Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant Example, J thermocouple with gain at 2 C of =51 µv/ C and time constant of approx.1 sec. k=5 [1 µv/ C] and T=.1 MATLAB program k=5; T=.1; num=[ k]; den=[t 1]; step(num,den);grid Fig. y(t) plot 5 Step Response 4.5 4 3.5 3 Amplitude 2.5 2 1.5 1.5.1.2.3.4.5.6 Time (sec)
Bode diagram MATLAB program >> bode(num,den); grid; 2 Bode Diagram 1 Magnitude (db) -1-2 -3 Phase (deg) -45-9 1 1 1 1 2 1 3 1 4 Frequency (rad/sec) Bandwidth Cutoff frequency, ω b,defining the bandwidth, is defined as the frequency ω f which the Magnitude 2 log G(jω) drops 3 db below its zero-frequency value 2 log G(j) 2 log G(jω) <2 log G(j) - 3 db f ω> ω b In the above case of a first der transfer function G(s)=k/(1+Ts), k=5, T =.1 s G(s)=5/(1+.1s) G(jω)=5/(1+.1jω) its zero-frequency value is G(j)=5 2 log G(j) =14 db Cutoff frequency, ω b,defining the bandwidth, is obtained from the equation 2 log G(jω b ) = 2 log G(j) - 3 =14 3 = 11 db ω b 1 rad/sec f b 1/(2π) 14 Hz
Static calibration of the sens F: Xn in unknown sens input to the sens and Ym is the measured output from the sens, static calibration uses steady state sens gain K c f calibration, i.e f estimating the unknown input Using limit value theem f unit step input X(s)=1/s f s tending towards zero K c = lim s G(s) /s= lim G(s) = G(j)=5 Consequently, unknown sens input to the sens is estimated as Y/ K c = Y/5 In practice, this is considered acceptable within bandwidth, i.e f input signal frequencies ω < ω b Sinusoidal Response of the sens x(t)=sin ωt k/(1+ts) y(t) 1/ K c x est (t) f X(s)= ω/(s 2 +ω 2 ) and unit impulse input δ(s)=1 δ(s)=1 ω/(s 2 +ω 2 ) k/(1+ts) y(t) 1/ K c x est (t) k=5; T=.1; K c =5 MATLAB Simulations 1) ω=1 rad/s f= 1.58 Hz period =.63 sec MATLAB program num=[ 5]; den=[.1 1 1 1]; impulse(num,den); Fig. y(t) plot
5 Impulse Response 4 3 2 1 Amplitude -1-2 -3-4 -5 1 2 3 4 5 6 Time (sec) 2) ω=1 rad/s f= 15.8 Hz period =.63 sec MATLAB program num=[ 5]; den=[.1 1 1 1]; impulse(num,den); Fig. y(t) plot
4 Impulse Response 3 2 1 Amplitude -1-2 -3-4 1 2 3 4 5 6 Time (sec) 3) ω=1 rad/s f= 158 Hz period =.63 sec MATLAB program num=[ 5]; den=[.1 1 1 1]; impulse(num,den); Fig. y(t) plot
1 Impulse Response.5 Amplitude -.5.1.2.3.4.5.6.7.8.9 1 Time (sec) Analytical Solutions (K. Ogata, Modern Control Engineering, 4th edition, Prentice Hall pp. 268-271), where L{}=Laplace transfm L{x(t)=sin ωt } G(s) =k/(1+ts) L{y(t)} 1/ K c L{ x est (t)} where static calibration constant is K c = G(j) F X(s)= ω/(s 2 +ω 2 ) and unit impulse input δ(s)=1 δ(s)=1 ω/(s 2 +ω 2 ) k/(1+ts) y(t) 1/ K c L{ x est (t)} F k=5; T=.1; results x(t) =1 x(t)=1 sin ωt G(s) =k/(1+ts) G(jω) =k/(1+t jω)=5/(1+.1 jω)= 5(1-j.1ω)/(1+.1 2 ω 2 )= 5(1-j.1ω)/(1+.1 2 ω 2 ) G(jω) = 5/(1+.1 2 ω 2 ) 1/2 Φ = tan -1 (-.1ω) K c = G(j)=k=5 y(t)=1 G(jω) sin (ωt+ Φ) x est (t)= y(t)/ K c =( G(jω) / K c ) sin (ωt+ Φ)= ( G(jω) /5) sin (ωt+ Φ)
