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© 2015 Technological Studies Higher Finalised Marking Instructions Scottish Qualifications Authority 2015 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. Part One: General Marking Principles for: Technological Studies 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) Marks 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. (b) 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 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 Part Two: Marking Instructions for each Question Question 1 Mark Allocation (a) (b) S 0 0 0 0 1 1 1 1 W 0 0 1 1 0 0 1 1 A 0 1 0 1 0 1 0 1 B = S.W.A + S.W.A + S.W.A B 0 0 0 0 1 0 1 1 Marks 3 marks for all B output 2 marks 7 or 6 B correct 1 mark 5 or 4 B correct 0 marks 3 or less correct 3 (1 mark) for each correct combination 4 OR B = S(W+A) (c) S W A B NOT AND OR OR S B W A Page 3 1 1 1 3 Question 1 (d) Mark Allocation Marks SW A B NOT equivalent AND equivalent OR equivalent Cancellation 1 1 1 1 4 (14) 2 (a) FH = 0 F × cos70 = 8·08 × cos80 F = 4·10 kN (b) 2 1 3 2 terms @1 answers (units not required) 2 1 3 2 moments @ 2 each answers including units 4 1 5 FV = 0 W = 8·08 × cos10 + 4·10 × cos20 W = 11·8 kN (c) 2 components answer including units MA = 0 8·08cos10 × 5 = 11·8 × L L = 3·37 m (11) Page 4 Question 3 Mark Allocation (a) Find b2 Number of degrees = 360 × 1·3 = 468 No of pulses = 468/1·8 = 260 Marks OR Find b2 360/1·8 = 200 steps/rev 200 × 1·3 = 260 No of loops = 260/4 b2 = 65 3 (b) Find b8 1 Rev requires 360/1·8 = 200 steps OR Find b8 6 rev/min = 1 rev in 10s no of steps per minute = 200 × 6 = 1200 10 time for 1 step = 60000/1200 b8 = 50ms b8 = 50ms extrudrev: for b0 = 1 to 25 pins = %01100000 pause 14 pins = %01010000 pause 14 pins = %10010000 pause 14 pins = %10100000 pause 14 next b0 return 200 0 05 s / steps 2 for calculating 25 answer 1 substitutions answer 1 1 substitutions answer 1 1 answer 1 answer 1 substitutions answer 1 1 1 for command calculate 25 – 180/1·8 = 100 100/4 = 25 3 pause 14 reversing step order all 4 pauses 1 1 1 label + return 1 9 7 (16) Page 5 Question 4 (a) Mark Allocation Closed loop (1) proportional (1) (b) 2 Tachogenerator set speed (c) Marks − + error error detector amplifier fuel regulator driver generator electrical output set speed (or desired speed) error detector & error amplifier (labels no essential) driver (transistor) fuel regulator generator & output (change of speed) tachogenerator 1 1 1 1 1 1 answer 1 Error = 2·8/10·9 = 0·257 substitution answer 1 1 V at non-inverting input = 6·8 + 0·257 = 7·06V substitution answer 1 1 substitution answer including units 1 1 Gain = 196/18 = 10·9 R = 20 × 8·94/7·06 = 25·3 k 6 7 (15) Page 6 Question 5 (a) Mark Allocation σ = F/A = 40/40 = 1 kN/mm2 ɛ = I/I = 0·51/100 = 0·0051 Marks substitution of value within elastic region answer 1 1 substitution of corresponding extension value answer 1 1 substitution answer including correct units 1 1 E = σ/ɛ = 1/0·0051 = 196 kN/mm2 (b) Mild steel/stainless