solution

for every value of k .
(2) From the second figure in Figure E.30, the signal is odd with T0 = 1, ω0 = 2π
1
T0
− jkω0t X[ k] = x( t) e dt
T ∫0
0
1/2
= ∫
1/2
2
1/2 ()
− jk πt
x te dt
−
1/2 1/2
∫ ∫
= x( t)cos(2 kπt) dt− j x( t)sin(2 kπt) dt
−1/2 −1/2
odd
even
So, x()cos(2 t k t) dt 0
1/2 π ∫ = , the harmonic function X[ k ] have a purely imaginary
−
value for every value of k .
Problem 4 (Chapter 8, P. 32)
Period of a sine wave is
1
6
10 ; period of a burst of “1” is 1
; period of a burst of “0”
5
10
1
is 5
10 ; period of a binary signal with alternating 1’s and 0’s is 1
T 0 = 2× and 5
10
ω π
5
0 = × 10 . The binary signal we want should be like:
x(t)
1
0.5
0
-0.5
-3 -2 -1 0 1 2 3
x 10 -5
-1
t
How can it come?
As we already know the basic sine wave is
x 1 (t)
1
0.5
0
-0.5
x t = × t , shown as
6
1 () sin(210 ) π
-3 -2 -1 0 1 2 3
x 10 -5
-1
t
In order to get the wanted signal, the signal shown above should multiply with a