Lecture #12:

INTERCONNECTED SYSTEMS AND FEEDBACK

Motivation:

Outline:

I. INTERCONNECTION OF SYSTEMS

Systems are interconnections of sub-systems. For example, consider the cascade of LTI systems shown below.

The presumption in such a cascade is that H1(s) and H2(s) do not change when the two systems are connected.

1/ Cascade of a lowpass and a highpass filter

Suppose H1(s)H1(s) size 12{H rSub { size 8{1} } \( s \) } {}and H2(s)H2(s) size 12{H rSub { size 8{2} } \( s \) } {}have the following form.

H 1 ( s ) = Y 1 ( s ) X 1 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 H 1 ( s ) = Y 1 ( s ) X 1 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 size 12{H rSub { size 8{1} } \( s \) = { {Y rSub { size 8{1} } \( s \) } over {X rSub { size 8{1} } \( s \) } } = { { { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } over {s+ { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } } } {}

H 2 ( s ) = Y 2 ( s ) X 2 ( s ) = s s + 1 R 2 C 2 H 2 ( s ) = Y 2 ( s ) X 2 ( s ) = s s + 1 R 2 C 2 size 12{H rSub { size 8{2} } \( s \) = { {Y rSub { size 8{2} } \( s \) } over {X rSub { size 8{2} } \( s \) } } = { {s} over {s+ { {1} over {R rSub { size 8{2} } C rSub { size 8{2} } } } } } } {}

2/ Loading

Now cascade H1(s)H1(s) size 12{H rSub { size 8{1} } \( s \) } {} and H2(s)H2(s) size 12{H rSub { size 8{2} } \( s \) } {}.

l

H 1 ( s ) = 1 R 1 C 1 s s 2 + s 1 R 1 C 1 + 1 R 2 C 2 + 1 R 2 C 1 + 1 R 1 R 2 C 1 C 2 H 1 ( s ) = 1 R 1 C 1 s s 2 + s 1 R 1 C 1 + 1 R 2 C 2 + 1 R 2 C 1 + 1 R 1 R 2 C 1 C 2 size 12{H rSub { size 8{1} } \( s \) = { { { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } s} over {s rSup { size 8{2} } +s left [ { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } C rSub { size 8{2} } } } + { {1} over {R rSub { size 8{2} } C rSub { size 8{1} } } } right ]+ { {1} over {R rSub { size 8{1} } R rSub { size 8{2} } C rSub { size 8{1} } C rSub { size 8{2} } } } } } } {}

Note that

H ( s ) ≠ H 1 ( s ) H 2 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 s s + 1 R 2 C 2 H ( s ) ≠ H 1 ( s ) H 2 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 s s + 1 R 2 C 2 size 12{H \( s \) <> H rSub { size 8{1} } \( s \) H rSub { size 8{2} } \( s \) = left [ { { { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } over {s+ { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } } right ] left [ { {s} over {s+ { {1} over {R rSub { size 8{2} } C rSub { size 8{2} } } } } } right ]} {}

3/ Isolation

With the use of an op-amp, the two systems can be isolated from each other or buffered so that the system function is the product of the individual system functions.

Note that

H ( s ) ≠ H 1 ( s ) H 2 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 s s + 1 R 2 C 2 H ( s ) ≠ H 1 ( s ) H 2 ( s ) = 1 R 1 C 1 s + 1 R 1 C 1 s s + 1 R 2 C 2 size 12{H \( s \) <> H rSub { size 8{1} } \( s \) H rSub { size 8{2} } \( s \) = left [ { { { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } over {s+ { {1} over {R rSub { size 8{1} } C rSub { size 8{1} } } } } } right ] left [ { {s} over {s+ { {1} over {R rSub { size 8{2} } C rSub { size 8{2} } } } } } right ]} {}

4/ Conclusion

When we draw block diagrams of the form

we assume that the individual systems are buffered or that the loading of system 1 by system 2 is taken into account in H1(s)H1(s) size 12{H rSub { size 8{1} } \( s \) } {}

II. FEEDBACK EXAMPLES

1/ Man-made system — robot car

2/ Robot car block diagram

To drive the robot car to the target we use the camera to compare the measured target position with the desired target position. The difference is an error signal whose value is used to change the wheel position. Therefore, the output variable, the angle of the wheels, is fed back to the input to control the new output variable.

