Signals and Systems - Electrical Engineering

1.4 Representation Using Basic Signals 93

% t: time support

% m: slope of ramp

% ad : advance (positive), delay (negative) factor

% USE: y = ramp(t,m,ad)

N = length(t);

y = zeros(1,N);

for i = 1:N,

if t(i)>= -ad,

y(i) = m * (t(i)+ad);

end

end

function y = ustep(t,ad)

% generation of unit step

% t: time

% ad : advance (positive), delay (negative)

% USE y = ustep(t,ad)

N = length(t);

y = zeros(1,N);

for i = 1:N,

if t(i)>= -ad,

y(i) = 1;

end

end

Analytically,

n y(t)=0 fort<−3 and fort>3, so the chosen support− 5 ≤t≤5 displays the signal in a

region where the signal occurs.

n For− 3 ≤t≤−1,y(t)is 3r(t+ 3 )= 3 (t+ 3 ), which is 0 att=−3 and 6 att=−1.

n For− 1 ≤t≤0,y(t)is 3r(t+ 3 )− 6 r(t+ 1 )= 3 (t+ 3 )− 6 (t+ 1 )=− 3 t+3, which is 6 att=

−1 and 3 att=0.

n For 0≤t≤3,y(t)is 3r(t+ 3 )− 6 r(t+ 1 )+ 3 r(t)=− 3 t+ 3 + 3 t=3.

n Fort≥3 the signal is 3r(t+ 3 )− 6 r(t+ 1 )+ 3 r(t)− 3 u(t− 3 )= 3 − 3 =0.

These coincide with the signal shown in Figure 1.8. n

nExample 1.17

Consider the following script that uses the functionsrampandustepto generate a signaly(t). Obtain

analytically the formula for the signaly(t). Write a function to compute and plot the even and odd

components ofy(t).

clear all; clf

t = -5:0.01:5;

y1 = ramp(t,2,2.5);

y2 = ramp(t,-5,0);