x est (t) / x(t) = G(jω) /5 F ω = 1 results G(jω) = 5/(1+.1 2 ω 2 ) 1/2 =5/(1+.1 2 1 2 ) 1/2 5 Φ = tan -1 (-.1ω) =-5.71 y(t)=1 G(jω) sin (ωt+ Φ) 5 sin ωt x est (t)= y(t)/k 1 sin ωt=x(t) x est (t) / x(t) 1 Summary of results: ω [rad/s] f[hz] Period [s] G(jω) Φ y(t) x est (t) x est (t) / x(t) 1 1.58.63 5 5sin ωt sin ωt 1 1 15.8.63 3.5-45 3.5sin(ωt-45).7sin(ωt-45).7 1 158.63 3.5-9.5sin(ωt-9).1sin(ωt-45).1 The result f ω = 1, gives x est (t) / x(t).7 2 log x est (t) / x(t) = - 3 db, i.e ω = 1 = ω b Consequently, f ω > ω b x est (t) / x(t) <.7 ω > >ω b x est (t) / x(t) <<.7 ω = 1 x est (t) / x(t).1 i.e the estimation is only 1% of the amplitude of the sens input signal Dynamic estimation (calibration) can use inverse problem solution, increasing gains, in this case 1/.7 f ω = 1 f and 1/.1=1 f ω = 1. (E. Doebelin, Measurement Systems, McGraw Hill, 199, Ch. 1.5 Dynamic Compensation, pp. 84-88) Obviously these gains would increase with ω and can lead to various difficulties (overflow in numerical computations, over-amplification of noise high frequency-low amplitude components in the y(t) signal etc), to be studied as part of inverse problem they, a topic of advanced mechatronics.
Bode diagram G -1 (s) = (1+Ts)/ k L{x(t)=sin ωt } G(s) k/(1+ts) L{y(t)} G -1 (s) = (1+Ts)/ k 1/ K c x est (t) MATLAB program f the dynamic compensat k=5; T=.1; num=[t 1]; den=[ k]; bode(num,den);grid; 3 Bode Diagram 2 Magnitude (db) 1-1 -2 9 Phase (deg) 45 1 1 1 1 2 1 3 1 4 Frequency (rad/sec) The magnitude of the inverse dynamic compensat G -1 (s) = (1+Ts)/ k shows 2 db/decade increase beyond bandwidth cutoff frequency, ω b =1, indicating growing computational difficulties as ω>> ω b. In general, inverse dynamic compensat f increasing senss bandwidth is subjet to various difficulties :
- computational difficulties as ω>> ω b due to increasing magnitude of the inverse dynamic compensat G -1 (s) = (1+Ts)/ k f ω> ω b ; digital wd length limitation can lead to overflow. - high frequency noise in the sens output is amplified by increasing magnitude of the inverse dynamic compensat G -1 (s) = (1+Ts)/ k f ω> ω b reducing signal to noise ratio - unmodelled dynamics and parametric uncertainty result in reduced effect of inverse dynamics compennsat - non-minimum phase sytems have unstable inverse dynamics Some solutions to the above difficulties are : - low pass filter f removing high frequency noise in the sens output L{x(t)} G(s) k/(1+ts) L{y(t)} Low Pass Filter G -1 (s)=(1+ts)/ k 1/ K c x est (t) - Modified Output Approach (MOA) applied to non-minimum phase sytems to avoid unstable inverse dynamics L{x(t)} G(s)k/(1+Ts) L{y(t)} Low Pass Filter G MOA -1 (s)=(1+ts)/ k 1/ K c x est (t)
B) Second der Instruments (E. Doebelin, Measurement Systems, McGraw Hill, 199, pp. 123-157) b) Second der transfer function (K. Ogata, Modern Control Engineering, Prentice Hall, 1997, Ch. 4-3, pp. 141-187) f example f a servo system (pp.142-146), a massspring-damper system (Example 4-7, pp. 174-175). G(s)=k /(s 2 +2 ζ ω n s + ω n 2 ) where k=gain, ω n = undamped natural frequency, ζ= damping ratio An example could be a fce f(t) transducer f(t) a fce measuring spring based instrument d(t) position measurement potentiometer v(t) where f(t) [N] is the input fce to measure d(t) [m] is the output displacement v(t) [V]is the output voltage of the potentiometer Assume the fce measuring spring based instrument measured by a mass-spring-damper M-B-K system moving hizontally (such that gravity effect can be igned in deriving motion equation) F(s)/ D(s)= 1/ (M s 2 +B s + K) where D(s) =L{d(t)} F(s) =L{f(t)} Assume the approximate transfer function of the position measurement potentiometer V(s)/D(s)= K p where V(s)= L{v(t)} D(s)=L{d(t)} K p [V/m] is the calibration constant of the potentiometer F(s) 1/ (M s 2 +B s + K) D(s) K p V(s) The overall transfer function is V(s)/F(s)= K p / (M s 2 +B s + K)= (K p / M)/ (s 2 +( B / M) s +( K /M) )