3/ Wheel position controller block diagram

The wheel controller system is itself a feedback system. A voltage proportional to the angular position of the motor shaft is subtracted from the desired value and the difference signal is used to drive the motor.

4/ Physiological control systems examples

Voluntary everyday activities

Involuntary everyday occurrences

5/ Spinal reflex

Tapping the patella stretches muscle receptors that, through a neural feedback system, results in muscle contraction. This reflex is used in the maintenance of posture.

III. SIMPLE LINEAR FEEDBACK SYSTEM

1/ Black’s formula

K(s) is called the open loop system function, and H(s) = Y (s)/X(s) is called the closed-loop system function. Note, when β(s) = 0, H(s) = K(s)

We can find H(s) by combining

E ( s ) = X ( s ) − β ( s ) Y ( s ) and Y ( s ) = K ( s ) E ( s ) E ( s ) = X ( s ) − β ( s ) Y ( s ) and Y ( s ) = K ( s ) E ( s ) size 12{E \( s \) =X \( s \) - β \( s \) Y \( s \) matrix { {} # {} } ital "and" matrix { {} # {} } Y \( s \) =K \( s \) E \( s \) } {}

to obtain

Y ( s ) = K ( s ) X ( s ) − β ( s ) Y ( s ) Y ( s ) = K ( s ) X ( s ) − β ( s ) Y ( s ) size 12{Y \( s \) =K \( s \) X \( s \) - β \( s \) Y \( s \) } {}

which can be solved to obtain Black’s formula,

H ( s ) = Y ( s ) X ( s ) = K ( s ) 1 + β ( s ) K ( s ) = forward transmission 1 − loop gain H ( s ) = Y ( s ) X ( s ) = K ( s ) 1 + β ( s ) K ( s ) = forward transmission 1 − loop gain size 12{H \( s \) = { {Y \( s \) } over {X \( s \) } } = { {K \( s \) } over {1+β \( s \) K \( s \) } } = { { ital "forward" matrix { {} # {} } ital "transmission"} over {1 matrix { {} # {} } - matrix { {} # {} } ital "loop" matrix { {} # {} } ital "gain"} } } {}

Two-minute miniquiz problem

Problem 8-1 Simple position control system

The objective of the position control system shown below is for the output position Y (s) to track the input signal X(s).

a) Determine the closed-loop system function H(s) = Y (s)/X(s).

b) For x(t) = u(t), a unit step, determine the steady state value of y(t).

Solution

H ( s ) = K ( s + 1 ) ( s + 100 ) 1 + K ( s + 1 ) ( s + 100 ) = K ( s + 1 ) ( s + 100 ) + K H ( s ) = K ( s + 1 ) ( s + 100 ) 1 + K ( s + 1 ) ( s + 100 ) = K ( s + 1 ) ( s + 100 ) + K size 12{H \( s \) = { { { {K} over { \( s+1 \) \( s+"100" \) } } } over {1+ { {K} over { \( s+1 \) \( s+"100" \) } } } } = { {K} over { \( s+1 \) \( s+"100" \) +K} } } {}

b) The steady-state response to a unit step is simply the response to the complex exponential x(t)= 1. e0.t = 1 x(t)= 1. e0.t = 1 size 12{x \( t \) =" 1" "." " e" rSup { size 8{0 "." "t "} } =" 1 "} {} which is y(t)= 1. H(0).e0.t = K/(100 + K)y(t)= 1. H(0).e0.t = K/(100 + K) size 12{y \( t \) =" 1" "." " H" \( 0 \) "." e rSup { size 8{0 "." "t "} } =" K/" \( "100 "+" K" \) } {}. The position error ε=K/(100+K)−1 = −100/(100 + Κ)ε=K/(100+K)−1 = −100/(100 + Κ) size 12{ε=K/ \( "100"+K \) - 1" = "-"100/" \( "100 + "Κ \) } {} Hence, this position controller (with proportional feedback) has an error that diminishes as the gain increases. However, the error is never zero no matter how large the gain.

Effect of feedback on system performance

Feedback is used to enhance system performance.

