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Unformatted text preview: ELEN417-Measurement Systems Homework 3 – Solution Chapter 4 Problem 1: Water, 100 °C 1 min Air, 20 °C Air, 20 °C Steady state sensitivity: K = 1 (Thermal exchange equilibrium) () = 1 1 + ℎ : = ℎ : = = 10 = 50 (0− ) = 20 ℃ () = 100 ℃; 0 ≤ < 60 ( , , () = 20 ℃; ≥ 60 ( , , 0 ≤ < 60: () = ∆() 1 80 = ; ℎ ∆ () = ∆ () 1 + 10 ELEN417-Measurement Systems University of Balamand Problem Solutions : () = ʆ−1 [()Δ ()] − Δ () () = ʆ−1 [ 80 − 80u() (1 + 10) 80 8 80 80 = = − (1 + 10) ( + 0.1) + 0.1 ℎ, () = 80u() − 80 −0.1 u() − 80u() = −80 −0.1 u() t (sec) 10 20 50 E(t) (°C) –29.4 –10.8 –0.54 ≥ 60: ∆ () = −80( − 60) () = ∆() 1 80−60 = ; ℎ ∆ () = − ∆ () 1 + 50 : () = ʆ−1 [()Δ ()] − Δ () () = ʆ − −1 80−60 [− + 80u( − 60) (1 + 50) 80−60 −1.6−60 80−60 80−60 = =− + (1 + 50) ( + 0.02) + 0.02 ℎ, () = −80u( − 60) + 80 −0.02(−60) u( − 60) + 80u( − 60) = 80 −0.02(−60) u( − 60) t (sec) 120 300 E(t) (°C) 24.1 0.66 ALTERNATIVE APPROACH Recall that: T t T T 0 1 exp t T 0 ; where MC UA In the first part: W 10 sec with T 0 20 °C and T 100 °C. ELEN417-Measurement Systems University of Balamand Problem Solutions T1 t 100 20 1 exp t 10 20 80 1 exp t 10 20 Error = T1(t) – T(∞) = T1(t) – 100 In the second part: A 50 sec T 2 t T T 60 1 exp t 60 T 60 with T 60 99 .8 °C and T 20 °C. T 2 t 20 99.8 1 exp t 60 50 99.8 79.8 1 exp t 50 99.80 Error = T2(t) – T(∞) = T2(t) – 20 ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 4.2: m = 0.5 Kg λ = 6.0 N·s/m KS = 200 N/m ) − : = 1 = 0.005 / : = √ ⁄ = 20 / : = ) 2√ ∙ (0− ) = 2 ( ) : (0− ) = ) = 0.3 2 = = 0.01 = 1 200 ∙ −1 Δ() = (3 − 2)() = (1)() () ( − ℎ ) : ∆() 1 1 = ∆() ( 1 2 + 2 + 1) 2 ∆() = 1 ()∆() ∆() = 1 −1 ʆ [()Δ()] ELEN417-Measurement Systems University of Balamand Problem Solutions ∆() = 1 1 − − ( ) + ( ) 2 √1 − { [} = √1 − 2 = 19.1 / ∆() = () − (0− ) → () = ∆() + (0− ) () = 1 + 0.5{1 − −6 [(19.1) + 0.314(19.1)]} ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 3: Overall steady-state sensitivity: K = 1 All linear elements : = 40 / : = 0.1 : ∆() 1 = ∆() ( 1 2 + 2 + 1) 2 − : () = 1 2 1 − ( ) + 2 1 − : |()| = 2 2 2 √(1 − ( ) ) + 4 2 ( ) ω (rad/sec) 10 30 50 ℎ − : () = −−1 ( ω (rad/sec) 10 30 50 ω/ωn 0.25 0.75 1.25 |()| 1.065 2.160 1.625 2 ⁄ ) 1 − (⁄ )2 ω/ωn 0.25 0.75 1.25 () –3.05° –18.9° –156° Dynamic Error: 1 () = 50 {[1.065(10 − 3.1) − (10)] + 3[2.16(30 − 18.9) − (30)] 1 + 5[1.625(50 − 156) − (50)]} ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 4: ℎ (): = 10 ) () = 1 1 + 10 : 0 = 1 = 0.1 / 3 − ℎ: = 0.1 / ) |()| ≥ 0.95, ℎ ≤ 1 |()| = 1 √1 + 100 2 = 0.95 1 = 0.9025 → 1 + 1002 = 1.108 → = 0.033 / 2 1 + 100 |()| ≥ 0.95, ℎ ≤ 0.033 / ) − : () = 1 1 + 10 1 × = 1 + 10 1+ 1+ ± 5 %: | ()| ≥ 0.95 1 = 0.9025 → 1 + 2 = 1.108 → = 0.33 / 1 + 2 ELEN417-Measurement Systems University of Balamand Problem Solutions 1 10s T ( s ) 0.5 s 0.1 1 2 Tin ( s ) 1 20s 1 s s 1.05 s 0.05 d) Bode Diagram 1 0.9 Magnitude (abs) 0.8 0.7 0.6 0.5 0.4 -3 10 -2 -1 10 10 0 10 Frequency (rad/s) The magnitude plot shows that the ±5% frequency range extends to about 0.01 rad/sec. This value is lower than the value reached in (c) above. ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 6: m = 0.5 Kg λ = 2.0 N·s/m KS = 100 N/m Displacement Transducer: sensitivity = 10 / ) − : = 1 = 0.01 / : = √ ⁄ = 14.14 / : = 2√ ∙ = 0.1414 = 2 −1 : = √1 − 2 = 14 / − ℎ: Δ = (Δ) = 0.5 × 9.81 −2 = 4.05 Transducer response to force step-change: ∆() = (10 ) (4.905 ) (0.01 ) {1 − −2 [(14) + 0.143(14)]} ∆() = (0.49 ){1 − −2 [(14) + 0.143(14)]} ) 60 → 1 ( 1 ) → = 6.28/ ELEN417-Measurement Systems University of Balamand Problem Solutions − : () = 1 2 1 − ( ) + 2 1 − : |()| = 2 2 2 √(1 − ( ) ) + 4 2 ( ) 1 |(6.28)| = 2 2 √(1 − ( 6.28 ) ) + 4(0.1414)2 ( 6.28 ) 14.14 14.14 = 1.231 2 At 6.28 rad/s, the response will undergo more than 23% overshoot. To remedy the low frequency response limitation, use negative feedback: check section 4.4 “Techniques for dynamic compensation”. ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 7: a) System transfer ratio: FM s Fin s FM s 50 20 10 3 0.1s 2 Fin s 1 0.1s s 50 0.4 s 50 1 1 0.1s 1 0.1s s 50 2 0.4 s 50 1 FM j Fin j 0.1 j 1 2 1 0.1 j 1 50 0.4 j 50 For ω = 10 rad/s: FM j10 Fin j10 F M j10 Fin j10 j j 1 1 1 j 1 0.04 j 0.08 1 j 0.96 j 0.08 1 11 1 1 1 0.734 0.9216 0.0064 0.08 40.24 0 . 96 j10 90 Tan 1 Tan 1 ELEN417-Measurement Systems University of Balamand Problem Solutions For ω = 30 rad/s: FM j 30 Fin j 30 FM j 30 j3 j3 1 1 1 j 3 1 0.36 j 0.24 1 j 3 0.64 j 0.24 Fin j 30 3 1 9 1 1.388 0.4096 0.0576 3 1 0.24 2.12 0.64 j30 90 Tan 1 Tan 1 For ω = 50 rad/s: FM j 50 Fin j 50 FM j 50 Fin j 50 j5 j5 1 1 12.5 1 j 5 1 1 j 0.4 1 j 5 j 0.4 1 j 5 12.5 2.45 1 25 5 1 j50 Tan 1 78.69 Therefore, FM t 50 0.735 sin 10 t 40.24 50 0.648 sin 30 t 2.12 3 50 1.471 sin 50 t 78.69 5 F M t 36 .75 sin 10 t 40.24 10.8 sin 30 t 2.12 14.71 sin 50 t 78.69 Error 36.75 sin 10 t 40.24 50 sin 10 t 10.8 sin 30 t 2.12 16.67 sin 30 t 14.71 sin 50 t 78.69 10 sin 50 t ELEN417-Measurement Systems University of Balamand Problem Solutions 50 40 30 20 10 0 -10 -20 -30 -40 -50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0.6 0.8 1 1.2 1.4 Fin (blue) and FM (red) waveforms 100 80 60 40 20 0 -20 -40 -60 -80 -100 0 0.2 0.4 Error waveform b) Use negative feedback to increase the bandwidth, significantly beyond 50 rad/sec, and thus be able to pass the 50 rad/sec frequency without the observed overshoot, while the lower frequencies would pass with less attenuation. ELEN417-Measurement Systems University of Balamand Problem Solutions ...
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  • Winter '15
  • Amir Alimohammad
  • Harshad number, University of Balamand

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