Frequency Response Analysis Sinusoidal Forcing of a First-Order Process For a first-order transfer function with gain K and time constant τ , the response to a general sinusoidal input, xt ( ) = A sin ω t is:
KA −t / τ yt()=−+(ωτ e ωτcosω tsin ω t) (5-25) ωτ22+1 Note that y(t) and x(t) are in deviation form. The long-time Chapter 13 Chapter response, yl(t), can be written as: KA ytA ()=+→∞sin()ωφ tfor t (13-1) ωτ22+1 where: φ =−tan−1 (ωτ)
1 Chapter 13 Chapter
Figure 13.1 Attenuation and time shift between input and output sine waves (K= 1). The phase angle φ of the output signal is given by φ =− Time shift / P × 360 D , where ∆ t is the (period) shift and P is the period of oscillation.
2 Frequency Response Characteristics of a First-Order Process ˆ Forxt ( )==+→∞ A sinω t , yA ( t) A sin (ωφ t)as t where : KA Aˆ ==−and φ tan−1 ()ωτ ωτ22+1 1. The output signal is a sinusoid that has the same frequency, ω, as the input.signal, x(t) =Asinωt. 2. The amplitude of the output signal, A ˆ , is a function of the Chapter 13 Chapter frequency ω and the input amplitude, A: KA Aˆ = (13-2) ωτ22+1
3. The output has a phase shift, φ, relative to the input. The amount of phase shift depends on ω. 3 Dividing both sides of (13-2) by the input signal amplitude A yields the amplitude ratio (AR)
AKˆ AR== (13-3a) A ωτ22+1 which can, in turn, be divided by the process gain to yield the
normalized amplitude ratio (ARN) 1 ARN = (13-3b) Chapter 13 Chapter ωτ22+1
4 Shortcut Method for Finding the Frequency Response The shortcut method consists of the following steps:
Step 1. Set s=jω in G(s) to obtain Gj () ω . Step 2. Rationalize G(jω); We want to express it in the form. G(jω)=R + jI where R and I are functions of ω. Simplify G(jω) by
Chapter 13 Chapter multiplying the numerator and denominator by the complex conjugate of the denominator. Step 3. The amplitude ratio and phase angle of G(s) are given by: AR =+RI22 Memorize ⇒ ϕ = tan−1 (IR / ) 5 Example 13.1 Find the frequency response of a first-order system, with 1 Gs()= (13-16) τs +1 Solution First, substitute sj = ω in the transfer function 11 Gj()ω == (13-17) τωjj++1 ωτ 1
Chapter 13 Chapter Then multiply both numerator and denominator by the complex conjugate of the denominator, that is, −+jωτ 1 − jjωτ +−+1 ωτ 1 Gj()ω == jjωτ()()+−1 ωτ +1 ωτ22+1 1 ()−ωτ =+jRjI =+ (13-18) ωτ22+1 ωτ 22+1 6 1 where: R = (13-19a) ωτ22+1 −ωτ I = (13-19b) ωτ22+1 From Step 3 of the Shortcut Method,
22 22 ⎛⎞⎛⎞1 −ωτ AR =+=RI ⎜⎟⎜⎟ + ⎝⎠⎝⎠ωτ22++1 ωτ 22 1 or
Chapter 13 Chapter 22 (1+ ωτ) 1 AR== (13-20a) 22 2 22 ()ωτ +1 ωτ +1 Also, −−11⎛⎞I − 1 φ ==−=−tan⎜⎟ tan ()ωτ tan ωτ ()(13-20b) ⎝⎠R
7 Complex Transfer Functions Consider a complex transfer G(s),
GsGsGsabc( ) ( ) ( )" Gs()=() () () (13-22) GsGsGs123" Substitute s=jω, GjGjGjabc( ωωω) ( )()" Gj()ω =()()() (13-23) GjG123ωωω jGj" From complex variable theory, we can express the magnitude and angle of Gj ω as follows:
Chapter 13 Chapter ()
Gjabc( ωωω) Gj( ) Gj( )" Gj()ω = (13-24a) Gj123()ωωω G() j Gj()"
∠=∠+∠+∠+Gj()ωωωω Gabc( j) G( j) G( j) " () () () −∠[ Gj123ωωω +∠ G j +∠ Gj +"] (13-24b) 8 Bode Diagrams • A special graph, called the Bode diagram or Bode plot, provides a convenient display of the frequency response characteristics of a transfer function model. It consists of plots of AR and φ as a function of ω. • Ordinarily, ω is expressed in units of radians/time. Bode Plot of A First-order System Recall: 1 −1
Chapter 13 Chapter ARN ==− and φ tan ()ωτ ωτ22+1
•→τAt low frequencies (ω 0andω 1) :
ARN =ϕ=0 1 and •→∞τAt high frequencies (ω and ω 1) : D ARN = 1/ ωτ and ϕ=−90 9 Chapter 13 Chapter
Figure 13.2 Bode diagram for a first-order process. 10 • Note that the asymptotes intersect at ωω = b = 1/ τ , known as the break frequency or corner frequency. Here the value of ARN from (13-21) is:
1 AR N ()ωω==b =0.707 (13-30) 11+
• Some books and software defined AR differently, in terms of
decibels. The amplitude ratio in decibels ARd is defined as Chapter 13 Chapter
ARd = 20 log AR (13-33)
11 Integrating Elements The transfer function for an integrating element was given in Chapter 5: Ys( ) K Gs()== (5-34) Us() s
KK AR ===Gj()ω (13-34) jωω
φω=∠Gj( ) =∠ K −∠( ∞) =−90D (13-35) Chapter 13 Chapter Second-Order Process A general transfer function that describes any underdamped, critically damped, or overdamped second-order system is K Gs()= (13-40) τ22ss++2ζτ 1 12 Substituting sj = ω and rearranging yields: K AR= (13-41a) 2 22 2 ()1−+ωτ ()2ωτζ ⎡⎤−2ωτζ φ = tan−1 (13-41b) ⎢⎥22 ⎣1− ωτ ⎦ Chapter 13 Chapter
Figure 13.3 Bode diagrams for second-order processes. 13 Time Delay Its frequency response characteristics can be obtained by substituting sj = ω , Gj()ω = e− jωθ (13-53) which can be written in rational form by substitution of the Euler identity, Gj()ω == e− jωθ cos ωθ − jsin ωθ (13-54) From (13-54) Chapter 13 Chapter AR ==Gj()ω cos22ωθ +=sin ωθ 1 (13-55)
−1 ⎛⎞sin ωθ φω=∠Gj() =tan ⎜⎟ − ⎝⎠cosωθ or φωθ= − (13-56)
14 Chapter 13 Chapter
Figure 13.6 Bode diagram for a time delay, e − θ s. 15 Chapter 13 Chapter
Figure 13.7 Phase angle plots for e − θ s and for the 1/1 and 2/2
Padé approximations (G1 is 1/1; G2 is 2/2).
