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bode_plot By DEV | PPT
Bode Plot 
NNaaffeeeess AAhhmmeedd 
Asstt. Professor, EE Deptt 
DIT, DehraDun
Poles & Zeros and Transfer Functions 
Transfer Function: A transfer function is defined as the ratio of the Laplace 
transform of the output to the input with all initial 
conditions equal to zero. Transfer functions are defined 
only for linear time invariant systems. 
Considerations: Transfer functions can usually be expressed as the ratio 
of two polynomials in the complex variable, s. 
Factorization: A transfer function can be factored into the following form. 
= + + + 
( ) ( )( )...( ) 
1 2 
m 
s p s p s p 
( )( ) ... ( ) 
1 2 
n 
G s K s z s z s z 
+ + + 
The roots of the numerator polynomial are called zeros. 
The roots of the denominator polynomial are called poles.
Poles, Zeros and S-Plane 
An Example: You are given the following transfer function. Show the 
poles and zeros in the s-plane. 
G s s s 
= + + 
s s s 
( ) ( 8)( 14) 
+ + 
( 4)( 10) 
S - plane 
origin 
o x o x x 
-14 -10 -8 -4 0 
s axis 
jw axis
Bode Plot 
• It is graphical representation of transfer function 
to find out the stability of control system. 
• It consists of two plots 
– Magnitude (in dB) Vs frequency plot 
– Phase angle Vs frequency plot
Bode Plot… 
• Consider following T.F 
= + + + 2 
( ) ( )( )...( ) 
s s p s p s p s 
• Put s=jw 
ö çè 
1 1 2 
( )( ) ... ( ) 1 2 
s 
= + + + 2 
( ) 1 ( 1 )( 2 
)...( ) 
jw jw p jw p jw p jw 
( ) ( )( ) ... ( ) 1 2 
• Arrange in following form 
ïþ 
ïý ü 
ïî 
ïí ì 
ö çè 
÷ø 
+ + + + + æ 
1 2 
jw 
n n 
n 
N 
m 
w 
w 
G jw K jw z jw z jw z 
x 
ïþ 
ïý ü 
ïî 
ïí ì 
÷ø 
+ + + + +æ 
1 2 
n n 
n 
N 
m 
w 
w 
G s K s z s z s z 
x 
= + + + 2 
( ) 1 (1 1 )(1 2 
)...(1 ) 
jw jwT jwT jwT jw 
G( jw) =|G( jw) |ÐG( jw) 
ïþ 
ïý ü 
ïî 
ïí ì 
ö çè 
÷ø 
+ + + + + æ 
( ) (1 )(1 ) ... (1 ) 1 2 
jw 
n n 
a b n 
N 
m 
w 
w 
G jw K jwT jwT jwT 
x
Bode Plot… 
• So 
G( jw) =|G( jw) |ÐG( jw) 
Magnitude Phase Angle 
20log | ( ) | 10 Magnitude in dB = G jw 
• Hence Bode Plot consists of two plots 
20log | ( ) | 10 jw G) 
– Magnitude ( dB) Vs frequency plot (w) 
– Phase angle ( ÐG( jw 
)Vs frequency plot (w)
Bode Plot… 
Magnitude in dB 
20log10 |G( jw) |=20log10 | K | +20log10 |1+ jwT1 | +20log10 |1+ jwT2 | ...+20log10 |1+ jwTn | 
+20log10 |1+ jwTa | +20log10 |1+ jwTb | ...+20log10 |1+ jwTm | +etc 
Phase Angle 
( ) ( ) ( ) ÐG( jw)=Ð(K) +Ð1+ jwT1 +Ð1+ jwT2 ...+Ð1+ jwTn 
-Ð(1+ jwTa ) -Ð(1+ jwTb )...-(1+ jwTm ) -etc 
( ) ( ) ( ) jwT jwT jwTn 1 
2 
=900 +tan-1 - 1 
- 
1 
+tan...+tan-tan-1( jwTa ) -tan-1( jwTb )...-tan-1( jwTm ) -etc
Bode Plot… 
Type of 
System 
Initial Slope Intersection with 
0 dB line 
0 0 dB/dec Parallel to 0 axis 
1 -20dB/dec =K 
2 -40dB/dec =K1/2 
3 -60dB/dec =K1/3 
. . . 