G(s)=V(s)/F(s)= k /(s 2 +2 ζ ω n s + ω n 2 ) where ω n 2 = K /M 2 ζ / ω n = B / M k= K p / M Time response of such second der instruments are significantly dependent on the value of ζ = B ω n / (2 M) ζ < 1 underdamped response ζ = 1 critically damped response ζ > 1 overdamped response MATLAB Simulations F k =1, ω n = 1 rad/s f n = 1.58 Hz period =.63 sec 1) Step response of G(s) = k /(s 2 +2 ζ ω n s + ω n 2 ) = k /(s 2 +b s + c) where b=2 ζ ω n c= ω n 2 Steady state of v(t) f unit step input f(t) is obtained using limit value theem f unit step input F(s)=1/s f s tending towards zero V = lim s G(s) /s= lim G(s) =lim k /(s 2 +2 ζ ω n s + ω n 2 )= 1/ω n 2 =.1 a) f ζ =, b=2 ζ ω n = MATLAB program f k=1; b=; c=1; num=[ k]; den=[1 b c]; step(num,den);grid Fig. v(t) plot shows an undamped oscillaty response, obviously not usefull in practical applications. Higher values f ζ are required.
.2 Step Response.18.16.14.12 Amplitude.1.8.6.4.2 1 2 3 4 5 6 b) f ζ =.1, b= 2 ζ ω n =2 MATLAB program k=1; b=2; c=1; num=[ k]; den=[1 b c]; step(num,den);grid Time (sec) The results show significant maximum overshoot of 7% and long 2% settling time of 4/(ζ ω n ) = 4/ (.1 x 1)= 4 sec.
.18 Step Response.16.14.12 Amplitude.1.8.6.4.2 1 2 3 4 5 6 Time (sec) c) f ζ =.6, b= 2 ζ ω n =12 MATLAB program k=1; b=12; c=1; num=[ k]; den=[1 b c]; step(num,den);grid The results show significant reduced overshoot of 5% and reduced 2% settling time of 4/(ζ ω n ) = 4/ (.6 x 1)= 1.7 sec.
.12 Step Response.1.8 Amplitude.6.4.2.1.2.3.4.5.6.7.8 Time (sec) d) f overdamped case, ζ = 1.2, b= 2 ζ ω n =24 MATLAB program k=1; b=24; c=1; num=[ k]; den=[1 b c]; step(num,den);grid The results show no overshoot but sluggish response.
.1 Step Response.9.8.7.6 Amplitude.5.4.3.2.1.2.4.6.8 1 1.2 Time (sec) Present day practice is to provide underdamped response f ζ = B ω n / (2 M) in the range of.6-.7, but selecting a damping coefficient of B = 2 M ζ / ω n (E. Doebelin, Measurement Systems, McGraw Hill, 199, pp. 131). Bode diagram f such a case ζ =.6, b= 2 ζ ω n =12 is given by MATLAB program k=1; b=12; c=1; num=[ k]; den=[1 b c]; bode(num,den); grid;
-2 Bode Diagram -4 Magnitude (db) -6-8 -1-12 -45 Phase (deg) -9-135 -18 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) Cutoff frequency, ω b,f the above case of a second der transfer function G(s) = k /(s 2 +2 ζ ω n s + ω n 2 ) = k /(s 2 +b s + c) where k=1; b=12; c=1; gives G(jω)=1/( -ω 2 +12jω + 1) its zero-frequency value is G(j)=1/1=.1 2 log G(j) =- 4 db Cutoff frequency, ω b,defining the bandwidth, is obtained from the equation 2 log G(jω b ) = 2 log G(j) - 3 =-4 3 = -43 db The Magnitude diagram gives ω b 1 rad/sec f b 1/(2π) 1.4 Hz Dynamic estimation (calibration) can use inverse problem solution, increasing gains, in this case Obviously these gains would increase with ω and can lead to various difficulties (overflow in numerical computations, over-amplification of noise high frequency-low amplitude components in the y(t) signal etc).
