solution

square wave signal 2 () x t looks like:
x 2 (t)
1
0.5
0
-0.5
-3 -2 -1 0 1 2 3
x 10 -5
-1
t
How to get this square wave signal? It’s just from a similar and basic function 3 () x t ,
as shown below. Then x2() t = x3() t × 2− 1.
x 3 (t)
1
0.5
-3 -2 -1 0 1 2 3
x 10 -5
0
t
The expression for this x3 () t is just a convolution of a rectangle function and
impulse train:
⎛ t ⎞
x3() t = rect ⎜ δ 5 5 () t 21/10 × 1/10
⎟∗
⎝ ⎠
⎛ t ⎞
x () 2 x () t − 1= 2rect ∗ () t − 1
⎝1/10 ⎠
⇒ 2 t = 3 ⎜ δ 5 ⎟
5 21/10 ×
6 ⎛ ⎛ t ⎞
⎞
xt () = x1() t x2() t = sin(2× 10 πt) ×⎜2rect ⎜ 5 δ 5 () t 1
21/10 × ⎟
1/10
⎟∗
−
⎝ ⎝ ⎠
⎠
1
From Fourier Series Pairs table, with T 0 = 2× 5
10
,
1
rectangular function is T = 5
10
,
ω π
5
0 = × 10 , width of