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Discretization of continuous-time state-space system.
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This question is from a Systems Theory test without answers or solutions.
Consider the folowing continuous-time state-space system
$dotx=Ax+Bu, quad y=Cx.$
The continuous-time system given above is sampled at times $kh, kin N$, with sampling time $h = 3$. We assume that the input function $u(t)$ is constant between two subsequent sampling times. To be more precise, $u(t)=u_k$, when $t in [kh, (k+1)h]$. The result of this exact discretization is given by:
$x_k+1=A_dx_k+B_dx_k$
$y_k=C_dx_k$
with matrices
$A_d = beginbmatrix e^3 & 3e^3 \ 0 & e^3 endbmatrix, quad B_d = beginbmatrix e^3-1 \ 0 endbmatrix, quad C_d = beginbmatrix e^3 & e^3 endbmatrix$
Which of the following continuous-time state-space systems yields the discrete time state space system after exact discretization is applied?
A) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 0 \
0 & 1 & 1 \
hline
1 & 1 \ endarray right]$
B) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
C) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
D) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 3 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
E) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & e^3-1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
linear-algebra control-theory linear-control
asked Aug 6 at 6:59
user463102
1288
add a comment |Â
up vote
0
down vote
favorite
This question is from a Systems Theory test without answers or solutions.
Consider the folowing continuous-time state-space system
$dotx=Ax+Bu, quad y=Cx.$
The continuous-time system given above is sampled at times $kh, kin N$, with sampling time $h = 3$. We assume that the input function $u(t)$ is constant between two subsequent sampling times. To be more precise, $u(t)=u_k$, when $t in [kh, (k+1)h]$. The result of this exact discretization is given by:
$x_k+1=A_dx_k+B_dx_k$
$y_k=C_dx_k$
with matrices
$A_d = beginbmatrix e^3 & 3e^3 \ 0 & e^3 endbmatrix, quad B_d = beginbmatrix e^3-1 \ 0 endbmatrix, quad C_d = beginbmatrix e^3 & e^3 endbmatrix$
Which of the following continuous-time state-space systems yields the discrete time state space system after exact discretization is applied?
A) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 0 \
0 & 1 & 1 \
hline
1 & 1 \ endarray right]$
B) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
C) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
D) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 3 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
E) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & e^3-1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
linear-algebra control-theory linear-control
asked Aug 6 at 6:59
user463102
1288
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question is from a Systems Theory test without answers or solutions.
Consider the folowing continuous-time state-space system
$dotx=Ax+Bu, quad y=Cx.$
The continuous-time system given above is sampled at times $kh, kin N$, with sampling time $h = 3$. We assume that the input function $u(t)$ is constant between two subsequent sampling times. To be more precise, $u(t)=u_k$, when $t in [kh, (k+1)h]$. The result of this exact discretization is given by:
$x_k+1=A_dx_k+B_dx_k$
$y_k=C_dx_k$
with matrices
$A_d = beginbmatrix e^3 & 3e^3 \ 0 & e^3 endbmatrix, quad B_d = beginbmatrix e^3-1 \ 0 endbmatrix, quad C_d = beginbmatrix e^3 & e^3 endbmatrix$
Which of the following continuous-time state-space systems yields the discrete time state space system after exact discretization is applied?
A) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 0 \
0 & 1 & 1 \
hline
1 & 1 \ endarray right]$
B) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
C) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
D) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 3 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
E) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & e^3-1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
linear-algebra control-theory linear-control
asked Aug 6 at 6:59
user463102
1288
This question is from a Systems Theory test without answers or solutions.
Consider the folowing continuous-time state-space system
$dotx=Ax+Bu, quad y=Cx.$
The continuous-time system given above is sampled at times $kh, kin N$, with sampling time $h = 3$. We assume that the input function $u(t)$ is constant between two subsequent sampling times. To be more precise, $u(t)=u_k$, when $t in [kh, (k+1)h]$. The result of this exact discretization is given by:
$x_k+1=A_dx_k+B_dx_k$
$y_k=C_dx_k$
with matrices
$A_d = beginbmatrix e^3 & 3e^3 \ 0 & e^3 endbmatrix, quad B_d = beginbmatrix e^3-1 \ 0 endbmatrix, quad C_d = beginbmatrix e^3 & e^3 endbmatrix$
Which of the following continuous-time state-space systems yields the discrete time state space system after exact discretization is applied?
A) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 0 \
0 & 1 & 1 \
hline
1 & 1 \ endarray right]$
B) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
C) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & 1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
D) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 3 & 1 \
0 & 1 & 0 \
hline
1 & 1 \ endarray right]$
E) $ left[ beginarrayc
A & B \
hline
C & \ endarray right]
=left[ beginarrayc
1 & 1 & e^3-1 \
0 & 1 & 0 \
hline
e^3 & e^3 \ endarray right]$
linear-algebra control-theory linear-control
asked Aug 6 at 6:59
user463102
1288
asked Aug 6 at 6:59
user463102
1288
asked Aug 6 at 6:59
user463102
1288
asked Aug 6 at 6:59
user463102
1288
1288
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Considering the architecture
$$
left[
beginarrayccc
A_d & vdots &B_d\
cdots & cdot& cdots\
C_d & vdots &
endarray
right] =
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] =
e^3left[
beginarraycccc
1 & 3 & vdots & 1 - e^-3\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
but $e^-3approx 0.05$ hence
$$
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] equiv
e^3left[
beginarraycccc
1 & 3 & vdots & 1\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
NOTE
If we have
$$
A_d = lambda A_d'\
B_d = lambda B_d'\
C_d = lambda C_d'
$$
we have
$$
x_k+1 = A_d x_k + B_d u_k\
y_k = C_d x_k
$$
or assuming null initial conditions
$$
Y(z) = C_dleft(I z - A_d right)^-1B_d U(z)
$$
or
$$
Y(z) =lambda C_d'left( I z - lambda A_d'right)^-1lambda B_d' U(z) = lambda C_d'left( I fraczlambda - A_d'right)^-1B_d' U(z)
$$
Resuming
$$
Y(lambdaxi) = lambda C_d'left( I xi - A_d'right)^-1B_d' U(lambdaxi)
$$
due to the variable change $xi = fraczlambda$. Now the $xi$-transform
$$
Y(lambdaxi) = sum_ky_k (lambdaxi)^k = sum_ky_k lambda^kxi^k = sum_ky'_k xi^k
$$
with $y'_k = lambda^k y_k, u'_k = lambda^k u_k$ so the answer is the transference matrix
$$
left[
beginarrayccc
A'_d & vdots &B'_d\
cdots & cdot& cdots\
lambda C'_d & vdots &
endarray
right];;;mboxor;;; left[
beginarrayccc
A'_d & vdots &lambda B'_d\
cdots & cdot& cdots\
C'_d & vdots &
endarray
right]
$$
answered Aug 6 at 8:21
Cesareo
5,8372412
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
add a comment |Â
up vote
0
down vote
When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
beginbmatrix
A_d & B_d \ 0 & 1
endbmatrix =
expleft(
beginbmatrix
A & B \ 0 & 0
endbmatrix h
right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
exp M = sum_n=0^infty frac1n! M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^3 - 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
and calculating the discretization for E) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^6 - 2,e^3 + 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & e^3 - 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
from this it can be concluded that C) is the answer.
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Considering the architecture
$$
left[
beginarrayccc
A_d & vdots &B_d\
cdots & cdot& cdots\
C_d & vdots &
endarray
right] =
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] =
e^3left[
beginarraycccc
1 & 3 & vdots & 1 - e^-3\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
but $e^-3approx 0.05$ hence
$$
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] equiv
e^3left[
beginarraycccc
1 & 3 & vdots & 1\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
NOTE
If we have
$$
A_d = lambda A_d'\
B_d = lambda B_d'\
C_d = lambda C_d'
$$
we have
$$
x_k+1 = A_d x_k + B_d u_k\
y_k = C_d x_k
$$
or assuming null initial conditions
$$
Y(z) = C_dleft(I z - A_d right)^-1B_d U(z)
$$
or
$$
Y(z) =lambda C_d'left( I z - lambda A_d'right)^-1lambda B_d' U(z) = lambda C_d'left( I fraczlambda - A_d'right)^-1B_d' U(z)
$$
Resuming
$$
Y(lambdaxi) = lambda C_d'left( I xi - A_d'right)^-1B_d' U(lambdaxi)
$$
due to the variable change $xi = fraczlambda$. Now the $xi$-transform
$$
Y(lambdaxi) = sum_ky_k (lambdaxi)^k = sum_ky_k lambda^kxi^k = sum_ky'_k xi^k
$$
with $y'_k = lambda^k y_k, u'_k = lambda^k u_k$ so the answer is the transference matrix
$$
left[
beginarrayccc
A'_d & vdots &B'_d\
cdots & cdot& cdots\
lambda C'_d & vdots &
endarray
right];;;mboxor;;; left[
beginarrayccc
A'_d & vdots &lambda B'_d\
cdots & cdot& cdots\
C'_d & vdots &
