Signals and Systems - Electrical Engineering

94 C H A P T E R 1: Continuous-Time Signals

FIGURE 1.8

Generation of

y(t)= 3 r(t+ 3 )− 6 r(t+ 1 )+ 3 r(t)−

3 u(t− 3 ),− 5 ≤t≤ 5 , and zero otherwise.

− 5 0 5

− 1

0

1

2

3

4

5

6

7

t(sec)

y(

t)

y3 = ramp(t,3,-2);

y4 = ustep(t,-4);

y = y1 + y2 + y3 + y4;

plot(t,y,’k’); axis([-5 5 -3 5]); grid

The signaly(t)=0 fort<−5 andt>5.

Solution

The signaly(t)displayed on Figure 1.9(a) is given analytically by

y(t)= 2 r(t+2.5)− 5 r(t)+ 3 r(t− 2 )+u(t− 4 )

Clearly,y(t)is neither even nor odd. To find its even and odd components we use the function

evenodd, shown in the following code with inputs as the signal and its support and outputs as the

even and odd components. The results are shown on the bottom plots of Figure 1.9. Adding these

two signals gives back the original signaly(t). The script used is as follows.

%%%%%%%%%%%%%%%%%%%

% Example 1.17

%%%%%%%%%%%%%%%%%%%

[ye, yo] = evenodd(t,y);

subplot(211)

plot(t,ye,’r’)

grid

axis([min(t) max(t) -2 5])

subplot(212)

plot(t,yo,’r’)