solution

x 1 (t)
x 2 (t)
x 20 (t)
3
2
1
0
N=1
-1
-3 -2 -1 0 1 2 3
6
4
2
0
-2
-3 -2 -1 0 1 2 3
60
40
20
0
t
N=2
-20
-3 -2 -1 0 1 2 3
Problem 3 (Chapter 8, P. 30)
t
N=20
(1) From the first figure in Figure E.30, the signal is even with T0 = 2, ω0 = π
So,
1
T0
− jkω0t X[ k] = x( t) e dt
T ∫0
0
1
=
2 ∫
1
−1
t
− jkπt x() te dt
1 1
= x( t)[cos( kπt) jsin( kπt)] dt
2 ∫ − + −
−1
1 1 j 1
= ()cos( ) ()sin( )
2∫ x t kπt dt− x t kπt dt
−1 2∫−1
even odd
j 1
()sin( ) 0
2 ∫ x t kπt dt = , the harmonic function X[ k ] have a purely real value
−1