Student
on
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RMS Values
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RMS Values
How can the root mean square of a
sinusoidally varying value be calculated
and how is it derived?
Contents
Initial Problem Statement 2
Narrative 3-19
Notes 20-21
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Appendices 22-24
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Student
on
page: 2 of 24
so an average value is required that gives an
a rate which is proportional to the power they
average heating effect. A commonly used
expend. This causes heating of components
average is the root mean square value and it
which may reduce their reliability or have
gives the effective average voltage to use when
consequences for the safe operation of the
calculating power dissipation.
lv
Electrically powered devices dissipate energy at
device; for example, you may burn yourself on
a handle that is in thermal contact with a hot
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ia
component.
The power used by a device is proportional to
the square of the current through a conductor
and the square of the voltage across the
conductor. This is unambiguous for DC but for
AC these quantities are continuously varying
How can the root mean square of a
sinusoidally varying value be calculated
and how is it derived?
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RMS Values
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Initial Problem Statement
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RMS Values
Student
Narrative
Introduction
The heating caused by a current is proportional to the power dissipated in the material that is
conducting it. For a purely resistive circuit one expression for the power, P, is given by
P=
V2
R
on
where V is the voltage drop across the circuit and R is the resistance in the circuit. For a steady
voltage (direct current or DC), the above expression is easy to evaluate. For an alternating current
(AC) the voltage changes continuously with time as shown below for a domestic supply. Note the
curve continues to the right for as long as the supply is connected.
Voltage, V (volts)
400
0
-100
-200
-300
-400
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0.055
0.06
Time (s)
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0
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100
RMS Values
200
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300
Figure 1.
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Discussion
How could you work out the heating effect of a voltage that varies as shown in
Figure 1? Does it matter that the value is sometimes positive and sometimes
negative?
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The variation of voltage with time for a domestic AC supply is shown in Figure 1, repeated below
Student
2. Heating due to a domestic AC supply
Voltage, V (volts)
400
300
200
100
0.01
0.015
0.02
0.025
0.03
0.035
-100
-300
-400
The equation for this curve is
0.05
0.055
0.06
Time (s)
er
Figure 1.
0.045
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-200
0.04
page: 4 of 24
0.005
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0
on
0
V = 325sin ωt
where ω is the angular frequency of the curve. This is given by
2π
T
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ω=
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where T is the period of the curve.
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Student
As heating is proportional to the square of the voltage through
V2
P=
R
you need to consider the graph of voltage-squared, as shown below.
Voltage-squared
120000
100000
on
80000
60000
Time (s)
0
0.005
0.01
0.015
0.02
0.025
Activity 1
0.035
0.04
0.045
0.05
er
Figure 2.
0.03
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0
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20000
What is the period of the domestic AC voltage shown in Figure 1 and how does
this compare with the period of the voltage-squared shown in Figure 2? What is the
angular frequency of the curve?
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Activity 2
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Write down an expression for the value of voltage squared as a function of time.
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RMS Values
40000
P=
dE
dt
Student
Average power is defined as the total work done (energy delivered) divided by the time taken. At any
instant the power is given by
If the power is constant the average power is equal to the instantaneous power. However, in the
case of alternating current the power is not constant; it varies through each cycle of the electrical
current.
on
To find the average power you have to find the energy delivered during one complete cycle and
divide it by the time the cycle takes. The total energy delivered is given by the area under a power
versus time graph. As P ∝ V 2 then it follows that the energy delivered must also be proportional to
the area under a voltage-squared versus time graph. This means that to establish a single average
value that gives an equivalent heating you need to find a value of the voltage squared such that the
areas under the two graphs are the same. To find this will require an average value for each cycle of
the voltage squared which may be scaled to provide an average over all values of time.
Voltage-squared
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See box-out “Power,
energy and time” on
page 20
Voltage-squared
120000
120000
100000
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100000
80000
60000
Time (s)
60000
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80000
Time (s)
average
40000
40000
20000
0
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20000
0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05
Figure 3.
What is the value of average that gives the same area under the graph?
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The average value will be called the mean sqaure value and denoted MS.
Discussion
Do you need to consider
the AC graph for all time
values?
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Discussion
How could you calculate the
area under the AC graph and
how would you use this to
find an average value that
gives an equivalent area?
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2
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Discussion
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Do you expect the estimated area to be greater or smaller than the true area?
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RMS Values
Figure 4.
