Unit - 4

Time and Frequency Domain Analysis

4.1 Time Domain Analysis

For a second order control system, the total time response is analysed by its transient as well as ready state response.

Any system which contains energy along elements (L1 c etc) if suffer any disturbance in the energy state effect at the input end as at this output or at both the ends takes source time to change in thus form one state other. This change its then is called as Transient times and the values of reverent and voltage during this period is Transient Response.

Fig:1 Transient Response

The above fig: A clearly shows steady state response is that part of response when the transient has died. If the steady star response of the output does not match with input, then the system has steady state errors.

Consider the following open loop system –

Fig 2 Open loop system block diagram

L-1[c(s)] = L-1[ G(s) R (S)]

C(t) = L-1[ G(S) Q (s)]

Fig 3 Closed loop system block diagram

E(s) = R(s) –c(s) [ error signal]

C(S)= [ R(s)- c(Q)] G(s)

C(s) = R(s) G(s)- c(s) G(s)

C(s) [ 1+G(s)] = R(s) G(s)

C(s)/R(s) = G(s)/1+G(s)

C(s)/R(s) = G(s)/1+G(s)

E(s) = R(s)- c(s)

=R(s)-R(s) G(s)/1+G (s0

E(s) = R(s) [1/1+G(s)

If no error then E(s) = 0, f(t) = 0 Hence, r(t) = c(t)

Key takeaway

The error signal is the difference between the unity feedback signal and input signal.

4.2 Unit step Response analysis via Transient response specifications (Percentage overshoot, Rise time, Delay time, Steady state error etc.)

Standard Test Inputs of Time Domain Analysis

The Impulse signal, Ramp signal, unit step and parabolic signals are used as the standard test signals. All these signals are explained below.

Impulse Signal:

This signal has zero amplitude everywhere except at the origin. Fig below shown the representation of Impulse signal.

Fig 4 Unit Impulse Signal

The mathematical representations

A(t) = 0 for t ≠0

dt = A                e

Where A represent energy or area of the Laplace Transform of Impulse signal is

L [A (t)] = A

Unit Impulse Signal:

If A = 1

(t) = 0 for t ≠0

L [(t)] = 1

The transfer function of a linear time invariant

System is the Laplace transform of the impulse response of the system. If a unit impulse signal is applied to system, then Laplace transform of the output c(s) is the transfer function G(s)

As we know G(s) = c(s)/R(S)

r(t) = (t)

R(s) = L [(t)] = 1

:. G(S) = C(s)

(b) Step signal:

Step signal of size A is a signal that change from zero level to A in zero time and stays there forever.

Fig 5 Unit Step Signal

r(t)= A     t >=0

=0      t<0

L[r(t)] = R(s) = A/s

Unit Step Signal:  If the magnitude of the slip signal is I then it is called unit step signal.

u(t) = 1  

t>=0

t<0

L[u(t)] = 1/s

(c) Ramp Signal:

The vamp signal increase linearly with time from initial value of zero at t= 0 as shown in fig is below

Fig 6 Ramp Signal

r(t) = At      t>=0

=0   t<0

A is the slope of the line The Laplace transform of ramp signal is

L[r(t)] = R(s) = A/s2

(d) Parabolic Signal:

The instantaneous value of a parabolic signal varies as square of the time from an initial value of zero t=0. The signal representation in fig 14 below.

Fig 7 Parabolic Signal

r(t) At2 t>=0

=0   t<0

Then Laplace Transform is given as

R(s) = L[At2] = 2A/s3

If no error then E(t) =0, R(t) = c(t), output is tracking the input.

Steady state Errors signal (ess): - (t )

Ess = t e(t)

Using final values, the theorem

= ess = S.E (s)

ess=    S[R(s)/1+G/(s)]

Key takeaway

If no error then E(t) =0, R(t) = c(t), output is tracking the input.

