Unit-5
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 a 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 1 Frequency Domain Specification
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 the resonant frequency is given as
Φr = - tan-1 [√1-2ξ2/ ξ] - - (7)
As we already know for the step response of the second-order system the value of damped frequency and peak overshoot is given as
Wd = Wn√1-ξ2 - - (8)
Mp = e- πξ2|√1-ξ2 - - (9)
The comparison of Mr and Mp is shown in figure(1). The two performance indices are correlated as both are functions of the damping factor ξ only. When subjected to step input the system with a 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(1) it is clear that for ξ> 1/2, the value of Mr does not exist.
Key takeaway
i) Mr and Mp are correlated as both are functions of the damping factor ξ only
When subjected to step input the system with a given value of Mr of its frequency response will exhibit a corresponding value of Mp.
The band of frequencies lying between -3db points.
Cut-off frequency –The frequency at which the magnitude is 3db below its zero frequency.
Cut-off Rate – It is the slope of the log magnitude curve near the cut off frequency.
Effect of addition of zeros
1) The root locus shifts away from the imaginary axis.
2) Stability of the system increases.
3) The settling time decreases.
4) The gain margin increases.
5) The system becomes less oscillatory.
Effect of Addition of Poles:
1) The root locus is shifted towards the imaginary axis.
2) The system becomes oscillatory.
3) The stability of the system decreases.
4) The settling time increases.
5) The range of k reduces.
It gives the frequency response of the system. If the transfer function is given, then from the plot number of poles and zeros can be calculated.
Polar plot of some standard functions:-
# TYPE ‘O’
Ex : 1T(S) = 1/S + 1
(1). For polar plot substitute S=jw.
TF = 1/1 + jw
(2). Magnitude M = 1 + 0j / 1 + jw = 1/√1 + w2
(3). Phase φ = tan-1(0)/ tan-1w = - tan-1w
W M φ
0 1 00
1 0.707 -450
∞ 0 -900
The plot is shown in fig. 3(a)
Fig 3(a)
Ex.2>. T(S) = 1/(S+1)(S+2)
(1). S = jw
TF = 1/(1+jw)(2+jw)
(2). M = 1/(1+jw)(2+jw) = 1/-w2 + 3jw + 2
M = 1/√1 + w2√4 + w2
(3). Φ = - tan-1 w - tan-1(w/2)
W M Φ
0 0.5 00
1 0.316 -71.560
2 0.158 -108.430
∞ 0 -1800
The plot is shown in fig3(b)
fig3(b)
Intersection of polar plot with imaginary axis will be when real part of Transfer function = 0
M = 1/(jw + 1)(jw + 2)
= 1/-w2 + j3w + 2
TYPE ‘1’
Ex.1 T(S) = 1/S
(1). S = jw
(2). M = 1/W
(3). Φ = -tan-1(W/O) = -900
W M φ
0 ∞ -900
1 1 -900
2 0.5 -900
∞ 0 -900
The plot is shown in fig.4(a)
Fig 4(a)
Ex.2 T(S) = 1/S2
(1). S = jw
(2). M = 1/w2
(3). Φ = -tan-1(W/O)-tan-1(W/O) = -1800
W M Φ
0 ∞ -1800
1 1 -1800
2 0.25 -1800
∞ 0 -1800
The plot is shown in fig.4(b)
Fig 4(b)
Key takeaway
TYPE 1 ORDER 2
Ex.1 T(S) = 1/S(S+1)
(1). M = 1/W√1+w2
(2). Φ = -900 - tan -1(W/T)
W M φ
0 ∞ -900
1 0.707 -1350
2 0.45 -153.40
∞ 0 -1800
The plot is shown in fig.5(a)
Fig 5(a)
Ex.2 TYPE 2 ORDER 3
T(s) = 1/S2(S+1)
(1). M = 1/w2√1+jw
(2). Φ = -1800 – tan-1W/T
The plot is shown in fig.5(b)
Fig 5(b)
It gives the frequency response as well as comments on the stability and the relative stability of the system.
Nyquist Stability Criteria:-
The Nyquist criteria is a semi graphical method that determines the stability of CL system investigating the properties of the frequency domain plot (Polar plot), the Nyquist plot of the OLTF G(S) H(S) is represented as L(S)
L(S) = G(S)H(S)
Especially the Nyquist plot of L(S) is a plot drawn by substituting S=jw and varying the value of w as per in polar plot. In the polar plot we take one-sided frequency response ( 0 - ∞) in the Nyquist plot we will vary the frequency in the entire range possible from ( -∞ to 0 ) and (0 to ∞ )
Nyquist Criteria also gives:-
(1). In addition to providing absolute stability like other plots, the Nyquist criteria also give information on the relative stability of a stable system and the degree of instability of an unstable system.
