A First Course on Modelling and Control by John Anthony Rossiter

Root-loci 19: Lead and lag design

methods in brief Assumptions

We will use a simple feedback loop of the following form, where the compensator is M(s)

and the process is G(s). Often it is convenient to write the compensator as M(s) = KH(s), to

separate out a gain which will be used in design.

Specifications

In order to design a control law, you must begin from the required specifications. These can

come in alternative forms and hence you must be flexible.

Settling time: This is more likely to be used in a root-loci design using the insight from time

constants that the settling time is roughly 3 times the dominant time constant where a

simple pole [1/(s+a)] has a time constant T=1/a. For complex roots only the real part is

taken to find the implied time constant.

Maximum allowable overshoot/damping/phase margin: The formulae for overshoot/decay

rate is 𝑒

−𝜁𝜋

√1−𝜁2 . Hence one can solve for the implied damping from the maximum

overshoot specification by solving the quadratic:

𝑜 = 𝑒

−𝜁𝜋

√1−𝜁2 (1 − 𝜁2)log(𝑜) = −𝜁𝜋

For a 2nd order system, there is also a relationship between the damping and the phase

margin (PM) using the formula:

𝑃𝑀 = 𝑡𝑎𝑛−1(2𝜁

√−2𝜁2+√1+4𝜁4).

Hence, given a maximum overshoot specification, you can find the implied damping and

from there, the implied PM requirement (approximately as based on the 2nd order

assumption).

Rise time/bandwidth: Rise time is difficult to quantify in terms of easily obtainable terms

when doing design and hence one is likely just to use bandwidth which is another measure

of speed of response. As closed-loop bandwidth is also not easily used in design, a pragmatic

solution is to approximate the closed-loop bandwidth by the open-loop gain cross over

frequency which is easy to use in design.

Offsets to steps and ramps and ramp error constant Kv: It is common to assume that zero

offsets to steps are a given, that is an integrator is included, although you may occasionally

encounter scenarios where this is not true. Offset to a step is the well-known formula:

𝑠𝑡𝑒𝑝𝑜𝑓𝑓𝑠𝑒𝑡 = 𝑙𝑖𝑚𝑠→01

1 + 𝐺(𝑠)𝑀(𝑠)

Often the designs in the text books use the offset to ramp criteria to help tie a design down

more precisely.

𝑟𝑎𝑚𝑝𝑜𝑓𝑓𝑠𝑒𝑡 = 𝑙𝑖𝑚𝑠→01

𝑠𝐺(𝑠)𝑀(𝑠);𝐾𝑣 = 𝑙𝑖𝑚𝑠→0𝑠𝐺(𝑠)𝑀(𝑠); 𝑜𝑓𝑓𝑠𝑒𝑡 =

1

𝐾𝑣

Typically one can cancel out the integrator with the s-term before doing this limit. So

assuming H(s)=sG(s), then 𝐾𝑣 = 𝐻(0)𝑀(0).

Assessment of specifications: In order to give tractable problems for assessment (the real

world will be more nuanced), it is common to assume you follow the procedures that deliver

a desired PM, cross-over frequency and/or offset. As the design formula above assume a

simple 2nd order model, the alignment to closed-loop damping and overshoot and closed-

loop bandwidth is not guaranteed in general, so some fine tuning may be required in a

practical scenario; this means the answers are not be unique and thus are hard to assess in

any absolute sense.

Design Methods

Next we summarise the following methods to some or all of these specifications using root-

loci and/or frequency response tools.

1. Proportional only design.

2. Lag compensator design

3. Lead compensator design.

Simple gain design with root-loci methods

Choose the maximum overshoot allowable and hence define the desired damping ratio.

Given the desired damping ratio find out the ratio of real to imaginary part in the poles:

𝑠2 + 2𝜁𝑤𝑛𝑠 + 𝑤𝑛2

𝑦𝑖𝑒𝑙𝑑𝑠→ 𝑠 = −𝜁𝑤𝑛 ± 𝑤𝑛√1 − 𝜁2

Hence the ratio of real part to imaginary part is: 𝑟𝑒𝑎𝑙

𝑖𝑚𝑎𝑔=

−𝜁

√1−𝜁2

This ratio can be used to draw the corresponding line in the complex plane.

