Title: Analysis of a closed loop feedback system
1Analysis of a closed loop feedback system
2Complex system
3Multi-variable system
4Problem Description
- Recall the following thermal system
- The inflow temperature is time varying
- The flowrate F m3/s is constant
- Heat losses from the tank are constant at the
operating conditions - Find the outflow temperature, Tout, given the
inflow temperature Tin - Power input from the electrical source is P Wd
W where W is the wattage of the heater and d is
the duty cycle set at the controller. D is a
variable. - The fluid is milk
- The Tank is well mixed
5Laplace Transformation
We found that
We can manipulate the duty cycle D in order to
compensate for changes in qout caused by changes
in qin. We must measure qout , compare it to
the desired temperature and adjust D.
6Set point and contoller
A strategy would be to make the manipulation
proportional to error, that is subtract qout
from qset (the desired temperature) and multiply
the difference by some constant to get D.
Proportional Controller
To measure qout recall our development for a
thermistor
7Combined block diagram
Combining these pieces in a block diagram. We can
calculate how well qout is controlled when qin
changes by solving for the transfer function
relating qout to qin
8Algebraic manipulation
Eliminating qout_measured
Input and output variables only
Eliminating D
Letting qset be at steady state (0) we can solve
for qin
9Closed loop equation by inspection
Notice that this is in the form of
Transfer functions for closed loop systems can be
written by inspection using this rule and the
sign of Loop is opposite the sign of the
feedback. For qout/ qset the transfer function is
For a step in qin we can calculate qout . We
would like qout to remain at 0 and any deviation
from 0 is undesirable.
10Proportional Offset
Rather than solving by inversion, we can
calculate the final value using the final value
theorem
If Kc is very large the error is negligible but
there is always some offset
This error is known as proportional offset. It
is a result of the need for some error to exist
to allow the controller to counter the load
caused by the load, qin , in this case.
11Simplifying the TF
- Let the time constants tm 1 and tp 10. Plot
qout for a step in qin.
12Matlab simulation - Low controller gain
n1 1 D10 11 5 Step(n,d)
13Matlab simulation - high controller gain
- Let KcK2 100
- Conclusion Turning up the controller gain
improves offset but affects stability negatively.
n1 1 D10 11 100 Step(n,d)