Signals and Systems - Electrical Engineering

8.2 Discrete-Time Signals 475

nExample 8.17

Consider how to generate a train of triangular, discrete-time pulsest[n], which is periodic of period

N=11. A period oft[n] is

τ[n]=







n 0 ≤n≤ 5

−n+10 6≤n≤ 10

0 otherwise

Find then an expression for its finite differenced[n]=t[n]−t[n−1].

Solution

The periodic signal can be generated by adding shifted versions ofτ[n], or

t[n]=···+τ[n+11]+τ[n]+τ[n−11]+···=

∑∞

k=−∞

τ[n− 11 k]

The finite differenced[n] is then

d[n]=t[n]−t[n−1]

=

∑∞

k=−∞

(τ[n− 11 k]−τ[n− 1 − 11 k])

The signald[n] is also periodic of the same periodN=11 ast[n]. If we let

s[n]=τ[n]−τ[n−1]=







1 0≤n≤ 5

−1 6≤n≤ 10

0 otherwise

then

d[n]=

∑∞

k=−∞

s[n− 11 k]

When sampled, these two signals look very much like the continuous-time train of triangular

pulses, and its derivative. n

nExample 8.18

Consider the discrete-time signal

y[n]= 3 r(t+ 3 )− 6 r(t+ 1 )+ 3 r(t)− 3 u(t− 3 )|t=0.15n