NOTE: It is commonplace to use 𝜁~0.7 or, in effect, real = -imag, so a line at 45o to the

horizontal.

Figure showing lines

drawn at 45o to the

horizontal. Any poles

lying on these lines, in

effect, have a

damping ratio of

about 0.7.

Illustration of

choosing closed-loop

poles.

Find intercept of the

45o lines with the

root-loci and then

find K such that the

closed-loop poles are

here.

)3)(2(

)4(

sss

sG

-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0-5

-4

-3

-2

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1-4

-3

-2

-1

0

1

2

3

4

Root Locus

Real Axis

Imagin

ary

Axis

Choose K to get CL poles here

To find K, simply exploit the fact that for a closed-loop pole to be at s=z, then:

|𝐺(𝑠)|𝐾 = 1 ⇒ 𝐾 = 1

|𝐺(𝑧)|

It is also worth noting, that for a factorised transfer function, G(s) is the product of the

factors, hence, for example:

𝐺(𝑠) =𝑃(𝑠 + 𝑎)

𝑠(𝑠 + 𝑏)(𝑠 + 𝑐)(𝑠 + 𝑑)⇒ |𝐺(𝑧)| =

𝑃|𝑧 + 𝑎|

|𝑧||𝑧 + 𝑏||𝑧 + 𝑐||𝑧 + 𝑑|

Illustration with G(s) = 1/(s+1)(s+3)(s+5)

What value of gain will give

closed-loop poles with a

damping ratio of about 0.7?

Solve the following, for s on

the 45o lines:

1)5)(3)(1(

sss

K

Hence, first GUESS that z=-

1.6+j1.6, then calculate:

531 sssK

Distances marked by double

arrows in the figure.

𝐾 = √(3.42 + 1.62)√(1.42 + 1.62)√(0.62 + 1.62)

Remark: using MATLAB tools, one can find K iteratively rather than doing these

computations explicitly. For example, if you know roughly what K should be, use:

p=pzmap(feedback(G*K,1)) with suitable values. Or, using drag of the closed-loop poles in

sisotool and then read the corresponding K from the compensator editor.

s = -1.6+j1.6

Simple gain design with frequency response methods

A simple gain design is typically premised on: (i) an integral already being present to ensure

offset free tracking and (ii) it is possible to achieve the desired PM simply by modifying gain

and this implying closed-loop stability. Your aim is to make the phase margin about 60

degrees; other PM values may be specified so read the specifications carefully. (In practice

plus or minus 2-3 degrees is fine as the behavior is not usually highly sensitive to the PM.)

1. Ensure you have cleared the compensator so that M=1 (called C in sisotool).

2. Drag the bode gain plot up or down (this is easy in SISOTOOL) as required until the PM is

as required. The corresponding compensator is your required gain as implicitly you have

solved arg(GK)=-180+PM and |GK|=1.

3. Note the implied gain cross over-frequency w with this compensation.

SUMMARY: There is only one d.o.f. and so typically this is used to achieve the desired PM

(which in design is taken as equivalent to damping/overshoot). Fine tuning can be used if

the specified PM does not deliver the desired behaviour, but this step is, to some extent,

arbitrary and therefore hard to assess.

Simple lag design with root-loci methods

A lag design begins with a simple proportional design. Hence, the achievable speed of

response and damping is largely determined by what can be achieved with a proportional

design – always do this first!

So, what is a lag for? In simple terms, a lag allows you to increase the low frequency gain

and therefore improve steady state offsets to steps and ramps. Examples in the text books

tend to focus on the offset to ramps (see specification section above) and thus are based on

a desirable Kv. To complete a lag design you need two specifications:

1. What is the desired damping ratio (or equivalently overshoot)? Do this with a

proportional design ‘K’ as above.