Properties of feedback

2/ Stabilize gain

overall gain = 10 Overall gain = 100 × 10 1 + 0 . 099 × 100 × 10 = 10 overall gain = 10 Overall gain = 100 × 10 1 + 0 . 099 × 100 × 10 = 10 size 12{ ital "overall" matrix { {} # {} } ital "gain"="10" matrix { {} # {} } ital "Overall" matrix { {} # {} } ital "gain"= { {"100" times "10"} over {1+0 "." "099" times "100" times "10"} } ="10"} {}

Note that both the open-loop and the same gain which equals 10.

But now suppose that the gain of the power amplifier is reduced to 5.

overall gain = 10 Overall gain = 100 × 5 1 + 0 . 099 × 100 × 5 = 9 . 9 overall gain = 10 Overall gain = 100 × 5 1 + 0 . 099 × 100 × 5 = 9 . 9 size 12{ ital "overall" matrix { {} # {} } ital "gain"="10" matrix { {} # {} } ital "Overall" matrix { {} # {} } ital "gain"= { {"100" times 5} over {1+0 "." "099" times "100" times 5} } =9 "." 9} {}

Note that a change in gain of the power amplifier of 50% leads to a change in gain of the feedback system of only 1%.

The stabilization of the gain resulting from feedback can be appreciated more generally from Black’s formula.

H ( s ) = K ( s ) 1 + β ( s ) K ( s ) H ( s ) = K ( s ) 1 + β ( s ) K ( s ) size 12{H \( s \) = { {K \( s \) } over {1+β \( s \) K \( s \) } } } {}

If K(s) is large so that |β(s)K(s)|>>1 then

H ( s ) ≈ 1 β ( s ) H ( s ) ≈ 1 β ( s ) size 12{H \( s \) approx { {1} over {β \( s \) } } } {}

So if H(s) has a gain then β(s) must have an attenuation. If the attenuation β(s) is determined precisely but the gain K(s) varies (with time, temperature, etc.), then we can make an amplifier whose gain is independent of K(s) and determined almost entirely by β(s).

But how can we make β(s) precise?

Consider the non-inverting amplifier with a non-ideal (finite gain) op-amp — network (left), block diagram (right).

For an op-amp (model 741) the gain is typically K≈107K≈107 size 12{K approx "10" rSup { size 8{7} } } {}. Hence, provided KR2/(R1+R2)>>1KR2/(R1+R2)>>1 size 12{"KR" rSub { size 8{2} } / \( R rSub { size 8{1} } +R rSub { size 8{2} } \) ">>1"} {}, we have

V o V i = 10 7 1 + 10 7 R 2 R 1 + R 2 ≈ R 1 + R 2 R 2 V o V i = 10 7 1 + 10 7 R 2 R 1 + R 2 ≈ R 1 + R 2 R 2 size 12{ { {V rSub { size 8{o} } } over {V rSub { size 8{i} } } } = { {"10" rSup { size 8{7} } } over {1+"10" rSup { size 8{7} } { {R rSub { size 8{2} } } over {R rSub { size 8{1} } +R rSub { size 8{2} } } } } } approx { {R rSub { size 8{1} } +R rSub { size 8{2} } } over {R rSub { size 8{2} } } } } {}

Conclusion — the gain of the feedback amplifier depends primarily on the values of the resistors and not on the gain of the op-amp which depends on parameters of transistors which change with time, temperature, etc.

3/ Reduce the effect of an output disturbance

The transfer functions for the input and the disturbance are

Y ( s ) X ( s ) = K ( s ) 1 + β ( s ) K ( s ) and Y ( s ) D ( s ) = 1 1 + β ( s ) K ( s ) Y ( s ) X ( s ) = K ( s ) 1 + β ( s ) K ( s ) and Y ( s ) D ( s ) = 1 1 + β ( s ) K ( s ) size 12{ { {Y \( s \) } over {X \( s \) } } = { {K \( s \) } over {1+β \( s \) K \( s \) } } matrix { {} # {} } ital "and" matrix { {} # {} } { {Y \( s \) } over {D \( s \) } } = { {1} over {1+β \( s \) K \( s \) } } } {}

Therefore, if β(s)K(s) is made arbitrarily large then

Y ( s ) / X ( s ) ⇒ 1 / β ( s ) and Y ( s ) / D ( s ) ⇒ 0 Y ( s ) / X ( s ) ⇒ 1 / β ( s ) and Y ( s ) / D ( s ) ⇒ 0 size 12{Y \( s \) /X \( s \) drarrow 1/β \( s \) matrix { {} # {} } ital "and" matrix { {} # {} } Y \( s \) /D \( s \) drarrow 0} {}

4/ Improve dynamic characteristics

Our model of an op-amp was simplistic. A more realistic model that takes the finite bandwidth of the op-amp into account is shown below.

