Frequency Response
EP314/323
PROCESS DYNAMICS &
CONTROLS
First Order System
𝑘𝑝
𝐺 𝑠 =
𝜏𝑠 + 1
Replacing s with 𝒋𝝎 in the 1st Order system TF
𝑘𝑝
𝑘𝑝
𝑘𝑝
−𝑘𝑝 𝜔𝜏
1 − 𝑗𝜔𝜏
𝐺 𝑗𝜔 =
=
.
=
+𝑗
2
2
𝜏 𝑗𝜔 + 1 1 + 𝑗𝜔𝜏 1 − 𝑗𝜔𝜏 1 + 𝜔 𝜏
1 + 𝜔2𝜏 2
𝐺 𝑗𝜔 =
𝑘𝑝
1 + 𝑗𝜔𝜏
AR
AR
∅
= tan−1 −𝜔𝜏
∡ tan−1 (−𝜔𝜏)
∅
First Order System Cont.…
Notes:
1.Magnitude
and phase lag arg
is
the amplitude ratio AR and phase lag ∅ .
obtained when the system has input with the
same sinusoidal wave and frequency
2. AR and ∅ change with frequency 𝛚
Second Order System
Let;
Dead Time
Let;
∅
Systems in series
for
𝑠 = 𝑗𝜔
𝐺1 𝑗𝜔 =
𝐺2 𝑗𝜔 =
𝑒𝑡𝑐. .
𝑘𝑝
1 + 𝜏12 𝜔 2
𝑘𝑝
1 + 𝜏22 𝜔 2
𝑎𝑛𝑑 ∅1 = ∡ tan−1 (−𝑗𝜔)
𝑎𝑛𝑑 ∅2 = ∡ tan−1 (−𝑗𝜔)
Systems in series Cont..
Overall
Overall
∅ = ∅1 + ∅2 + … + ∅𝑁
Integrating System
𝑘𝑝
𝐺 𝑠 =
𝑠
𝑠 = 𝑗𝜔
𝑘𝑝
𝐺 𝑗𝜔 =
𝑗𝜔
𝑘𝑝
AR = 𝐺 𝑗𝜔 =
𝜔
∅
−1
0
∅ = tan −∞ = −90
Controllers
P Controller:
∅
PI Controller:
∅
Controllers Cont.…
PD Controller:
∅
Controllers Cont.…
PID Controller:
∅
Can be phase lag or lead (Depending on 𝝉D, 𝝉I, 𝛚)
BODE DIAGRAM
Convenient graphical representation of
dependence on AR and ∅ with frequency.
(Log-Log graph)
a.k.a Asymptotic plot
It consists of 2 graphs
1. log AR vs log 𝛚
2. phase ∅ vs log 𝛚
Important parameters
(Semi-Log graph)
a.k.a Phase lag plot
•
•
•
•
Low-frequency Asymptote
Corner frequency
High-frequency Asymptote
True curve
Example of Bode Diagram
1st Order System
In Laplace domain:
𝐻 𝑠
𝑅
1
=
= 𝐾𝑝
𝑄 𝑠
𝜏𝑠 + 1
𝜏𝑠 + 1
𝐺 𝑠
Let G(s) = 1/(𝜏𝑠 + 1) ,
thus in frequency domain:
1
𝐺 𝑗𝜔 =
𝜏𝑗𝜔 + 1
Asymptotic plot can be obtained from Log AR vs Log 𝝎
1
1
𝐴𝑅 = 𝐺 𝑗𝜔 =
log 𝐴𝑅 = − log 𝜔𝜏
2
2
2
1+𝜔 𝜏
∅ = tan−1 −𝜔𝜏
2
+1
Asymptotic plot
𝜔𝜏
1
a) IF 𝜔𝜏 → 0 we use: log 𝐴𝑅 = − log 𝜔𝜏
2
Thus, 𝜔𝜏 = 0; 𝒕𝒉𝒆𝒏 𝐴𝑅 = 1
2
+1
b) log 𝐴𝑅 = − log 𝜔𝜏 , is when asymptotic with gradient -1
c)𝜔𝑐 = 1, 𝒕𝒉𝒆𝒏 𝐴𝑅 = 1
Phase lag plot
𝜔𝜏
a. 𝜔𝜏 → 0 𝑡ℎ𝑒𝑛 ∅ → − tan−1 𝜔𝜏 = 0o
b. 𝜔𝜏 → ∞ 𝑡ℎ𝑒𝑛 ∅ → − tan−1 𝜔𝜏 = −90o
c. 𝜔𝜏 → 1 𝑡ℎ𝑒𝑛 ∅ → − tan−1 𝜔𝜏 = −45o
Summary of Bode Diagram
(First Order System)
Bode diagram for first-order system
Example of Bode Diagram
2nd Order System
2nd Order System
a) 𝒊𝒇 𝜔𝜏 → 0 𝒕𝒉𝒆𝒏 𝐴𝑅 → 1 𝒂𝒏𝒅 ∅ → 0
b) 𝒊𝒇 𝜔𝜏 → ∞
1
𝒕𝒉𝒆𝒏 𝑙𝑜𝑔 𝐴𝑅 → − log
2
1
→ − log
2
1 − 𝜔2 𝜏 2
𝜔2 𝜏 2
2
+ 2𝜏ζ𝜔
2
→ −2 log 𝜏 − 2 log 𝜔
𝒂𝒏𝒅 asymptote with gradient −2 where, ∅ = −180o
c) 𝒊𝒇 𝜔𝑐 =
1
𝜏
; ∅ = − tan−1 ∞ = −90o
2
Summary of Bode Diagram
(Second Order System)
Example of Bode Diagram
1st Order System in series
∅
𝐺 𝑠 =
1
𝜏1 𝑠 + 1
1
𝜏2 𝑠 + 1
Example 15.6 in textbook
For a 1st Order System in series with transfer function
We know that,
Thus
Bode Diagram
(First Order System in series as Example 15.6 )
Example of Bode Diagram: P controller
A proportional controller transfer function is
𝐺 𝑠 = 𝐾𝑐
has amplitude ratio, AR = 𝐾𝑐 and phase angle
zero at all frequencies, ∅ = ∀𝜔𝜏 .
Thus, No Bode diagram is necessary for this
component.
Example of Bode Diagram: PI controller
1
𝜏𝐼 𝑠
𝐺 𝑠 = 𝐾𝑐
1
1+
𝜏𝐼 𝑠
Example of Bode Diagram: PD controller
𝐺 𝑠 = 𝐾𝑐 𝜏𝐷 𝑠 + 1
𝜏𝐷 𝑠 + 1
Example 15.7 (textbook)
The open loop transfer function is:
Example 15.7 (textbook) Answer