Steady state Errors signal (ess): - (t )

Ess = t e(t)

Using final values, the theorem

= ess = S.E (s)

Unit step Response analysis via Transient response specifications

OLTF G(s) = k/s(1+TS)

CLTFC(s)/R(s) = k/k+s(1+Ts)

= k/s2+ s/T +k/T

Comparing above equation with standard 2nd order eqn

CLTF = wn2/s2+2rs wns+wn2    

Standard 2nd order equation

CEs2+2 wn s+wn2= 0

S= -2wn±/2

=2wn±2wn/2

S= wn±wn

S= -wn±wn

S= -wn ±wn

S= -wn ±gwn

Standard eqn:

T(s) = wn2/s2+2wns+wn2

Our eqn T(s) = K/T/s2+1/T s+ K/T

Wn2 = k/T

2 wn = 1/T

2k/T = 1/T

k/T = 1/2T

= 1/2T T/K

= 12KT

Graphically showing the position of loops s1, s2 for offered of

As the characteristic Eqn (location of poles) is dependent only on (wn constant for a given system)

S1 = -wn + jwn 1-2

S2= - wn - jwn1-2

CASE 1: (=0)

S1=jwn, S2, = -jwn

Fig 8 Location of poles for =0

Undamped

CASE 2:(0<<1)

S1= -wn + jwn 0.75

=-wn/2 +jwn (0.26)

S2 = -wn/2 – jwn (0.86)

Fig 9 Location of poles for <1

CASE:3 (=1)

S1= S2 = -wn = -wn

Fig 10 Location of poles for =1

CASE4: (=-2)

S1 = -2wn +jwn -3

=-2wn – jwn (1.73)

S1  = -3.73wn

S2 = -2wn –jwn -3

= -2wn + jwn (1.73)

= -0.27 wn

Fig 11 Location of poles for >1

Overdamped

All practical systems are 2 order to, if R(e) = (t) R(s) = 1: CLTF = 2nd order st eqn and hence already the system is possible.

CLTF = e(s)/ R(s) = wn2/s2+2wns+wn2

Now calculating c(s), c(t) for different values of input

1)     Impulse I/p

R(t) = (t)

R(s) = 1

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2/s2 +2wn+wn2

Under this i/p (R(t) = (t)) the output varies with different values of . So,

CONDITION 1: ( =0)

C(s) =wn2/s2 +wn2

Os2 +wn2 =0

S=+-jwn

C(t) = wn sinwnt

Sinwnt

Fig 12 Undamped oscillations

As there in no damping i.e., oscillations at t= 0 are some at t so, called UNDAMPED

CONDITION 2:  0<<1

c(s) = R(s) wn2/s22wn+wn2 R(t) = (t)

R(s) =1

C(s) = wn2/s2+2wns+wn2 

CE

S2+2wns +wn2 =0

S2, S1 = - wn ±jwn 1-2

C(t) = e-wnt   sin (wn )t

Wd=wn)

C(t)=e-wnt sin (wdt)

Fig 13 Underdamped oscillations

The oscillations are present but at t- infinity the Oscillations are 0 so, it is UNDERDAMPED

CONDITION 3: =1

C(s) = R(s) wn2/s2+2wns+wn2

C(s)=wn2/S2 +2wns+wn2

=wn2/ (s +wn)2

CE    S= -Wn

Diagram

C(t)=w2n / ()2

C(t)=

Fig 14 Critically damped oscillations

No damping obtained at so is called CRITICALLY DAMPED.

CONDITIONS  4:>1

C(s) = wn2/s22wnS+Wn2

S1, s2 = WN+-jwn

Fig 15 location of poles for over damped oscillations

a)     UNIT Step Input:

R(s) = 1/s

C(s)/R(s) = wn2/s2+2wns+wn2

C(s) = R(s) wn2 /s2 +2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = R(s) wn2/s2+2wns+wn2

C(s) = wn2s(s2+2wns+wn2)

C(t) = 1- ewnt/1-es2 sin (wdt + ø)

Wd = wn1-2

Ø=

Where

Wd = Damping frequency of oscillations

Wn = natural frequency of oscillations

wn = damping coefficient.

