PI controller: Changes system type by 1 Bode Plots: For 2nd order equation: (closed loop)

• Given a transfer function G(s)

( ) ( )
• Identify poles and zeros
PID Controller: • Transform into Bode forms (normalized Peak Time:
forms)
( )
√
PD Controller: Overshoot:
( )
√
• List the breakpoints in ascending order
Steady State Error: needs to be as small as – for magnitude plots, breakpoints are ( )
possible. (open loop systems)
( ) √ ( )
( ) ( ) ( ) list them in ascending order,
Take limit as e goes to 0. For each sign – for phase plots, they are
change there are 2 zeros. Setting Time:
Step: R(s)=1/s
Ramp: R(s)=1/s2
list them in ascending order,
Parabola: R(s)=1/s3
• Draw the first line
System Type: Number of poles at origin
– for magnitude, the first line is drawn for
{
ω<first BP
( ) Rise time:
• If there is any sp term, the first line has a
( ) slope 20 dB/dec
( ) • If not, the first line is flat until the first
breakpoint Linearization:
input Type 0 Type 1 Type 2 • This line should be such that it passes 1. Take linear terms to the left
Step 0 0 20logKdB point at ω=1 2. Non-linear terms to the right
– for phase, the first line is drawn for 3. Linear=f(x1,x2, x3)= the non-linear part
Ramp 0 ω<first 0.1BP 4.
• If there is any sp term, the first line is flat ( ) ( )
parabola Means marginally stable. at 90p
• If not, the first line is flat at 0 degrees
Block Diagram Simplification: • Continue plotting for subsequent 5. Replace 4 into 3
breakpoints
– at each breakpoint, change the slope
accordingly
• Magnitude if a breakpoint is for the 1st
order one, change the slope by 20db/dec
• if it is for the 2nd order one, change it by
40db/dec
• the change is positive for zeros and
negative for poles
• the last line must have a slope of -20(n-
m)db/dec
– Phase at each point(0.1x , 10x ), change
the slope accordingly
• for each 0.1x point,
if it is for the 1st order term, change the
Root Locus Steps: 1. Open loop poles and slope by 45 deg/dec
zeros. 2. Mark them on diagram and draw root if it is for the 2nd order term, change the
FVT: loci on axis. Draw line from 1st, 3rd, 5th …pole slope by 90deg/dec
or zero to the one after it. 3. When n>m there the change is positive for zeros and
exist n-m asymptotes for negative for poles
∑ • For each 10x point, do it in opposite way,
( )
e.g., (flip sign)
Stability: system is stable if all roots of the • The final line must be flat with angle
denominator are in the LHP 4. breakaway/break 4 in points when closed Phase -90(n-m)
( )
loops have multiple roots at Arrows
Routh Howitz:if all elements in the first Gain Margin: magnitude below 0 db at the
going out of poles into zeros.
column are positive the system is stable.
5. Departure angle, which way complex poles frequency where the phase becomes -180.
are heading Phase margin: phase angle above -180 at
the frequency where the gain becomes 0 db.
∑ ( ) ∑ ( ) Time delay delta t ω=phase margin

( )
6. jω crossing: a) fin value of k to switch from
stable to unstable. B) Insert G(s) into
1+KG(s)=0 (characteristic eqn) and solve for
k=0 from routh. Replace k in characteristic
eqn andwith s=jω then solve for ω.
CH6 Frequency Response:
Bandwidth: usually frequency at which
magnitude2=0.5 DC Gain
( ) ( )
For second order sk eqn TF
√( ) √
DC Gain: take s=0 for G(s)
√
( )
√
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Resonant Peak: Max value for frequency

response magnitude
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