ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
1
ECE 301 Solution to Homework Assignment 2
1. Indicate whether the following systems are causal, invertible, linear, memoryless,
and/or time invariant by circling the correct options. (A system may have more
than one of these properties.) Justify your answer.
(a) y(t) = x(t−2)+x(2−t)
(causal, invertible, linear , memoryless, time invariant )
Causality: The system is NOT causal because for any t < 1, the output
depends on a future input, e.g. y(0) = x(−2) + x(2).
Invertibility: The system is NOT invertible because, if w(t) is even, then the
output y(t) corresponding to x1 (t) = w(t)u(t) is indistinguishable from
that corresponding to x2 (t) = w(t)u(−t).
Linearity: The system is linear because if
y1 (t) = x1 (t − 2) + x1 (2 − t),
y2 (t) = x2 (t − 2) + x2 (2 − t), and
x(t) = αx1 (t) + βx2 (t),
(1)
(2)
(3)
then the output y(t) corresponding to the input x(t) is
y(t) =
=
=
=
=
x(t − 2) + x(2 − t)
(αx1 + βx2 )(t − 2) + (αx1 + βx2 )(2 − t)
αx1 (t − 2) + βx2 (t − 2) + αx1 (2 − t) + βx2 (2 − t)
α (x1 (t − 2) + x1 (2 − t)) + β (x2 (t − 2) + x2 (2 − t))
αy1(t) + βy2(t).
(4)
(5)
(6)
(7)
(8)
Memorylessness: The system is NOT memoryless because the output at
time t depends on input values at times other than t.
Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa (t) = x(t − a)1 . Then
the output ya (t) corresponding to the input signal xa (t) is
ya (t) = xa (t − 2) + xa (2 − t)
= x((t − a) − 2) + x(2 − (t − a))
= y(t − a).
(b) y(t) =
1
R2
−∞
x(τ )dτ
(9)
(10)
(11)
(causal, invertible, linear , memoryless, time invariant)
The character “a” on the left hand side of the equals sign is a label. The variable a on the right
hand side represents an arbitrary real number. It may be easier to think of a specific number, say, 3, so
long as you note that for the system to be time invariant, the test must work for any real value, not just
3.
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
2
Causality: The system is NOT causal because for any t < 2, the output
depends on a future input.
Invertibility: The system is NOT invertible because only the integral and
not the input function itself determines the output, e.g. the output corresponding to input x1 (t) = u(t) − u(t − 1) is the same as that corresponding
to input x2 (t) = 2[u(t) − u(t − 1/2)].
Linearity: The system is linear because if
Z
y1 (t) =
2
−∞
2
Z
y2 (t) =
x1 (τ )dτ, and
(12)
x2 (τ )dτ,
(13)
−∞
and x(t) = αx1 (t) + βx2 (t), then the output y(t) corresponding to the
input x(t) is
y(t) =
=
Z
2
(αx1 + βx2 )(τ )dτ
−∞
2
Z
αx1 (τ )dτ +
−∞
= αy1 (t) + βy2(t).
Z
2
βx2 (τ )dτ
−∞
(14)
(15)
(16)
Memorylessness: The system is NOT memoryless because the output at
time t depends on input values at times other than t.
Time Invariance: The system is NOT time invariant. To see this, we let
x(t) = u(t) − u(t − 1) so that
(17)
y(t) =
(18)
=
Z
2
x(τ )dτ
−∞
Z 1
0
dτ
(19)
= 1,
(20)
i.e. constant for all t, so in particular, y(t − 3) = 1 for any value of t2 .
However, if we let
x3 (t) = x(t − 3) = u(t − 3) − u(t − 4)
. Then
y3 (t) =
=
Z
2
x3 (τ )dτ
−∞
Z 2
3
x(τ )dτ
= 0
6
=
y(t − 3).
2
(21)
(22)
(23)
(24)
We need only a single delay value for which the property does not hold to show that the system is
not time invariant, so 3 will do.
ECE 301 Signals and Systems
(c) y(t) = (dx/dt)(t)
Solution to Assignment 2
3
September 7, 2006
( causal , invertible, linear , memoryless , time invariant )
Causality: The system is memoryless, hence causal.
Invertibility: The system is NOT invertible because any two inputs that differ
by a constant yield the same output.
Linearity: The system is linear because if
y1 (t) = (dx1 /dt)(t), and
y2 (t) = (dx2 /dt)(t),
(25)
(26)
and x(t) = αx1 (t) + βx2 (t), then the output y(t) corresponding to the
input x(t) is
y(t) = (d(αx1 + βx2 )/dt)(t)
= α(dx1 /dt)(t) + β(dx2 /dt)(t)
= αy1 (t) + βy2 (t).