The finite bandwidth of the op-amp is modeled as

K ( s ) = K s + α K ( s ) = K s + α size 12{K \( s \) = { {K} over {s+α} } } {}

Using Black’s formula we obtain

H ( s ) = K ( s ) 1 + βK ( s ) = K s + α 1 + β K s + α = K s + ( α + βK ) H ( s ) = K ( s ) 1 + βK ( s ) = K s + α 1 + β K s + α = K s + ( α + βK ) size 12{H \( s \) = { {K \( s \) } over {1+βK \( s \) } } = { { { {K} over {s+α} } } over {1+β { {K} over {s+α} } } } = { {K} over {s+ \( α+βK \) } } } {}

Therefore, the frequency response is

H ( jω ) = K / ( α + βK ) jω / ( α + βK ) + 1 H ( jω ) = K / ( α + βK ) jω / ( α + βK ) + 1 size 12{H \( jω \) = { {K/ \( α+βK \) } over {jω/ \( α+βK \) +1} } } {}

the impulse response is

h ( t ) = Ke − ( α + βK ) t u ( t ) h ( t ) = Ke − ( α + βK ) t u ( t ) size 12{h \( t \) = ital "Ke" rSup { size 8{ - \( α+βK \) t} } u \( t \) } {}

and the step response is

s ( t ) = K α + βK ( 1 − e − ( α + βK ) t ) u ( t ) s ( t ) = K α + βK ( 1 − e − ( α + βK ) t ) u ( t ) size 12{s \( t \) = { {K} over {α+βK} } \( 1 - e rSup { size 8{ - \( α+βK \) t} } \) u \( t \) } {}

We compare the open-loop with the closed-loop characteristics. We use typical parameters of a model 741 op-amp, K = 8×106K = 8×106 size 12{"K "=" 8" times "10" rSup { size 8{6} } } {} and α = 40 rad/sec.

Effect on pole-zero diagram

As the loop gain is increased from 0 the pole moves out along the negative real axis from −40 rad/s and when β = 0.1 the pole reaches −8×105rad/s−8×105rad/s size 12{ approx - 8 times "10" rSup { size 8{5} } ital "rad"/s} {}. This type of diagram which shows the trajectory of the closed loop poles in the complex s-plane as the gain is changed is called a root locus plot.

Effect of feedback on frequency response.

B

Effect of feedback on step response.

As the loop gain is increased from 0 the time constant of the step response decreases, i.e., the system responds faster.

Summary of effect of feedback on dynamic characteristics.

5/ Reduce the effect of nonlinear distortion

A common power amplifier configuration found in many electronic systems (e.g., stereo amplifiers) is the push-pull emitter follower amplifier.

This configuration has an inherent nonlinearity. When vi> 0.6 Vvi> 0.6 V size 12{v rSub { size 8{i} } >" 0" "." "6 V"} {}, T1 conducts and when vi< 0.6 Vvi< 0.6 V size 12{v rSub { size 8{i} } <" 0" "." "6 V"} {}, T2 conducts. Thus, there is a dead zone between −0.6 and +0.6 V where neither transistor conducts.

The transfer characteristic, which contains an idealized dead zone, is shown below.

In an audio amplifier, the type of distortion caused by the dead zone is called crossover distortion.

We examine the use of feedback to reduce cross-over distortion with the aid of MATLAB’s block diagram language SIMULINK.

6/ Effect of cross-over distortion on simple signals and on music

The figure shows the effect of crossover distortion on a sinusoid obtained with Simulink. A demo will show the effect of cross-over distortion in an audio amplifier on simple signals and on music and the role of feedback to minimize this distortion.

IV. DYNAMIC PERFORMANCE OF FEEDBACK SYSTEMS

1/ Simple position control system with proportional controller

The objective of the position control system shown below is for the output position Y (s) to track the input signal X(s).

