2011 Technological Studies. Advanced Higher. Finalised Marking Instructions

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1 0 Technological Studies Advanced Higher Finalised Marking Instructions Scottish ualifications Authority 0 The information in this publication may be reproduced to support SA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SA s N Delivery: Exam Operations Team. Where the publication includes materials from sources other than SA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SA s N Delivery: Exam Operations Team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes. Page

2 Section A Marks (a) f = ( + )C T = + 50 = C T T = T = 5760 = ( ) = 880 Ω (b) See Worksheet 5 (8) main: btfss POTB,0 goto main movlw d 6 movwf COUNTE loop: bsf POTB,7 movlw d call pause bcf POTB,7 movfw DELAY call pause rrf DELAY,F decfsz COUNTE,F goto loop bsf btfsc POTB,6 POTB, bsf POTB,5 movlw d 0 ) call wait ) bcf POTB,6 movlw d 50 movwf COUNTE check: btfsc POTB, goto release movlw d O movlw d 00 call decfsz wait COUNTE,F call pause goto check bsf POTB, end release: bsf POTB, end () Page

3 (a) Wein bridge (b) This waveform is used due to: rapid transition of state; only two states; useful for providing clock pulse (c) V 8 k k 7 V 0 V V out = P P + 7 = = / p =/ /000 p = 800Ώ = = 9Ω = 0Ω Page

4 continued (d) V out = Vin dt C 8 = Vint 6000C 8 = t 6000C 8 = C C = C = F = 5.0 μf (9) Page

5 (a) 000N 500 N 500 N 60 N/m 0 6 m 5 m 0 m 0 5 m Take moments about ΣCWM = ΣACWM (60 5) + ( ) + (500 ) + (500 5) = = 770 = 57 N Σ F V = 0 = [ ( 60)] 57 = 9 N (b) (i) See Worksheet (ii) From LHE 0m and m = 0 Nm m : (0 6 9) 60 = Nm m : ( 9) (0 000) 60 = 09 Nm 6 6m : ( 6 9) ( 000) m : ( 9) ( 5 000) 60 = 000 Nm 60 = 906 Nm 5 5m : ( 5 9) ( 000) 60 = 6 Nm for plotting points 6 () Page 5

6 5 (a) See Worksheet 5 7 (b) S Cross-coupled NOs Correct labels (9) Page 6

7 6 main: bsf POTB,5 movfw MAK call pause bcf POTB,5 movfw SPACE ) call pause ) call adcread movlw d 60 subwf DATA,W btfss STATUS,C goto incmk movfw DATA sublw d 50 btfss STATUS,C goto main movfw DATA addwf MAK goto main incmk: movfw DATA addwf SPACE goto main () Page 7

8 7 (a) (i) More area resisting bending depth material depth (or any acceptable answer) Greater resistance to air flow, wind (ii) Better strength to weight ratio (b) (i) I = mm, y = 5mm, σ = 00N/mm my σ = I σi M = y M = 5 = 6. knm FL M = M M = L 6. 0 = 500 = 969N (ii) FL deflection = 8 EI E(mild steel) = 96 0 N/mm d = = 6.8mm Page 8

9 7 (c) (continued) FL M = M = M = 75000Nmm my σ = I my I = = σ 90 I = 508mm I = I = πd I = 908mm BD I = I = B D B=5.mm 5 (d) M = I E EI = M = = 970mm Page 9

10 7 (continued) (e) range is 0 6 6v = 08V (msb) = = = 8 kω 08 5 = 8 = 65 6 kω = 8 = kω = 8 8 = 6 5 kω = 6 8 = 55 kω = 8 = 050 kω Sketch on Worksheet 5 (f) 7 0 = 7 = 6 = 8V 6 (0) Page 0

11 8 (a) grab: bsf POTB, ) bcf POTB,5 ) movlw d 50 * call wait * bcf POTB, movlw d 0 * call wait * bsf POTB,5 movlw d 50 * call wait * * delays return (b) dump: bsf POTB,7 ) call adcread ) check: movfw ACTUALPOS sublw d 50 btfss STATUS,Z ) goto check ) bcf POTB,7 ) movlw d 0 ) call wait ) bsf POTB, ) movlw d 0 ) call wait ) bcf POTB, ) return ) (c) main: movlw d ) movwf COUNTE ) btfss POTB, ) goto main ) movlw d 00 ) movwf DESIEDPOS ) loop: bsf POTB,6 ) call adcread ) movfw DESIEDPOS ) subwf ACTUALPOS ) btfss STATUS,Z ) goto loop ) call grab ) call dump ) movlw d 50 ) addwf DESIEDPOS ) decfsz COUNTE ) goto loop ) goto main ) (+ labels) Page