steel/nickel alloy (c) UTS = 430N/mm2 1 (from data book) 1 WS = 430/8 = 53·8N/mm2 substitution answer 1 1 A = F/σ = 6000/53·8 = 112mm2 substitution answer 1 1 substitution answer including correct units 1 d (4A / ) (4 112 / ) = 12mm 6 1 7 (14) Page 7 Question 6 Mark Allocation (a) Protects transistor / from back e.m.f. (b) R = V/I = (5·8 – 0·7)/(4·4 × 10−3) = 1·16 k (c) Marks 2 Ic = P/V = 20/6 = 3·33 A hFE = Ic/Ib = 3·33/(4·4 × 10−3) = 757 Use of – 0·7 substitution answer including unit 1 1 1 substitution answer, unit not required 1 1 substitution answer, no unit 1 1 4 1 1 1 3 1 1 1 1 1 5 3 (d) two npn transistors base connections collector connections (e) Ib = 3·33/1020 = 3·26 × 10−3 substitution answer use of 1·4 substitution answer including units R = V/1 = (5·8 – 1·4)/3·26 × 10−3 = 1·35 k (f) Draws negligible (no) current from input OR high input (gate) resistance (impedance). No base resistor required. Can provide high output current. High switching speed. Low power consumption. any 2 @ 1 Voltage controlled 2 (19) Page 8 Question 7 (a) Mark Allocation main: heater: onfull: 7 (b) if pin0 = 1 then heater low 5 low 4 goto main high 4 if pin1 = 1 then onfull gosub adcread DATA = DATA/10 high 5 pause DATA low 5 pause 12 goto main high 5 goto main Marks 2 1 (mark awarded below) high 4 low 4 1 1 2 1 1 (mark awarded below) high 5 low 5 including label Pulse Width Modulation (PWM) 1 1 1 1 1 1 15 1 (16) Page 9 Question 8 Mark Allocation Op-amp A – Summing Amplifier Op-amp B – Inverting Amplifier (a) 1 1 V V V 1 R 1 3 R R 1 3 f V1 = −8 (5/100 + 5/50) = −1·2 V Vout = −1 × −1·2 = 1·2 V (b) 3·6 = −8(5/200 + 5/50 + 5/R) × −1 3·6 = 0·2 + 0·8 + (40/R) 3·6 – 1 = 40/R (c) R = 40/2·6 R = 15·4 k (d) Marks (i) Max hot drink output – tea/milk/sugar (ii) Vout = 1·2 + 2·6 = 3·8V Supply = 3·8 × 100/85 = 4·47V all substitutions 3 answer including units 1 3 substitutions 3 answer including units 1 2 4 4 1 substitutions answer 1 1 use of “saturation” answer including units 1 1 4 (15) Page 10 Question 9 (a) (b) Mark Allocation Voltage at non-inverting input of op-amp 2 increases. Non-inverting input becomes greater than inverting input. Op-amp output goes high and transistor switches on. Relay switches boiler on. T1 = 200k T2 = 800k R/800 = 200/18 R = 200 × 800/18 = 8·89 M (c) (d) 1 1 1 1 1 1 ratios 1 answer 1 4 1 1 1 1 1 5 Voltage at inverting input = 24 × 18/64 = 6·75 substitution answer 1 1 Voltage at non-inverting input = 24 × 18/48 = 9V substitution answer 1 1 answer 1 substitution answer 1 1 answer including units 1 1 Av = 2·3/2·25 = 1·02 Rf = 1·02 × 320 Rf = 326 k Mh = 0 3 terms @ 2 each (Acos10 × 600) + (90cos15 × 100) – (240sin30 × 920) = 0 A × 591 = 110400 − 8693 A = 172 N answer, including unit Page 11 4 from data book from data book Role of V1 Resistance of Th1 decreases / V3 increases When V+ve > V−ve, op-amp switches on Output greater than VGS Pump on. Error = 9 − 6·75 = 2·25V (e) Marks 9 6 1 7 Question 9 (f) Mark Allocation Fv = 0 (↑+ve) +Hv – 90cos15 – 172cos10 + 240sin30 = 0 HV = 86·9 + 169·3 – 120 HV = 136 N (↑) FH = 0 (→+ve) +HH – 90sin15 + 172sin10 – 240cos30 = 0 HH = 23·3 – 29·9 + 207·8 HH = 201 N (→) H (1362 2012 ) = 243 N tan 136 Marks three components @ 1 each 3 answer (units not necessary) 1 three components @ 1 each 3 answer (units not necessary) 1 substitution 1 answer including unit 1 answer 1 201 θ = 34·1° 11 (40) Page 12 Question 10 (a) Mark Allocation For AC: FV = 0 FAC sin27° = 1·18 FAC = 2·6 kN (TIE) For AB: FH = 0 FAB = 2·6cos27 FAB = 2·32 kN (STRUT) For CB: FV = 0 FBC = 1·18 kN (TIE) FCE FCD 27° Strain Sensor 1 1 1 substitution answer with units correct nature 1 1 1 answer with units correct nature 1 1 3 terms answer with units correct nature 3 1 1 2 terms answer with units correct nature 2 1 1 17 1 1 1 1 1 1 6 27° 1·18 kN 2·6 kN For CE: FH = 0 FCE = 2·32 + 5·2cos27 FCE = 6·95 kN (TIE) Wind Sensor substitution answer with units correct nature Node C For CD: FV = 0 FCDcos63 = 1·18 + 1·18 FCD = 5·2 kN (STRUT) (b) Marks Signal Conditioning Signal Conditioning Multi plexer A to D Micro controller Data logger Two sensors Two signal conditioners Multiplexer ADC Microcontroller Datalogger Page 13 Question 10 Mark Allocation (c) Marks sample rate MPX low windspeed Is DATA <70? Y N b3 = 600 Y Is DATA <120? N b3 = 300 b3 = 60 MPX high readstrain sample rate & return MPX low/MPX high all other elements with associated links 9 @ 1 pause b3 1 1 9 looped 10 times? return 11 (d) Non-inverting amp Vin = 2V from graph 1 1 Vout = 5 × 186/255 = 3·65V answer 1 Gain = 3·65/2 = 1·83 For non-inverting amplifier, gain = 1 + Rf/Ri = 1·83 Rf/Ri = 0·83 Rf = 83 k, Ri = 100 k (or other suitable pair in k range) answer 1 answer 1 1 6 (40) Page 14 Question 11 (a) Mark Allocation steer: loop1: loop2: subtrim: addtrim: adddata: subdata: (b) (c) Marks if pin1 = 1 then subtrim if pin2 = 1 then addtrim low 7 gosub adcread if TRIM > 128 then adddata if TRIM < 128 then suddata gosub steerangle return 2 2 1 1 2 2 1 TRIM = TRIM − 1 high 7 pause 100 goto steer (mark allocated below) (mark allocated below) TRIM = TRIM + 1 high 7 pause 100 goto loop1 (both high 7) (both pause 100) 1 1 RESULT = TRIM – 128 DATA = DATA + RESULT goto loop2 1 1 (mark allocated below) RESULT = 128 – TRIM DATA = DATA – RESULT goto loop2 (both goto loop2) Sub A is summing amp; produces 2 voltages for fwd & back Sub B inverts signal for A to +ve Sub C produces a ref voltage for Sub D Sub D: when +ve > −ve input, T1 on; car forward when –ve > +ve input, T2 on; car backward Vref = 6 × 8/28 = 1·71V substitutions answer with unit Page 15 1 1 1 1 1 1 1 22 1 1 1 2 1 6 1 1 2 Question 11 (d) (i) Mark Allocation Diff amp: Summing amp: (d) (ii) Diff amp V0 = + 6·7V 300 6·7 = (1·71 – V1) 20 V1 = 1·26V 6 8 1 26 10 R1 6 8 0·126 = R1 R1 = 54 k 300 (1·71 – V1) 20 V1 = 2·16V 6 8 6 8 Summing amp: 2 16 10 R2 54 6 8 0·216 = 0·126 + R2 R2 = 75·6 k − 6·7 = Marks substitutions 1 answer 1 substitutions 1 1 answer 1 substitutions 1 answer 1 substitutions 1 5 1 1 5 (40) [END OF MARKING INSTRUCTIONS] Page 16