Example 3 - Waveform Representation

Example:
Find equations representing the waveform above over two full cycles (or periods)

Solution:
Note: for the vertical line mathematically:



we can define the upper point as defined and the lower point as undefined

or we can define the upper point as undefined and the lower point as defined

or we can define the upper point as undefined and the lower point as undefined (we will use this approach in this example)



In reality, there is a rapid but not instantaneous transition

The period is the time of one full cycle
T = 8 milliseconds = 8 ms = 0.008 s

To represent first period, need equations for line segments (1) 0 ≤ t ≤ 2;
(2) 2 ≤ t ≤ 6; (3) 6 ≤ t ≤ 8; ...

Equation for line (1); (in equations below v in Volts, t in ms)
\  v = -4; (0 ≤ t ≤  2)

Equation for line (2); (2 ≤ t ≤ 6)
v(t) = m t + b; m = ∆v / ∆t = (v2 - v1) / (t2 - t1)
v(t) = (7 - (-4)) / (6 - 2) = 11 / 4 = 2.75 volts/ms
\  v = 2.75 t + b;
Calculate b ... use point (2; 4):
-4 = 2.75 (2) + b; --> b = -9.5
\  v = 2.75 t - 9.5; (2 ≤ t ≤  6)

Equation for line (3); 6 ≤ t ≤ 8
\v  = 7; (6 ≤ t ≤ 8)

Using shifting conception, we extend to second period or cycle (8 ≤ t ≤ 16)...
v = -4; (8 ≤ t ≤  10)
v = 2.75 (t - 8) - 9.5; or v = 2.75 t - 31.5 (10 ≤ t ≤  14)
v = -7; (14 ≤ t ≤ 16)

Note: If t in seconds, we'll find (check it...)
v = -4; (0 ≤ t ≤ 0.002 s)
v = 2750 t - 9.5; (0.002 s ≤ t ≤ 0.006 s)
v = 7; (0.006 s ≤ t ≤ 0.008 s)

Extending to second cycle...
v = -4; (0.008 s ≤ t ≤ 0.010 s)
v = 2750 (t - 0.008) - 9.5; or v = 2750 t - 31.5; (0.010 s ≤ t ≤ 0.014 s)
v = 7; (0.014 s ≤ t ≤ 0.016 s)