2007 Technological Studies
Higher
Finalised Marking Instructions
 Scottish Qualifications Authority 2007
The information in this publication may be reproduced to support SQA qualifications only on a
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Materials Team at Dalkeith 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.
Section A
1
Question
Mark Allocation
(a)
C = D + (P . L) OR
C = (D.P.L) +(D.P.L) +(D.P.L) +(D.P.L) +(D.P.L)
(b)
Marks
Brackets optional
1
D
C
P
L
NOT gates and correct input connections @ ½ each
AND gate and correct input connections
OR gate and correct input connections
(c)
1
½
½
2
D
C
P
L
NOT equivalents and correct input connections
AND equivalent and correct input connections
OR equivalent and correct input connections
Cancellation of redundant gates
Page 2
½
1
1
½
3
(6)
Question
2
(a)
(b)
(c)
(d)
•
•
•
•
•
•
•
•
•
Mark Allocation
Base current too low
Collector current too low
Thermistor resistance too high
Current gain (hFE) too low
Fixed resistor R too low
Lamp current required too high
Lamp resistance too low
Transistor unsuitable
Base voltage < 0.7 V
V = IR = 140 mA × 20 Ω
= 2.8 V
Marks
(any one answer)
1
calculation
answer including units
½
½
1
answer (units not necessary)
answer including units
½
½
1
Thermistor resistance @ -10 0C = 3.2 MΩ (accept 3.2 - 3.8 MΩ) (from Data Booklet)
R/3.2 = 7/5
formula, stated or implicit
R = 3.2 × 7/5 
calculation
= 4.48 MΩ
answer including units
½
½
½
½
2
(5)
Lamp current @ 12 V: I = V/R = 12/20
= 0.6 A (600 mA)
Minimum gate voltage = 7 V
Page 3
Question
3
(a)
Mark Allocation
(i) Gain = 2.55/0.34 = 7.5
(no units)
(ii) For non-inverting amplifier, gain = 1 + Rf /Ri = 7.5
Rf /Ri = 6.5
calculation (must be shown)
Rf = 650 kΩ, Ri = 100 kΩ (or other suitable pair, higher value in the kΩ range)
650 kΩ
Correct non inverting amplifier circuit
with resistor values in correct places
(Vin and Vout do not have to be shown)
_
+
100 kΩ
Vin
(b)
Vout
(i) Gain = 2.55/4.25 = 0.6
(no units)
(ii) Use two inverting amplifiers, one with a gain of -0.6, and one with a gain of -1
(or other suitable pair.)
-0.6 × -1 = 0.6
(calculation must be shown)
For first amplifier, Rf = 60 kΩ, Ri = 100 kΩ (or other suitable pair - higher in the kΩ range)
For second amplifier, Rf = 100 kΩ, Ri = 100 kΩ (or other suitable pair- higher in kΩ range)
60 kΩ
(c)
100 kΩ
_
Vin
+
Water
Temperature
Oil
Pressure
(optional)
100 kΩ
100 kΩ
_
+
Temperature
Sensor
Signal
Conditioning
Signal
Pressure
Conditioning
Sensor
½
½
(both sensors) (both signal
conditioners)
Vout
Multiplexer
Two correct inverting
amplifier circuits with
resistor values in correct
places @ ½ each
(Vin and Vout do not have to
be shown)
A to D
Microcontroller
½
½
½
(signals from microcontroller
to multiplexer &ADC not
penalised if included)
Marks
½
½
½
½
2
½
½
½
½
1
3
non volatile
memory
½
3
(8)
Page 4
Question
4
(a)
•
•
•
Mark Allocation
Additional loading due to weather – wind/snow etc
Possible harm to people
Cost of failure
Marks
(any one answer)
(b)
(i) σ = E × ε
= 110 × 103 × 0.0009
= 99.0 N/mm2
(formula, stated or implicit)
½
answer (including units) ½
(ii) UTS = 1000 N/mm2
FOS = Ultimate Tensile Stress/Safe Working Stress
= 1000/99
= 10.1
(c)
1
A = π d2 / 4 = 100π / 4 = 78.5 mm2
F=σ×A
= 99.0 × 78.5
= 7.77 kN
(from Data Booklet)
formula (stated or implicit)
calculation
answer (with no units)
calculation (units not necessary)
formula, stated or implicit
calculation
answer including units
Page 5
½
½
½
½
½
½
½
½
3
2
(6)
Question
5
(a)
(b)
Mark Allocation
D to A converter
To convert to analogue
To give output voltage proportional to values of binary input
(Summing and inverting amplifiers - ½)
(To add voltages from microcontroller - ½)
(i) V1 = -Rf (V1/R1 + V2/R2 + V3/R3....)