(2). It also gives indications on how the system stability can be improved.
(3). The Nyquist plot of G(S) H(S) is the polar plot of G(S) H(S) drawn with a wider range of frequency ( -∞ to ∞ ) and along the Nyquist path.
(4). The Nyquist plot of G(S) H(S) gives information on frequency domain characteristics such as B.W, gain margin, and phase margin.
Construction of Nyquist Plot
Encircled: A point or region in a complex function phase i.e. S-plane is said to be encircled by a closed path if it is found inside the path.
Assumption:-
In this example point, A is encircled by the closed path Y. Since, A is inside the closed path point B is not encircled by y. it is outside the path. Furthermore, when the closed path Y, has a direction assign to it, encirclement, if made can be in the clockwise direction or the anti-clockwise direction.
Point A is encircled by Y inan anticlockwise direction. We can say that the region inside the path is encircled in the prescribed direction and the region outside the path is not encircled.
Enclosed:-
A point or region is said to be enclosed by a closed path if it is encircled in the counter clockwise direction, or the point or region lies to the left of the path (always) when the path is traveling in the prescribed direction.
The concept of enclosure is particularly useful if only a portion of a closed path is shown.
In this example the shaded region is
Considered to be enclosed by the closed path Y. In other words, point A is enclosed by Y in fig a. but is not enclosed by Y in fig b. and for point B it is viceversa.
No of encirclements and enclosures:-
A line is cut once
For B line is cut twice
As it’s overlapping but 2 times in
Same direction
When a point is encircled by a closed path Y, a no. N can be assigned to the no. of times it is encircled. The magnitude of N can be determined by drawing an arrow around the closed path Y.
Taking an arbitrary point S, and moving around in clockwise direction and anti-clockwise direction respectively. We are getting a direction.
The path followed by S1 gives us the direction and this path which covers the total number of revolution travelled by this point S1 is N or the net angle is ‘ 2 π N ’.
For B = 2 = N for A = 1 = N
In this eg. point A is encircled ones (or 2 π radians) by function Y and point B is encircled twice (or 4 π radians) all in a clockwise direction.
In diagram b again A and B are encircled but in a counter-clockwise direction thus for this diagram A is enclosed one’s and B is enclosed twice.
By definition M is +ve for anticlockwise(direction) encirclement and –ve for clockwise encirclement.
OLTF G(S) = (S + Z1)(S + Z2)/(S + P1)(S + P2) H(S) = 1 - - (1)
CLTF G(S)/1 + G(S)
CE = 1 + G(S)
= 1 + (S + Z1)(S + Z2)/(S + P1)(S + P2)
CE = (S + P1)(S + P2) + (S + Z1)(S + Z2)/ (S + P1)(S + P2) - - (2)
Key takeaway:
# OLTF poles are equal to CE poles.
CE = (S + Z’1)( S + Z’2)/( S + P1)( S + P2) - - (3)
CLTF = G(S) –(1) / 1 + G(S) –(3)
= (S + Z1)( S + Z2) / (S + Z’1)( S + Z’2) - - (4)
Key takeaway:
# Zeros of characteristic equation is poles of CLTF (3 and 4).
For the closed-loop system to be stable zeros of CE(i.e. poles of CLTF) should not be located at the right half of the S-plane.
Consider a contour, which covers the entire right half of the S-plane.
P1 W(0 - ∞)
P2RejR ∞
ϴ - π/2 to 0 to + π/2
P3 W(∞ to 0 )
If every point along the boundary of the contour is mapped in q(S) where q(S) is 1+G(S)H(S)[CE]. The CE is drawn in S-domain. Now, as the CE : q(s) = 1 + G(S)H(S) contour is drawn into S-plane.
This q(S) contour may encircle the origin. Thus, the number of encirclement of q(S) contour with respect to the origin is given by
N = Z – P
Where: Z1P zeros and poles of q(S)[CE]
N Total no of encirclement of origin
** for the CL system to be stable Z=0 always.
Important:-
# Open loop System(stable):- When the OL system is stable P=0 i.e. no of poles on the right half
N = Z – P
If P = 0
N = Z
# Open loop system(unstable) :-Let P = 1 i.e. one OL pole is located in right half of S-plane i.e. OLTF is unstable.
As N = Z – P
N = Z – 1
For the CL system to be stable the only criteria are (Z=0) i.e.