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Estimate the area under the graph of V = 105 625 sin ωt between t = 0 seconds
and t = 0.01 seconds by constructing rectangles of width 0.001 seconds and that
are always just below the curve, as shown below. Recall that ω = 100π.
Student
Activity 3 (Group 1)
2
2
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Do you expect the estimated area to be greater or smaller than the true area?
RMS Values
Discussion
Figure 5.
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Estimate the area under the graph of V = 105 625 sin ωt between t = 0 seconds
and t = 0.01 seconds by constructing rectangles of width 0.001 seconds and that
are always just above the curve, as shown below. Recall that ω = 100π.
Student
Activity 3 (Group 2)
You now need to work out an average voltage-squared value that gives the
equivalent area. Recall that this average value will be called the mean square value
and denoted MS. Use you result from the above area estimate to calculate MS.
Discussion
The area estimates made using rectangles vary considerable. How could you
improve the accuracy of the estimate?
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Student
Activity 4
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the curve
Student
3. Improving the estimate of the area under
Figure 8 rectangle width 0.0005
2
2
er
Multimedia
si
Estimate the area under the graph of V = 105 625 sin ωt between t = 0 seconds
and t = 0.01 seconds by constructing rectangles of width 0.0005 seconds and that
are always just below the curve. Repeat the activity for rectangles of width 0.00025
seconds. Recall that ω = 100π.
The resource RMS Values Interactive is available to automate these calculations.
lv
Optionally you can perform the calculations by hand or by using a spreadsheet. Remember that you
only have to work out half the values – the whole area estimate can be found using symmetry.
Rectangle
t lower
t upper
V
V2
Area
1
0.0000
0.0005
0.00
0.00
0.00
2
0.0005
0.0010
50.84
2584.83
1.29
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Rectangle width = 0.0005
3
0.0010
0.0015
4
0.0015
0.0020
5
0.0020
0.0025
6
0.0025
0.0030
7
0.0030
0.0035
8
0.0035
0.0040
9
0.0040
0.0045
10
0.0045
0.0050
Total for half graph
Total for whole graph
You could use a
spreadsheet to fill
in these values.
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page: 10 of 24
Activity 5 (Group 1)
RMS Values
Figure 7 rectangle width 0.001
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In the previous activity you estimated the area under the curve using rectangles and found different
values depending on whether you used rectangles that were just above the curve or just below
the curve. After discussion it was decided that the estimate for both could be improved by using
narrower rectangles, shown below for the case where the rectangles are just below the curve.
t lower
t upper
V
V2
Area
1
0.00000
0.00025
0.00
0.00
0.00
2
0.00025
0.00050
25.50
650.21
0.16
3
0.00050
0.00075
4
0.00075
0.00100
5
0.00100
0.00125
6
0.00125
0.00150
7
0.00150
0.00175
8
0.00175
0.00200
9
0.00200
0.00225
10
0.00225
0.00250
11
0.00250
0.00275
12
0.00275
0.00300
13
0.00300
0.00325
14
0.00325
0.00350
15
0.00350
0.00375
16
0.00375
0.00400
17
0.00400
0.00425
18
0.00425
0.00450
19
0.00450
0.00475
20
0.00475
0.00500
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Rectangle
RMS Values
Total for half graph
Total for whole graph
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You could use a
spreadsheet to fill
in these values.
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Student
Rectangle width = 0.00025
Supported by
2
2
Estimate the area under the graph of V = 105 625 sin ωt between t = 0 seconds
and t = 0.01 seconds by constructing rectangles of width 0.0005 seconds and that
are always just above the curve. Repeat the activity for rectangles of width 0.00025
seconds. Recall that ω = 100π.
Student
Activity 5 (Group 2)
Multimedia
The resource RMS Values Interactive is available to automate these calculations.
Optionally you can perform the calculations by hand or by using a spreadsheet. Remember that you
only have to work out half the values – the whole area estimate can be found using symmetry.
t lower
t upper
V
V2
Area
1
0.0000
0.0005
50.84
2584.83
1.29
2
3
0.0005
0.0010
100.43
10086.29
5.04
0.0010
0.0015
4
0.0015
0.0020
5
0.0020
0.0025
6
0.0025
0.0030
7
0.0030
0.0035
8
0.0035
0.0040
9
0.0040
0.0045
10
0.0045
0.0050
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Rectangle
on
Rectangle width = 0.0005
Total for half graph
Total for whole graph
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You could use a
spreadsheet to fill
in these values.