Bode diagram G -1 (s) = (s 2 +b s + c) / k F(s) G(s) D(s) G -1 (s) V(s) MATLAB program k=1; b=12; c=1; den =[ k]; num =[1 b c]; bode(num,den); grid; 12 Bode Diagram 1 Magnitude (db) 8 6 4 2 18 135 Phase (deg) 9 45 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) The magnitude of the inverse dynamic compensat G -1 (s) = (1+Ts)/ k shows 4 db/decade increase beyond bandwidth cutoff frequency, ω b =1, indicating growing computational difficulties as ω>> ω b, even me significant the in the case of first der instruments. N-der instrument will have 2 N db/decade increase beyond bandwidth cutoff frequency.
Some solutions to the above difficulties were outlined f the case of first eder instruments: - low pass filter f removing high frequency noise in the sens output - Modified Output Approach (MOA) applied to non-minimum phase sytems to avoid unstable inverse dynamics In practice the increase beyond bandwidth cutoff frequency, ω b is nmally limited up to a maximum useful frequency component ω useful which might be achieved by inverse dynamic compensat approach without reachng unacceptable high magnitudes of the inverse dynamic compensat, making this approach advantageus in computer based instrumentation.
Sinusoidal Response of the second der sens with ζ =.6 F F(s) 1/ (M s 2 +B s + K) D(s) V(s) the overall transfer function is V(s)/F(s)= K p / (M s 2 +B s + K)= (K p / M)/ (s 2 +( B / M) s +( K /M) ) G(s)=V(s)/F(s)= k /(s 2 +2 ζ ω n s + ω n 2 ) where ω n 2 = K /M 2 ζ ω n = B / M k= K p / M Time response of such second der instruments are significantly dependent on the value of ζ = B ω n / (2 M) where ζ =.6 MATLAB simulation of x(t)=sin ωt k/( s 2 +2 ζ ω n s + ω n 2 ) y(t) 1/K c x est (t) where static calibration constant is K c = G(j) F X(s)= ω/(s 2 +ω 2 ) and unit impulse input δ(s)=1 can be achieved with MATLAB instruction impulse (num, den) f δ(s)=1 ω/(s 2 +ω 2 ) k/( s 2 +2 ζ ω n s + ω n 2 ) y(t) 1/ K c x est (t) The transfer function f the g eneral MATLAB program is k ω/[(s 2 +ω 2 )( s 2 +2 ζ ω n s + ω n 2 )] = k ω/(s 4 + 2 ζ ω n s 3 +(ω n 2 +ω 2 )s 2 +2 ζ ω n ω 2 s + ω n 2 ω 2 ) = k ω /( s 4 +as 3 + (b +ω 2 )s 2 +a ω 2 s + bω 2 ) where a= 2 ζ ω n b= ω n 2 Let us assume k =1, ζ =.6 ω n = 1 rad/s such that a= 2 ζ ω n =12 b= ω n 2 =1
a=12; b=1; num=[ ω]; den=[1 a b+ω*ω a*ω*ω b*ω*ω]; impulse(num,den); MATLAB Simulations 1) ω=1 rad/s f=.158 Hz period = 6.3 sec MATLAB program is a=12; b=1; num=[ 1]; den=[1 a b+1 a b]; impulse(num,den);.15 Impulse Response.1.5 Amplitude -.5 -.1 -.15 5 1 15 2 25 3 Time (sec) The result agrees to Bode diagram amplitude f ω=1of -4 db.1 2) ω=5 rad/s