N = -1
which means q(S) contour should encircle the original ones in the CW direction.
Key takeaway:
(1). When the OL system is unstable then the corresponding CL will be stable only when q(S) contour will encircle origin in CW direction.
(2). The no of encirclements should be equal to no of open-loop poles located in the right half of the S-plane.
** the no of encirclements(N) can also be calculated by using G(S) contour (instead of q(S) contour) but the reference is -1+j0 instead of 0+j0 i.e. the no of encirclements should be considered w.r.t -1+j0 and not with the origin.
Explanation Mapping
q(S) = 1 + G(S)
G(S) is always given to us, so we can relate G(S) with q(S).
G(S) = q(S) – 1
But q(S) can be drawn by adding 1 real part to the q(S).
G(S) given then q(S) shift to the right side.
Q1. For the transfer function below plot the Nyquist plot and also comment on stability?
G(S) = 1/S+1
Sol:- N = Z – P ( No pole of the right half of S plane P = 0 )
P = 0, N = Z
NYQUIST PATH:-
P1 = W – (0 to - ∞)
P2 = ϴ( - π/2 to 0 to π/2 )
P3 = W(+∞ to 0)
Substituting S = jw
G(jw) = 1/jw + 1
M = 1/√1+W2
Φ = -tan-1(W/I)
for P1 :- W(0 to -∞)
W M φ
0 1 0
-1 1/√2 +450
-∞ 0 +900
Path P2:-
W = Rejϴ R ∞ϴ -π/2 to 0 to π/2
G(jw) = 1/1+jw
= 1/1+j(Rejϴ) (neglecting 1 as R ∞)
M = 1/Rejϴ = 1/R e-jϴ
M = 0 e-jϴ = 0
Path P3:-
W = -∞ to 0
M = 1/√1+W2 , φ = -tan-1(W/I)
W M φ
∞ 0 -900
1 1/√2 -450
0 1 00
The Nyquist Plot is shown in fig 6
Fig 6 Nyquist Plot
From the plot we can see that -1 is not encircled so, N = 0
But N = Z, Z = 0
So, the system is stable.
Q.2. for the transfer function below plot the Nyquist Plot and comment on stability G(S) = 1/(S + 4)(S + 5)
Soln :- N = Z – P , P = 0, No pole on right half of S-plane
N = Z
NYQUIST PATH
P1 = W(0 to -∞)
P2 = ϴ(-π/2 to 0 to +π/2)
P3 = W(∞ to 0)
Path P1 W(0 to -∞)
M = 1/√42 + w2 √52 + w2
Φ = -tan-1(W/4) – tan-1(W/5)
W M Φ
0 1/20 00
-1 0.047 25.350
-∞ 0 +1800
Path P3 will be the mirror image across the real axis.
Path P2:ϴ(-π/2 to 0 to +π/2)
S = Rejϴ
G(S) = 1/(Rejϴ + 4)( Rejϴ + 5)
R∞
= 1/ R2e2jϴ = 0.e-j2ϴ = 0
The plot is shown in fig 7. From plot N=0, Z=0, system stable.
Fig 7 Nyquist Plot
Q.3. For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?
Soln: As the poles exist at the origin. So, the first time we do not include poles in the Nyquist plot. Then check the stability for the second case we include the poles at the origin in the Nyquist path. Then again check the stability.
PART – 1: Not including poles at the origin in the Nyquist Path.
Fig 8(a) Nyquist Paths
P1 W(∞ Ɛ) where Ɛ 0
P2 S = Ɛejϴ ϴ(+π/2 to 0 to -π/2)
P3 W = -Ɛ to -∞
P4 S = Rejϴ, R ∞, ϴ = -π/2 to 0 to +π/2
For P1
M = 1/w.w√102 + w2 = 1/w2√102 + w2
Φ = -1800 – tan-1(w/10)
W M Φ
∞ 0 -3 π/2
Ɛ ∞ -1800
Path P3 will be a mirror image of P1 about a Real axis.
G(Ɛejϴ) = 1/( Ɛejϴ)2(Ɛejϴ + 10)
Ɛ 0, ϴ = π/2 to 0 to -π/2
= 1/ Ɛ2 e2jϴ(Ɛejϴ + 10)
= ∞. e-j2ϴ [ -2ϴ = -π to 0 to +π ]
Path P2 will be formed by rotating through -π to 0 to +π
Path P4 S = Rejϴ R ∞ ϴ = -π/2 to 0 to +π/2
G(Rejϴ) = 1/ (Rejϴ)2(10 + Rejϴ)
= 0
N = Z – P
No poles on the right half of the S plane so, P = 0
N = Z – 0
But from the plot shown in fig 8(a). it is clear that many encirclements in Anticlockwise direction. So,
N = 2
N = Z – P
2 = Z – 0
Z = 2
Hence, the system unstable.