2. What is the desired steady-state gain characteristic?

a. Offset to step if no integrator.

b. Offset to ramp if integrator included.

The 2nd step is where the lag is introduced.

Choose the lag compensator to be:

𝑀(𝑠) = 𝐾𝑠+𝑎

𝑠+𝑎

𝑟

; 𝑟 > 1 where K is the value from the proportional design.

We have two parameters to choose, that is a and r. With a suitable specification of steady-state requirements, neither of these is a free choice and in fact they are both fixed.

Choosing r: The steady-state offset to a step (if there is no integrator):

𝑠𝑡𝑒𝑝𝑜𝑓𝑓𝑠𝑒𝑡 = 𝑙𝑖𝑚𝑠→01

1 + 𝐺(𝑠)𝑀(𝑠)=

1

1 + 𝐺(0)𝐾𝑟

Hence, if the maximum offset requirement is known, then r can be computed.

The offset to a ramp criteria where an integrator is included is:

𝑟𝑎𝑚𝑝𝑜𝑓𝑓𝑠𝑒𝑡 = 𝑙𝑖𝑚𝑠→01

𝑠𝐺(𝑠)𝑀(𝑠);𝐾𝑣 = 𝑙𝑖𝑚𝑠→0𝑠𝐺(𝑠)𝑀(𝑠) = 𝑙𝑖𝑚𝑠→0𝑠𝐺(𝑠)𝐾𝑟;

𝑜𝑓𝑓𝑠𝑒𝑡 = 1

𝐾𝑣⇒ 𝑟 =

1

𝑜𝑓𝑓𝑠𝑒𝑡 × 𝑙𝑖𝑚𝑠→0𝑠𝐺(𝑠)𝐾

Choosing a: It is important that a is not large compared to the dominant poles or the lag would cause a significant movement to the right in the asymptotes/centroid of the root-loci, thus reducing speed of response and possibly negating the original proportional design.

A rule of thumb is to choose a to be about a factor of 10 less than the real part of the dominant poles, that is, the pole positions used in the proportional design.

Illustration lag design with G(s) = (s+3)/[s(s+1)(s+2)(s+4)] Steady-state requirement is offset to ramps less than 50% (i.e. Kv=2)

The proportional design giving

a damping ratio of about 0.7 is

K=0.9.

Kv =(3/8)Kr = 0.3375r

Kv = 2 r=5.9

Real part of dominant pole is

at -0.4, so choose a = 0.04.

Hence:

𝑀(𝑠) = 0.9𝑠 + 0.04

𝑠 + 0.04 5.9⁄

Note, the compensated root-

loci looks almost the same as

the uncompensated one,

except of course of the

additional pole and zero very

close to the origin. This

pole/zero pair embeds a very

slow mode into the

asymptotic behaviour and is

the price paid for the increase

in low frequency gain.

As expected, note the

transient behaviour is similar

with and without the lag, but

the slow mode dominates

asymptotically.

The real advantage would be

evident when tracking a ramp.

s = -0.4+j0.4

Simple lag design with frequency response methods

Choose the lag compensator to be:

𝑀(𝑠) = 𝐾𝑠 +

𝑤10

𝑠 +𝑤10𝛼

where the values of K, w are both taken from the simple gain design above (so the first step

is to do a simple gain design) and thus, implicitly, K, w are not free parameters but

determined solely by the required PM used in the simple gain design. [Remark: most books

will advise you begin from a simple gain design based on a PM of about 5 degrees more than

what you want as you will lose some PM when the lag pole and zero are added.]

The parameter 𝛼 is the only parameter you can choose (typically between 2 and 10) and

determines how much increase in low frequency gain is achieved. Obviously 𝛼 is often

defined implicitly by a requirement on the low frequency gain or equivalently the ramp

error constant Kv.

