Design via frequency response

Transient response via gain adjustment

Consider a unity feedback system, where G(s) =ω2

ns(s+2ζωn)

. The closed loop transfer function is

T(s) =ω2

n

s2 + 2ζωs + ω2n

Figure above; The time response of the second

order underdamped system

1

The percentage overshoot, %OS, is given by

%OS =cmax − cfinal

cfinal

Note that %OS is a function only of the damp-

ing ratio, ζ.

%OS = e−(ζπ/√

1−ζ2) × 100

The inverse is given by

ζ =− ln(%OS/100)

√

π2 + ln2(%OS/100)

There is also a relationship between damping

ratio and phase margin. The phase margin is

obtained by solving |G(jω)| = 1 to obtain the

frequency as

ω1 = ωn

√

−2ζ2 +

√

1 + 4ζ4

The phase margin is

ΦM = arctan2ζ

√

−2ζ2 +√

1 + 4ζ4

2

Thus if we can vary the phase margin, we can

vary the percent overshoot, via a simple gain

adjustment.

Figure above; Bode plots showing gain adjust-

ment for a desired phase margin.

3

Problem: For the position control system shown

below, find the value of preamplifier gain, K,

to yield a 9.5% overshoot in the transient re-

sponse for a step input. Use only frequency

response methods.

Figure above; Bode plots for the example above.

4

Solution: 1. G(s) = 100Ks(s+36)(s+100)

. Choose

K = 3.6 to start the magnitude plot at 0dB.

2. Use

ζ =− ln(%OS/100)

√

π2 + ln2(%OS/100)

to find ζ = 0.6 for %OS/100 = 0.095, and then

use

ΦM = arctan2ζ

√

−2ζ2 +√

1 + 4ζ4

to find ΦM = 59.2◦ for ζ = 0.6.

3. Locate on the phase plot the frequency that

yields a 59.2◦ margin. This means −120.8◦

phase angle at frequency of 14.8 rad/s.

4. At 14.8 rad/s on the magnitude plot, the

gain is found to be -44.2dB. This magnitude

has to be raise to 0dB to yield the required

phase margin. 44.2dB increase is 162.2 in

gain, Thus K = 3.6 × 162.2 = 583.9.

5

Lag compensation

The steady error constants are:

position constant

Kp = lims→0

G(s),

velocity constant

Kv = lims→0

sG(s),

and acceleration constant

Ka = lims→0

s2G(s).

The value of the steady-state error decreases

as the steady error constants increases.

The function of the lag compensator is to (i)

improve the steady error constant by increas-

ing only the low-frequency gain without any

resulting instability, and (2) increase the phase

margin of the system to yield the desired tran-

sient response.

6

Visualizing lag compensation

The transfer function of the lag compensator

is

Gc(s) =s + 1

T

s + 1αT

where α > 1.

Figure above; Frequency response of a lag com-

pensator s+0.1s+0.01

7

In the figure below, the uncompensated system

is unstable since the gain at 180◦ is greater

than 0dB. The lag compensator, while not

changing the low-frequency gain, does reduce

the high frequency gain. The magnitude curve

can be shaped to go through 0dB at the de-

sired phase margin to obtain the desired tran-

sient response.

Figure above; visualizing lag compensation

8

Design procedure

1. Set the gain,K, to the value that satis-

fies the steady-state error specification and

plot the Bode plots based on this gain.

2. Find the frequency where the phase margin

is 5◦ to 12◦ greater than the phase margin

that yields the desired transient response.

3. Select a lag compensator whose magnitude

response yields a composite Bode diagram

that goes 0dB at the frequency found in

step 2; (For detail see the next example)

4. Reset the system gain, K, to compensate

for any attenuation in the lag network in

order to keep the static error constant the

same as that found in step 1.

9

Problem: For the same previous position con-

trol system, use Bode Diagrams to design a lag

compensator to yield a a ten-fold improvement

in steady-state error over the gain-compensated

system while keeping the percent overshoot at

9.5%.

Solution: 1. The gain-compensated system is

G(s) =58390

s(s + 36)(s + 100)

Thus Kv = 16.22, a ten-fold improvement changes

Kv = 162.2. Thus K needs to be set at K =

5839, and the open loop transfer function

G(s) =583900

s(s + 36)(s + 100)

10

Figure above; Bode plots showing lag compen-

sator design.

2. The phase margin required for %OS/100 =

0.095 (ζ = 0.6) is ΦM = 59.2◦. We increase

this value by 10◦ to ΦM = 69.2◦ (i.e. phase

angle −110.8◦). The frequency is 9.8rad/s.

At this frequency the magitude plot must go

through 0dB. The magnitude at 9.8rad/s is

24dB. Thus the lag compensator must provide

-24dB attenuation at 9.8rad/s.

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3.&4. We now design the compensator. First

draw the high frequency asymptote at -24dB.

Arbitrarily select the higher break frequency to

be about one decade below the phase margin

frequency, i.e. at 0.98rad/s, to start drawing a

-20dB/decade until 0dB is reached. The lower

break frequency is found to be 0.062rad/s.

Hence the lag compensator’s transfer function

is

Gc(s) =0.063(s + 0.98)

(s + 0.062)

where the gain 0.063 of the compensator is set

to achieve a dc gain of unity.

The compensated system’s forward transfer func-

tion is thus

Gc(s)G(s) =36786(s + 0.98)

s(s + 36)(s + 100)(s + 0.062)

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