1.4 Representation Using Basic Signals 91where the first termx 1 a(t)is continuous and the secondx 1 b(t)is discontinuous. The derivative is
dx 1 (t)
dt=− 2 πsin( 2 πt)[u(t)−u(t− 1 )]+(cos( 2 πt)− 1 )[δ(t)−δ(t− 1 )]+δ(t)−δ(t− 1 )=− 2 πsin( 2 πt)[u(t)−u(t− 1 )]+δ(t)−δ(t− 1 )
since
(cos( 2 πt)− 1 )[δ(t)−δ(t− 1 )]=(cos( 2 πt)− 1 )δ(t)−(cos( 2 πt)− 1 )δ(t− 1 )
=(cos( 0 )− 1 )δ(t)−(cos( 2 π)− 1 )δ(t− 1 )
= 0 δ(t)+ 0 δ(t− 1 )= 0The termδ(t)in the derivative indicates that there is a jump from 0 to 1 inx 1 (t)att=0 and that
in−δ(t− 1 )there is a jump of−1 (from 1 to 0) att=1. See Figure 1.7.0 0.2 0.4 0.6 0.8 1− 202x^1(t
)0 0.2 0.4 0.6 0.8 1− 202x^1(a
t)0 0.2 0.4 0.6 0.8 1− 202x^1(b
t)t
(c)t
(b)t
(a)FIGURE 1.7
(a) Decomposition ofx 1 (t)=cos( 2 πt)[u(t)−u(t− 1 )]into (b) a continuous and (c) a discontinuous signal
(a pulse).