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t lower
t upper
V
V2
Area
1
0.00000
0.00025
25.50
650.21
0.16
2
0.00025
0.00050
50.84
2584.83
0.65
3
0.00050
0.00075
4
0.00075
0.00100
5
0.00100
0.00125
6
0.00125
0.00150
7
0.00150
0.00175
8
0.00175
0.00200
9
0.00200
0.00225
10
0.00225
0.00250
11
0.00250
0.00275
12
0.00275
0.00300
13
0.00300
0.00325
14
0.00325
0.00350
15
0.00350
0.00375
16
0.00375
0.00400
17
0.00400
0.00425
18
0.00425
0.00450
19
0.00450
0.00475
20
0.00475
0.00500
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Rectangle
RMS Values
Total for half graph
Total for whole graph
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lv
You could use a
spreadsheet to fill
in these values.
MEI ©2011
Student
Rectangle width = 0.00025
Supported by
Student
Activity 6
Collate your results to fill in the following table
Rectangle
width
0.001
0.0005
0.00025
Group 1 Area
(Rectangles
below curve)
Group 2 Area
(Rectangles
above curve)
Plot your results on the following graph
on
Estimated area
650
625
600
575
500
450
425
400
0
0.0001
0.0002
0.0003
er
475
0.0004
0.0005
0.0006
0.0007
0.0008
0.
0009
0.001
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525
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550
0.0011
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Rectangle width (s)
Figure 9.
Discussion
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What do you notice
about the two lines as the
rectangle width decreases?
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Discussion
Do you expect the estimate
to be closer to the actual area
as the width of the rectangles
reduces? Why?
Supported by
Student
4. Finding the limit
Recall the result from the previous activity
Rectangle
width
0.001
0.0005
0.00025
Group 1 Area
(Rectangles
below curve)
422.50
475.31
501.72
Group 2 Area
(Rectangles
above curve)
633.75
580.94
554.53
Estimated area
650.00
on
625.00
600.00
Group 2
575.00
550.00
500.00
475.00
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525.00
400.00
0
0.0001
0.0002
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425.00
0.0003
0.0004
0.0005
0.0006
0.0007
0.0008
0.
0009
0.001
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Group 1
450.00
0.0011
Rectangle width (s)
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Figure 10.
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It is observed that a halving of the rectangle width (which doubles the number of rectangles used)
produces a new estimate for the area which falls on a straight line. As the width of the rectangles
reduces the expected error decreases as the rectangles more closely fit the actual curve as shown
below.
Figure 6 rectangle width 0.001
Figure 7 rectangle width 0.0005
The lines for the two different approaches (over-estimating and under-estimating) appear to converge
to a single value when the rectangle width is zero.
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Find the equations for the two lines and determine the y-intercept value. Give values
to 1 d.p. as the estimate areas given do not retain all the calculated decimal places.
Student
Activity 7
Activity 8
Use your result to find the average value, MS, that gives an equivalent area over the
time interval 0.01 seconds.
Discussion
Discussion
Discuss what happens to the
area of one of the rectangles
used to estimate the area
as its width is reduced. How
does this affect the sum of all
the areas?
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Do the two lines converge
to the same point?
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In the previous activity you determined the area under the curve by summing area of a number of
rectangles and found that it was possible to extrapolate results to zero thickness and still produce a
finite, non-zero estimate of area.
Student
5. Using integration
If the width of an individual rectangle is a small value dt then the small area contribution of rectangle
i, dAi, for the function V 2 = 105 625 sin 2 ωt at time ti is given approximately by
2
dAi = dt ×105 625 sin ωti
The total area is given by summing all the small areas
A = ∑ dAi
You found that
A = ∑ dAi → true area
as dt → 0
i
on
i
This is true of many functions; V 2 = 105 625 sin 2 ωt is not a special case. For the general function
f(t) this limit is known as the integral of f(t) and is written as
0.01
A = lim ∑ f ( t ) × dt =
∫ f ( t ) dt
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i
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dt →0
0
The limits on the integral represent the start and end times over which the area is calculated.
