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whichimpliesthefollowingassessment:
δ
[]
y
~
Ś
O
+
x
2
+
x
2
B
Ś
(
|
k
1
+
x
2
+
x
2
N
|
)
1%
±
Fx
()
1%
with
Fx
()
=
3
x
x
+
+
2
2
Thefunction
Fxisincreasingfor
()
xE
[]
0,1
because:
dFx
dx
()
=
(
x
+
4
2
)
2
>
0
Therefore:
δ
[]
y
~
Ś
F
()
11%
|
±
32
12
+
+
1%1.67%
d
Solution#2:Thesameresultmaybeobtainedbydirectlycomputingthecoefficientsofdataerrorspropaga-
tion:
Tx
a
()
±
ady
yda
b
a
±
±
1
2
±
axb
(
ba
–
+
)
2
da
d
(
xb
ba
+
–
)
2
b
a
±
±
2
1
±
ba
–
–
a
a
b
±
±
1
2
±–
1
Tx
b
()
±
bdy
ydb
a
b
±
±
1
2
±
bxb
(
ba
–
+
)
2
db
d
(
xb
ba
+
–
)
2
b
a
±
±
1
2
±
ba
bx
–
+
xb
2
+
ab
–
b
a
±
±
1
2
±
x
2
+
x
2
andsubstitutingthemtotheerrorformula:
δ
[]
y
~
±
Tx
a
()
O
+
Tx
b
()
B
±–+
O
x
2
+
x
2
B
δ
[]
y
~
Ś
Tx
a
()
1%
+
Tx
b
()
1%
±
(
|
k
1
+
x
2
+
x
2
N
|
)
1%
±
(
|
k
3
x
x
+
+
2
2
N
|
)
1%
for
xE
[]
0,1
ThismethodhasbeenappliedinthefollowingMATLABscript:
%RELATIVEERROROFCOMPUTINGTHEDERIVATIVE
%OFTHEFUNCTION:f(x)=(x+a)/(x+b)FOR0<=x<=1,
%CAUSEDBYERRORSCORRUPTINGaANDb
dxmax=1;
symsxab
f=(x+a)/(x+b);
y=diff(f,x);
Ta=a*diff(y,a)/y;%coefficientoferrorpropagationfora
Tb=b*diff(y,b)/y;%coefficientoferrorpropagationforb
T=abs(Ta)+abs(Tb);%coefficientforerrorassessment
dymax=subs(T,[a,b],[1,2])*dxmax;%maximumerrorin%
fplot(dymax,[0,1]);
xlabel('x');ylabel('dymax[%]');grid
%maximumerrorcorruptingthedatain%
%f(x)
%formulatobeanalysed
Problem1.6:DeterminethefunctionT(x),characterisingthepropagationofthedataerrorthroughthefol-
lowingformula:
y
±|
(
k
1
1
+x
–
x
N
|
)
x
for
xE
(
0,1
)
Solution:TheproblemmaybeconsiderablysimplifiedbycomputingthenaturallogarithmoftheLHSand
RHSoftheanalysedformula:
ln
()
y
±
x
ln
(
|
k
1
1
+x
–
x
N
|
)
±f
x
L
ln1
(
–
x
)
–
ln1
(
+
x
)
1
J
Thenthedifferentiationofbothsidesofthetransformedfunctionyields:
dx
d
ln
()
y
±f
L
ln1
(
–
x
)
–
ln1
(
+
x
)
1+
J
x
f
|
L
–
1
–
1
x
–
1
+
1
x
1
|
J
17