= -80(5/800 + 5/400)
= -80 × 5 × 3/800
= -1.5 V
(ii) Vout = -(Rf/Ri) × Vin
Vout = -(400/300) × V1
= -(4/3) × -1.5 V
=2V
(c)
Marks
(any one answer)
formula, stated or implicit
correct substitution
calculation
answer, negative, including units
1
½
½
½
½
formula, stated or implicit, and substitution
½
answer including units
½
By proportion: maximum output voltage = 2 × 15/3
formula, stated or implicit
= 10 V calculation ½ answer including units ½
Alternative method:
Vout = -Rf (V1/R1 + V1/R1 + V1/R1....) × - Rf/Ri
formula, stated or implicit
= -80 (5/800 + 5/400 + 5/200 + 5/100) × -4/3
correct substitution
= -80 × 75/800 × -4/3
calculation
= 10 V
answer including units
1
1
½
½
½
½
2
1
2
(d)
Vout
continuous final level - ½
12 steps - ½
offset - ½
axes including labels - ½
Time
(i)
2
(ii) For the 'soft start' of the motor; for starting the motor gently; for accelerating the
motor to full speed gradually.
any one answer
1
(iii) This is the lowest value that will cause the motor to turn; any value lower than
this and the motor will not start to turn.
either answer
Page 6
1
(10)
Question
6
(a)
Mark Allocation
displaytemp: for b0 = 1 to 5
½ (or any other variable from b1 to b11)
serout 7, T2400,(254,1)
½
gosub adcread
½
serout 7, T2400,(254,128)
½
OR combine lines- see * below
serout 7, T2400,(“TEMP”) ½
serout 7, T2400,(254,192)
½
OR combine lines- see ** below
serout 7, T2400,(#data)
½
pause 1000
½
next b0
½
return
½ (return command & 'displaytemp' label)
Marks
}
}
* serout 7, T2400,(254, 128, "TEMP") (1)
** serout 7, T2400,(254, 192, #data) (1)
Page 7
(5)
Question
7
Mark Allocation
Marks
1
(a)
Comparator
(b)
(i) Resistance of thermistor at 125 0C = 10 kΩ (accept 10.1 kΩ)
½
R1/Rth = 5/1
R1 = Rth × 5/1
= 10 kΩ × 5
= 50 kΩ
formula (stated or implicit)
½
calculation
answer including units
½
½
Alternative method:
Highest reference voltage = 9 × 5/6 = 7.5 V
R1 = 10 × 7.5/1.5
= 50 kΩ
2
formula (stated or implicit) ½
(ii) Rth/R1 = 5/1
Rth = 5/1 × R1
= 5 × 50 kΩ
= 250 kΩ
(c)
½
formula (stated or implicit) ½
answer including units ½
answer including units
Vout = 9 × 85/100
= 7.65V
calculation (units not necessary
R3 = V/I
formula, stated or implicit
-3
calculation
= (7.65 – 1.5) / 30 × 10 (deduct ½ if 1.5 V not subtracted)
= 205Ω
(255 Ω if 1.5 V not subtracted)
answer including units
Page 8
½
1
½
½
½
½
2
(6)
Question
8
(a)
(b)
Mark Allocation
ΣFv = 0
Weight of log = 10.5 Cos 300 + 8.2 Cos 400
= 15.4 kN
ΣFv = 0
FLV = 10.5Cos30 + 8.17Cos 40 – 4Cos75
= 14.3 kN
Marks
(½ if no other mark allocated below)
1
two terms @ ½ each
1
calculation ½ answer including unit ½
(½ if no other mark allocated below)
3 components @ ½ each 1½
answer (units not necessary) ½
ΣFH = 0
(½ if no other mark allocated below)
FLH = 10.5Sin30 + 4Sin75 – 8.17Sin40
3 components @ ½ each
= 3.86 kN
answer (units not necessary)
OR
ΣFH = 0
½ if no other mark allocated below
FLH = 4Sin75
The other two forces (10.5 kN and 8.17 kN) when added give a vertical resultant
(with no horizontal component) to balance weight of log
stated or implicit
FLH = 3.86 kN
answer (units not necessary)
FL = √ (14.32 + 3.862)
= 14.8 kN
θ = Tan-1 (3.86/14.8)
= 15.10 (74.90) only if marked correctly on a diagram)
Page 9
2
formula and calculation
answer including units
formula and calculation
answer including units
1½
½
1
½
½
½
½
½
½
6
(8)
9
(a)
(b)
(c)
(d)
(e)
A – voltage follower
B – difference amplifier
½
½
Greater change in voltage for a given change in strain (signal amplification)
Temperature compensation.