We can use Black’s formula to find H(s) as follows

H ( s ) = K ( s + 1 ) ( s + 100 ) 1 + K ( s + 1 ) ( s + 100 ) = K ( s + 1 ) ( s + 100 ) + K H ( s ) = K ( s + 1 ) ( s + 100 ) 1 + K ( s + 1 ) ( s + 100 ) = K ( s + 1 ) ( s + 100 ) + K size 12{H \( s \) = { { { {K} over { \( s+1 \) \( s+"100" \) } } } over {1+ { {K} over { \( s+1 \) \( s+"100" \) } } } } = { {K} over { \( s+1 \) \( s+"100" \) +K} } } {}

How good is this system at controlling the motor position?

Let us examine the steady-state step response first. The system function is

H ( s ) = K ( s + 1 ) ( s + 100 ) + K H ( s ) = K ( s + 1 ) ( s + 100 ) + K size 12{H \( s \) = { {K} over { \( s+1 \) \( s+"100" \) +K} } } {}

The steady-state response to a unit step is simply the response to the complex exponential x(t)= 1.e0.t= 1x(t)= 1.e0.t= 1 size 12{x \( t \) =" 1" "." e rSup { size 8{0 "." t} } =" 1"} {} which is y(t) = 1.H(0) e0.t= K/(100+K)e0.t= K/(100+K) size 12{e rSup { size 8{0 "." t} } =" K/" \( "100"+K \) } {}. The position error ε=K/(100+K)−1=−100/(100+K). Hence, this position control system (with proportional controller) has an error that can be made arbitrarily small, although not zero, as the gain is made arbitrarily large.

Two-minute miniquiz problem

Problem 9-1

For K = 1000 determine the unit step response y(t) of the position control system.

[ H int: the polynomial s 2 + 101 s + 1100 ≈ ( s + 88 . 5 ) ( s + 12 . 5 ) ] [ H int: the polynomial s 2 + 101 s + 1100 ≈ ( s + 88 . 5 ) ( s + 12 . 5 ) ] size 12{ \[ H"int:" matrix { {} # {} } ital "the" matrix { {} # {} } ital "polynomial" matrix { {} # {} } s rSup { size 8{2} } +"101"s+"1100" approx \( s+"88" "." 5 \) \( s+"12" "." 5 \) \] } {}

Solution

Substituting K = 1000 into the system function yields

Y ( s ) = X ( s ) H ( s ) = 1 s 1000 ( s + 1 ) ( s + 100 ) + 1000 Y ( s ) = X ( s ) H ( s ) = 1 s 1000 ( s + 1 ) ( s + 100 ) + 1000 size 12{Y \( s \) =X \( s \) H \( s \) = left [ { {1} over {s} } right ] left [ { {"1000"} over { \( s+1 \) \( s+"100" \) +"1000"} } right ]} {}

The denominator polynomial can be factored and expanded in a partial fraction expansion as follows

Y ( s ) = 1000 s ( s + 88 . 5 ) ( s + 12 . 5 ) = 1000 / ( 88 . 5 ) ( 12 . 5 ) s + 100 / ( − 88 . 5 ) ( − 76 ) s + 88 . 5 + 1000 / ( − 12 . 5 ) ( 76 ) s + 12 . 5 = 0 . 9 s + 0 . 15 s + 88 . 5 + 1 . 05 s + 12 . 5 Y ( s ) = 1000 s ( s + 88 . 5 ) ( s + 12 . 5 ) = 1000 / ( 88 . 5 ) ( 12 . 5 ) s + 100 / ( − 88 . 5 ) ( − 76 ) s + 88 . 5 + 1000 / ( − 12 . 5 ) ( 76 ) s + 12 . 5 = 0 . 9 s + 0 . 15 s + 88 . 5 + 1 . 05 s + 12 . 5 alignl { stack { size 12{Y \( s \) = { {"1000"} over {s \( s+"88" "." 5 \) \( s+"12" "." 5 \) } } } {} # size 12{ matrix { {} # {} } = { {"1000"/ \( "88" "." 5 \) \( "12" "." 5 \) } over {s} } + { {"100"/ \( - "88" "." 5 \) \( - "76" \) } over {s+"88" "." 5} } + { {"1000"/ \( - "12" "." 5 \) \( "76" \) } over {s+"12" "." 5} } } {} # size 12{ matrix { {} # {} } = { {0 "." 9} over {s} } + { {0 "." "15"} over {s+"88" "." 5} } + { {1 "." "05"} over {s+"12" "." 5} } } {} } } {}

