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Use of the function H(X) with LDEC, LAP, or ILAP, is not allowed in the calculator.
You have to use the main results provided earlier when dealing with the
Heaviside step function, i.e., L{H(t)} = 1/s, L
-1
{1/s}=H(t),
L{H(t-k)}=e
–ks
⋅L{H(t)} = e
–ks
⋅(1/s) = ⋅(1/s)⋅e
–ks
and L
-1
{e
–as
⋅F(s)}=f(t-a)⋅H(t-a).
Example 2 – The function H(t-t
o
) when multiplied to a function f(t), i.e., H(t-t
o
)f(t),
has the effect of switching on the function f(t) at t = t
o
. For example, the solution
obtained in Example 3, above, was y(t) = y
o
cos t + y
1
sin t + sin(t-3)⋅H(t-3).
Suppose we use the initial conditions y
o
= 0.5, and y
1
= -0.25. Let’s plot this
function to see what it looks like:
Θ Press „ô, simultaneously if in RPN mode, to access to the PLOT SETUP
window.
Change
TYPE to FUNCTION, if needed
Change EQ to ‘0.5*COS(X)-0.25*SIN(X)+SIN(X-3)*H(X-3)’.
Make sure that
Indep is set to ‘X’.
H-VIEW: 0 20, V-VIEW: -3 2.
Press @ERASE @DRAW to plot the function.
Press @EDIT L @LABEL to see the plot.
The resulting graph will look like this:
Notice that the signal starts with a relatively small amplitude, but suddenly, at
t=3, it switches to an oscillatory signal with a larger amplitude. The difference
between the behavior of the signal before and after t = 3 is the “switching on”
of the particular solution y
p
(t) = sin(t-3)⋅H(t-3). The behavior of the signal before
t = 3 represents the contribution of the homogeneous solution, y
h
(t) = y
o
cos t +
y
1
sin t.
The solution of an equation with a driving signal given by a Heaviside step
function is shown below.
Example 3
– Determine the solution to the equation, d
2
y/dt
2
+y = H(t-3),