For systems with no integrator, the offset to steps requirement is given as:

𝑒𝑠𝑠 =1

1 + 𝐺(0)𝐾𝛼;𝑒𝑠𝑠 ≤ 𝑃 ⇒ 𝛼 ≥ (

1

𝑃− 1)

1

𝐺𝐾(0)

For systems with an integrator, the offset to ramps requirement using the lag compensator

above is given as:

𝐺𝑀 =𝑀(𝑠)𝐻(𝑠)

𝑠; 𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 =

1

𝐻(0)𝐾𝛼;𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 ≤ 𝑃 ⇒ 𝛼 ≥

1

𝑃𝐾𝐻(0)

Alternatively, if the requirement is specified in terms of the desired ramp offset constant Kv,

then:

𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 ≤ 𝑃 ⇒ 𝐾𝑣 =1

𝑃⇒ 𝐻(0)𝐾𝛼 ≥ 𝐾𝑣 ⇒ 𝛼 ≥

𝐾𝑣𝐻(0)𝐾

SUMMARY: A lag design is premised on the PM requirement being satisfied by a simple gain

design; if you cannot do a simple gain design, then you cannot do a lag compensation. If the

required alpha is bigger than 10, the requirement is likely to be unrealistic or to lead to an

unreliable implementation or need multiple lag compensators. You could try reducing the

PM requirement but clearly this also has negative impacts such as loss of dynamic

performance (under-damping).

Lead design based root-loci methods

A lead design with root-loci does not lend itself to a simple mechanistic solution although

some good insights are possible which allows you to do an intuitive design. Hence, in these

notes we focus primarily on the core concepts rather than a detailed design suggestion.

1. Lead is a high gain strategy, the aim

is to obtain a faster response than

possible with a simple proportional

design. A typical ‘lead specification’

is based on the desired speed of

response (or equivalently

dominant pole -b).

2. In order to speed up the dominant

poles, the centroid/asymptotes of

the root-loci need to be moved to

the left. Ultimately, you need to

decide how far left it is reasonable

to move them. Obviously, if your

specification sets a desired dominant pole -b, then you know roughly where your

centroid needs to be.

Choose the lead compensator to be:

𝑀(𝑠) = 𝐾𝑠+𝑎

𝑠+𝑟𝑎; 𝑟 > 1 where K is a value to be determined last.

Observation: The centroid of the asymptotes will move to the left by: (r-1)a/(k-m)

We have 2 degrees of freedom, a and r, and it may be unclear how to choose these to

achieve the desired centroid movement. A rule of thumb would bridge the pole and zero

around the desired closed-loop pole positions of -b, that is:

𝑟𝑎 > 𝑏 > 𝑎

A simplification to this would be to choose a=b so that only r remains to be selected.

Increasing r will move the centroid to the left and change the shape of the loci.

Once you have selected the pole and zero to achieve a desirable loci shape, K can be

selected using a method similar to the proportional design method to ensure the poles lie in

the correct place on the loci.

Remark: This form of iterative and conceptual design is well supported in sisotool where the

compensator poles and zeros can easily be dragged in the root-loci plot.

0 2 4 6 8 10 12 14 160

0.2

0.4

0.6

0.8

1

1.2

K=1

Klead

Step Response

Time (seconds)

Am

plit

ud

e

Lead enables you to speed up

transient behaviour, but with

similar damping.

Lead design based on bandwidth and PM specification

Define the lead compensator as:

M (s) = (K b )

s+ wb

s+w b

In this case the values K,w are very different from those used in the lag.

1. Choose w to be the desired gain cross over frequency (in effect this is almost equivalent

to the desired closed-loop bandwidth and so for practical purposes to make the design

easier, take the two to be the same).

The target bandwidth should logically be higher than the gain cross over frequency achieved

by the lag compensator because, if not, the lead is not required and will do nothing.

Realistically you can probably increase bandwidth compared to the lag by about 2 to 5 times

but this requirement should be provided.

2. Determine a K such that this w becomes the gain cross over frequency, that is find K such

that |G(jw)K|=1. This may give you very poor margins, but do not worry about that for now.

