MECHATRONICS Quiz 1

1) Consider a first order instrument with output y(t), given , the input x(t)=1 sin ωt. a) Obtain the analytical solution for y(t), xest(t) and |xest(t)|/|x(t)| b) Calculate y(t), xest(t) and |xest(t)|/|x(t)|, where k=5 and T=0.01, for ω = 10, 100 and1000 c) Obtain the analytical solution the cutoff frequency, ωb d) Calculate the value of ωb for k=5 and T=0.01 Analytical Solutions (K. Ogata, Modern Control Engineering, 4th edition, Prentice Hall pp. 268-271), where L{}=Laplace transform a)|x(t)|=1 x(t)=1 sin ωt G(s) =k/(1+Ts) G(jω) =k/(1+T jω) y(t)=1 |G(jω)| sin (ωt+ Φ) xest(t)= y(t)/k=(|G(jω)|/k) sin (ωt+ Φ)= (|G(jω)|/5) sin (ωt+ Φ) |xest(t)|/|x(t)|= |G(jω)|/k b) For k=5; T=0.01; results 5/(1+0.01 jω)= 5(1-j0.01ω)/(1+0.012ω2)= 5(1-j0.01ω)/(1+0.012ω2) |G(jω)|= 5/(1+0.012ω2) ½ Φ = tan -1 (-0.01ω) xest(t)= y(t)/k=(|G(jω)|/k) sin (ωt+ Φ)= (|G(jω)|/5) sin (ωt+ Φ) |xest(t)|/|x(t)|= |G(jω)|/k= |G(jω)|/5 For ω = 10 results |G(jω)|= 5/(1+0.012ω2) 1/2 =5/(1+0.012102) 1/2 ≈5 Φ = tan -1 (-0.01ω)=-5.71 ≈0 y(t)=1 |G(jω)| sin (ωt+ Φ) ≈5 sin ωt xest(t)= y(t)/k ≈1 sin ωt=x(t) |xest(t)|/|x(t)| ≈1 c) Exact calculation of the cutoff frequency, ωb ,defining the bandwidth, obtained from the equation 20 log|G(jωb)|= 20 log|G(j0)| - 3 or log|G(j0)| - log|G(jωb)| = log(|G(j0)| / |G(jωb)| )=3/20 or |G(j0)| /|G(jωb)| = log-1 3/20 = 1.4125 or |G(jωb)| =|G(j0)| /1.4125 =0.0709 |G(j0)| For the above first order instrument G(jω) =k/(1+T jω) |G(jω)|= k/((1+T2ω2) 1/2 |G(j0)|=k and |G(jωb)| = k/((1+T2ωb

2) 1/2 such that |G(jωb)=|G(j0)| /1.4125 gives k/((1+T2ωb

2) 1/2 =k /1.4125 d)) For the above first order instrument and for k=5 and T=0.01 |G(jω)|= 5/((1+0.012ω2) 1/2 |G(j0)|= 5 The cutoff frequency, ωb can be obtained from |G(jωb)| =|G(j0)| /1.4125 For obtaining ωb the equation to solve is then 5/((1+0.012ωb

2) 1/2=5 /1.4125 or 1+0.012ωb

2 = 1.41252 1+0.0001 ωb

2 =1.99516 The solution for ωb

2 is ωb

2 =9951.6 Only the positive solution gives ωb =√9951.6 ≈ 100 [rad/s]

2) Consider a second order instrument with output y(t), the input x(t)=1 sin ωt, a) Obtain the analytical solution for y(t) b) Calculate y(t), where k=1, ζ = 0.7 and ωn =10 rad/s, for ω = 1, 10 and1000 c) Obtain the analytical solution the cutoff frequency, ωb d) Calculate the value of ωb for k=1, ζ = 0.7 and ωn =10 rad/s HINTS a) the overall transfer function is V(s)/F(s)= Kp / (M s2+B s + K)= (Kp / M)/ (s2+( B / M) s +( K /M) ) or G(s)=V(s)/F(s)= k /(s2+2 ζ ωn s + ωn

2) where ωn

2 = K /M 2 ζ ωn = B / M k= Kp / M Time response of such second order instruments are significantly dependent on the value of ζ = B ωn / (2 M) where ζ =0.7 For |x(t)|=1 x(t)=1 sin ωt G(jω) =1/( ωn

2 -ω2+ j 2 ζ ωnω) |G(jω)| =1/( (ωn

2 -ω2) 2+ (2 ζ ωωn) 2 ) 1/2

Φ = tan -1 (-2 ζ ωnω /( ωn2-ω2) )

y(t)=1 |G(jω)| sin (ωt+ Φ) b) for ζ = 0.7 ωn = 10 rad/s such that a= 2 ζ ωn =14 b= ωn

2=100 G(s) = k/( s2+14s + 100) G(jω) =1/( - ω2+100+14 jω) |G(jω)|= 1/((100-ω2)2+142ω2) 1/2 =1/((100-ω2)2+142ω2) 1/2

For ω = 1 |G(j1)|= 1/((100-1)2+142) 1/2 ≈1/100

Φ = tan -1 (-14 /(100-1) ) ≈ -8.05 y(t)=1 |G(j1)| sin (t+ Φ)= 1/10000 sin (t-8.05)

c) The cutoff frequency, ωb can be obtained from |G(jωb)| =|G(j0)| /1.4125 where |G(jωb)|=1/((100-ωb

2)2+142ωb2) 1/2

|G(j0)|=1/100 d) 1/((100-ωb

2)2+142ωb2) 1/2 =(1/100)/1.4125

or ((100-ωb2)2+142ωb

2) 1/2 =141.25 or (100-ωb

2)2+142ωb2 =141.25 2

100 2 -200ωb2 +ωb

4 +142ωb2- 141.25 2=0

-200ωb2 +ωb

4 +142ωb2 =141.25 2-100 2

(-200 + ωb2 +142 )ωb

2 -9951.6=0 ωb

4 - 4ωb2 -9951.6=0

The solution is ωb

2=2 ± 99.8 i.e. ωb2 = -97.8 and 101.8

For ωb2 = 101.8

ωb ≈± 10.1 Only the positive solution gives ωb =10.1≈10 = ωn