T= Time constant

Condition 1= 0

C(s) = wn2 /s(s2+wn2)

C(t) = 1- e° sin wdt +ø

C(t)= 1- sin (wn +90)

C(t) = 1+cos wnt

Constant

C(t) = 1+constant

Fig 16 C(t) = 1+cos wnt

Condition 2:  0<<1

C(s) = /s2+wns +wn2

C(s) =1/s –    s+ wn/s2+ wns +wn2

=1/s –    s+wn/(s+wn)2+wd2- wn/(s+wn2) +wd2

Wd = wn 1-2

Taking Laplace inverse of above equation

L --1 s+wn/(s+wn) +wd2= e-wnt cos wdt

L-1 s+wn/(s+wn)2+wd2 = e-wnt sinWdt

C(t) = 1-e-wnt [coswdt +/1-2 sinwdt]

= 1-ewnt /1-2 sin [wdt + 1-2/]    t>=0

C(t) = 1-ewnt/1-2 sin(wdt+ø)

Ø = 1+2/

Fig 17. Transient Response of second order system

Specifications:

1)     Rise Time (tp): The time taken by the output to reach the already status value for the first time is known as Rise time.

C(t) = 1-e-wnt/1-2 sin (wdt+ø)

Sin (wd +ø) = 0

Wdt +ø = n

tr =n-ø/wd

For first time so, n=1.

tr = -ø/wd

T=1/

2)     Peak Time (tp)

The peak value attained by the output is called peak time. The time required by the output to reach this value is lp.

d(cct) /dt = 0 (maxima)

d(t)/dt = peak value

tp = n/wd   for n=1

tp = wd

3) Peak Overshoot Value:

Maximum deviation of output from steady state value is called peak overshoot value (Mp).

(ltp) = 1 = Mp

(Sin (Wat + φ)

(Sin (Wd∏/Wd + φ)

Mp = e-∏ξ / √1 –ξ2

Condition 3 ξ = 1

C (S) = R (S) Wn2 / S2 + 2ξWnS + Wn2

C (S) = Wn2 / S (S2 + 2WnS + Wn2) [ R(S) = 1/S]

C (S) = Wn2 / S (S2 + Wn2)

C (t) = 1 – e-Wnt + tWne-Wnt

The response is critically damped.

4) Settling Time (ts):

ts = 3 / ξWn (5%)

ts = 4 / ξWn (2%)

Key takeaway

Rise Time tr = -ø/wd

Peak Time tp = n/wd   for n=1

tp = wd

Peak Overshoot Value

Mp% = e-∏ξ / √1 –ξ2

Settling Time

ts = 3 / ξWn (5%)

ts = 4 / ξWn (2%)

Examples

Q.1. The open loop transfer function of a system with unity feedback gain G (S) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.

Sol: Finding closed loop transfer function,

C (S) / R (S) = G (S) / 1 + G (S) + H (S)

As it is unity feedback so, H(S) = 1

C(S)/R(S) = G(S)/1 + G(S)

= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4

C(S)/R(S) = 20/S2 + 5S + 24

Standard equation for second order system,

S2 + 2ξWnS + Wn2 = 0

We have,

S2 + 5S + 24 = 0

Wn2 = 24

Wn = 4.89 rad/sec

2ξWn = 5

(a). ξ = 5/2 x 4.89 = 0.511

(b). Mp% = e-∏ξ / √1 –ξ2 x 100

= e-∏ x 0.511 / √1 – (0.511)2 x 100

Mp% = 15.4%

(c). tr = ∏ - φ / Wd

φ = tan-1√1 – ξ2 / ξ

φ= tan-1√1 – (0.511)2 / (0.511)

φ = 1.03 rad.

tr = ∏ - 1.03/Wd

Wd = Wn√1 – ξ2

= 4.89 √1 – (0.511)2 

Wd = 4.20 rad/sec

tr = ∏ - 1.03/4.20

tr = 502.34 msec

(d). tp = ∏/4.20 = 747.9 msec

Q.2. A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Settling time (v) Peak overshoot.