3. Find the new PM and hence compute the value = PMd-PM, which is the phase rotation

required to get a desired phase margin PMd.

4. The maximum phase rotation of a lead depends upon the ratio of pole to zero using the

formulae:

𝜙 = 𝑡𝑎𝑛−1√𝛽 − 𝑡𝑎𝑛−11

√𝛽= 𝑠𝑖𝑛−1 (

𝛽 − 1

𝛽 + 1)

For a given rotation , we can solve for beta as follows:

𝛽 = (𝑡𝑎𝑛𝜙 + √𝑡𝑎𝑛2𝜙 + 1)2

=sin𝜑 + 1

1 − sin𝜑

5. Substitute the values K, w and into the formulae above to define the lead compensator.

You will know it is correct if the PM is PMd and the gain cross over freq. is w.

SUMMARY: With this design, the USER has no influence on the implied low frequency gain

as the design is fixed by the two choices of: (i) desired gain cross-over frequency and (ii)

desired PM.

Remark: The user can alternatively formulate a design which fixes: (i) cross-over and low-

freq gain or (ii) PM and low freq gain. These are given at the end as popular in text books

although in my view these would not be used in practice. If you want to fix the low

frequency gain (ramp offset), better to use a lead-lag compensator given next.

Lead-lag design

If you really want to fix low frequency gain while also aiming for a good bandwidth, then it is

better to use a lead-lag compensator. The procedure is straightforward as follows.

STEP 1: Do a lead design which gives the desired PM and desired cross over-frequency.

However, you should choose to start with a PM about 5 degrees above what you want as

you will lose this amount when adding the lag.

STEP 2: Assuming the lead design is in place, do a standard lag design to recover the

required low frequency gain.

Illustration of lead lag design: Take the system G= (s+10)/[s(s+1)(s+2)(s+3)] and design a

lead-lag compensator to give a cross-over frequency of 0.9, a PM of 60 degrees and a

steady-state offset to a ramp of less than 1.

STEP 1: A standard lead design

Find K such that |G(jw)K|=1 and find the corresponding PM. Then compute the value =

60-PM and use this to find 𝛽 from: 𝛽 = (𝑡𝑎𝑛𝜙 + √𝑡𝑎𝑛2𝜙 + 1)2=sin𝜑+1

1−sin𝜑

w=0.9, K=0.83, PM=12.2, phi=47.8, 𝛽 =6.72

Hence define:

ws

ws

KsK

)()(

Substitute the values K, w and into the formulae above to define the lead compensator.

𝐾𝑙𝑒𝑎𝑑 = 0.83 × √6.72 × (𝑠 + 0.9/√6.72

𝑠 + 0.9 × √6.72)

You will know it is correct if the PM is 60 degrees and the gain cross over freq. is w.

STEP 2: Now focus on the low frequency gain using a lag which should only have a minor

impact on the achieved margins and bandwidth.

The current offset to ramps is given from

𝐺𝐾 =𝐾(𝑠)𝐻(𝑠)

𝑠; 𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 =

√𝛽

𝐻(0)𝐾=

√6.72

10 × 0.83/6= 4.855

Therefore, gain needs to be increased by a factor of 4.855 to reduce the offset to one.

Therefore, noting the current cross over frequency is 0.9, the lag component is given as:

𝐾𝑙𝑎𝑔 =𝑠 + 0.9 10⁄

𝑠 + 0.9 48.55⁄

Define the final control law as:

𝐾𝑙𝑒𝑎𝑑𝑙𝑎𝑔 = 0.83 × √6.72 × (𝑠 + 0.9 10⁄

𝑠 + 0.9 48.55⁄)(

𝑠 + 0.9/√6.72

𝑠 + 0.9 × √6.72)

As expected, the PM has dropped by 5 degrees (which could be avoided if step 1 used 65

rather than 60), but otherwise the specification has been met.