16 Process Zeros
Consider a process zero term, Gs( )= Ks( τ +1) Substituting s=jω gives
Gj( ω)= Kj( ωτ +1) Thus: 22
Chapter 13 Chapter AR ==Gj()ωωτ K +1 φω=∠Gj() =+tan−1 ωτ ()
Note: In general, a multiplicative constant (e.g., K) changes the AR by a factor of K without affecting φ .
17 Frequency Response Characteristics of Feedback Controllers
Proportional Controller. Consider a proportional controller with positive gain
Gscc( )= K (13-57)
In this case G cc ( j ω ) = K , which is independent of ω. Therefore, Chapter 13 Chapter
ARcc= K (13-58) and D φc = 0 (13-59)
18 Proportional-Integral Controller. A proportional-integral (PI) controller has the transfer function (cf. Eq. 8-9),
⎛⎞⎛⎞1 τI s +1 Gscc()=+= K⎜⎟⎜⎟1 K c (13-60) ⎝⎠⎝⎠ττIIss Substitute s=jω:
⎛⎞⎛⎞⎛11jω+τI 1 ⎞ Gjcc()ω= K⎜⎟⎜⎟⎜11 + = K c = K c − j⎟ ⎝⎠⎝⎠⎝τττIIIjjω ωω⎠
Chapter 13 Chapter Thus, the amplitude ratio and phase angle are:
2 1 ()ωτI +1 AR ==+=Gj()ω K1 K (13-62) cc c2 cωτ ()ωτI I −11− D φωcc=∠Gj() =tan( − 1/ ωτ I) =tan (ωτ I) −90 (13-63) 19 Chapter 13 Chapter
⎛10s + 1⎞ Figure 13.9 Bode plot of a PI controller, Gs c () = 2⎜ ⎟ ⎝ 10s ⎠
20 Ideal Proportional-Derivative Controller. For the ideal proportional-derivative (PD) controller (cf. Eq. 8-11)
GsccD( )= K(1+ τ s) (13-64) The frequency response characteristics are similar to those of a LHP zero: 2 AR cc=+K ()ωτ D1 (13-65)
−1 φ = tan (ωτD ) (13-66) Chapter 13 Chapter Proportional-Derivative Controller with Filter. The PD controller is most often realized by the transfer function
⎛⎞τDs +1 Gscc()= K⎜⎟ (13-67) ⎝⎠ατDs +1
21 Figure 13.10 Bode plots of an ideal PD controller and a PD controller with derivative filter.
Idea: Gsc ( )= 24( s+ 1) With Derivative Chapter 13 Chapter Filter: ⎛ 41s + ⎞ Gsc ()= 2⎜ ⎟ ⎝ 0.4s + 1⎠
22 PID Controller Forms
Parallel PID Controller. The simplest form in Ch. 8 is ⎛⎞1 Gscc()=++ K⎜⎟1 τ D s ⎝⎠τ1s
Series PID Controller. The simplest version of the series PID controller is
⎛⎞τ1s +1 Gscc()=+ K⎜⎟()τ D s1 (13-73) Chapter 13 Chapter ⎝⎠τ1s
Series PID Controller with a Derivative Filter.
⎛⎞⎛⎞τ1ss++1 τD 1 Gscc()= K⎜⎟⎜⎟ ⎝⎠⎝⎠ττ1ssα D +1 23 Figure 13.11 Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 1). Idea parallel: ⎛ 1 ⎞ Gsc ()=++21⎜ 4 s⎟ Chapter 13 Chapter ⎝ 10s ⎠
Series with Derivative Filter: ⎛⎞⎛⎞10ss+ 1 4+ 1 Gsc ()= 2⎜⎟⎜⎟ ⎝⎠⎝⎠10ss 0.4+ 1 24 Nyquist Diagrams
Consider the transfer function
1 Gs()= (13-76) 21s + with 1 AR ==Gj()ω (13-77a) 2 ()2ω +1
Chapter 13 Chapter and φω=∠Gj( ) =−tan−1 ( 2ω) (13-77b)
25 Chapter 13 Chapter
Figure 13.12 The Nyquist diagram for G(s) = 1/(2s + 1) plotting Re () G ( j ω ) and Im(Gj( ω)). 26 Chapter 13 Chapter
Figure 13.13 The Nyquist diagram for the transfer function in Example 13.5: 5(8se+ 1) −6s Gs()= (201)(41)ss+ +
27