0
\$\begingroup\$

Say that a system (circuit, pump, population model, etc.) has a exponential output of the type $$y = ae^{bx}$$ What do control engineers do in order to achieve control objectives such as tracking and stabilization around a set point?

\$\endgroup\$
  • \$\begingroup\$ It depends on how much performance you want. The non-linearity will just degrade the response as you'll have to design the controller for the point of highest plant gain, which will make it sluggish when the plant gain is low. In many situations this may be good enough. Otherwise you basically try to model the non-linearity and compensate for it (apply the inverse to the controller output). \$\endgroup\$ – Jon Jul 18 '15 at 10:08
  • \$\begingroup\$ Give us something a little more specific to work with. Currently, the question is too broad. Non-linear control is a huge field due to the diversity of non-linear functions and their interaction with linear system components, in addition to the dependence on signal types and magnitudes. \$\endgroup\$ – Chu Jul 18 '15 at 10:44
  • \$\begingroup\$ Exponential response can be linearized to the extent necessary. It may not even be necessary if the setpoint is fixed. Gain scheduling can work with moderate ranges. \$\endgroup\$ – Spehro Pefhany Jul 18 '15 at 13:28
3
\$\begingroup\$

It's usually no problem at all. Take as a simple electronic example an op-amp precision rectifier. It has to cope with severe non-linearities i.e. the diode in the output circuit. It has a formula shown below: -

enter image description here

Where I is the current thru the device and V is the applied voltage to the terminals. It's exponential like in the question but the circuit below copes well: -

enter image description here

Vout has is produced and is perfectly linear with respect to the peak input voltage i.e. negative feedback has overcome the difficulties. The same is ture for a push pull amplifier with negative feedback. You can make one without final transistor biasing (to overcome the distortion arising when changing from one transistor to another) by choosing an op-amp that is high-speed.

enter image description here

In effect, when the transistors are at the crossover-point the op-amps's high speed almost perfectlyy compensates and "rushes" the input voltage to the transistors past this cross-over point. Again, negative feedback comes to the rescue: -

enter image description here

Real difficulties arise in control systems with complex frequency responses, hysterisis and deadband. Now these can cause major headaches.

\$\endgroup\$
  • \$\begingroup\$ Dead time (eg. Transport Delay) is a real killer. Especially if it's similar or longer than the response time of the plant. \$\endgroup\$ – Spehro Pefhany Jul 18 '15 at 13:24
  • \$\begingroup\$ @Spehro Pefhany, SPC can deal with time delay if a plant model is avaiable \$\endgroup\$ – Chu Jul 18 '15 at 14:42
  • \$\begingroup\$ Oops. Better check. That's a precision rectifier, not a log amp. \$\endgroup\$ – WhatRoughBeast Jul 18 '15 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.