Digital Signal Processing Reference
In-Depth Information
Table 6.3: Sinusoids that simplify things for us.
cos(=2) = 0; sin(=2) = 1
cos() =1; sin() = 0
cos(2) = 1; sin(2) = 0
cos(3=2) = 0; sin(3=2) =1
x[1] =
1
4
((x
0
+ x
1
+ x
2
+ x
3
)+
(x
0
(j) + x
1
+ x
2
(j) + x
3
(1j0))+
(x
0
(1 + j0) + x
1
+ x
2
(1j0) + x
3
(1j0))+
(x
0
(j(1)) + x
1
+ x
2
(j(1)) + x
3
(1j0)))
Simplifying:
x[1] =
1
4
((x
0
+ x
1
+ x
2
+ x
3
)+
(x
0
(j) + x
1
+ x
2
(j) + x
3
(1))+
(x
0
(1) + x
1
+ x
2
(1) + x
3
(1))+
(x
0
(j) + x
1
+ x
2
(j) + x
3
(1))):
Grouping terms:
x[1] =
1
4
(x
0
(1 + j1j) + x
1
(1 + 1 + 1 + 1) + x
2
(1j1 + j) + x
3
(11 + 11))
x[1] =
1
4
(x
0
(0) + x
1
(4) + x
2
(0) + x
3
(0))
x[1] = x
1
:
Thus, the inverse transform gives us back the data that we had before the forward
transform. The
1
N
term appears in the inverse transform to scale the outputs back
to the original magnitude of the inputs. We only show this for one of the original
values, but the interested reader should be able to verify that this works for all four
of the original values.
