Signals.7
Forced Response
ECE 311
Forced Response
Objective:
Find the natural response for a differential equation
Discussion
The forced response is the response due to the input, x(t), assuming the initial conditions are zero. This works
somewhat well for simple inputs where the LaPlace transform is known. It doesn't work well for arbitrary signals.
Example: Find the forced response the following differential equation
d3y
dt 3
d2y
dy
+ 6 dt 2 + 11 dt + 6y = 5 dx
+ 3x
dt
Assume
⎧0
x(t) = u(t) = ⎨
⎩1
t<0
t>0
Since x(t) = 0 for t<0, y(t) and its derivatives will also be zero for t=0. This lets you replace derivatives with
powers of 's':
d3y
dt 3
d2y
dy
+ 6 dt 2 + 11 dt + 6y = 5 dx
+ 3x
dt
s 3 Y + 6s 2 Y + 11sY + 6Y = 5sX + 3X
Note that for zero initial conditions, sY means 'the derivative of y'. 's' is an operator meaning 'the derivative of'.
Solve for Y:
⎞
Y = ⎛⎝ s 3 +6s5s+3
2 +11s+6 ⎠ X
Replace X with it's LaPlace transform:
⎞ ⎛1⎞
Y = ⎛⎝ s 3 +6s5s+3
2 +11s+6 ⎠ ⎝ s ⎠
Use partial fractions:
5s+3
⎞ ⎛1⎞
Y = ⎛⎝ (s+1)(s+2)(s+3)
⎠ ⎝s⎠
⎞ ⎛ 1 ⎞ ⎛ −3.5 ⎞ ⎛ 2 ⎞
Y = ⎛⎝ 0.5
s ⎠ + ⎝ s+1 ⎠ + ⎝ s+2 ⎠ + ⎝ s+3 ⎠
Take the inverse LaPlace transform:
y(t) = (0.5 + 1e −t − 3.5e −2t + 2e −3t )u(t)
JSG
1
rev April 13, 2009