Sol: The closed loop transfer function is

C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

= (5)2 / S2 + 2 x 0.7 x S + (5)2

C(S)/R(S) = 25 / S2 + 7s + 25

(ii). tp = ∏ / Wd

Wd = Wn√1 - ξ2

= 5√1 – (0.7)2

= 3.571 sec

(iii). tr = ∏ - φ/Wd

φ= tan-1√1 – ξ2 / ξ = 0.795 rad

tr = ∏ - 0.795 / 3.571

tr = 0.657 sec

(iv). For 2% settling time

ts = 4 / ξWn = 4 / 0.7 x 5

ts = 1.143 sec

(v). Mp = e-∏ξ / √1 –ξ2 x 100

Mp = 4.59%

Q.3. The open loop transfer function of a unity feedback control system is given by

G(S) = K/S (1 + ST)

Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?

Sol: C(S)/R(S) = G(S) / 1 + G(S)H(S) H(S) = 1

C(S)/R(S) = K/S (1 + ST) / 1 + K/S (1 + ST)

C(S)/R(S) = K/S (1 + ST) + K

C(S)/R(S) = K/T / S2 + S/T + K/T

For second order system,

S2 + 2ξWnS + Wn2

2ξWn = 1/T

ξ = 1/2WnT

Wn2 = K/T

Wn =√K/T

ξ = 1 / 2√K/T

ξ = 1 / 2 √KT

Forξ1 = 0.2, for ξ2 = 0.4

ξ1 = 1 / 2 √K1T

ξ2 = 1 / 2 √K2T

ξ1/ ξ2 = √K2/K1

K2/K1 = (0.2/0.4)2

K2/K1 = 1 / 4

K1 = 4K2

Q.4. Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2

Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?

Sol:

Mp = 5% = 0.05

Mp = e-∏ξ / √1 –ξ2

0.05 = e-∏ξ / √1 –ξ2

Cn 0.05 = - ∏ξ / √1 –ξ2

-2.99 = - ∏ξ / √1 –ξ2

8.97(1 – ξ2) = ξ2∏2

0.91 – 0.91 ξ2 = ξ2

0.91 = 1.91 ξ2

ξ2 = 0.69

(ii). ts = 4/ ξWn

4 = 4/ ξWn

Wn = 1/ ξ = 1/ 0.69

Wn = 1.45 rad/sec

4.3 Frequency Domain Analysis

1>. Resonant Peak (Mr): The maximum value of magnitude is known as Resonant peak. The relative stability of the system can be determined by Mr. The larger the value of Mr the undesirable is the transient response.

2>. Resonant Frequency (Wr): The frequency at which magnitude has maximum value.

3>. Bandwidth: The band of frequencies lying between -3db points.

4>. Cut-off frequency –The frequency at which the magnitude is 3db below its zero frequency.

5>. Cut-off Rate – It is the slope of the log magnitude curve near the cut off frequency.

Fig 18 Frequency Domain Specification

Correlation between time domain and frequency domain specifications

The transfer function of second order system is shown as

C(S)/R(S) = W2n / S2 + 2ξWnS + W2n - - (1)

ξ = Ramping factor

Wn = Undamped natural frequency for frequency response let S = jw

C(jw) / R(jw) = W2n / (jw)2 + 2 ξWn(jw) + W2n

Let U = W/Wn above equation becomes

T(jw) = W2n / 1 – U2 + j2 ξU

So,

| T(jw) | = M = 1/√ (1 – u2)2 + (2ξU)2 - - (2)

T(jw) = φ = -tan-1[ 2ξu/(1-u2)] - - (3)

For sinusoidal input the output response for the system is given by

C(t) = 1/√(1-u2)2 + (2ξu)2Sin [wt - tan-1 2ξu/1-u2] - - (4)

The frequency where M has the peak value is known as Resonant frequency Wn. This frequency is given as (from eqn (2)).

DM/du|u=ur = Wr = Wn√(1-2ξ2) - - (5)

From equation (2) the maximum value of magnitude is known as Resonant peak.

Mr = 1/2ξ√1-ξ2 - - (6)

The phase angle at resonant frequency is given as

Φr = - tan-1 [√1-2ξ2/ ξ] - - (7)

As we already know for step response of second order system the value of damped frequency and peak overshoot are given as

Wd = Wn√1-ξ2 - - (8)

Mp = e- πξ2|√1-ξ2 - - (9)

The comparison of Mr and Mp is shown in figure. The two performance indices are correlated as both are functions of the damping factor ξ only. When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

Similarly, the correlation of Wr and Wd is shown in fig (2) for the given input step response [ from eqn (5) & eqn (8)]

Wr/Wd = √ (1- 2ξ2)/(1-ξ2)

Mp = Peak overshoot of step response

Mr = Resonant Peak of frequency response

Wr = Resonant frequency of Frequency response

Wd = Damping frequency of oscillation of step response.