12 8 (d) (continued) Take moments about A ΣCWM = ΣACWM (8 5) + (0 5) = B B = 75 kn ( 5 + 5) 75 ΣFV = A = = 5 kn (e) N 5 M 0 M M N 6 m m 5 kn Take moments about N M and M = 0 ΣCWM = ΣACMW 6 5 = M M M 6 5 = = 875kN (TIE) Page

13 8 (continued) (e) Take moments about N ΣCWM M = M = = ΣACMW x =.7 sin 8.5 = m = 99kN (STUT) Σ Horizontal forces = M horiz 99cos5 = 0 M horiz = 99cos5 875 M horiz =.0kN.0 M = sin 6.6 M = 6kN (TIE) 9 (0) Page

14 9 (a) Time based each step ends after a set time event based A sensor being actuated ends step. (b) See Worksheet 9(a) (c) V 0 = V dt in C Vout = C t dt V 0 = t = C t C =.5 0 C 9 t V 0 = t C (d) The voltage divider sets a reference for the inverting input. As the capacitor charges the non-inverting input increases when it is greater than the inverting input the transistor switches the relay. The relay then discharges the capacitor and the process repeats. The larger the input voltage the faster the capacitor charges and the faster the frequency. Page

15 9 (continued) (e) See 9(b) Worksheet (f) delay: movlw d movwf COUNT loop: movfw TIMEVAL call wait decfsz COUNT goto loop return (g) main: movlw d 5 movwf COUNTE check: btfss POTB, goto check loop: bsf POTB,7 movfw TIMEVAL ) call delay ) * bcf POTB,7 movfw TIMEVAL ) call delay ) * movlw d subwf TIMEVAL decfsz COUNTE,F goto loop goto main (0) Page 5

16 Worksheet Signal from display switch Signal from run switch S 5V D D D D D Clock output from 555 Vcc k 7 8 for 5 AND CKK for 5 AND for 5 6ANDS for pin AND for 5 AND 88 k In multiple connections for + µf 6 5 0V for 555 circuit Page 6

17 Worksheet kn 500 N 500 N 60 N /m 0 6 m 5 m 0 m 0 5 m B(ii) 6 marks (b) (i) marks Shear-force D iagram (N ) Bending moment D iagram (N m) N m 00 m 60 m 0 m 50 m 00 m 57 N D istance (m) mark each highlighted point apart from and mark for plot D istance (m) DISTANCE FOM LEFT END (m) BENDING MOMENT (Nm) () () () () () () Mark for both Page 7

18 Worksheet 5 A c 5 V 0 V B C D LT BI EL b a g f e d v- a b c 7- Segment Display for 7 BCD Seg for ABCD for 5 v JK for AND for CLK for fire for fire S 5 V Signal from fire button J K S J K S J K S J K S 9 V V 0 V Hz S Igniter 0 V Page 8

19 Worksheet 7 0 k lsb 0 k 0 k 5 6 msb 0 V 0 V Analogue Signal EN A B C D for summing for correct counter connections EN A B C D 5 V Page 9

20 Worksheet 9(a) to valve A Vcc 6 CLOCK 5 A 0 to valve B CLOCK 7 B +Vcc 5 D 0 0 V 5 9 C 6 B D 7 ESET C A 0 BINAY COUNTE 8 0 V 8 9 DECODE to valve C clock pin pin pin ABCD ABCD B&D ESET 6,7,8,9 O VALVE A 0, O INV VALVE B 9 INV VALVE C Page 0

21 Worksheet 9(b) for pattern for repeating pattern for inversion ( A ) A B for pattern for repeating pattern for shift for inversion ( B ) [END OF MAKING INSTUCTIONS] Page

2013 Technological Studies. Advanced Higher. Finalised Marking Instructions

2013 Technological Studies. Advanced Higher. Finalised Marking Instructions 03 Technological Studies Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 03 The information in this publication may be reproduced to support SQA qualifications only on