©
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.
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should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for
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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