From fig it is clear that for ξ> 1/2, value of Mr does not exist.

Key takeaway

i)                   Mr and Mp are correlated as both are functions of the damping factor ξ only

Ii)                 When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

4.4 Frequency Domain Parameters - Natural Frequency, Damping Frequency and Damping Factor

Natural frequency

The natural frequency is defined as the frequency at which a system oscillates when not subjected to a continuous or repeated external force. This frequency can be calculated simply by the following relationship: wn=sqrt(k/m) where k is the stiffness and m is the mass of the system.

Damping frequency

The damped frequency or ringing frequency is the frequency at which a DAMPED system oscillates when not subjected to a continuous or repeated external force. Damped frequency is lower than natural frequency and is calculated using the following relationship: wd=wn*sqrt(1-z) where z is the damping ratio and is defined as the ratio of the system damping to the critical damping coefficient, z=C/Cc where Cc, the critical damping coefficient, is defined as: Cc=2*sqrt(km).

Damping factor

Technically, the damping factor of a system refers to the ratio of nominal loudspeaker impedance to the total impedance driving it (amplifier and speaker cable). In practice, damping is the ability of the amplifier to control speaker motion once signal has stopped. A high damping factor means that the amplifier’s impedance can absorb the electricity generated by speaker coil motion, stopping the speaker’s vibration.

Other points:

• Damping varies with frequency. Some manufacturers publish a damping curve for their amps.
• The effects of damping are most apparent at low frequencies, in the range of the woofer’s resonance. Well damped speakers sound “tighter” in the low end. Low damping factors result in mushy or indistinct bass.
• Speakers connected in series or parallel will experience the same damping factor from the amp. Impedance determines damping factor, not speaker wiring.
• Higher impedance speakers increase system damping factor.
• The damping factors you see published as amp specs are for the amp only, not referenced to an entire system. Higher is better, and you’ll often see quite high numbers, 200, 300, even 3000 or higher.
• System damping factors over 10 are generally acceptable. The higher the better.

Key takeaway

The natural frequency is defined as the frequency at which a system oscillates when not subjected to a continuous or repeated external force.

4.5 Mapping of Pole Zero plot with damping factor, Natural frequency and unit step response

Second Order Systems

Fig 19 Second order system

The order of a differential equation is the highest degree of derivative present in that equation. A system whose input-output equation is a second order differential equation is called Second Order System.

There are a number of factors that make second order systems important. They are simple and exhibit oscillations and overshoot. Higher order systems are based on second order systems. In case of mechanical second order systems, energy is stored in the form of inertia whereas in case of electrical systems, energy can be stored in a capacitor or inductor.

Standard form of second order system is given by:

Where:

Fig 20 Pole plot for an underdamped second order system

• ωn Is the natural frequency
• Introduction to Second Order Systems 3is the damping ratio
• If 0< Introduction to Second Order Systems 3 <1, system is named as Damped System
• If < Introduction to Second Order Systems 3 =1, system is named as Critically Damped System
• If < Introduction to Second Order Systems 3 >1, system is named as Over Damped System

Fig 21 Second order response as a function of damping ratio

Fig 22 Second order underdamped response for different damping ratio

Fig 23 Unit step response of second order system (a) with constant real part (b) imaginary part (c) with constant damping ratio

Let us move the poles to the right or left. Since the imaginary part is now constant, movement of the poles yields the responses of Fig (b). Here the frequency is constant over the range of variation of the real part. As the poles move to the left, the response damps out more rapidly.

Moving the poles along a constant radial line yields the responses shown in Fig(c). Here the percent overshoot remains the same. Notice also that the responses look exactly alike, except for their speed. The farther the poles are from the origin, the more rapid the response.