A = lim ∑ f ( x ) × dx = ∫ f ( x ) dx
dx →0
.
i
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Of course this is valid for any variables so
You also found that the mean value required to give an equivalent area over the time, MS, is
calculated by
A
t
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MS × t = A ⇒ MS =
Activity 9
Using the definitions above write down an equation for the mean square, MS, in the
form
A
, where A is expressed as an integral.
t
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MS =
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Solve the equation and compare your mean square (MS ) value with the result from
the previous activity. (Recall MS = 52812.66 and 52812.33 for the rectangles under
the curve and rectangles over the curve results respectively. Note, these values are
derived based on rounded values of area estimates.)
Student
Activity 10
Discussion
The value of MS is intended to give a rectangle of equal area to the curve
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V 2 = 105 625 sin 2 ωt over the specified time. Compare MS with the peak value
2
2
of V = 105 625 sin ωt .
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Recall the introduction to this problem:
Student
6. Mean square and the root mean square
Electrically powered devices dissipate energy at a rate which is proportional to the power
they expend. This causes heating of components which may reduce their reliability or have
consequences for the safe operation of the device; for example, you may burn yourself on a
handle that is in thermal contact with a hot component.
on
The power used by a device is proportional to the square of the current through a conductor and
the square of the voltage across the conductor. This is unambiguous for DC but for AC these
quantities are continuously varying so an average value is required that gives an average heating
effect. A commonly used average is the root mean square value and it gives the effective average
voltage to use when calculating power dissipation.
You have considered a domestic AC supply where the voltage at a given time is given by the function
V = 325sin ωt
ω = 100π.
page: 19 of 24
T
and calculated that an average mean square value, MS, of the V 2 function to given an equal area
under the graph over a given time is given by
2
RMS Values
ω=
si
where ω is the angular frequency of the curve. This is given by
2π
where T is the period of the curve, 0.02 seconds for a domestic 50 Hz supply so that
You have also determined that heating effects are proportional to V 2, i.e.
2
2
2
peak V
105625
=
2
2
lv
MS =
er
P ∝ V ⇒ P ∝ ( 325 sin ωt ) = 105625 sin ωt
Discussion
The above expression gives the mean square value, i.e. the mean value of the
voltage squared. How can you use this to produce a mean value for the voltage
that gives an equivalent heating?
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Activity 11
Calculate the RMS value of the voltage in a domestic AC supply to the nearest
integer.
Discussion
Is the result familiar?
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Student
Notes
Power, energy and time
For a general curve of power against time the area under the curve between
any two time points gives the total energy between those two times.
Power
The above follows from the definition of power as energy per unit time,
variables gives
dE
dt
Integrating both sides
t2
t2
E
t1
o
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∫ Pdt = ∫ dE = E
∫ Pdt represents the area under the P curve between time t
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The expression
dE
. Separating the
dt
P=
⇒ Pdt = dE
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P=
time
t2
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Figure 11.
1
and t2.
t1
The above is analogous to the more familiar fact that the area under a velocity/time graph gives the
total distance travelled. Denoting the velocity as v and distance as x, the definition of velocity as the
rate of change of distance gives v =
dx
.
dt
Separating the variables and integrating gives
dx
v=
t2
The integral
dt
⇒ vdt = dx
t2
x
t1
0
∫ vdt = ∫ dx = x
∫ vdt represents the area under the velocity/time graph between time points t
1
and t2.
t1
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RMS Values
t1
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on
Area = energy
Student
Separating the variables and integrating gives
v=
dx
⇒ vdt = dx
dt
t2
x
t1
0
∫ vdt = ∫ dx = x
t2
The integral
∫ vdt represents the area under the velocity/time graph between
t1
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time points t1 and t2.
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Student
Appendix 1
using the interactives
Multimedia
RMS Values Interactive is available to automate these calculations of areas under a graph
RMS Values
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by using rectangles of different sizes.
Figure 14.
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The selection choices at the left of the screen determine whether the rectangles are bound under the
curve, as shown above, or above the curve, as shown below.
Figure 15.
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Figure 16.
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The slider at the bottom reduces the width of the rectangles so that they more closely fit the curve.
Student
The box at the bottom right hand side of the screen gives the estimated area based on the sum of
the areas of the rectangles.
Figure 17.
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Student
Appendix 2
mathematical coverage
ASL Outcomes:
Understand the use of trigonometry to model situations involving oscillations
4.
Undertsand the mathematical structure of a range of functions and be familiar with their graphs.
6.
Know how to use differentiation and integration in the context of engineering analysis and problem solving.
7.
Understand the methods of linear algebra. Know how to use algebraic processes.
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3.
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