(27)
(28)
(29)
Memorylessness: The system is memoryless because the output at time t
depends on input values at only time t.
Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa (t) = x(t − a). Then
the output ya (t) corresponding to the input signal xa (t) is
ya (t) = (dxa /dt)(t) = (d(x)/dt)(t − a) = y(t − a).
(d) y(t) = x(t/3)
(30)
(causal, invertible , linear , memoryless, time invariant )
Causality: The system is NOT causal because if t < 0, then the output
depends on future values of the input, e.g. for t = 3 we have y(−3) =
x(−1).
Invertibility: The system is invertible by applying the function w(t) = y(3t).
Linearity: The system is linear because if
y1 (t) = x1 (t/3), and
y2 (t) = x2 (t/3)
(31)
(32)
and x(t) = αx1 (t) + βx2 (t), then the output y(t) corresponding to the
input x(t) is
y(t) = (αx1 + βx2 )(t/3)
= αx1 (t/3) + βx2 (t/3)
= αy1 (t) + βy2 (t)
(33)
(34)
(35)
Memorylessness: The system is NOT memoryless because the output at
time t depends on input values at times other than t.
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
4
Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa (t) = x(t − a). Then
the output ya (t) corresponding to the input signal xa (t) is
ya (t) = xa (t/3) = x((t − a)/3) = y(t − a)
(e) y(t) = cos(x(t))
(36)
( causal , invertible, linear, memoryless , time invariant )
Causality: The system is memoryless, hence causal.
Invertibility: The system is NOT invertible, e.g. suppose that x1 (t) = (π/2)u(t)
and x2 (t) = −(π/2)u(t). Then y1 (t) = cos(x1 (t)) = 0 = cos(x2 (t)) =
y2 (t), ∀t.
Linearity: The system is NOT linear because if
y1 (t) = cos(x1 (t)), and
y2 (t) = cos(x2 (t)),
(37)
(38)
and x(t) = αx1 (t) + βx2 (t), then the output y(t) corresponding to the
input x(t) is
y(t) = cos(αx1 (t) + βx2 (t))
= cos(αx1 (t)) cos(βx2 (t)) − sin(αx1 (t)) sin(βx2 (t))
6= α cos(x1 (t)) + β cos(x2 (t))
(39)
(40)
(41)
in general. (We used an identity from Chapter B of the texbook to get the
second equality above.)
Memorylessness: The system is memoryless because the output at time t
depends on input values at only time t.
Time Invariance: The system is time invariant. To see this, we let y(t) be
the output corresponding to the input x(t) and let xa (t) = x(t − a). Then
the output ya (t) corresponding to the input signal xa (t) is
ya (t) = cos(xa (t)) = cos(x(t − a)) = y(t − a).
(42)
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
5
2. For the system described by the differential equation
(D 2 + αD + 6)y(t) = (D + 4)x(t)
(43)
(a) Suppose α = 5.
i. Find the eigenvalues and modes.
Solution: Solving Q(λ) = λ2 + 5λ + 6 = (λ + 2)(λ + 3) = 0, we obtain the
eigenvalues λ1 = −2 and λ2 = −3. The corresponding modes are c1 e−2t
and c2 e−3t .
ii. Find the zero input response y0 (t) corresponding to the initial conditions
y(0) = 1, ẏ(0) = 3.
Solution: The ZIR is
y0 (t) = c1 e−2t + c2 e−3t , so
ẏ0 (t) = −2c1 e−2t − 3c2 e−3t .
(44)
(45)
y0 (0) = 1 = c1 + c2 and
ẏ0 (0) = 3 = −2c1 − 3c2
(46)
(47)
Solving
for c1 and c2 yields
c1 = 6 and c2 = −5
so
y0 (t) = 6e−2t − 5e−3 t
iii. Find the impulse response h(t).
Solution: First we solve for yn (t) by applying the initial conditions y(0) =
0 and ẏ(0) = 1 to the linear combination of the modes
yn (t) = c1 e−2t + c2 e−3t .
Solving
yn (0) = 0 = c1 + c2 and
ẏn (0) = 1 = −2c1 − 3c2
(48)
(49)
for c1 and c2 yields
c1 = 1 and c2 = −1
so
yn (t) = e−2t − e−3t , and
ẏn (t) = −2e−2t + 3e−3t .
(50)
(51)
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
6
The impulse response will be given by
h(t) = b0 δ(t) + [P (D)yn (t)]u(t).