. . . 
. . . 
N -20NdB/dec =K1/N
Bode Plot Procedure 
• Steps to draw Bode Plot 
1. Convert the TF in following standard form & put s=jw 
= + + + 2 
( ) 1 (1 1 )(1 2 
)...(1 ) 
jw jwT jwT jwT jw 
2. Find out corner frequencies by using 
ïþ 
ïý ü 
ïî 
ïí ì 
ö çè 
÷ø 
+ + + + + æ 
( ) (1 )(1 ) ... (1 ) 1 2 
jw 
n n 
a b n 
N 
m 
w 
w 
G jw K jwT jwT jwT 
x 
1 , 1 , 1 ... 1 , 1 , 1 / sec 
1 2 3 
Rad etc 
T T T Ta Tb Tc
Bode Plot Procedure … 
3. Draw the magnitude plot. The slope will change at each 
corner frequency by +20dB/dec for zero and -20dB/dec for 
pole. 
 For complex conjugate zero and pole the slope will change 
by 
±40dB/ dec 
4. Starting plot 
i. For type Zero (N=0) system, draw a line up to first (lowest) 
corner frequency having 0dB/dec slope of magnitude 
(height) 20log10K 
ii. For type One (N=1) system, draw a line having slope 
-20dB/dec from w=K and mark first (lowest) corner 
frequency. 
iii. For type One (N=2) system, draw a line having slope 
-40dB/dec from w=K1/2 and mark first (lowest) corner 
frequency.
Bode Plot Procedure … 
5. Draw a line up to second corner frequency by 
adding the slope of next pole or zero to the 
previous slope and so on…. 
i. Slope due to a zero = +20dB/dec 
ii. Slope due to a pole = -20dB/dec 
5. Calculate phase angle for different value of ‘w’ 
and draw phase angle Vs frequency curve
Bode Plot GM & PM 
• Gain Margin: It is the amount of gain in db that can be 
added to the system before the system become 
unstable 
– GM in dB = 20log10(1/|G(jw|) = -20log10|G(jw| 
– Gain cross-over frequency: Frequency where magnitude plot 
intersect the 0dB line (x-axis) denoted by wg 
• Phase Margin: It is the amount of phase lag in degree 
that can be added to the system before the system 
become unstable 
– PM in degree = 1800+angle[G(jw)] 
– Phase cross-over frequency: Frequency where phase plot 
intersect the 1800 dB line (x-axis) denoted by wp 
– Less PM => More oscillating system
Bode Plot GM & PM
Bode Plot & Stability 
Stability by Bode Plot 
1. Stable 
If wg<wp => GM & PM are +ve 
2. Unstable 
If wg>wp => GM & PM are –ve 
3. Marginally stable 
If wg=wp => GM & PM are zero
Bode Plot Examples 
Example 1:Sketech the Bode plot for the TF 
Determine 
(i) GM 
(ii) PM 
(iii) Stability 
( ) 1000 
s s 
(1 0.1 )(1 0.001 ) 
G s 
+ + 
=
Bode Plot Examples… 
Solution: 
1. Convert the TF in following standard form & put s=jw 
( ) 1000 
jw jw 
+ + 
(1 0.1 )(1 0.001 ) 
G jw 
= 
2. Find out corner frequencies 
1 = 10 
1000 
0.1 
1 = 
0.001 
So corner frequencies are 10, 1000 rad/sec
Bode Plot Examples… 
• How to draw different slopes
Bode Plot Examples… 
• Magnitude Plot
Bode Plot Examples… 
• Phase Plot 
S.N 
o 
W Angle (G(jw) 
1 1 ---- 
2 100 -900 
3 200 -980 
4 1000 -134.420 
5 2000 -153.150 
6 3000 -161.360 
7 5000 -168.570 
8 8000 -172.790 
9 Infi -1800
Bode Plot Examples… 
• Phase Plot
Bode Plot Examples… 
• Phase Plot …
Bode Plot Examples… 
• So Complete Bode Plot
References 
• Automatic Control System By Hasan Saeed 
– Katson Publication
Questions?
