input output equations - 6.4

d

x( 0)

Given,

----x + 5x

=

3t

= 10

dt

Dx + 5x = 3t

3t

x =

-------------

D + 5

–5t

∫e

5t

x = e

3tdt + C

–5t

te5t

– e5t

x =

e

3

--------------------

+ C

5

Initial conditions,

( 0) ( 1) – ( 1)

x( 0)

( 1)

=

3

-----------------------------

+ C =

10

C = 10.6

5

te5t

– e5t

–5t

x =

e

3

--------------------

+ 10.6

5

x= 0.6( t – 1) + 10.6e–5t

x= 0.6t – 0.6 + 10.6e–5t

d

5t

5t

guess,

----( te

– e

)

dt

= 5te5t + e5t – e5t

d

= 5te5t

5t

5t

5t

----( te

– e

)

=

5e

t

dt

te5t

– e5t

∫e

5t

--------------------

=

tdt

5

Figure 6.4 An example of a solution for a first-order system

6.3 INPUT-OUTPUT EQUATIONS

A typical system will be described by more than one differential equation. These equations can be solved to find a single differential equation that can then be integrated. The basic technique is to arrange the equations into an input-output form, such as that in Figure 6.5. These equations will have only a single output variable, and these are always shown on the left hand side. The input variables (there can be more than one) are all on the right hand side of the equation, and act as the non-homogeneous forcing function.

input output equations - 6.5

e.g.,	···	··		·	·	+ u1	··	+ u3
2y1		+ y1		+ y1	+ 4y1 = u1		+ 3u2 + u3
··		·	+ y2		··	·	·
y2	+ 6y2				= u1 + 3u2	+ u2	+ 0.5u2 + u3

where,

y = outputs u = inputs

Figure 6.5 Developing input-output equations

A sample derivation of an input-output equation from a system of differential equations is given in Figure 6.6. This begins by replacing the differential operator and combining the equations to eliminate one of the output variables. The solution ends by rearranging the equation to input-output form.

Given the differential equations,

y·1

=

– 3y1 + 2y2 + u1 + 2u·2

(1)

y·2

=

2y1 + y2 + u·1

(2)

Find the input-output equations.

(1) Dy1 = – 3y1 + 2y2 + u1 + 2Du2

y1( D + 3) = 2y2 + u1 + 2Du2

										y				= y							D + 3							– 0.5u									– Du
												2							1			-------------													1								2
																								2
				(2) Dy2									= 2y1 + y2 + Du1
									y2( D – 1) = 2y1 + Du1
										y						D + 3																							( D – 1) = 2y
																-------------								– 0.5u							– Du																										+ Du
													1			2														1								2																	1										1
													D2 + 2D – 3																																												2
										y				-----------------------------														– 2				–			0.5Du									+ 0.5u								– D							u				+ Du									=				Du
												1									2																					1								1											2									2									1
										0.5D2y										1	+ Dy					1	– 3.5y						1			= Du							1		+ 0.5Du								1			– 0.5u									1		+ D2u								2	– Du						2
										0.5y1'' + y1' – 3.5y1																					=				u1' + 0.5u1' – 0.5u1 + u2' – u2'

Figure 6.6 An input output equation example

input output equations - 6.6

Find the second equation for the example in Figure 6.6 for the output y2.

Figure 6.7 Drill problem: Find the second equation in the previous example

6.3.1 Converting Input-Output Equations to State Equations

In some instances we will want to numerically integrate an input-output equation. The example starting in Figure 6.8 shows the development of an input-output equation for two freely rolling masses joined by a spring. The final equation has a derivative on the right hand side that would prevent it from being analyzed in many cases. In particular if the input force ’F’ was a step function the first derivative would yield an undefined (infinite) value that could not be integrated.

input output equations - 6.7

F

x1

Ks

x2

M1

M2

Equations of motion can be derived for these masses.

Ks( x2 – x1)

∑Fx

= F + Ks( x2 – x1)

2

F

M1

= M1D

x1

x

( M

D2

+ K ) = F + K

x

2

1

1

s

s

Ks( x2 – x1)

∑Fx = –Ks( x2 – x1) = M2D2x2

M2

K

x

= x

( M

D2

+ K )

s

1

2

2

s

The equations can be combined to eliminate x2.

2

Ksx1

( M

D + K ) = F + K

--------------------------

x

1

1

s

s M

D2

+ K

2

s

x

( ( M

1

D2 + K ) ( M

D2 + K ) – K2) = F( M

D2

+ K )

1

s

2

s

s

2

s

x

( D4M

M

+ D2K

( M

+ M ) + K2 – K2) = F( M

D2

+ K )

1

( D4M

1

2

+ D2K

s

1

2

s

s

D2

2

s

x

1

M

2

s

( M

1

+ M ) ) = F( M

+ K )

1

2

2

s

d

4

d

2

x K ( M + M )

d

2

x M M +

=

FM + FK

----

----

----

dt

1

1

2

dt

1

s 1

2

dt

2

s

Figure 6.8 Writing an input-output equation as a differential equation

The equation is then converted to state variable form, including a step to calculate a second derivative of the input, as shown in Figure 6.9.

input output equations - 6.8

This can then be written in state variable form by creating dummy variables for integrating the function ’F’.

d

x

=

v

----

1

1

dt

d

----

v

1

=

a

1

dt

d

----

a

1

=

d

1

dt

d

( M

+ M ) = a

----

d

1

M

1

M

2

+ a

1

K

1

F

M

2

+ FK

s

dt

s

2

d

d

=

a

–Ks

( M1 + M2)

+ a

1

+ F

Ks

----

-----------------------------------

------

--------------

1

1

F

dt

M1M2

M1

M1M2

The approximate value of the second derivative of a unit time step can be calculated using the time step.

1

(when the timestep, step, it is turned on)

a

F0

=

-----

T

2

1

a

F1

=

–

-----

(after the first timestep)

T

2

1

1

v

F

=

Ta

F0

+ Ta

F1

=

T

-----

+ T

–-----

=

0

(to verify)

T

2

T

2

T2

T2

T2

1

T2

1

1 1

xF

=

-----aF

+ -----aF

=

-----

-----

+

-----

-----

=

-- + -- = 1

2

2

2

T

2

2

T

2

2 2

These equations can then be written in matrix form.

																						0		1						0									0
														x1																															x1
																						0		0						1									0
												d		v		1																													v		1
										----								=				0 0								0									1											+
											dt			a1												–Ks				( M1				+ M2)											a1
														d1								0		0		-----------------------------------													0						d1
																														M1M2

		0			0
	0				0
													aF
	0				0
														F
	1					Ks
	------				--------------
	M1 M1M2

Figure 6.9 Writing state equations for equations with derivatives