2014 Technological Studies. Advanced Higher. Finalised Marking Instructions

Transcription

1 204 Technological Studies Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 204 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Assessment team. Where the publication includes materials from sources other than SQA (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. SQA s NQ Assessment team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Part One: General Marking Principles for: Technological Studies Advanced Higher This information is provided to help you understand the general principles you must apply when marking candidate responses to questions in this Paper. These principles must be read in conjunction with the specific Marking Instructions for each question. (a) (b) for each candidate response must always be assigned in line with these general marking principles and the specific Marking Instructions for the relevant question. If a specific candidate response does not seem to be covered by either the principles or detailed Marking Instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader/Principal Assessor. Marking should always be positive ie, marks should be awarded for what is correct and not deducted for errors or omissions. GENERAL MARKING ADVICE: Technological Studies Advanced Higher The marking schemes are written to assist in determining the minimal acceptable answer rather than listing every possible correct and incorrect answer. The following notes are offered to support Markers in making judgements on candidates evidence, and apply to marking both end of unit assessments and course assessments. Page 2

3 Part Two: Marking Instructions for each Question Section A Q (a) (i) Wein-bridge oscillator (ii) Sine wave (b) f = 2 RC R = 2 fc = 6 3 subs 3 R =59 Answer plus unit 4 (c) Output fed into a Schmitt trigger () which would produce a square wave () which is suitable as a falling edge. 2 (8) Page 3

4 Q2 main: btfss PORTB,0 ;start switch goto main water: btfss PORTB,3 ;float switch goto ALARM bsf PORTB, ;poll mux for quantity dial call adcread movwf QUANTITY bsf PORTB,7 ;heat water bcf PORTB, ;poll mux for temperature HEAT: call adcread movwf TEMPERATURE movlw d 80 subwf TEMPERATURE,W btfss STATUS,C ;check temp reached 80 goto HEAT ;no, so keep heating bcf PORTB,7 ;heater off rrf QUANTITY,F ;quantity 2 movfw QUANTITY bsf PORTB,6 ;valve on call wait ;time delay bcf PORTB,6 ;valve off goto main ALARM: bsf PORTB,5 ;bleep on movlw d 0 call wait bcf PORTB,5 ;bleep off goto main 26 Page 4

5 Q3 (a) ƩM A = (0 6) + (5 2) + ( 2 3) (R B 2) = R B = 0 26 R B = 0 5kN 2 V = R A ( 2) = 0 R A = + 6 5kN kN (b) 6 (c) Page 5

6 Q3 (c) (cont) 0m & 2mBM = 0kNm 2m BM = + (6 5 2) = + 3kNm 6 5kN kn 3m 2 5m BM = + (6 5 3) ( 0 5) 2 = = +9kNm 6 5kN 2kN 4m 3m BM = + (6 5 4) (2 ) = = + 24kNm 2 6 5kN kn/m 6m BM = + (6 5 6) ( 2 3) 2 = = +33kNm 6 5kN Plotting line Nature 0 (20) Page 6

7 Q4 (a) 5 (b) 6 (c) 3 falling-edge JKs 5V J & K( 3) +5V to A & C Set +5V to B Reset Q 2 3 Q AND 7 AND RESET (b) (8) Page 7

8 Q5 (a) Neutral Axis (b) b d I = 3 3 bd 2 = = mm 4 2 (c) M I σ = = y σ = My I M = 40kNm = Nmm y = 00mm I = Nmm σ = 20N/mm 2 4 (7) Page 8

9 Q6 (a) Binary coded decimal to decimal decoder () converts binary to decimal () 2 Binary coded decimal to 7-segment display () decoder converts binary to segment signals. () 2 4 (b) When OR gate is high, Counter is enabled Counter counts up on each rising edge BCD to D decodes binary to decimal When BCD to D reaches 3 counter A disabled, buzzer sounds 2 When BCD to D reaches 3 AND enabled When clock goes high low 2 nd 4-bit counter B counts up. The 4-bit output is decoded to drive 7-segment display. 7-segment display shows a digit equivalent to count when count reaches 8: signal is sent to OR and counter A is re-enabled Decoder continues to 5 and resets counter B to 0. Counter A resets at 0 Any points@ (5) Page 9

10 Q7 (a)(i) (a) (i) gain (k) too low 2 (a)(ii) (a) (ii) gain (k) too high 2 (b)(i) Removes offset (b)(ii) Removes hunting / reduces settling time (c) (c) ideal PID response 2 (d) A substitutions Position A pulse-time (mark) = 25ms answer B Position B pulse-time (mark) = 75ms 3 (e) = 0 5 ms = 50 steps 2 Page 0