endarray
right]
$$
answered Aug 6 at 8:21
Cesareo
5,8372412
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
add a comment |Â
up vote
0
down vote
accepted
Considering the architecture
$$
left[
beginarrayccc
A_d & vdots &B_d\
cdots & cdot& cdots\
C_d & vdots &
endarray
right] =
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] =
e^3left[
beginarraycccc
1 & 3 & vdots & 1 - e^-3\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
but $e^-3approx 0.05$ hence
$$
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] equiv
e^3left[
beginarraycccc
1 & 3 & vdots & 1\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
NOTE
If we have
$$
A_d = lambda A_d'\
B_d = lambda B_d'\
C_d = lambda C_d'
$$
we have
$$
x_k+1 = A_d x_k + B_d u_k\
y_k = C_d x_k
$$
or assuming null initial conditions
$$
Y(z) = C_dleft(I z - A_d right)^-1B_d U(z)
$$
or
$$
Y(z) =lambda C_d'left( I z - lambda A_d'right)^-1lambda B_d' U(z) = lambda C_d'left( I fraczlambda - A_d'right)^-1B_d' U(z)
$$
Resuming
$$
Y(lambdaxi) = lambda C_d'left( I xi - A_d'right)^-1B_d' U(lambdaxi)
$$
due to the variable change $xi = fraczlambda$. Now the $xi$-transform
$$
Y(lambdaxi) = sum_ky_k (lambdaxi)^k = sum_ky_k lambda^kxi^k = sum_ky'_k xi^k
$$
with $y'_k = lambda^k y_k, u'_k = lambda^k u_k$ so the answer is the transference matrix
$$
left[
beginarrayccc
A'_d & vdots &B'_d\
cdots & cdot& cdots\
lambda C'_d & vdots &
endarray
right];;;mboxor;;; left[
beginarrayccc
A'_d & vdots &lambda B'_d\
cdots & cdot& cdots\
C'_d & vdots &
endarray
right]
$$
answered Aug 6 at 8:21
Cesareo
5,8372412
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Considering the architecture
$$
left[
beginarrayccc
A_d & vdots &B_d\
cdots & cdot& cdots\
C_d & vdots &
endarray
right] =
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] =
e^3left[
beginarraycccc
1 & 3 & vdots & 1 - e^-3\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
but $e^-3approx 0.05$ hence
$$
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] equiv
e^3left[
beginarraycccc
1 & 3 & vdots & 1\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
NOTE
If we have
$$
A_d = lambda A_d'\
B_d = lambda B_d'\
C_d = lambda C_d'
$$
we have
$$
x_k+1 = A_d x_k + B_d u_k\
y_k = C_d x_k
$$
or assuming null initial conditions
$$
Y(z) = C_dleft(I z - A_d right)^-1B_d U(z)
$$
or
$$
Y(z) =lambda C_d'left( I z - lambda A_d'right)^-1lambda B_d' U(z) = lambda C_d'left( I fraczlambda - A_d'right)^-1B_d' U(z)
$$
Resuming
$$
Y(lambdaxi) = lambda C_d'left( I xi - A_d'right)^-1B_d' U(lambdaxi)
$$
due to the variable change $xi = fraczlambda$. Now the $xi$-transform
$$
Y(lambdaxi) = sum_ky_k (lambdaxi)^k = sum_ky_k lambda^kxi^k = sum_ky'_k xi^k
$$
with $y'_k = lambda^k y_k, u'_k = lambda^k u_k$ so the answer is the transference matrix
$$
left[
beginarrayccc
A'_d & vdots &B'_d\
cdots & cdot& cdots\
lambda C'_d & vdots &
endarray
right];;;mboxor;;; left[
beginarrayccc
A'_d & vdots &lambda B'_d\
cdots & cdot& cdots\
C'_d & vdots &
endarray
right]
$$
answered Aug 6 at 8:21
Cesareo
5,8372412
Considering the architecture
$$
left[
beginarrayccc
A_d & vdots &B_d\
cdots & cdot& cdots\
C_d & vdots &
endarray
right] =
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] =
e^3left[
beginarraycccc
1 & 3 & vdots & 1 - e^-3\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
but $e^-3approx 0.05$ hence
$$
left[
beginarraycccc
e^3 & 3 e^3 & vdots & e^3 - 1\
0 & e^3 & vdots & 0\
cdots & cdots & cdot& cdots\
e^3 & e^3 & vdots &
endarray
right] equiv
e^3left[
beginarraycccc
1 & 3 & vdots & 1\
0 & 1 & vdots & 0\
cdots & cdots & cdot& cdots\
1 & 1 & vdots &
endarray
right]
$$
NOTE
If we have
$$
A_d = lambda A_d'\
B_d = lambda B_d'\
C_d = lambda C_d'
$$
we have
$$
x_k+1 = A_d x_k + B_d u_k\
y_k = C_d x_k
$$
or assuming null initial conditions
$$
Y(z) = C_dleft(I z - A_d right)^-1B_d U(z)
$$
or
$$
Y(z) =lambda C_d'left( I z - lambda A_d'right)^-1lambda B_d' U(z) = lambda C_d'left( I fraczlambda - A_d'right)^-1B_d' U(z)
$$
Resuming
$$
Y(lambdaxi) = lambda C_d'left( I xi - A_d'right)^-1B_d' U(lambdaxi)
$$
due to the variable change $xi = fraczlambda$. Now the $xi$-transform
$$
Y(lambdaxi) = sum_ky_k (lambdaxi)^k = sum_ky_k lambda^kxi^k = sum_ky'_k xi^k
$$
with $y'_k = lambda^k y_k, u'_k = lambda^k u_k$ so the answer is the transference matrix
$$
left[
