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With our understanding of the functions
e
x
e
x
,
e
j
Θ
e
j
Θ
, and the quadratic equation
z
2
+
b
a
z
+
c
a
=
0
z
2
+
b
a
z+
c
a
=0, we can undertake a rudimentary study of differential and difference equations.
Differential Equations. In your study of circuits and systems you will encounter the homogeneous differential equation
d2dt2x(t)+a1ddtx(t)+a2=0.d2dt2x(t)+a1ddtx(t)+a2=0.
(1)Because the function estest reproduces itself under differentiation, it is plausible
to assume that x(t)=estx(t)=est is a solution to the differential equation. Let's
try it:
d
2
d
t
2
(
e
s
t
)
+
a
1
d
d
t
(
e
s
t
)
+
a
2
(
e
s
t
)
=
0
(
s
2
+
a
1
s
+
a
2
)
e
s
t
=
0
.
d
2
d
t
2
(
e
s
t
)
+
a
1
d
d
t
(
e
s
t
)
+
a
2
(
e
s
t
)
=
0
(
s
2
+
a
1
s
+
a
2
)
e
s
t
=
0
.
(2)If this equation is to be satisfied for all
t
t, then the polynomial in
s
s must be
zero. Therefore we require
s2+a1s+a2=0.s2+a1s+a2=0.
(3)As we know from our study of this quadratic equation, the solutions are
s1,2=-a12±12a12-4a2.s1,2=-a12±12a12-4a2.
(4)This means that our assumed solution works, provided s=s1s=s1 or
s
2
s
2
. It is a fundamental result from the theory of differential equations that the most general solution for x(t)x(t) is a linear combination of these assumed solutions:
x(t)=A1es1t+A2es2t.x(t)=A1es1t+A2es2t.
(5)If a12-4a2a12-4a2 is less than zero, then the roots
s
1
s
1
and
s
2
s
2
are complex:
s1,2=-a12±j124a2-a12.s1,2=-a12±j124a2-a12.
(6)Let's rewrite this solution as
s
1
,
2
=
σ
±
j
ω
s
1
,
2
=
σ
±
j
ω
(7)where σ and ω are the constants
σ
=
-
a
1
2
σ
=
-
a
1
2
(8)ω=124a2-a12.ω=124a2-a12.
(9)With this notation, the solution for x(t)x(t) is
x(t)=A1eσtejωt+A2eσte-jtωt.x(t)=A1eσtejωt+A2eσte-jtωt.
(10)If this solution is to be real, then the two terms on the right-hand side must
be complex conjugates. This means that A2=A1*A2=A1* and the solution for x(t)x(t) is
x
(
t
)
=
A
1
e
σ
t
e
j
ω
t
+
A
1
*
e
σ
t
e
-
J
ω
t
=
2
Re
{
A
1
e
σ
t
e
j
ω
t
}
.
x
(
t
)
=
A
1
e
σ
t
e
j
ω
t
+
A
1
*
e
σ
t
e
-
J
ω
t
=
2
Re
{
A
1
e
σ
t
e
j
ω
t
}
.
(11)The constant
A
1
A
1
may be written as A1=|A|ejφA1=|A|ejφ. Then the solution for x(t)x(t)
is
x(t)=2|A|eσtcos(ωt+φ).x(t)=2|A|eσtcos(ωt+φ).
(12)This “damped cosinusoidal solution” is illustrated in Figure 1.
Find the general solutions to the following differential equations:
-
d
2
d
t
2
X
(
t
)
+
2
d
d
t
x
(
t
)
+
2
=
0
d
2
d
t
2
X
(
t
)
+
2
d
d
t
x
(
t
)
+
2
=
0
;
-
d
2
d
t
2
x
(
t
)
+
2
d
d
t
x
(
t
)
-
2
=
0
d
2
d
t
2
x
(
t
)
+
2
d
d
t
x
(
t
)
-
2
=
0
;
-
d
2
d
t
2
x
(
t
)
+
2
=
0
d
2
d
t
2
x
(
t
)
+
2
=
0
.
Difference Equations. In your study of digital filters you will encounter homogeneous difference equations of the form
xn+a1xn-1+a2xn-2=0.xn+a1xn-1+a2xn-2=0.
(13)What this means is that the sequence{xn}{xn} obeys a homogeneous recursion:
x
n
=
-
a
1
x
n
-
1
-
a
2
x
n
-
2
.
x
n
=
-
a
1
x
n
-
1
-
a
2
x
n
-
2
.
(14)A plausible guess at a solution is the geometric sequence xn=znxn=zn. With this guess, the difference equation produces the result
z
n
+
a
1
z
n
-
1
+
a
2
z
n
-
2
=
0
(
1
+
a
1
z
-
1
+
a
2
z
-
2
)
z
n
=
0
.
z
n
+
a
1
z
n
-
1
+
a
2
z
n
-
2
=
0
(
1
+
a
1
z
-
1
+
a
2
z
-
2
)
z
n
=
0
.
(15)If this guess is to work, then the second-order polynomial on the left-hand side must equal zero:
1
+
a
1
z
-
1
+
a
2
z
-
2
=
0
1
+
a
1
z
-
1
+
a
2
z
-
2
=
0
1
+
a
1
z
-
1
+
a
2
z
-
2
=
0
z
2
+
a
1
z
+
a
2
=
0
.
1
+
a
1
z
-
1
+
a
2
z
-
2
=
0
z
2
+
a
1
z
+
a
2
=
0
.
(16)The solutions are
z
1
,
2
=
-
a
1
2
±
j
1
2
4
a
2
-
a
1
2
=
r
e
j
θ
.
z
1
,
2
=
-
a
1
2
±
j
1
2
4
a
2
-
a
1
2
=
r
e
j
θ
.
(17)The general solution to the difference equation is a linear combination of the
assumed solutions:
x
n
=
A
1
z
1
n
+
A
2
(
z
1
*
)
n
=
A
1
z
1
n
+
A
1
*
(
z
1
*
)
n
=
2
Re
{
A
1
z
1
n
}
=
2
|
A
|
r
n
cos
(
θ
n
+
φ
)
.
x
n
=
A
1
z
1
n
+
A
2
(
z
1
*
)
n
=
A
1
z
1
n
+
A
1
*
(
z
1
*
)
n
=
2
Re
{
A
1
z
1
n
}
=
2
|
A
|
r
n
cos
(
θ
n
+
φ
)
.
(18)This general solution is illustrated in Figure 2.
Find the general solutions to the following difference equations:
- xn+2xn-1+2=0xn+2xn-1+2=0;
- xn-2xn-1+2=0xn-2xn-1+2=0;
- xn+2xn-2=0xn+2xn-2=0.
(What does "Endorsed by" mean?)
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