either answer
1
1
RG2 is on top.
answer
The top surface is under tension, and the strain gauge on this surface will experience
a rise in resistance
reason
½
½
1
V2 = 9 x 120.15/240
= 4.505625 V
½
½
1
formula, stated or implicit
answer to 6 decimal places, including units
Vout = Rf/Ri (V2 - V1)
formula, stated or implicit
Rf/Ri = Vout / (V2 - V1)
= 6/(4.505625 - 4.5)
substitution of values
= 1067 (accept 1070)
answer
Rf = 107 kΩ, Ri = 0.1 kΩ or other suitable pair, higher value in kΩ range
Page 10
½
½
½
½
2
(6)
Section B
Question
10
(a)
(b)
Mark Allocation
Marks
(i) If a request is received from the third floor, and the lift is not at the third floor.
If a request is received from the second floor, and the lift is at the first floor.
(ii) If the lift is not at the first or third floor (OR if the lift is at the second floor)
(no half marks)
label 'secondfloor' ½
secondfloor: if pin1 = 1 then up
1
if pin3 = 1 then down 1
goto fin (OR return)
½
high 6
½
if pin2 = 0 then up (or test) ½
goto stop (OR low 6) ½
(return)
down:
test2:
high 5††
if pin2 = 0 then down
(or test2)
fin:
label 'secondfloor' ½
if pin 1 = 0 then button 1
high 6†
½
OR
†
up:
test:
stop:
}
1
1
low 5
low 6**
return
}
(†accept pins = %01000000)
test: if pin2 = 1 then stop
½
goto test
½
button: if pin3 = 0 then fin
1½
½
high 5††
½
1 OR test2: if pin2 = 1 then stop * ½
goto test2
* ½
}
½ OR
½
fin:
pins = %00000000
return
½
½
7
††
(accept pins = %00100000)
*(OR goto test - 1)
** (may be omitted if used higher up in program)
(c)
(d)
-Potentiometer gives an adjustable reference voltage
-Reference voltage is applied to inverting input of op. amp.
-As light level falls, resistance of LDR rises.
-As light level falls, voltage across LDR rises.
-Voltage across LDR is applied to non-inverting input of op. amp.
-If light level is below threshold, op. amp. output is high
-If voltage at non-inverting input is greater than voltage at inverting input, op. amp.
output is high (or door opens).
any five answers @ ½ each 2½
-If switch is pressed, voltage across LDR rises (to Vcc)
-If switch is pressed, voltage at non-inverting input rises ( to Vcc)
-If switch is pressed, op-amp output is high (or door opens)
any one answer @ ½ ½
(i) Taking moments about point D:
ΣMc/w = ΣMac/w
(½ if no other marks awarded)
2.8R1 = 12 × 0.8 + 6 × 2
3 terms @½ each 1½
R1 = 7.71 kN
answer including units ½
Page 11
3
2
Question
10
(d)
Mark Allocation
(ii) Analysing Node A:
FAB
7.71 kN
500
Marks
Taking vertical components
ΣFup = ΣFdown
(½ if no other mark awarded)
7.71 = FAECos50
substitution
FAE= 7.71/Cos50
= 12.0 kN tension (tie) magnitude, including units
nature
FAE
½
½
½
Alternative method: closed vector diagram
7.71 kN
0
FAE/Sin90 = 7.71/sin40
formula, stated or implicit
FAE = 7.71/sin40
FAE = 12.0 kN tension (tie) magnitude, including units
nature
FAE
50
½
½
½
FAB
Taking horizontal components
ΣFleft = ΣFright
FAB= FAECos40
FAB = 12.0Cos40
FAB = 9.19 kN compression (strut)
(½ if no other mark awarded)
substitution
½
magnitude, including units
nature
½
½
Taking vertical components
ΣFup = ΣFdown
(½ if no other mark awarded)
FBE = 6 kN compression (strut)
magnitude, including units
nature
½
½
(Can be solved also by closed vector method)
Analysing node B
6 kN
FAB
FBC
FBE
Analysing node E
6 kN
12.0 kN
500
FCE
29.20
Taking vertical components
ΣFup = ΣFdown (½ if no other mark awarded)
12 Cos50 = 6 + FCECos 60.8 (deduct ½ for