4.7 Introduction to Bode Plot

What is Bode Plot

A Bode plot is a graph commonly used in control system engineering to determine the stability of a control system. A Bode plot maps the frequency response of the system through two graphs – the Bode magnitude plot (expressing the magnitude in decibels) and the Bode phase plot (expressing the phase shift in degrees).

A Bode plot is a standard format for plotting frequency response of LTI systems. Becoming familiar with this format is useful because:

1. It is a standard format, so using that format facilitates communication between engineers.

2. Many common system behaviour’s produce simple shapes (e.g., straight lines) on a Bode plot

Advantages

1. By looking at bode plot we can write the transfer function of system

Q. G(S) =

1. Substitute S = j

G(j) = 

M =

= tan-1 = -900

Magnitude varies with ‘w’ but phase is constant.

MdB = +20 log10

MdB = -20 log10

Decade frequency:

W present = 10 past

Then present is called decade frequency of past

2 = 10 1

2 is decade frequency of 1

 MdB

0.01 40

0.1 20

1 0 (shows pole at origin)

0 -20

10 -40

100 -60

Slope = (20db/decade)

Fig 24 MAGNITUDE PLOT

Fig 25 Phase Plot

Que. G(S) =

Sol: G(j) =

M =   ;  = -1800 (-20tan-1)

MdB = +20 log -2

MdB = -40 log10

 MdB

0.01 80

0.1 40

1 0 (pole at origin)

10 -40

100 -80

Slope = 40dbdecade

Que. G(S) = S

Sol: M= W

= 900

MdB = 20 log10

 MdB

0.01 -40

0.1 -20

1 0 

10 20

100 90

1000 60

Fig 26 Bode Plot G(S) = S

Que. G(S) = S2

Sol: M= 2   

MdB = 20 log102 = 40 log10

= 1800

W MdB

0.01 -80

0.1 -40

1 0 

10 40

100 80

Fig 27 Magnitude Plot G(S) = S2

Que. G(S) =

Sol: G(j) =

M =

MdB = 20 log10 K-20 log10

= tan-1() –tan-1()

= 0-900 = -900

	K=1		K=10
	MDb		MdB
	=-20 log10	=20 -20 log10
0.01	40		60
0.1	20		40
1	0		20
10	-20		0
100	-40		-20

Fig 28 Bode Plot G(S) =

Fig 29 All bode plots in one plot

Fig 30 Variation in K shifts magnitude plot by +20db

As we vary K then plot shift by 20 log10K i.e., adding a d.c. to a.c. Quantity

Approximation of Bode Plot:

If poles and zeros are not located at origin

G(S) =

TF =

M =

MdB = -20 log10 (

= -tan-1

Approximation: T >> 1. So, we can neglect 1.

MdB = -20 log10

MdB = -20 log10T.; = -tan-1(T)

Approximation: T << 1. So, we can neglect T.

MdB= 0dB, = 00

At a point both meet so equal i.e., a time will come hence both approx. become equal

-20 log10T= 0

T= 1

  corner frequency

At this frequency both the cases are equal

MdB = -20 log10

Now for

MdB = -20 log10

= -20 log10

= -10 log102

MdB = 10

Fig 31 Approximation in bode plot

When we increase the value of in app 2 and decrease the of app 1 so a RT comes when both cases are equal and hence for that value of where both app are equal gives max. Error we found above and is equal to 3dB

At corner frequency we have max error of -3Db

Que. G(S) =

Sol: TF =

M =

MdB = -20 log10 (     at T=2

  MdB

1  -20 log10

10  -20 log10

100            -20 log10

  MdB  = =

0.1  -20 log10 = 1.73 10-3

0.1  -20 log10 = -0.1703

0.5  -20 log10 = -3dB

1  -20 log10 = -6.98

10  -20 log10 = -26.03

100 -20 log10 = -46.02

Fig 32 Magnitude Plot with approximation

Without approximation

For second order system

TF =

TF =

=

=

=

M=

MdB=

Case 1 <<

<< 1

MdB= 20 log10 = 0 dB

Case 2  >>

>> 1

MdB = -20 log10

= -20 log10

= -20 log10

< 1 is very large so neglecting other two terms

MdB = -20 log10

= -40 log10

Case 3. When case 1 is equal to case 2

-40 log10 = 0

= 1

The natural frequency is our corner frequency

Fig 33 Magnitude Plot

Max error at i. e at corner frequency

MdB = -20 log10

For

MdB = -20 log10

error for

Completely the error depends upon the value of (error at corner frequency)

