《信号与系统 Signals and Systems》课程教学资料（英文版）lecture20 A Typical Feedback System

H AA Signals and systems Fall 2003 Lecture#20 0Noⅴ ember2003 Feedback Systems 2. Applications of Feedback Systems

Signals and Systems Fall 2003 Lecture #20 20 November 2003 1. Feedback Systems 2. Applications of Feedback Systems

A Typical feedback system To Be Designed Reference or-- Command Input ? Plant To Be Designed Why use Feedback? Reducing effects of nonidealities Reducing Sensitivity to Uncertainties and variability Stabilizing Unstable Systems Reducing effects of Disturbances Tracking Shaping system response Characteristics(bandwidth/speed)

Why use Feedback? • Reducing Effects of Nonidealities • Reducing Sensitivity to Uncertainties and Variability • Stabilizing Unstable Systems • Reducing Effects of Disturbances • Tracking • Shaping System Response Characteristics (bandwidth/speed) A Typical Feedback System

One motivating example 0 k ki[0D-8(t) Potentia meter Comparator Amplifie gain a(t) e(t) Potentia v(t) Motor e(t k Motor e(t) Platform voltage angular Open-Loop System Closed-Loop Feedback System

One Motivating Example Open-Loop System Closed-Loop Feedback System

Analysis of( Causal!LTI Feedback Systems: Black's Formula CT System (t)- e H(s) Loop r G(s) Y H X()1+G(s)H( Blacks formula(1920s) orward gain Closed-loop system function loop gain Forward gain-total gain along the forward path from the input to the output Loop gain-total gain around the closed loop

Analysis of (Causal!) LTI Feedback Systems: Black’s Formula CT Syste m Black’s formula (1920’s) Closed -loop system function = forward gain 1 - loop gain Forward gain — total gain along the forward path from the input to the output Loop gain — total gain around the closed loop

Applications of Blacks Formula Example: A(s) d(t A(s) C(s) Y(s Forward gain A B X(s 1-loop gain 1+ A' BC AB 1+A X() 1+A+ABC Forward gain B B(1+ -loop gain 1+A'BC 1+A+ABC

Applications of Black’s Formula Example: 1) 2)

The Use of feedback to Compensate for nonidealities () K P(s) G(s) Assume kPO is very large over the frequency range of interest In fact assume KP(j)G(0)>>1 Qiu) Y(jw kP(w) X u) 1+kPlwG(w G(jw Independent of P(s)!!

The Use of Feedback to Compensate for Nonidealities Assume KP (j ω) is very large over the frequency range of interest. In fact, assume — Independent of P(s)!!

Example of Reduced Sensitivity IThe use of operational amplifiers 2)Decreasing amplifier gain sensitivity Example (a) Suppose KP(O)=1000,G(j01)=0.099 1000 Q(O1) =10 1+(1000.099 (b)Suppose KP(@))=500, G(@))=0.099 (50% gain change) 500 Q(02)= =9.9(1% gain change) 1+(500(0.099

Example of Reduced Sensitivity 10 1 (1000)( 0 099 ) 1000 ( ) ( ) 1000 ( ) 0 099 1 1 1 = + = = = . Q j KP j , G j . ω ω ω 1)The use of operational amplifiers 2)Decreasing amplifier gain sensitivity Example: (a) Suppose (b) Suppose (50% gain change) 9 9 1 (500)( 0 099 ) 500 ( ) ( ) 500 ( ) 0 099 2 2 2 . . Q j KP j , G j . ≅ + = = = ω ω ω (1% gain change)

Fine, but why doesnt GG@) fluctuate Note QGu) For amplification, GG@)must attenuate, and it is much easier to build attenuators(e. g. resistors )with desired characteristics There is a price KPG(1)>>1→KP()>> G(w) Needs a large loop gain to produce a steady(and linear) gain for the whole system Consequence of the negative(degenerative) feedback

Fine, but why doesn’t G(jω) fluctuate ? Note: Needs a large loop gain to produce a steady (and linear) gain for the whole system. ⇒ Consequence of the negative (degenerative) feedback. For amplification, G(jω) must attenuate, and it is much easier to build attenuators (e.g. resistors) with desired characteristics There is a price:

Example: Operational amplifiers KAV ()X/+E=△y K E=△V|K y(t) G(s) R+R2 If the amplitude of the loop gain KG(s>> 1 - usually the case, unless the battery is totally dead Th en Y 1R1+B2 Steady st 1 The closed-loop gain only depends on the passive components (R,& r2), independent of the gain of the open-loop amplifier K

Example: Operational Amplifiers If the amplitude of the loop gain |KG(s)| >> 1 — usually the case, unless the battery is totally dead. The closed-loop gain only depends on the passive components (R1 & R2), independent of the gain of the open-loop amplifier K. Then Steady State

The Same Idea works for the compensation for nonlinearities Example and demo: Amplifier with a deadzone (t) (t) EK, y(t) The second system in the forward path has a nonlinear input-output relation (a deadzone for small input), which will cause distortion if it is used as an amplifier. However, as long as the amplitude of the "loop gain is large enough, the input-output response= 1/K

The Same Idea Works for the Compensation for Nonlinearities Example and Demo: Amplifier with a Deadzone The second system in the forward path has a nonlinear input-output relation (a deadzone for small input), which will cause distortion if it is used as an amplifier. However, as long as the amplitude of the “loop gain” is large enough, the input-output response ≅ 1/K2