page 113
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Proportional (P) |
Gc |
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K |
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Proportional-Integral (PI) |
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1 + ---- |
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Proportional-Derivative (PD) |
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K( 1 + τ s) |
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Proportional-Integral-Derivative (PID) |
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1 + ---- |
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τ s |
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Gc |
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Lead |
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1 + τ s |
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1 + τ s |
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Gc |
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Lag |
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1 + ατ |
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Lead-Lag |
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1 + τ |
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1 + ατ |
1s |
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τ 1 > τ 2 |
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7.2 ROOT-LOCUS PLOTS
•Consider the basic transform tables. A superficial examination will show that the denominator (bottom terms) are the main factor in determining the final form of the solution. To explore this further, consider that the roots of the denominator directly impact the partial fraction expansion and the following inverse Laplace transfer.
•When designing a controller with variable parameters (typically variable gain), we need to determine if any of the adjustable gains will lead to an unstable system.
•Root locus plots allow us to determine instabilities (poles on the right hand side of the plane), overdamped systems (negative real roots) and oscillations (complex roots).
•Note: this procedure can take some time to do, but the results are very important when designing a control system.
•Consider the example below,
page 114
Note: This controller has adjustable gain. After this design is built we must anticipate that all values of K will be used. It is our responsibility to make sure that none of the possible K values will lead to instability.
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G( s) |
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H( s) = 1 |
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First, we must develop a transfer function for the entire control system.
GS( s) |
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1 + G( s) |
H( s) |
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= -------------------------- |
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s + K |
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Next, we use the characteristic equation of the denominator to find the roots as the value of K varies. These can then be plotted on a complex plane. Note: the value of gain ’K’ is normally found from 0 to +infinity.
s + K = 0 K |
root |
jω |
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Note: because all of the roots for all values of K are real negative this system will always be stable, and it will always tend to have a damped response. The large the value of K, the more stable the system becomes.
• Consider the previous example, the transfer function for the whole system was found, but then only the denominator was used to determine stability. So in general we do not need to find the transfer function for the whole system.
page 115
Consider the general form for a negative feedback system.
GS( s) |
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G( s) |
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1--------------------------------+ G( s) H( s) |
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Note: two assumptions that are not often clearly stated are that we are assuming that the control system is a negative feedback controller, and that when not given the feedback gain is 1.
The system response is a function of the denominator, and it’s roots.
1 + G( s) H( s) = 0
It is typical, (especially in textbook problems) to be given only G(s) or G(s)H(s).
The transfer function values will often be supplied in a pole zero form.
G( s) H( s) |
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K( s + z0) ( s + z1)… |
( s + zm) |
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------------------------------------------------------------------( s + p0) ( s + p1)… |
( s + pn) |
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• Consider the example, |
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page 116
Given the system elements (you should assume negative feedback),
G( s) = |
K |
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s-------------------------2 + 3s + 2 |
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First, find the characteristic equation,. and an equation for the roots,
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K |
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s + 3s + 2 |
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s2 + 3s + 2 + K = 0 |
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roots |
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– 3 ± 9 – 4( 2 + K) |
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1 – 4K |
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Next, find values for the roots and plot the values, |
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******CALCULATE AND PUT IN NUMBERS
7.2.1 Approximate Plotting Techniques
•The basic procedure for creating root locus plots is,
1.write the characteristic equation. This includes writing the poles and zeros of the equation.
page 117
1 + G( s) H( s) |
( s + z1) ( s + z |
2)… |
( s + zm) |
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= 1 + K( s + p ) ( s + p )… |
( s + p ) |
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2.count the number of poles and zeros. The difference (n-m) will indicate how many root loci lines end at infinity (used later).
3.plot the root loci that lie on the real axis. Points will be on a root locus line if they have an odd number of poles and zeros to the right. Draw these lines in.
4.determine the asymptotes for the loci that go to infinity using the formula below. Next, determine where the asymptotes intersect the real axis using the second formula. Finally, draw the asymptotes on the graph.
β ( k) |
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± 180°( 2k + 1) |
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k [ 0, n – m – 1] |
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n – m |
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( p1 + p2 + … + pn) ( z1 |
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+ zm) |
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5. the breakaway and breakin points are found next. Breakaway points exist between two poles on the real axis. Breakin points exist between zeros. to calculate these the following polynomial must be solved. The resulting roots are the breakin/breakout points.
A = ( s + p1) ( s + p2)… ( s + pn) |
B = ( s + z1) ( s + z2)… ( s + zm) |
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B – A |
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6. Find the points where the loci lines intersect the imaginary axis. To do this substitute the fourier frequency for the laplace variable, and solve for the frequencies. Plot the asymptotic curves to pass through the imaginary axis at this point.
( jω + z1) ( jω + z2)… |
( jω + zm) |
1 + K(--------------------------------------------------------------------------jω + p1) ( jω + p2)… |
( jω + pn) = 0 |
• Consider the example in the previous section,
page 118
Given the system elements (you should assume negative feedback),
G( s) = |
K |
H( s) = 1 |
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2----+-----3---s----+-----2- |
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Step 1: (put equation in standard form)
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K |
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1 + G( s) H( s) = 1 + |
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Step 2: (find loci ending at infinity)
m = 0 |
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(from the poles and zeros of the previous step) |
n – m = |
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(loci end at infinity) |
Step 3: (plot roots) |
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jω |
σ
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Step 4: (find asymptotes angles and real axis intersection)
β ( k) |
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180°( 2k + 1) |
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k I[ 0, 1] |
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β ( 0) |
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page 119
Step 5: (find the breakout points for the roots) |
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A = 1 |
B = s2 + 3s + 2 |
jω |
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d |
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dsA = 0 |
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dsB = 2s + 3 |
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A |
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-1 |
1( 2s + 3) – ( s2 + 3s + 2) ( 0) = 0 |
-1.5 |
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2s + 3 = 0
s = –1.5
Note: because the loci do not intersect the imaginary axis, we know the system will be stable, so step 6 is not necessary, but we it will be done for illustrative purposes.
Step 6: (find the imaginary intercepts)
1 + G( s) H( s) = 0
1 |
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1 + Ks-------------------------2 + 3s + 2 |
= 0 |
s2 + 3s + 2 + K = 0
( jω ) 2 + 3( jω ) + 2 + K = 0
–ω 2 + 3jω + 2 + K = 0
ω 2 + ω ( –3j) + ( – 2 – K) = 0 |
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ω |
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3j ± ( –3j) 2 – 4( – 2 – K) |
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3j ± – 9 + 8 + 4K |
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3j ± 4K – 1 |
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------------------------------- |
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In this case the frequency has an imaginary value. This means that there will be no frequency that will intercept the imaginary axis.
• Plot the root locus diagram for the function below,