18.03 Class 14, March 10, 2006
Exponential signals, higher order equations, operators
[1] Exponential signals
x" + bx' + kx = A e^{rt}
(*)
We want to find some solution.
Try for a solution of the form
xp = B e^{rt} :
k] xp = B e^{rt}
b] xp' = B r e^{rt}
xp" = B r^2 e^{rt}
_____________________
A e^{rt} = B ( r^2 + br + k ) e^{rt}
or
xp = ( A / p(r)) e^{rt}
where p(s) = s^2 + bs + k is the characteristic polynomial.
Eg
x" + 2x' + 2x = 4 e^{3t}
p(3) = 3^2 + 2(3) + 2 = 17 so
xp = ( 4 / 17 ) e^{3t} .
The general solution is given by
x = (4/17) e^{3t} + e^{t} ( a cos(t) + b sin(t) )
Of course this will let us solve (*) for sinusoidal signals as well:
Eg
y" + 2y' + 2y = sin(3t)
This is the imaginary part of
z" + 2z' + 2z = e^{3it}
p(3i) = (3i)^2 + 2(3i) + 2 = 7 + 6i
so
zp = ( 1 / (7 + 6i) ) e^{it}
We want the imaginary part. Lets do it by writing out real and imaginary
parts:
zp = ( (7  6i) / 85 ) ( cos(3t) + i sin(3t) )
and find the imagnary part of the product:
yp =  (7/85) cos(3t) + (6/85) sin(3t)
The general solution is
y
=
 (7/85) cos(3t) + (6/85) sin(3t) + e^{t} ( a cos t + b sin t
).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
The work shows:
The Exponential Response Formula: a solution to
x" + bx' + kx = A e^
{rt}
is given by
xp = ( A / p(r) ) e^{rt}
provided p(r) is not zero.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 Staff
 Equations, LTI system theory, higher order equations

Click to edit the document details