0
\$\begingroup\$

I read the following line in a control system book:

Higher bandwidth corresponds to better command following, disturbance rejection and sensitivity to parametric variations. On the other hand, achievable bandwidth is limited by presence of noise and dynamic uncertainties.

Can you please explain how " Higher bandwidth corresponds to better command following, disturbance rejection and sensitivity to parametric variations"

\$\endgroup\$
  • \$\begingroup\$ You are talking about PID loops, and how fast (bandwidth) the hardware in the loop responds to user or PC input as well as motor RPM, motor current, etc. PID are settings using RC networks to trim the loop to match the motor response time and other parameters. \$\endgroup\$ – Sparky256 Oct 22 '18 at 5:49
  • \$\begingroup\$ It is a general statement,wherein it is given that choice of bandwidth is an important design criteria for controller design because Higher bandwidth corresponds to better command following, disturbance rejection and sensitivity to parametric variations \$\endgroup\$ – ShiS Oct 22 '18 at 5:54
1
\$\begingroup\$

Every system has a time and frequency spectrum of inputs and outputs determined mass & energy storage in the system, choice of inputs, sensors and tolerance vs resolution vs time response. The ideal system matches the signal bandwidth to the desired spectrum to get the desired response with many choices of criteria;

Maximum SNR, minimum group delay, minimum overshoot, minimum step error, maximum disturbance rejection, linear phase, min. static error, dynamic error, min. Integrated squared error, optimal anticipated min. error, maximal interference jitter rejection, min BER, max fault tolerance, max efficiency, max climatic stability, min. FIT rate or max MTBF with redundancy, optimal fault detection/correction/recovery.

You must define the mission, control system with a hierarchical-input-process-output or HIPO design with full design specs. Then the spectral and time response are just necessary parameters to specify for the solution.

The best simple explanation is the step response rise time from 10 to 90% is inversely proportional to the loop bandwidth at half signal power output as a function of frequency.
Tr=0.35/f
for f = max freq. BW @ -3dB. But the dampening factor or it’s inverse , Q must be low to reduce ringing or overshoot with compensation to ensure proper stability.

Road Analogy;

too much delay in feedback might look like a drunk driver wandering off centre or , hyperactive might kill someone in the next lane to avoid a sitting duck, or a tired or lazy driver might have more risk to accidents sampling the road traffic every few seconds rather than being attentive with a visual 25 frames per second.

\$\endgroup\$
  • \$\begingroup\$ Is it possible to give any reason (apart from mathematical formula), why bandwidth is inversely proportional to rise time, i.e., why would increase in bandwidth lead to decrease in rise time. I want to explain it physically not mathematically. Also why higher bandwidth lead to better disturbance rejection \$\endgroup\$ – ShiS Oct 23 '18 at 4:49
  • 1
    \$\begingroup\$ T=1/f ?? The more BW use to correct error, the less error due to delay \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 23 '18 at 13:51
  • \$\begingroup\$ Got it. Thanks. And what about this "why higher bandwidth lead to better disturbance rejection " \$\endgroup\$ – ShiS Oct 23 '18 at 19:49
  • 1
    \$\begingroup\$ The faster the feedback (slew rate) the tighter control of error \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 23 '18 at 22:09
  • 1
    \$\begingroup\$ Think of faster sampling rate and error result driving on curved road and measure your cm error centred from both edges. Then shut your eyes every second and increase and see the error increase. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 23 '18 at 22:13

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.