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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 ) 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 ) 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 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

  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 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 . e 0 . t = 1 size 12{x \( t \) =" 1" "." " e" rSup { size 8{0 "." "t "} } =" 1 "} {} which is y ( t ) = 1 . H ( 0 ) . e 0 . 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 ) 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

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 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 ) 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 ) 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 10 7 size 12{K approx "10" rSup { size 8{7} } } {} . Hence, provided KR 2 / ( R 1 + R 2 ) >>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 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 ) 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

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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