11 Q7 (cont) Assembler Code (f) drive: movlw d 25 movwf MARK movlw d 50 movwf COUNTER loop: bsf PORTB,4 movfw MARK call short bcf PORTB,4 movlw d 20 call pause incf MARK,F decfsz COUNTER,F goto loop movlw d 50 movwf COUNTER loop2 bsf PORTB,4 movfw MARK for repeat code call short bcf PORTB,4 movlw d 20 call pause decf MARK,F decfz COUNTER,F goto loop2 return 3 (26) Page

12 Q8 (a) obstruction: call adcread movlw d 8 subwf DATA,W btfsc STATUS,C goto finish bsf PORTB,6 movlw d 26 call wait bcf PORTB,6 finish: return 7 (b) main: call navdata loop: bsf PORTB,3 bsf PORTB,7 movlw d 0 call pause call obstruction decfsz FORWARDTIME goto loop (incl. goto main below) movlw d subwf TURN btfsc STATUS,Z goto main clrf TURNTIME ACW: bsf PORTB,4 movlw d 8 call pause bcf PORTB,4 movfw d 9 call pause incf TURNTIME movfw TURNTIME subwf TURNSTOP btfss STATUS,Z goto ACW bcf PORTB,7 (incl. goto loop above) goto main 20 Page 2

13 Q8 (c) 3 2 BD bd ID = = D I= 64 4 (2 36) 2 64 I D = I total = I total = mm 4 6 (d) yield stress aluminium = 30 N/mm 2 my I σi m= y = FL 4m m= F= 4 L = 3 87kN 7 (40) Page 3

14 Q9 (a) volt increments k 5v R v 0v V R V f out Ri R 0 R k R = 300kΩ R = 650kΩ 2 R = 325kΩ 3 R = 62.5kΩ 4 R = 8.3kΩ 5 R = 40.6kΩ 6 R = 20 3kΩ 7 Page 4

15 Q9 (a) (cont) 0 Page 5

16 Q9 (b) (i) Voltage controlled Oscillator 2 (ii) SubA = RC circuit,charging time prop to V in 2 SubB = volt divider, provides Volt ref 2 SubC = comparator compares Volt ref with Cap volt V cap > volt then +ve 2 output SubD = npn trans responds to +ve, switches on DPDT relay briefly, disch cap + switches on output 2 SubE = lamp and buzz, freqency of outputs prop to current drawn by pump motor 2 0 (iii) V in must be greater than volt, limited frequency of relay, mech life of relay 2 (c) main: bsf PORTB,4 movlw d 5 call wait bcf PORTB,4 loop: call adcread sublw d 40 btfsc STATUS,Z ( for status Z, for clear) 2 goto wet btfss STATUS,C ( for status C, for set) 2 goto wet ( for both goto wet) bcf PORTB,4 call delay goto main wet: bsf PORTB,4 goto loop 6 (40) Page 6

17 Q0 (a) B (i) + ƩM Q =0 (R B ) + (7 5 ) = 0 2 substitutions 2 R B = +7 5kN answer 3 (ii) ƩM Q = 0+ + (4. 09 perp) + (F ) (7 5 2) = 0 3 substitutions 3 + ( ) + F 5 = 0 calculation F = 0kN answer perp 44cos30 224m 6 Page 7

18 Q0 (b) ƩM P = 0 + (M 3 ) + (2 5 ) = 0 M 3 = + 2 5kN (TIE) 3 ƩM Q = 0 + +(2 5 2) + (M perp) = 0 perp = 44cos30 = 224m 5 M = M 4 08kN (STRUT) 5 perp 2 = 44sin30 ƩM R = 0 + = 0 707m (M perp 2) (M2 44) 0 OR : Fh 0 ve + ( ) + (M 2 44) = 0 + M cos5 + M 2 cos45 + M 3 = M2 4 08cos5 + M 2 cos = 0 44 M 2 = M 2 cos45 = 44 M 2 =2 04kN (TIE) M 2 = 2 04kN (TIE). 5 3 Page 8

19 Q0 (c) 0 32 V out = t 2 V out = 0 6t 2 (d) V out = Vin dt RC 0 6t = V 6 6 in dt substitutions 2 V in 0 6t t 8 36 integration V in = = 2 94 volts 4 (e) V out = 0 6t 2 = 0 6t substitutions 2 t= 0 6 t = 75 seconds answer 2 Page 9

20 Q0 (f) When V in = 5V: lower threshold, output already low. correct state Rp = 2 78k substitutions & answer 2 R 35 substitution R 5 P 35 R R 6 49k 6 (g) Switch-off voltage: Output already high correct state Rp = 2 73k substitution 68 V = 5 = 3 57 volts V 2 = 3 57 volts 4 (40) [END OF MARKING INSTRUCTIONS] Page 20