Thanks

bode_plot By DEV

  • 1.
    Bode Plot NNaaffeeeessAAhhmmeedd Asstt. Professor, EE Deptt DIT, DehraDun
  • 2.
    Poles & Zerosand Transfer Functions Transfer Function: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial conditions equal to zero. Transfer functions are defined only for linear time invariant systems. Considerations: Transfer functions can usually be expressed as the ratio of two polynomials in the complex variable, s. Factorization: A transfer function can be factored into the following form. = + + + ( ) ( )( )...( ) 1 2 m s p s p s p ( )( ) ... ( ) 1 2 n G s K s z s z s z + + + The roots of the numerator polynomial are called zeros. The roots of the denominator polynomial are called poles.
  • 3.
    Poles, Zeros andS-Plane An Example: You are given the following transfer function. Show the poles and zeros in the s-plane. G s s s = + + s s s ( ) ( 8)( 14) + + ( 4)( 10) S - plane origin o x o x x -14 -10 -8 -4 0 s axis jw axis
  • 4.
    Bode Plot •It is graphical representation of transfer function to find out the stability of control system. • It consists of two plots – Magnitude (in dB) Vs frequency plot – Phase angle Vs frequency plot
  • 5.
    Bode Plot… •Consider following T.F = + + + 2 ( ) ( )( )...( ) s s p s p s p s • Put s=jw ö çè 1 1 2 ( )( ) ... ( ) 1 2 s = + + + 2 ( ) 1 ( 1 )( 2 )...( ) jw jw p jw p jw p jw ( ) ( )( ) ... ( ) 1 2 • Arrange in following form ïþ ïý ü ïî ïí ì ö çè ÷ø + + + + + æ 1 2 jw n n n N m w w G jw K jw z jw z jw z x ïþ ïý ü ïî ïí ì ÷ø + + + + +æ 1 2 n n n N m w w G s K s z s z s z x = + + + 2 ( ) 1 (1 1 )(1 2 )...(1 ) jw jwT jwT jwT jw G( jw) =|G( jw) |ÐG( jw) ïþ ïý ü ïî ïí ì ö çè ÷ø + + + + + æ ( ) (1 )(1 ) ... (1 ) 1 2 jw n n a b n N m w w G jw K jwT jwT jwT x
  • 6.
    Bode Plot… •So G( jw) =|G( jw) |ÐG( jw) Magnitude Phase Angle 20log | ( ) | 10 Magnitude in dB = G jw • Hence Bode Plot consists of two plots 20log | ( ) | 10 jw G) – Magnitude ( dB) Vs frequency plot (w) – Phase angle ( ÐG( jw )Vs frequency plot (w)
  • 7.
    Bode Plot… Magnitudein dB 20log10 |G( jw) |=20log10 | K | +20log10 |1+ jwT1 | +20log10 |1+ jwT2 | ...+20log10 |1+ jwTn | +20log10 |1+ jwTa | +20log10 |1+ jwTb | ...+20log10 |1+ jwTm | +etc Phase Angle ( ) ( ) ( ) ÐG( jw)=Ð(K) +Ð1+ jwT1 +Ð1+ jwT2 ...+Ð1+ jwTn -Ð(1+ jwTa ) -Ð(1+ jwTb )...-(1+ jwTm ) -etc ( ) ( ) ( ) jwT jwT jwTn 1 2 =900 +tan-1 - 1 - 1 +tan...+tan-tan-1( jwTa ) -tan-1( jwTb )...-tan-1( jwTm ) -etc
  • 8.
