PROCESS CONTROL SYSTEM DESIGN Process Control System Design LECTURE 2: FREQUENCY DOMAIN ANALYSIS

Transcription

1 05444 PROCESS CONTROL SYSTEM DESIGN Process Control System Design LECTURE : FREQUENCY DOMAIN ANALYSIS Daniel R. Lewin Department of Chemical Engineering Technion, Haifa, Israel - Objectives On completing this section, you should be able to: Draw Bode and Nyquist plots of an arbitrary SISO linear system Both by hand and using MATLAB Sketch the temporal response of a SISO linear system, given its Bode plot Both by hand and using MATLAB/SIMULINK Determine an appropriate transfer function given the Bode plot of a SISO linear system. Both by hand and using MATLAB -

2 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Usin s= iω p( iω) p s ω t P(s) Ysin( ω t+ φ) φ ω Amplitude ratio, AR = Y U = p iω { ( ω) } { ( ω) } Phase shift, arg{ p( i )} tan Im p i φ= ω = Re p i -3 Frequency Response Example Response of dead sea pond to cyclic perturbation: dc π = 0.05c + 0.0E,c 0 = 0;E( t) = 3sin ( ωt ), ω= rad/h dt Solution: 0.05t c( t) = 0.05e ( sin ( ωt.383) ) p( ω=π ) = = φ ω=π =.383 rad (phase lag) φ ω -4

3 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response o System K p = τ s + K Kp ( iτω) p Kp ( iτω) piω = i τω+ ( + i τω ) ( i τω ) + τω The ultimate response has the following characteristics: Kp AR = p( iω ) = +τ ω Im { p( iω )} φ( ω ) = arg{ p( iω )} = tan = tan ( τω) Re{ p( iω) } Example (Dead Sea Pond): = = s s AR = p( i ω ) =. For ω= π,ar = ω φ ω = tan 0 ω. For ω= π, φ = Log (AR) Freq. Response Plots o System = s + Bode Plot AR(0. 0) = AR(0.0) =.000 AR(.00) = AR(0.0) = AR(00) = 0.00 In the Bode magnitude plot, log (AR) is plotted against log (ω) ( ω ) = pi + ω Phase φ(.00) = -45 o φ(0.0) = -84. o φ(00) = o In the Bode phase plot, Phase (in rad or degrees) is plotted against log (ω) φ(0.0) = o φ(0.0) = -5.7 o φ( ω ) = tan ( ω) -6 3

4 05444 PROCESS CONTROL SYSTEM DESIGN Log (AR) Freq. Response Plots o System = s + lim ω 0 Bode Plot K - p = K p +τ ω Kp Asymptotes join at breakpoint located at ω = /τ rad/min lim ω Asymptotes +τ ω = K τω Gradient of high frequency asymptote is - on a log-log scale p Phase Asymptotes join at break point lim tan ( τω ) = 0 ω 0 o lim tan ( τω ) = 90 ω o -7 Freq. Response Plots o System Generating Bode Plots with MATLAB w=logspace(-,,00);s=i*w; p=./(s+); AR=abs(p); ph=80*phase(p)/pi; subplot(,,) loglog(w,ar,'-r ) subplot(,,) semilogx(w,ph,'-r') -8 4

5 05444 PROCESS CONTROL SYSTEM DESIGN Freq. Response Plots o System In the Nyquist plot, p(iω), which is a complex number, is plotted directly, as a locus from ω = 0, to ω =. Nyquist Plot ω 0.0 AR.000 φ ( o ) w=logspace(-,,00);s=i*w; p=./(s+); plot(p) -9 Frequency Response o System Bode Plot Nyquist Plot -0 5

6 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Integrator K p = s K p Kp iω ikp pi ( ω ) = i ω + i ω ω ω ( i ) The ultimate response has the following characteristics: Kp AR = p( iω ) = ω Im { p( iω )} π φ( ω ) = arg{ p( iω )} = tan = tan ( ) = Re{ p( iω) } - Frequency Response Integrator Bode Plot Nyquist Plot - 6

7 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Interpretation Consider two first order systems: p( s ) = and = 5s + s + These two systems are excited by a series of steps (a square wave, approximately) as shown below. Note that p responds more sluggishly than p, because its time constant is 5 times larger. Note also that the exciting signal is only approximately a square wave Frequency Response Interpretation This is because it is actually made up of a series of harmonic terms. In fact, it can be shown that a true square wave can be expressed as the infinite series: sin ( ( i ) + t ) u ( t) = = sin( t) + sin( 3t) + sin( 5t) + 3 i i 5 = + More accurate approximations of the true square wave are obtained by using more terms in the above expansion. Any signal can be expressed as a Fourier expansion of the form: u( t) = α( ω) sin( ωt) ω where α(ω) is the amplitude of the signal at ω. -4 7

