] =    1 0 0    (2) (a) Is the system controllable? Solution: C = h A 2 B AB B i =    0 0 0 0 2 0 2 0 2    (3) Since the controllability matrix C only has rank 2, the system is not controllable. We would need it to be rank 3 here to span the full space R 3 . (b) Show that we can write the i th state as ⃗ x [ i ] =    2 i − 3 x 1 [ i − 1 ] + x 3 [ i − 1 ] x 2 [ i − 1 ] + 2 u [ i − 1 ]    (4) Is it possible to reach ⃗ x [ ℓ ] =    − 2 4 6    for some ℓ ? If so, for what input sequence u [ i ] up to i = ℓ − 1 ? Solution: We can write: ⃗ x [ i ] =    2 0 0 − 3 0 1 0 1 0    ⃗ x [ i − 1 ] +    0 0 2    u [ i − 1 ] (5) =    2 0 0 − 3 0 1 0 1 0       x 1 [ i − 1 ] x 2 [ i − 1 ] x 3 [ i − 1 ]    +    0 0 2    u [ i − 1 ] (6) =    2 x 1 [ i − 1 ] − 3 x 1 [ i − 1 ] + x 3 [ i − 1 ] x 2 [ i − 1 ] + 2 u [ i − 1 ]    (7) 1

EECS 16B Discussion 9B 2023-10-26 12:35:36-07:00 =    2 i x 1 [ 0 ] − 3 x 1 [ i − 1 ] + x 3 [ i − 1 ] x 2 [ i − 1 ] + 2 u [ i − 1 ]    (8) =    2 i − 3 x 1 [ i − 1 ] + x 3 [ i − 1 ] x 2 [ i − 1 ] + 2 u [ i − 1 ]    (9) Note that in this expression for ⃗ x [ i ] , x 1 [ i ] = 2 i is decoupled from all other states and inputs. From this expression we also see that there is no choice of inputs us to get to x 1 [ ℓ ] = − 2. Therefore, we will never be able to reach ⃗ x [ ℓ ] =    − 2 4 6    for any ℓ . (c) Is it possible to reach ⃗ x [ ℓ ] =    2 − 3 − 2    for some ℓ ? For what input sequence u [ i ] for i = 0 to i = ℓ − 1 ? HINT: Use the result for ⃗ x [ i ] from the previous part. Solution: We need ℓ = 1 since x 1 [ i ] = 2 i x 1 [ 0 ] = 2 i . Hence, ⃗ x [ 1 ] =    2 − 3 2 u [ 0 ]    (10) We realize that the first two entries of ⃗ x [ 1 ] are exactly what we want. Thus, we have to choose u [ 0 ] so the third entry is − 2. If we choose u [ 0 ] = − 1, then we have reached our desired state. Thus we see that a system being uncontrollable does not mean we are unable to reach anything at all, but just that the range that can be reached is limited. (d) Find the set of all ⃗ x [ 2 ] , given that you are free to choose any u [ 0 ] and u [ 1 ] . Solution: ⃗ x [ 1 ] =    2 − 3 2 u [ 0 ]    (11) ⃗ x [ 2 ] =    2 0 0 − 3 0 1 0 1 0       2 − 3 2 u [ 0 ]    +    0 0 2    u [ 1 ] (12) =    4 − 6 + 2 u [ 0 ] − 3 + 2 u [ 1 ]    (13) Since we can set u [ 0 ] and u [ 1 ] arbitrarily, we can reach any state of the form    4 0 0    + span       0 1 0    ,    0 0 1       after two timesteps. This means that we can reach any value for ⃗ x [ 2 ] ; contrast this with how the © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 2

dis09B_sol

Browse Popular Homework Q&A