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Lecture 12:Interconnected Systems And Feedback

Module by: Thanh Dinh Vu, Truc Pham-Dinh, Anh Tuan Hoang, Tam Huynh-Ngoc

Summary: Signals normally pass through interconnections of subsystems. Feedback provides an opportunity to use and to integrate material we have learned (Laplace transform, frequency response, step response) in an important application area. Stability is an important issue with feedback systems. Unstable systems can be stabilized with feedback.

Lecture #12:
INTERCONNECTED SYSTEMS AND FEEDBACK
Motivation:
  • Signals normally pass through interconnections of subsystems.
  • Feedback is widely used in both man-made and natural systems to enhance performance.
  • Feedback provides an opportunity to use and to integrate material we have learned (Laplace transform, frequency response, step response) in an important application area.
  • Stability is an important issue with feedback systems
  • Unstable systems can be stabilized with feedback
Outline:
  • Interconnection of systems
  • Simple feedback system — Black’s formula
  • Effect of feedback on system performance
  • Review properties of feedback
  • Dynamic performance of feedback systems
  • BIBO stability
  • Roots of second-order and third-order polynomials
  • Root locus plots of position control systems
  • Stabilization of unstable systems
  • Conclusion
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.
Figure 1
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} } } } } } } {}
Figure 2
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
Figure 3
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.
Figure 4
4/ Physiological control systems examples
Voluntary everyday activities
  • Driving a car
  • Filling a glass with water
Involuntary everyday occurrences
  • Pupil reflex
  • Blood glucose control
  • Spinal reflex
5/ Spinal reflex
Figure 5
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).
Figure 6
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
  1. 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} } } {}
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.
  • Stabilize gain
  • Reduce the effect of an output disturbance
  • Improve dynamic characteristics — increase bandwidth, improve response time
  • Reduce noise
  • Reduce nonlinear distortion
Properties of feedback
  • Increase input impedance
  • Decrease output impedance
2/ Stabilize gain
Figure 7
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.
Figure 8
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 )