8 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Interpretation Let s display the magnitudes of the terms in the expansion to seven terms: = u t sin t sin 3t sin 5t sin 3t on the Bode magnitude plot. This is equivalent to a step (integrator) -5 Frequency Response Interpretation Equivalent to the square wave (step) -6 8

9 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Interpretation -7 Frequency Response Lead/Lag = ( τ + ) p s K s p p iω = K iτω+ p = ( τ + ) p ( iω ) = K ( τω+ i ) p s K s p p The ultimate response has the following characteristics: AR = p( iω ) = p( iω ) = Kp +τ ω Im{ p ( iω )} φ( ω ) = arg{ p( iω )} = tan = tan ( τω) Re{ p ( iω) } Im { p ( iω )} φ( ω ) = arg{ p( iω )} = tan = tan ( τω) Re{ p ( iω) } -8 9

10 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response Lead/Lag Bode Plot + Nyquist Plot Gradient of high frequency asymptote is + on a log-log scale -9 Frequency Response o System Kp = τ s + ξτ s + piω = The ultimate response has the following characteristics: AR = p iω = piω = ( p τω ) + iξτω Kp (( τ ω ) iξτω) (( τ ω ) + iξτω) (( τ ω ) iξτω) Kp ( τ ω ) Kp ( ξτω) i ( τ ω ) + ( ξτω) ( τ ω ) + ( ξτω) piω = K ( τ ω ) + ( ξτω) { ( ω )} { } { ( ω) } p Im p i ξτω Re p i τ ω φ ω = arg p iω = tan = tan K -0 0

11 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response o System Bode Plot - lim ω K p = K p τω ( +τ ω ) Gradient of high frequency asymptote is - on a log-log scale ξτω lim tan = 80 ω τ ω o - Frequency Response Complex p(s) To construct asymptotes in Bode plots for: ( as+ )( as+ ) ( a s+ ) = Kp e bs+ bs+ bs+ m θs n On the Bode log AR plot, the amplitudes of each component add up: log = logkp + log as + + log as + + log bs + log bs + On the Bode linear phase plot scale, the phase of each component add up: φ = arg{ } = arg{ as + } + arg{ as + } + arg b s + arg b s + θs { } { } For large ω, log AR(ω) vs. log ω has an asymptotic slope of (n m) For large ω, for MP p(s) [no positive zeros or delay terms], φ approaches asymptotic value of (π/) (n m) -

12 05444 PROCESS CONTROL SYSTEM DESIGN Frequency Response MP/NMP MP = Minimum phase, NMP = Non-minimum phase. NMP systems are those that feature phase lags greater than anticipated based on the system s AR alone. Those whose phase lag corresponds to the systems AR are MP systems. NMP components are either: (a) Right-half plane (RHP) zeros, or (b) Dead time RHP zeros: It is convenient to factor the RHP zero with its LHP mirror image: pz ( s) = ( zs+ ) ( zs+ ). The AR of this component is AR =, which has a phase lag of zero. In contrast, for large ω, p z has a phase lag of: limarg{ pz ( s) } = limarg{ zs+ } limarg{ zs+ } ω ω ω = π π = π s Dead time: The phase lag of pd ( s) = e θ is θω. The AR of this component is AR =, which has a phase lag of zero. -3 Class Exercise - Sketch p(iω) Generate a Bode plot for the following transfer function: s+ = ( 0s + ) e 0.5s Solution. The Bode plot is plotted by combining the contributions of the components: s+, and e ( 0s + ) 0.5s -4

13 05444 PROCESS CONTROL SYSTEM DESIGN Class Exercise - Sketch p(iω) Solution (Cont d). s+ = e 0s + 0.5s s + ( 0s+ ) ω φ (e -0.5s ) s+ ( 0s+ ) e 0.5s Class Exercise - Determine p(s) Determine the transfer function, p(s) for the process whose Bode diagram is given above. -6 3

14 05444 PROCESS CONTROL SYSTEM DESIGN Class Exercise - Solution ω = ω = 0. K p = ( 0s+ ) ( 0s+ ) p( s ) = or ( s+ ) ( s+ ) -7 Class Exercise - Solution High ω asymptote -70 o (a) No delay (b) NMP system ( 0s+ ) ( 0s+ ) p( s ) = or ( s+ ) ( s+ ) -8 4

15 05444 PROCESS CONTROL SYSTEM DESIGN Summary On completing this section, you should be able to: Draw Bode and Nyquist plots of an arbitrary SISO linear system Compute each component separately and combine in Bode plot Sketch the response of a SISO linear system, given its Bode plot Based on interpretation of frequency response Determine an appropriate transfer function given the Bode plot of a SISO linear system. Uses skills developed in -9 5