each mistake)
FCE = 1.71/ Cos 60.8
FCE = 3.51 kN compression (strut)
magnitude, including units
nature
Page 12
1
½
½
6
(20)
Question
11
Mark Allocation
(a)
Marks
position sensor
desired
position
_
+
error
detector
driver
motor
rotational
motion
error
amplifier
desired position signal & correct connection to error detector
actual position signal (from position sensor) & correct connection to error detector
('+' and '-' can be swapped on error detector)
error detector & error amplifier
driver
motor and output
position sensor and correct feedback loop
(b)
(c)
(d)
(e)
Vout = Rf/Ri (V2 - V1)
0.85 × 12 = 500/10 (V2 - V1)
V2 - V1 = (0.85 × 12) / (500/10)
V2 - V1 = 0.204 V
½
½
½
½
½
½
3
formula (stated or implicit) & correct substitution
½
answer including units
½
1
(i) Change in V2 = 12 × 2.7/360 = 0.09 V
V2 =1.72 V + 0.09 V
V2 = 1.81 V
calculation
calculation
answer including units
1
½
½
2
(ii) V3 = Rf/Ri (V2 - V1)
V3 = 500/10 × 0.09
V3 = 4.5 V
calculation
answer including units
½
½
1
(iii) V4 = V3 - 0.7 V
V4 = 4.5 - 0.7 V
V4= 3.8 V
calculation
answer including units
½
½
1
As the dish rotates towards the desired position:• The difference between actual and desired position signals decreases (OR the
voltage difference between inputs to difference amplifier decreases).
• The output voltage from the difference amplifier decreases.
• The voltage to the motor decreases.
• The motor slows down.
• The satellite dish stops at (or near) the desired position.
any four answers @ ½ each
2
(i) Op-amp based control system: closed loop
Microcontroller based control system: open loop
open loop, closed loop
correct way around
Page 13
½
½
1
Question
11
(e)
(f)
Mark Allocation
Marks
(ii) -Infinite number of possible positions for the dish.
-Uses closed loop control and so does not rely on previous position being correct.
-Microcontroller system is open loop - if stepper motor rotated out of position
(say by wind) every position thereafter will be incorrect.
any one answer
(Less 'jerky' than stepper motor - ½)
Sets the actual position value to the desired position value.
(because) actual position will be the same as desired position after the subroutine
rotate has been executed.
(g)
(i) 115-37 = 78
(ii) Logic 1 (high)
(h)
rotate: for b3 = 1 to b2
for b4 = 1 to 4
high 5
pause 5
low 5
pause 5
next b4
next b3
return
1
½
½
1
1
1
('b3' can be any variable from b3 to b11)
('b4' can be any variable different from 'b3', from
b3 to b11)
(½ in total for pauses adding to 10 ms)
(completion of inner loop)
(completion of outer loop)
('return' command & 'rotate' label at start)
(inner loop and outer loop can be interchanged)
½
½
½
½
½
½
½
½
4
(Multiplying b2 by 4 to give a new variable such as b3 [b3 = b2*4] and then using
one loop [such as for b4 = 1 to b3] would not work as b3 cannot exceed 255. In this
case, award 1½ marks for the loops instead of 2 marks.)
(i)
100/4 = 25 degrees per second
= (25/360) × 60 rev/min
= 4.17 rev/min
calculation
answer and units
Page 14
½
½
1
(20)
Question
Mark Allocation
Marks
12
(a)
Input subsystem X:
When S1 is pressed, a low signal (logic 0) is sent to input A of the logic subsystem;
When S2 is pressed, a low signal (logic 0) is sent to input B of the logic subsystem.
OR
When S1 is released, a high signal (logic 1) is sent to input A of the logic subsystem;
When S2 is released, a high signal (logic 1) is sent to input B of the logic subsystem.