The step response is

y ( t ) = ( 0 . 9 + 0 . 15 e − 88 . 5t − 1 . 05 e − 12 . 5t ) u ( t ) y ( t ) = ( 0 . 9 + 0 . 15 e − 88 . 5t − 1 . 05 e − 12 . 5t ) u ( t ) size 12{y \( t \) = \( 0 "." 9+0 "." "15"e rSup { size 8{ - "88" "." 5t} } - 1 "." "05"e rSup { size 8{ - "12" "." 5t} } \) u \( t \) } {}

Note that y(∞) = 0.9 which fits with the result obtained from the steady-state analysis which gives 1000/1100 ≈ 0.9.

A plot of the step response for K = 1000 along with those for several values of K are shown next.

How does the step response change as K is increased?

Thus, we cannot achieve an arbitrarily small position error without causing damped oscillations with this controller design.

2/ Simple position control system with zero position error

Consider a new design in which the error is integrated.

We can use Black’s formula to find H(s) as follows

H ( s ) = K s ( s + 1 ) ( s + 100 ) 1 + K s ( s + 1 ) ( s + 100 ) = K s ( s + 1 ) ( s + 100 ) + K H ( s ) = K s ( s + 1 ) ( s + 100 ) 1 + K s ( s + 1 ) ( s + 100 ) = K s ( s + 1 ) ( s + 100 ) + K size 12{H \( s \) = { { { {K} over {s \( s+1 \) \( s+"100" \) } } } over {1+ { {K} over {s \( s+1 \) \( s+"100" \) } } } } = { {K} over {s \( s+1 \) \( s+"100" \) +K} } } {}

The steady-state response to a unit step is simply the response to the complex exponential x(t)= 1.e0.t= 1x(t)= 1.e0.t= 1 size 12{x \( t \) =" 1" "." e rSup { size 8{0 "." t} } =" 1"} {} which is y(t)= 1.H(0).e0.t= 1y(t)= 1.H(0).e0.t= 1 size 12{y \( t \) =" 1" "." H \( 0 \) "." e rSup { size 8{0 "." t} } =" 1"} {}. The position error ε = 1− 1 = 0. Hence, it appears that this position control system (with integral controller) has no position error for any value of K. So how do we pick K?

3/ Simple position control system with zero position error — step response

The step response is shown for several values of K.

4/ BIBO stability

There are many ways one can define stability of a system. We shall use the following. A system for which every bounded input yields a bounded output is called BIBO stable. A feedback system with closed loop system function

H ( s ) = K ( s ) 1 + β ( s ) K ( s ) H ( s ) = K ( s ) 1 + β ( s ) K ( s ) size 12{H \( s \) = { {K \( s \) } over {1+β \( s \) K \( s \) } } } {}

is BIBO stable if its poles (the natural frequencies of the closed loop system) are in the left half of the s-plane. Thus, determining the conditions for which a system is stable reduces to finding whether the zeros of 1+β(s)K(s) = 0 are in the left-half s-plane. When β(s)K(s) is a rational function, this condition is tested by determining whether the roots of the numerator polynomial of 1+β(s)K(s) are located in the left-half s-plane.

V. ROOTS OF SECOND-ORDER AND THIRD-ORDER POLYNOMIALS

We consider conditions that second- and third-order polynomials have roots in the left half of the complex s-plane.

