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In-Depth Information
Example: Compute the quadratic periodic basis functions.
Since the degree = 2, we have
m
= 3. Also, let us choose,
n
= 3 in equation
(5.1); i.e., we consider four control points:
V
0
,
V
1
,
V
2
,and
V
3
. Hence, we need
to compute four basis functions, namely
B
0
,
3
,
B
1
,
3
,
B
2
,
3
,and
B
3
,
3
.
We can compute using equation (5.6)
B
0
,
3
B
1
,
3
B
2
,
3
B
3
,
3
B
0
,
2
B
1
,
2
B
2
,
2
B
3
,
2
B
4
,
2
B
0
,
1
B
1
,
1
B
2
,
1
B
3
,
1
B
4
,
1
B
5
,
1
.
The inverse functions can be written using equation (5.7):
B
0
,
3
B
1
,
3
B
2
,
3
B
3
,
3
B
4
,
3
B
5
,
3
B
0
,
2
B
1
,
2
B
2
,
2
B
3
,
2
B
4
,
2
B
0
,
1
B
1
,
1
B
2
,
1
B
3
,
1
.
Let us now find out the knot vector. From equation (5.2), it is clear that
B
5
,
1
needs the knots
t
5
and
t
6
. The total number of knots is, therefore,
n
+
m
+1 =
3 + 3 + 1 = 7. Hence, the knot vector is
t
= [0123456], i.e.,
t
0
=0,
···
t
n
+
m
=
t
3+3=
t
6
.
5.3.1 Computation of Uniform Periodic B-spline Basis
The blending functions for a uniform periodic B-splines are also periodic.
This means for all values of
n
and
m
, all the blending functions have the same
shape. Each successive blending function is a shifted version of the previous
function. Hence,
B
i,m
(
u
)=
B
i
+1
,m
(
u
+
u
)
(5.8)
=
B
i
+2
,m
(
u
+2
u
)
,
where
u
is the interval between adjacent knot values.
Example: Computation of a uniform quadratic B-spline basis functions.
Let us now compute the blending functions corresponding to a uniform
quadratic B-spline. For a quadratic B-spline, the order is 3. Hence we choose
n=3. This means we have four control points. We, therefore, have four basis
or blending functions, e.g.,
B
0
,
3
,
B
1
,
3
,
B
2
,
3
,and
B
3
,
3
. Each of these blending
functions is defined over
m
subintervals. The total number of knots is
n
+
m
+ 1, which is 7 in the present case [0
,
1
,
2
,
3
,
4
,
5
,
6
,
7]. The total number of
subintervals is, therefore, 6. The parameter
u
ranges from 0 to
n
+
m
or 6.
Now,
B
0
,
3
(
u
)=
u
−
t
0
t
3
−u
t
3
−
B
0
,
2
+
B
1
,
2
t
2
−
t
0
t
1
1
2
uB
0
,
2
+
2
(3
=
−
u
)
B
1
,
2
1
2
u
u−t
0
t
0
B
0
,
1
+
2
u
t
2
−
u
=
t
1
B
1
,
1
t
1
−
t
2
−
+
2
(3
u
)
u−t
1
t
2
−
t
1
B
1
,
1
+
2
(3
u
)
t
3
−
u
t
3
−
−
−
t
2
B
2
,
1
(5.9)
1
2
u
2
B
0
,
1
+
1
=
{
2
u
(2
−
u
)
+
2
(3
−
u
)(
u
−
1)
}
B
1
,
1
+
2
(3
u
)
2
B
2
,
1
.
−
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