Help Session 4
PROBLEM 1
For the electromechanical system shown below, identify the state variables and obtain a statespace model. The output is y(t) (displacement) and the input is e(t) (source voltage).
The back voltage generated through the coil, as the core moves in it, is proportional to the
core velocity, whereas the force developed on the core is proportional to the current i(t)
applied on the coil. Assume back-emf and force constants ke and kf, respectively, and the
core massless and always vertical. The displacements are considered from the equilibrium
position. Ignore the acceleration of gravity and the horizontal bar mass.
1
Following the assumptions in the problem statement and the figure, this problem only has one
mass and one degree-of-freedom. The angular displacement of the bar is written as the
length of the bar,
multiplied by the vertical displacement y, such that
. Since very
large power is required to displace the core downward, it is assumed that mass M will only
move vertically, thus neglecting horizontal motion in the analysis.
2
(7)
3
PROBLEM 2
Consider the fluid system shown. Derive the mathematical model for the system and put them
in state-space form. The output is ‘q4’.
From the lecture notes, in particular lecture 7, it is clear that the sum of the flow rates at a
junction must be zero and that the sum of the pressure drops around a loop must be zero.
These are the hydraulic equivalents of Kirchhoff’s Current and Voltage laws.
Also,
the pressure drop across a fluid resistor is
the pressure drop across an inertor is
the pressure drop across a capacitor is
the general expression for pressure is
4
Assume there is a variable speed pump, which can be considered an ideal flow source, and
delivers a constant flow rate . This pump is connected to “ground” like the three tanks
which in fluidic systems means that at that point the pressure is equal to atmospheric
pressure.
To find an expression for the amount of fluid in a tank it is as simple as setting up a control
volume in thermodynamics, the volumetric rate of change of flow is equal to the flow in –
flow out, and the difference is then added to the previous amount:
or in electrical terms
Charge = initial Charge + (Current coming in – Current going out)*Time
The hydraulic expression involving flow rate and fluid height is:
Net flow rate = Cross-sectional Area of Tank * Rate of Change of Fluid Height
Using these fundamental relationships it is possible to write the governing differential
equations of any hydraulic system.
Equivalent electrical circuit:
Note: The circuit is split in two to demonstrate that the pressure across R2 is independent of
the pressure in tank 2. In other words, there is no direct connection between the tanks 1 and 2.
5
For tank 1
Looking at the node directly above capacitor (tank) 1, we can sum up the flow rates to find an
expression for the flow through the capacitor. Their sum will equal 0, just like according to
Kirchhoff’s Circuit Law.
Performing a loop on the left side of the circuit with R2 and C1 we can easily see that using
KVL the voltage across the capacitor will be equal to the voltage drop of the resistor.
Similarly, R2 and C1 have the same pressure (voltage) drop.
First we write an expression for the pressure drop (voltage) across the capacitor:
(1)
And now we write the pressure drop across the resistor:
(2)
(3)
Plugging these equations into the expression for the pressure drop across the capacitor will
give us the governing differential equation for tank 1.
Substituting (1), (2) for the flow rates in (3) yields:
(4)
For tank 2
As for the first tank, we start the analysis by observing the net flow at the pressure node
above the capacitor, in this case node P2. A quick “KCL” summation yields:
(5)
We know what q2 is, but we don’t have an expression for q3. By observing the right circuit,
we can draw a KVL loop relating q3, P2 and P3 which are the pressures at tanks 2 and 3,
respectively.
6
Using the given relationships this becomes,
(6)
From the pressure (voltage) drop formula for capacitors we can find the pressure drop across
tank 2:
(7)
And now substituting the flow rates (currents)
the second governing differential equation:
from (3) and (6) into equation (7) yields
(8)
For tank 3:
Repeating the same exercise for capacitor 3 gives us the following relationships:
(9)
(10)
(11)
Writing the pressure (voltage) drop across capacitor 3 by substituting (9) and (10) into (11):
(12)
Equations (4), (8) and (12) are the governing differential equations of the system.
Setting up the state space form:
7
8