The maximum error will be

MdB = -20 log10

M = -20 log10

= 0

is resonant frequency and at this frequency we are getting the maximum error so the magnitude will be

M = -+

=

Mr =

MdB = -20 log10

MdB = -20 log10 

= tan-1

Mr =

Gain Cross Over Frequency

The frequency at which the bode plot culls the 0db axis is called as Gain Cross Over Frequency.

Fig 34 Gain cross over frequency

Phase Cross Over Frequency

The Frequency at which the phase plot culls the -1800 axis.

Fig 35 Phase Margin and Gain Margin

GM=MdB= -20 log [ G (jw)]

                   .:    

               .:

Key takeaway

i)                   More the difference b/WPC and WGC core are the stability of system

Ii)                 If GM is below 0dB axis than take +ve and stable. If GM above 0dB axis, that is take -ve

GM= ODB - 20 log M 

Iii)              The IM should also lie above -1800 for making the system (i.e., pm=+ve

Iv)               For a stable system GM and PM should be -ve

v)                 GM and PM both should be +ve more the value of GM and PM more the system is stable.

Vi)               If Wpc and Wgc are in same line Wpc= Wgc than system is marginally stable as we get GM=0dB.

Vii)            When gain cross over frequency is smaller than phase curves over frequency the system is stable and vice versa.

4.8 Gain Margin

The greater the Gain Margin (GM), the greater the stability of the system. The gain margin refers to the amount of gain, which can be increased or decreased without making the system unstable. It is usually expressed as a magnitude in dB.

We can usually read the gain margin directly from the Bode plot (as shown in the diagram above). This is done by calculating the vertical distance between the magnitude curve (on the Bode magnitude plot) and the x-axis at the frequency where the Bode phase plot = 180°. This point is known as the phase crossover frequency.

4.9 Phase Margin

The greater the Phase Margin (PM), the greater will be the stability of the system. The phase margin refers to the amount of phase, which can be increased or decreased without making the system unstable. It is usually expressed as a phase in degrees.

We can usually read the phase margin directly from the Bode plot (as shown in the diagram above). This is done by calculating the vertical distance between the phase curve (on the Bode phase plot) and the x-axis at the frequency where the Bode magnitude plot = 0 dB. This point is known as the gain crossover frequency.

Sketch of Bode Plot

Q.1 Sketch the bode plot for transfer function

G(S) =

1. Replace S = j

G (j=

This is type 0 system. So initial slope is 0 dB decade. The starting point is given as

20 log10 K = 20 log10 1000

= 60 dB

Corner frequency 1 = = 10 rad/sec

2 = = 1000 rad/sec

Slope after 1 will be -20 dB/decade till second corner frequency i.e., 2 after 2 the slope will be -40 dB/decade (-20+(-20)) as there are poles

2. For phase plot

= tan-1 0.1 - tan-1 0.001

For phase plot

100  -900

200  -9.450

300  -104.80

400  -110.360

500  -115.420

600  -120.00

700  -124.170

800  -127.940

900  -131.350

1000 -134.420

The plot is shown in figure below

Fig 36 Magnitude Plot for G(S) =

Q.2 For the given transfer function determine

G(S) =

Gain cross over frequency phase cross over frequency phase mergence and gain margin

Initial slope = 1

N = 1, (K)1/N = 2

K = 2

Corner frequency

1 = = 2 (slope -20 dB/decade

2 = = 20 (slope -40 dB/decade

Phase

= tan-1  - tan-1 0.5 - tan-1 0.05

= 900- tan-1 0.5 - tan-1 0.05

1  -119.430

5  -172.230

10  -195.250

15  -209.270

20  -219.30

25  -226.760

30  -232.490

35  -236.980

40  -240.570

45  -243.490

50  -245.910

Finding gc (gain cross over frequency

M =

4 = 2 ( (

6 (6.25104) + 0.2524 +2 = 4

Let 2 = x

X3 (6.25104) + 0.2522 + x = 4

X1 = 2.46

X2 = -399.9

X3 = -6.50

For x1 = 2.46

gc = 3.99 rad/sec (from plot)