beginarrayccc
A'_d & vdots &B'_d\
cdots & cdot& cdots\
lambda C'_d & vdots &
endarray
right];;;mboxor;;; left[
beginarrayccc
A'_d & vdots &lambda B'_d\
cdots & cdot& cdots\
C'_d & vdots &
endarray
right]
$$
answered Aug 6 at 8:21
Cesareo
5,8372412
edited Aug 7 at 6:50
answered Aug 6 at 8:21
Cesareo
5,8372412
answered Aug 6 at 8:21
Cesareo
5,8372412
answered Aug 6 at 8:21
Cesareo
5,8372412
5,8372412
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
add a comment |Â
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
That's a lot easier than I expected. Thanks a lot Cesareo.
– user463102
Aug 6 at 8:44
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
You seem to imply that D) is the answer, which is incorrect.
– Kwin van der Veen
Aug 6 at 19:58
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
@KwinvanderVeen Note that $A_d,B_d,C_d$ are given.
– Cesareo
Aug 6 at 20:04
add a comment |Â
up vote
0
down vote
When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
beginbmatrix
A_d & B_d \ 0 & 1
endbmatrix =
expleft(
beginbmatrix
A & B \ 0 & 0
endbmatrix h
right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
exp M = sum_n=0^infty frac1n! M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^3 - 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
and calculating the discretization for E) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^6 - 2,e^3 + 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & e^3 - 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
from this it can be concluded that C) is the answer.
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
add a comment |Â
up vote
0
down vote
When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
beginbmatrix
A_d & B_d \ 0 & 1
endbmatrix =
expleft(
beginbmatrix
A & B \ 0 & 0
endbmatrix h
right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
exp M = sum_n=0^infty frac1n! M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^3 - 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
and calculating the discretization for E) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^6 - 2,e^3 + 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & e^3 - 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
from this it can be concluded that C) is the answer.
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
add a comment |Â
up vote
0
down vote
up vote
0
down vote
When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
beginbmatrix
A_d & B_d \ 0 & 1
endbmatrix =
expleft(
beginbmatrix
A & B \ 0 & 0
endbmatrix h
right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
exp M = sum_n=0^infty frac1n! M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^3 - 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
and calculating the discretization for E) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^6 - 2,e^3 + 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & e^3 - 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
from this it can be concluded that C) is the answer.
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
When the input is constant during one time period is also known as zero-order hold discretization and can be calculated with
$$
beginbmatrix
A_d & B_d \ 0 & 1
endbmatrix =
expleft(
beginbmatrix
A & B \ 0 & 0
endbmatrix h
right),
$$
and $C_d = C$. The matrix exponential can be calculated with
$$
exp M = sum_n=0^infty frac1n! M^n.
$$
In am not sure if the question is also considering similarity transformations, otherwise every option where $C neq C_d$ can already be omitted. When assuming that similarity transformations are not used then one only has to check two options, namely C) and E).
Calculating the discretization for C) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^3 - 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
and calculating the discretization for E) gives
$$
beginbmatrix
e^3 & 3,e^3 & e^6 - 2,e^3 + 1 \
0 & e^3 & 0 \
0 & 0 & 1
endbmatrix =
expleft(
beginbmatrix
1 & 1 & e^3 - 1 \
0 & 1 & 0 \
0 & 0 & 0
endbmatrix 3
right),
$$
from this it can be concluded that C) is the answer.
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
answered Aug 6 at 19:57
Kwin van der Veen
4,3792826
4,3792826
add a comment |Â
add a comment |Â
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