PART 2 Including poles at the origin in the Nyquist Path.
P1 W(∞ to Ɛ) Ɛ 0
P2 S = Ɛejϴ Ɛ 0 ϴ(+π/2 to +π to +3π/2)
P3 W(-Ɛ to -∞) Ɛ 0
P4 S = Rejϴ, R ∞, ϴ(3π/2 to 2π to +5π/2)
M = 1/W2√102 + W2 , φ = - π – tan-1(W/10)
P1 W(∞ to Ɛ)
W M φ
∞ 0 -3 π/2
Ɛ ∞ -1800
P3( mirror image of P1)
P2 S = Ɛejϴ
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10 + Ɛejϴ)
Ɛ 0
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10)
= ∞. e-j2ϴϴ(π/2 to π to 3π/2)
-2ϴ = (-π to -2π to -3π)
P4 = 0
The plot is shown in fig. from the plot it is clear that there is no encirclement of -1 in the Nyquist path. (N = 0). But the two poles at origin lies to the right half of the S-plane in the Nyquist path.(P = 2)[see path P2]
N = Z – P
0 = Z – 2
Z = 2
Hence, the system is unstable.
Path P2 will be formed by rotating through -π to -2π to -3π
NOTE: The sign convention for angles is shown
Angles are considered –ve for anticlockwise directions and +ve for clockwise directions.
In polar plot any point gives the magnitude phase of the transfer function in bode we split magnitude and 
plot.
Advantages
Q. G(S) = 
1. substitute S = j
G(j
) = 
M = 
= tan-1
= -900
Magnitude varies with ‘w’ but the phase is constant.
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)
MAGNITUDE PLOT
PHASE PLOT
Q.G(S) = 
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
Q. G(S) = S
M= W
= 900
MdB = 20 log10
MdB
0.01 -40
0.1 -20
1 0
10 20
100 90
1000 60
Q. G(S) = S2
M= 
2 MdB = 20 log10
2
= 1800 = 40 log10
W MdB
0.01 -80
0.1 -40
1 0
10 40
100 80
Q. (S) = 
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
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 the origin
G(S) = 
TF = 
M = 
MdB = -20 log10 (
= -tan-1
Approximation: 
T >> 1. so, we can neglect 1.
MdB = -20 log10
MdB = -20 log10
T . ; 
= -tan-1(
T)
Approximation: 
T << 1. so, we can neglecting 
T.
MdB= 0dB, 
= 00
At a point both meet so equallyi.e a time will come hence both approx become equal
-20 log10
T= 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
When we increase the value of 
in app 2 and decrease the 
of app 1 so an RT comes when both cases are equal and hence for that value of 
where both apps are equal gives max. the error we found above and is equal to 3dB
At corner frequency, we have a max error of -3dB
Q. G(S) = 
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
Without approximation
For the 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 the 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
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 the 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 = 
Type of system | Initial slope | Intersection |
0 | 0 dB/decade | Parallel to 0 axis |
1 | -20 dB/decade | =K1 |
2 | -40 dB/decade | =K1/2 |
3 | -60 dB/decade | =K1/3 |
. | . | 1 |
. | . | 1 |
. | . | 1 |
N | -20N dB/decade | =K1/N |
The frequency at which the bode plot culls the 0db axis is called Gain Cross Over Frequency.
The frequency at which the phase plot culls the -1800 axis.
GM=MdB= -20 log [ G (jw)]
.: 
.: 
1) When gain cross over frequency is smaller than phase curves over frequency the system is stable and vice versa.
Key takeaway:
i) More the difference between b/WPC and WGC core is the stability of the system
ii) If GM is below 0dB axis then take ilb +ve and stable. if GM above 0dB axis, that takes -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 the same line Wpc= Wgcthen the system is marginally stable. as we get GM=0dB.
References:
1) B. C. Kuo, “Automatic Control System”, Wiley India, 8th Edition, 2003.
2) Richard C Dorf and Robert H Bishop, “Modern control system”, Pearson Education, 12th edition, 2011.
3) D. Roy Choudhary, "Modern Control Engineering", PHI Learning Pvt. Ltd., 2005.