For phase margin

PM = 1800  -

= 900 – tan-1 (0.5×gc) – tan-1 (0.05 × gc)

= -164.50

PM = 1800  - 164.50

= 15.50

For phase cross over frequency (pc)

= 900 – tan-1 (0.5 ) – tan-1 (0.05 )

-1800 = -900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

-900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

Taking than on both sides

Tan 900 = tan-1

Let tan-1 0.5 pc = A, tan-1 0.05 pc = B

= 00

= 0

1 =0.5 pc  0.05pc

pc = 6.32 rad/sec

The plot is shown in figure below

Fig 37 Magnitude Plot G(S) =

Q3. For the given transfer function

G(S) =

Plot the rode plot find PM and GM

Sol: T1 = 0.5  1 = = 2 rad/sec

Zero so, slope (20 dB/decade)

T2 = 0.2  2 = = 5 rad/sec

Pole, so slope (-20 dB/decade)

T3 = 0.1 = T4 = 0.1

3 = 4 = 10 (2 pole) (-40 db/decade)

1. Initial slope 0 dB/decade till 1 = 2 rad/sec

2. From 1 to2 (i.e., 2 rad /sec to 5 rad/sec) slope will be 20 dB/decade

3. From 2 to 3 the slope will be 0 dB/decade (20 + (-20))

4. From 3,4 the slope will be -40 dB/decade (0-20-20)

Phase plot

= tan-1 0.5 - tan-1 0.2 - tan-1 0.1 - tan-1 0.1

500  -177.30

1000  -178.60

1500  -179.10

2000  -179.40

2500  -179.50

3000  -179.530

3500  -179.60

GM = 00

PM = 61.460

The plot is shown in figure below

Fig 38 Magnitude and phase Plot for G(S) =

Q 4. For the given transfer function plot the bode plot (magnitude plot) G(S) =

Sol: Given transfer function

G(S) =

Converting above transfer function to standard from

G(S) =

=

1. As type 1 system, so initial slope will be -20 dB/decade

2. Final slope will be -60 dB/decade as order of system decides the final slope

3. Corner frequency

T1 = , 11= 5 (zero)

T2 = 1, 2 = 1 (pole)

4. Initial slope will cut zero dB axis at

(K)1/N = 10

i.e., = 10

5.  Finding n and

T(S) =

T(S)=

Comparing with standard second order system equation

S2+2ns +n2

n = 11 rad/sec

n = 5

11 = 5

= = 0.27

6. Maximum error

M = -20 log 2

= +6.5 dB

As K = 10, so whole plot will shift by 20 log 10 10 = 20 dB

The plot is shown in figure below

Fig 39 Magnitude Plot for G(S) =

Q5. For the given plot determine the transfer function

Fig 40 Magnitude Plot

Sol: From figure above, we can conclude that

1. Initial slope = -20 dB/decade so type -1

2. Initial slope all 0 dB axis at = 10 so

K1/N     N = 1

(K)1/N = 10.

3. Corner frequency

1 = = 0.2 rad/sec

2 = = 0.125 rad/sec

4. At = 5 the slope becomes -40 dB/decade, so there is a pole at = 5 as slope changes from -20 dB/decade to -40 dB/decade

5. At = 8 the slope changes from -40 dB/decade to -20 dB/decade hence is a zero at = 8 (-40+(+20) = 20)

6. Hence transfer function is T(S) =

References:

1. Alciatore & Histand, Introduction to Mechatronics and Measurement system, 4th Edition, Mc-Graw Hill publication, 2011
2. Bishop (Editor), Mechatronics – An Introduction, CRC Press, 2006
3. Mahalik, Mechatronics – Principles, concepts and applications, Tata Mc-Graw Hill publication, New Delhi
4. C. D. Johnson, Process Control Instrumentation Technology, Prentice Hall, New Delhi