-150

-100

-50

0

50

100

Ma

gn

itu

de

(d

B)

10-3 10-2 10-1 100 101 102-315

-270

-225

-180

-135

-90

Ph

ase

(d

eg

)

Bode Diagram

Gm = 9.26 dB (at 1.92 rad/s) , Pm = 55.1 deg (at 0.906 rad/s)

Frequency (rad/s)

OTHER LEAD DESIGNS FOR COMPLETENESS ONLY

Lead design with a low frequency gain requirement and desired PM

Readers will realise that if we want to fix the low frequency gain, then we cannot fix both

the PM and the gain cross over frequency, something else will be outside our direct

influence.

Also, as a lead compensator is a high frequency/high gain approach, it is necessary in

general to include an integrator to ensure the low frequency has sufficient gain. We will

assume that here.

Below is close to a classic text book approach but students should note that a typical text

book approach is iterative, that is, one does not automatically arrive at a good solution and

may not achieve close to the original requirements where those are not explicitly the PM,

gain or cross-over frequency.

However, if we choose to fix: (i) low freq. gain and PM or (ii) low freq. gain and cross-over

frequency, then we can still use an algorithmic approach to find the design. Unfortunately,

this insight is not the text books which, in my view, use a rather clumsy approach which fails

to achieve the stated specifications.

To make the design more systematic when aiming for a low frequency gain requirement,

assume the lead has a slightly different structure, that is

M (s) = (Kb)

s+ wb

s+w b

where the core point is that the low frequency gain is given solely by K so we can meet our

low frequency objective by reference only to that value!

1.  Calculate required loop gain to satisfy error constants. This constraint does not depend

on beta due to the assumed structure above.

𝐺(𝑠)𝑀(𝑠) =𝑀(𝑠)𝐻(𝑠)

𝑠; 𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 =

1

𝐻(0)𝐾;𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 ≤ 𝑃 ⇒ 𝐾 ≥

1

𝑃𝐻(0)

2.  Let 𝛽 =1 and use K alone to satisfy the gain requirement.

3. Next we add some phase lift to achieve the desired PM by adding the dynamic part of the

compensator. We know that the gain of the dynamic part at the point of maximum phase is

given as: |𝛽𝑗𝑤𝑐+

𝑤𝑐√𝛽⁄

𝑗𝑤𝑐+𝑤𝑐√𝛽| = √𝛽 and phase uplift is 𝜙 = 𝑡𝑎𝑛−1√𝛽 − 𝑡𝑎𝑛−1

1

√𝛽

Hence, assuming that the NEW gain cross-over frequency will coincide with the frequency of

maximum phase lift, the effective cross over frequency on the graph of G(s)K is taken from

the intercepts with lines of the gain plot based on 1√𝛽⁄ .

4.  Specifically, the choice of beta now fixes the gain cross over frequency and thus for any

given beta, we can determine the implied PM.

Here it is probably easiest to do a form of trial and error until we get a match with the

requirements. A table can help and is based on the bode of just G(s) with the K from step 2:

(i) try values of 𝛽 from 2 to 10; (ii) use the implied gain lift of the lead to find the implied

gain cross over frequency and hence find the current phase of G(jw)M(jw); (iii) form the

implied PM when the pole and zero are added; (iv) select the beta which gives the PM

closest to what you want.

Try beta 2 3 4 6 8 10

Find cross over frequency wc at xx dB

-3dB -4.8dB -6dB -7dB -9dB -10dB

wc TBC TBC

Current PM for given wc

a b c d e f

Phase rotation provided at wc

19 30 37 46 51 55

PM achieved at wc a+19 30+b 37+c 46+d 51+e 55+f

SUMMARY: There is no point messing about with decimal places as the choice of

compensator is rarely that sensitive. Often integer values of beta are close enough except

perhaps between 2-4 where you might try 2.5 and 3.5. This approach is still somewhat

iterative and clumsy and in general I would not use it.

Lead design with a low frequency gain requirement and set cross over frequency:

Students will realise that as an alternative and quicker procedure which avoids the need for

the table is if you fix the cross over frequency rather than PM. You can select your desired

gain cross-over frequency and then use the Bode plot of GK to determine the required beta.