Input subsystem Y:
(i) The variable resistor adjusts the reference signal (threshold) at the inverting input
of the operational amplifier;
(ii) When the light level falls the resistance of the LDR rises;
(iii) When the light level falls the voltage across the LDR rises;
(iv) When the voltage at the non-inverting input of the operational amplifier rises
above the voltage on the inverting input, the output of the operational amplifier goes
high. OR When the voltage across the LDR rises above the reference voltage the
output of the operational amplifier goes high.
(v)When the voltage on the non-inverting input of the comparator falls below the
voltage on the inverting input the output of the comparator goes low. OR When the
voltage across the LDR falls below the reference voltage the output of the
operational amplifier goes low.
any four answers @ ½ each
Logic Subsystem:
When Inputs A and B are both low (at logic 0) and input C is high (at logic 1) the
output Z goes high (logic 1)
(no half marks)
OR
When S1 and S2 are both pressed and the light level is below the threshold the logic
system gives out a high signal (logic 1).
(no half marks)
(b)
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
½
½
2
1
1
4
Z
0
1
0
0
0
0
0
0
no half marks
Page 15
½
½
1
Question
12
Mark Allocation
(c)
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
D
1
1
0
0
0
0
0
0
Marks
E
0
0
1
1
1
1
1
1
F
1
0
1
0
1
0
1
0
Z
0
1
0
0
0
0
0
0
Column 'Z' is the same in both truth tables, therefore logic systems are equivalent.
Column D (no half marks)
Column E (no half marks)
Column F (no half marks)
Column Z (no half marks)
Statement regarding column 'Z'
(d)
(e)
I = V/R
= 5-0.7 /1 × 103
= 4.3 × 10-3 A (4.3 mA)
Ic = V/R = (16 - 0.2) / 14
= 1.13 A
1
½
½
½
½
3
calculation (deduct ½ for 0.7 V not subtracted) ½
(5 × 10-3 A if 0.7 V not subtracted)
answer and units ½
1
calculation (deduct ½ for 0.2 V not subtracted) ½
answer and (units not required) ½
hFE = Ic/Ib
= 1.13 / 4.3 × 10-3
= 262
formula and calculation ½
answer with no units ½
2
(f)
BC108
TIP31A
0V
(g)
(i) ΣMH = 0
(450 × Sin70 × F) = (70 × 150) + (60 × 350)
F = 31500 / 422.86
= 74.5 kN
Page 16
1
Darlington arrangement
(0V rail does to have to be shown)
correct transistor types
correct positions for transistor types
½
½
( ½ if no marks awarded below)
3 moments @ ½ each
1½
answer including units
½
2
2
Question
12
(g)
Mark Allocation
Marks
(ii) Treat beam as Horizontal
Σ Fv = 0 (components perpendicular to beam)
Rh(V) + F Sin70 = 70 + 60
equation (deduct ½ for each error)
Rh(V) = 130 – 74.5 Sin70
= 60 kN
answer (unit not necessary)
ΣFh = 0 (components parallel to beam)
Rh(h) = F Cos70
= 74.5Cos70
= 25.5 kN
1
½
equation and substitution
answer (unit not necessary)
1
½
formula
answer and units
½
½
θ = Tan-1 60/25.5
formula
= 67.0º (angle to beam)
answer and units
angle to horizontal = 67º + 30º = 97º (from opposite direction = 83º)
½
½
Rh = √ (602 + 25.52)
= 65.2 kN
60 kN
θ
25 kN
(There must be a statement or diagram making
clear the direction the angle is measured from either 670 from beam or 830 from horizontal,
otherwise deduct ½ mark.)
0
830 67
Alternative method: components perpendicular and horizontal to ground:
Σ Fv = 0
Rh(V) + 74.5Cos50 = 70Cos30 + 60Cos30
equation (deduct ½ for each error)
Rh(V) = 64.7 kN
answer (unit not necessary)
1
½
ΣFH = 0
Rh(H) = 74.5Sin50 – (70Sin30 + 60Sin30)
= 7.93 kN
equation (deduct ½ for each error)
answer (unit not necessary)
1
½
FH = √ (64.72 + 7.932)
= 65.2 kN
formula
answer and units
½
½
angle to horizontal = Tan-1 64.7/7.93
= 83º
formula
answer and units
½
½
(There must be a statement or diagram making clear the direction the angle is
measured from - either 670 from beam or 830 from horizontal, otherwise deduct ½
mark.)
[END OF MARKING INSTRUCTIONS]
Page 17
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