1/ Second-order polynomials

Second-order polynomials with real coefficients have either real or complex roots of the form

( s + a ) ( s + b ) = 0 or ( s + a + jc ) ( s + a - jc ) = 0 ( s + a ) ( s + b ) = 0 or ( s + a + jc ) ( s + a - jc ) = 0 size 12{ \( s+a \) \( s+b \) =" 0 " matrix { {} # {} } "or" matrix { {} # {} } \( s+a+"jc" \) \( s+"a - jc" \) " = 0"} {}

where a > 0 and b > 0. The polynomials can be expressed as

s 2 + ( a + b ) s + ab or s 2 + 2 as + a 2 + c 2 = 0 s 2 + ( a + b ) s + ab or s 2 + 2 as + a 2 + c 2 = 0 size 12{s rSup { size 8{2} } + \( a+b \) s+ ital "ab" matrix { {} # {} } ital "or" matrix { {} # {} } s rSup { size 8{2} } +2 ital "as"+a rSup { size 8{2} } +c rSup { size 8{2} } =0} {}

Thus, both polynomials have the form

s 2 + αs + β = 0 s 2 + αs + β = 0 size 12{s rSup { size 8{2} } +αs+β=0} {}

where α > 0 and β > 0. These conditions are both necessary and sufficient.

2/ Third-order polynomials

Third-order polynomials must have one real root and either a pair of real or complex roots of the form

( s + a ) ( s + b ) ( s + c ) = 0 or ( s + b ) ( s + a + jd ) ( s + a - jd ) = 0 ( s + a ) ( s + b ) ( s + c ) = 0 or ( s + b ) ( s + a + jd ) ( s + a - jd ) = 0 size 12{ \( s+a \) \( s+b \) \( s+c \) =" 0 " matrix { {} # {} } "or" matrix { {} # {} } \( s+b \) \( s+a+"jd" \) \( s+"a - jd" \) " = 0"} {}

where a > 0, b > 0, and c > 0. The polynomials can be expressed as

s 3 + ( a + b + c ) s 2 + ( ab + ac + bc ) s + abc = 0 or s 3 + ( 2a + b ) s 2 + ( a 2 + d 2 + 2 ab ) s + ( a 2 + d 2 ) b = 0 s 3 + ( a + b + c ) s 2 + ( ab + ac + bc ) s + abc = 0 or s 3 + ( 2a + b ) s 2 + ( a 2 + d 2 + 2 ab ) s + ( a 2 + d 2 ) b = 0 alignl { stack { size 12{s rSup { size 8{3} } + \( a+b+c \) s rSup { size 8{2} } + \( ital "ab"+ ital "ac"+ ital "bc" \) s+ ital "abc"=0 matrix { {} # {} } } {} # ital "or" matrix { {} # {} } s rSup { size 8{3} } + \( 2a+b \) s rSup { size 8{2} } + \( a rSup { size 8{2} } +d rSup { size 8{2} } +2 ital "ab" \) s+ \( a rSup { size 8{2} } +d rSup { size 8{2} } \) b=0 {} } } {}

Thus, both can be put in the form

s 3 + αs 2 + βs + γ = 0 s 3 + αs 2 + βs + γ = 0 size 12{s rSup { size 8{3} } +αs rSup { size 8{2} } +βs+γ=0} {}

Note that α > 0, β > 0, and γ > 0. But in addition, β > γ/α. These conditions are both necessary and sufficient.

VI. ROOT LOCUS PLOTS FOR POSITION CONTROL SYSTEMS

1/ Proportional controller

H ( s ) = K s 2 + 101 s + 100 + K H ( s ) = K s 2 + 101 s + 100 + K size 12{H \( s \) = { {K} over {s rSup { size 8{2} } +"101"s+"100"+K} } } {}

Recall the step response of the position control system with proportional controller.

The poles of the closed-loop system function are at

s 1,2 = − 101 2 ± 101 2 2 − 100 − K 1 / 2 s 1,2 = − 101 2 ± 101 2 2 − 100 − K 1 / 2 size 12{s rSub { size 8{1,2} } = - { {"101"} over {2} } +- left [ left [ { {"101"} over {2} } right ] rSup { size 8{2} } - "100" - K right ] rSup { size 8{1/2} } } {}

The root locus plot is

2/ Integral controller

H ( s ) = K s 3 + 101 s 2 + 100 s + K H ( s ) = K s 3 + 101 s 2 + 100 s + K size 12{H \( s \) = { {K} over {s rSup { size 8{3} } +"101"s rSup { size 8{2} } +"100"s+K} } } {}

Recall the step response of the position control system with integral controller.