So, if the desired gain cross-over is w1, then:

√𝛽 =1

|𝐺(𝑗𝑤1)𝐾|; 𝐾(𝑠) = 𝐾𝛽

𝑠 +𝑤1√𝛽⁄

𝑠 + 𝑤1√𝛽

ILLUSTRATION of: (i) fixing low frequency gain and PM followed by (ii) fixing low freq.

gain and cross-over freq. (see leaddesigniterative.m)

Take the system G= (s+10)/[s(s+1)(s+2)(s+3)].

STEP 1: For a steady-state offset to a ramp of less than 2, solve:

𝐺𝑀 =𝑀(𝑠)𝐻(𝑠)

𝑠; 𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 =

1

𝐻(0)𝐾;𝑒𝑠𝑠,𝑟𝑎𝑚𝑝 ≤ 𝑃 ⇒ 𝐾 ≥

1

𝑃𝐻(0)

with H(s) = (s+10)/[(s+1)(s+2)(s+3)] so H(0)=10/6 and thus K=6/20 = 0.3.

STEPs 2,3: Plot the bode diagram of G(s)K and read off the phases and cross over

frequencies corresponding to lines -3dB, -4.8dB, -6dB etc (that is corresponding to beta

=2,3,4,5,…).

Tabulate the results.

Try 𝛽 2 3 4 6 8 10

Find cross over frequency wc at xx dB

-3dB -4.8dB -6dB -7dB -9dB -10dB

wc 0.57 0.68 0.73 a b c

Current PM for given wc

37 28 25 d e f

Phase rotation provided at wc

19 30 37 46 51 55

PM achieved at wc 56 58 62 46+d 51+e 55+f

Try 𝛽 =3 as this will give close to a 60 degree PM. Hence the compensator will be:

𝐾 × 3𝑠+𝑤𝑐

√3⁄

𝑠+√3𝑤𝑐; 𝑤𝑐 = 0.68, 𝐾 = 0.3

-12

-10

-8

-6

-4

-2

0

Ma

gn

itu

de

(d

B)

0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1-180

-150

-120

Ph

ase (

de

g)

Bode of G(s)K without dynamic part of lead

Frequency (rad/s)

-3dB, -142o

-4.8dB, -152o

Remark: This design is somewhat clumsy and is done this

way simply to aid insight into what is

happening. The combination of PM

and low frequency gain

requirements means that the

design is not easily automated or

done on pen and paper as it

requires a table as above.

(ii) fixing low frequency gain and

cross over frequency: If you wish to

set the gain cross-over frequency to be

say 0.75, and accept whatever phase margin arises, then repeat STEP 1, form the bode

diagram of G(s)K and then use the formula:

√𝛽 =1

|𝐺(𝑗0.75)𝐾|; 𝐾(𝑠) = 𝐾𝛽

𝑠 + 0.75√𝛽⁄

𝑠 + 0.75√𝛽

This gives 𝛽 =4.24. As expected, the required cross- over

freq. is met exactly as seen below.

HOWEVER, select some other desired

cross-over frequency and you may find

the resulting PM is far from ideal!

Published with Creative Commons License by J.A. Rossiter, University of Sheffield

-100

-50

0

Ma

gn

itu

de

(d

B)

10-1 100 101 102-315

-270

-225

-180

-135

-90

-45

Ph

ase

(d

eg

)

Bode Diagram

Gm = 9.92 dB (at 1.71 rad/s) , Pm = 61 deg (at 0.75 rad/s)

Frequency (rad/s)

-150

-100

-50

0

Ma

gn

itu

de

(d

B)

10-1 100 101 102-315

-270

-225

-180

-135

-90

Ph

ase

(d

eg

)

Bode Diagram

Gm = 10.2 dB (at 1.54 rad/s) , Pm = 59.5 deg (at 0.662 rad/s)

Frequency (rad/s)