a=12; b=1; num=[ 5]; den=[1 12 125 3 25]; impulse(num,den);.15 Impulse Response.1.5 Amplitude -.5 -.1 -.15 1 2 3 4 5 6 7 8 9 1 Time (sec) The result agrees to Bode diagram amplitude f ω=5of -4 db.1 Analytical Solutions (K. Ogata, Modern Control Engineering, 4th edition, Prentice Hall pp. 268-271), where L{}=Laplace transfm L{x(t)=sin ωt } G(s)= k/( s 2 +2 ζ ω n s + ω 2 n ) L{y(t)} 1/ K c L{ x est (t)} f X(s)= ω/(s 2 +ω 2 ) and unit impulse input δ(s)=1 δ(s)=1 ω/(s 2 +ω 2 ) k/( s 2 +2 ζ ω n s + ω 2 n ) y(t) 1/ K c L{ x est (t)} F x(t) =1 x(t)=1 sin ωt ζ =.6 ω n = 1 rad/s such that a= 2 ζ ω n =12 b= ω 2 n =1 G(s) = k/( s 2 +12 s + 1) G(jω) =1/( -ω 2 +1+12 jω)= ( -ω 2 +1-12 jω)/ [( -ω 2 +1+12 jω) ( -ω 2 +1-12 jω)] = (1-ω 2-12 jω)/( (1-ω 2 ) 2 +12 2 ω 2 ) G(jω) = 1/((1-ω 2 ) 2 +12 2 ω 2 ) 1/2
Φ = tan -1 (-12 ω/(1-ω 2 ) ) y(t)=1 G(jω) sin (ωt+ Φ) G(j) = 1/((1) 2 ) 1/2 =1/1 K c =G(j)=1/1 x est (t)= y(t)/ K c =( G(jω) / K c ) sin (ωt+ Φ)= ( G(jω) /(1/1)) sin (ωt+ Φ) x est (t) / x(t) = G(jω) / K c = G(jω) / G(j) F ω = 1 results G(jω) = 1/((1-ω 2 ) 2 +12 2 ω 2 ) 1/2 =1/((1-1 2 ) 2 +12 2 1 2 ) 1/2 =1/12 Φ = tan -1 (-12 ω/(1-ω 2 ) )= tan -1 (-12/(1-1 2 ) ) = tan -1 (-12/)=-9 y(t)=1 ( G(jω) / K c )sin (ωt+ Φ) ((1/12)/(1/1)) sin( ωt-9) G(j1) / G(j) = (1/12)/(1/1)=1/12 = x est (t) / x(t).83 F ω = 1, Bode diagram gives appr -42 db and -9 which agrees with the above result. Exact calculation of the cutoff frequency, ω b,defining the bandwidth, obtained from the equation 2 log G(jω b ) = 2 log G(j) - 3 log G(j) - log G(jω b ) = log( G(j) / G(jω b ) )=3/2 G(j) / G(jω b ) = log -1 3/2 = 1.4125 G(jω b ) = G(j) /1.4125 =.79 G(j) which shows that, at cutoff frequency, ω b, the amplitude G(jω b ) drops to.79 of the G(j) F the above second der instrument G(jω) = 1/((1-ω 2 ) 2 +12 2 ω 2 ) 1/2 results G(j) = 1/((1-2 ) 2 +12 2 2 ) 1/2 = 1/1 The cutoff frequency, ω b can be obtained from G(jω b ) = G(j) /1.4125 =(1/1)/ 1.4125=1/141.25 F obtaining ω b the equation to solve is then 1/((1-ω b 2 ) 2 +12 2 ω b 2 ) 1/2 =1/141.25 ((1-ω b 2 ) 2 +12 2 ω b 2 ) 1/2 =141.25 (1-ω b 2 ) 2 +12 2 ω b 2 =141.25 2 =19952.6 ω b 4-2 ω b 2 +1-19952.6+ 144ω b 2 = ω b 4-56 ω b 2-9952.6 = The solution f ω b 2 is ω b 2 =28± (28 2 +9952.6) =28±13.6 131.6 and 75.6 Only the positive solution gives ω b = 131.6= 11.47 [rad/s] that can be recognized on the second der Bode diagram f the magnitude of -4-3 =-43 db