The poles are the roots of the polynomial

s 3 + 101 s 2 + 100 s + K = 0 s 3 + 101 s 2 + 100 s + K = 0 size 12{s rSup { size 8{3} } +"101"s rSup { size 8{2} } +"100"s+K=0} {}

The root locus plot is

Note that all the poles are in the lhp for K >0 and 100 > K/101 or 0 < K < 10100. For K>10100 two poles move into the rhp and the system is unstable.

VII. STABILIZATION OF UNSTABLE SYSTEMS

1/ Many common systems are unstable

Some common systems are annoyingly unstable (adapted from Figure 11.7 in Oppenheim & Willsky, 1983).

We can model the audio feedback system with SIMULINK as follows.

Another example of an unstable system is an inverted pendulum. For example, balancing a broomstick in your hand is an example of an inherently unstable system that is stabilized by your motor control system.

Figure adapted from Figure 11.2 in Oppenheim & Willsky, 1983.

2/ Inverted pendulum

We will analyze an inverted pendulum attached to a cart. A schematic diagram is shown on the left and a free-body diagram showing the forces on the pendulum is shown on the right.

The forces of attachment of the cart and pendulum on the pendulum are obtained from the equations of rectilinear motion.

The equation of rotational motion about the center of mass is

J d 2 θ ( t ) dt 2 = mgl sin θ ( t ) − ml cos θ ( t ) d 2 x ( t ) dt 2 J d 2 θ ( t ) dt 2 = mgl sin θ ( t ) − ml cos θ ( t ) d 2 x ( t ) dt 2 size 12{J { {d rSup { size 8{2} } θ \( t \) } over { ital "dt" rSup { size 8{2} } } } = ital "mgl""sin"θ \( t \) - ital "ml""cos"θ \( t \) { {d rSup { size 8{2} } x \( t \) } over { ital "dt" rSup { size 8{2} } } } } {}

where J is the moment of inertia of the mass about the mass, m is the mass, and g is acceleration of gravity. For small θ, this differential equation is linearized by noting that sinθ≈θ and cos θ ≈ 1. Therefore,

J d 2 θ ( t ) dt 2 − mgl sin θ ( t ) = − ml d 2 x ( t ) dt 2 J d 2 θ ( t ) dt 2 − mgl sin θ ( t ) = − ml d 2 x ( t ) dt 2 size 12{J { {d rSup { size 8{2} } θ \( t \) } over { ital "dt" rSup { size 8{2} } } } - ital "mgl""sin"θ \( t \) = - ital "ml" { {d rSup { size 8{2} } x \( t \) } over { ital "dt" rSup { size 8{2} } } } } {}

The system function is

H ( s ) = θ ( s ) X ( s ) = − ( ml / J ) s 2 s 2 − ( mgl / J ) H ( s ) = θ ( s ) X ( s ) = − ( ml / J ) s 2 s 2 − ( mgl / J ) size 12{H \( s \) = { {θ \( s \) } over {X \( s \) } } = { { - \( ital "ml"/J \) s rSup { size 8{2} } } over {s rSup { size 8{2} } - \( ital "mgl"/J \) } } } {}

Therefore, the poles occur at

s 1,2 = ± mgl J 1 / 2 s 1,2 = ± mgl J 1 / 2 size 12{s rSub { size 8{1,2} } = +- left [ { { ital "mgl"} over {J} } right ] rSup { size 8{1/2} } } {}

Hence, the system is inherently unstable since one of its poles (natural frequencies) is in the rhp.

3/ Stabilization of the inverted pendulum

To stabilize the inverted pendulum, a rotary potentiometer is used to measure θ(t). A current proportional to θ(t) − θo, where θo is the desired angle, drives a motor so as to increase x(t) (adapted from Figure 6.4-1, Siebert, 1986).

This inverted pendulum is connected in a feedback configuration as follows.

M(s) represents the system function of the motor dynamics, 1 + (a/s) is the system function of the proportional plus integral controller for θ(t), and c + bs is the system function of the proportional plus derivative controller for x(t).

Demo of stabilization of an inverted pendulum.

4/ Stabilize by pole cancellation?

Why not cascade an unstable system with another system that cancels the unstable pole?

VIII. CONCLUSIONS

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This work is licensed by Truc Pham-Dinh, Tam Huynh-Ngoc, Anh Tuan Hoang, and Thanh Dinh Vu under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by vocw on Jul 7, 2009 4:08 am -0500.