    Bode Plot… Typeof System Initial Slope Intersection with 0 dB line 0 0 dB/dec Parallel to 0 axis 1 -20dB/dec =K 2 -40dB/dec =K1/2 3 -60dB/dec =K1/3 . . . . . . . . . N -20NdB/dec =K1/N
  • 9.
    Bode Plot Procedure • Steps to draw Bode Plot 1. Convert the TF in following standard form & put s=jw = + + + 2 ( ) 1 (1 1 )(1 2 )...(1 ) jw jwT jwT jwT jw 2. Find out corner frequencies by using ïþ ïý ü ïî ïí ì ö çè ÷ø + + + + + æ ( ) (1 )(1 ) ... (1 ) 1 2 jw n n a b n N m w w G jw K jwT jwT jwT x 1 , 1 , 1 ... 1 , 1 , 1 / sec 1 2 3 Rad etc T T T Ta Tb Tc
  • 10.
    Bode Plot Procedure… 3. Draw the magnitude plot. The slope will change at each corner frequency by +20dB/dec for zero and -20dB/dec for pole.  For complex conjugate zero and pole the slope will change by ±40dB/ dec 4. Starting plot i. For type Zero (N=0) system, draw a line up to first (lowest) corner frequency having 0dB/dec slope of magnitude (height) 20log10K ii. For type One (N=1) system, draw a line having slope -20dB/dec from w=K and mark first (lowest) corner frequency. iii. For type One (N=2) system, draw a line having slope -40dB/dec from w=K1/2 and mark first (lowest) corner frequency.
  • 11.
    Bode Plot Procedure… 5. Draw a line up to second corner frequency by adding the slope of next pole or zero to the previous slope and so on…. i. Slope due to a zero = +20dB/dec ii. Slope due to a pole = -20dB/dec 5. Calculate phase angle for different value of ‘w’ and draw phase angle Vs frequency curve
  • 12.
    Bode Plot GM& PM • Gain Margin: It is the amount of gain in db that can be added to the system before the system become unstable – GM in dB = 20log10(1/|G(jw|) = -20log10|G(jw| – Gain cross-over frequency: Frequency where magnitude plot intersect the 0dB line (x-axis) denoted by wg • Phase Margin: It is the amount of phase lag in degree that can be added to the system before the system become unstable – PM in degree = 1800+angle[G(jw)] – Phase cross-over frequency: Frequency where phase plot intersect the 1800 dB line (x-axis) denoted by wp – Less PM => More oscillating system
  • 13.
  • 14.
    Bode Plot &Stability Stability by Bode Plot 1. Stable If wg<wp => GM & PM are +ve 2. Unstable If wg>wp => GM & PM are –ve 3. Marginally stable If wg=wp => GM & PM are zero
  • 15.
    Bode Plot Examples Example 1:Sketech the Bode plot for the TF Determine (i) GM (ii) PM (iii) Stability ( ) 1000 s s (1 0.1 )(1 0.001 ) G s + + =
  • 16.
    Bode Plot Examples… Solution: 1. Convert the TF in following standard form & put s=jw ( ) 1000 jw jw + + (1 0.1 )(1 0.001 ) G jw = 2. Find out corner frequencies 1 = 10 1000 0.1 1 = 0.001 So corner frequencies are 10, 1000 rad/sec
  • 17.
    Bode Plot Examples… • How to draw different slopes
  • 18.
    Bode Plot Examples… • Magnitude Plot
  • 19.
    Bode Plot Examples… • Phase Plot S.N o W Angle (G(jw) 1 1 ---- 2 100 -900 3 200 -980 4 1000 -134.420 5 2000 -153.150 6 3000 -161.360 7 5000 -168.570 8 8000 -172.790 9 Infi -1800
  • 20.
    Bode Plot Examples… • Phase Plot
  • 21.
    Bode Plot Examples… • Phase Plot …
  • 22.
    Bode Plot Examples… • So Complete Bode Plot
  • 23.
    References • AutomaticControl System By Hasan Saeed – Katson Publication
  • 24.
  • 25.