Ask question

Ask question

All categories

1 year ago

12

a) find the state-space representation of the system. b) is the system controllable? why? c) is the system observable? why

1 answer:

mart [117]1 year ago

5 0

If a controlled input can transfer (alter) the control system's initial states to some other desired states in a finite amount of time, the control system is said to be controllable.

Using Kalman's test, we can determine whether a control system is controllable. The evolution model for the state variables (time-varying unknowns) and the observation model, which connects the observations to the state variables, make up the state space representation of a dynamical system. The capacity to move a system about in its full configuration space using just specific permitted actions is generally referred to as controllability. The precise definition changes slightly depending on the model type or framework used.

Learn more about control here-

brainly.com/question/28540307

#SPJ4

You might be interested in

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0

3 years ago

A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well

ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = $\frac{Q}{4\pi T} * W (u)$

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0

3 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

b)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

c)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0

3 years ago

What type of oil pressure gauge should be used when

liq [111]

Answer:

The mechanical gauge would be the one for the job

Explanation:

6 0

3 years ago

A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in

DochEvi [55]

Answer:

$a=5.515\frac{m}{s^{2} }$

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

$1 mile=1609.34 mts.$

$1 hora=3600 segundos$

Performing the operations

$60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}$

Now, we will use the acceleration formula

$a=\frac{v}{t}$

Where v = speed and t = time

Substituting the values of $t=5.2s$

$a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }$

7 0

3 years ago