(52)
The degree of Q(D) being one greater than the degree of P (D), the value
of b0 is zero. We obtain
h(t) = [(D + 4)yn (t)]u(t)
= [−2e−2t + 3e−3t + 4(e−2t − e−3t )]u(t)
= [2e−2t − e−3t ]u(t).
(53)
(54)
(55)
(b) Now suppose α = 1.
i. Find the eigenvalues and modes.
2
Solution: Solving Q(λ)
where
√ = λ +λ+6 = (λ+α+jβ)(λ+α−jβ) = 0,√
α = −1/2 and β = √ 23/2. we obtain the eigenvalues λ1 = −1/2 + j 23/2
and λ2 = −1/2 − j 23/2. The corresponding modes are c1 eλ1 t and c2 eλ2 t ,
or ceαt cos(βt + θ), depending upon which representation one prefers.
ii. Find the zero input response y0 (t) corresponding to the initial conditions
y(0) = 1, ẏ(0) = 3.
Solution: For readability, I’ll write the intermediate equations in terms
of α and β and substitute for them at the end. The ZIR is
y0 (t) = c1 e(α+jβ)t + c2 e(α−jβ)t , so
ẏ0 (t) = (α + jβ)c1 e(α+jβ)t + (α − jβ)c2 e(α−jβ)t .
(56)
(57)
y0 (0) = 1 = c1 + c2 and
ẏ0 (0) = 3 = (α + jβ)c1 + (α − jβ)c2
(58)
(59)
Solving
for c1 and c2 yields
c1 =
β − j(3 − α)
β + j(3 − α)
and c2 =
2β
2β
(60)
so
β − j(3 − α) (α+jβ)t β + j(3 − α) (α−jβ)t
e
+
e
2β
2β
!
!
jβt
j(3 − α)eαt −ejβt + e−jβt
+ e−jβt
αt e
+
= e
2
β
2
αt
(3 − α)e
sin(βt)
= eαt cos(βt) +
β
√
√
7e−t/2
sin( 23t)
= e−t/2 cos( 23t/2) + √
23
y0 (t) =
(61)
(62)
(63)
(64)
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
7
iii. Find the impulse response h(t).
Solution: First we solve for yn (t) by applying the initial conditions y(0) =
0 and ẏ(0) = 1 to the linear combination of the modes
yn (t) == c1 e(α+jβ)t + c2 e(α−jβ)t
Solving
yn (0) = 0 = c1 + c2 and
ẏn (0) = 1 = (α + jβ)c1 + (α − jβ)c2
for c1 and c2 yields
c1 = −j
(65)
(66)
1
1
and c2 = j
2β
2β
so
1
(−e(α+jβ)t + e(α−jβ)t )
2β
!
αt sin(βt)
, and
= e
β
!
αt sin(βt)
ẏn (t) = αe
+ eαt cos(βt)
β
yn (t) = j
(67)
(68)
(69)
The impulse response is then
h(t) = b0 δ(t) + [P (D)yn (t)]u(t).
(70)
The degree of Q(D) being one greater than the degree of P (D), the value
of b0 is zero. We obtain
h(t) = [(D + 4)yn (t)]u(t)
"
!
#
αt sin(βt)
αt
= (α + 4)e
+ e cos(βt) u(t)
β
√
!
√
−t/2 sin( 23t)
√
+ e−t/2 cos( 23t).
= 7e
23
√
(c) Finally, suppose α = 2 6.
(71)
(72)
(73)
i. Find the eigenvalues and modes. √
√
Solution: Solving Q(λ) =√λ2 + 2 6λ + 6 = (λ + 6)2 = 0, we obtain
the
√
− 6t
and
eigenvalues
λ1 = λ2 = − 6. The corresponding modes are c1 e
√
c2 te− 6t .
ECE 301 Signals and Systems
Solution to Assignment 2
September 7, 2006
8
ii. Find the zero input response y0 (t) corresponding to the initial conditions
y(0) = 1, ẏ(0) = 3.
Solution: The ZIR is
√
√
y0 (t) = c1 e− 6t + c2 te− 6t , so
√
√
√
√
ẏ0 (t) = − 6(c1 e− 6t + c2 te− 6t ) + c2 e− 6t .
(74)
(75)
Solving
y0 (0) = 1 = c1 and
√
ẏ0 (0) = 3 = − 6c1 + c2
for c1 and c2 yields
c1 = 1 and c2 = 3 +
so
y0 (t) = (1 + (3 +
√
√
6)t)e−
(76)
(77)
6
√
6t
.
iii. Find the impulse response h(t).
Solution: First we solve for yn (t) by applying the initial conditions y(0) =
0 and ẏ(0) = 1 to the linear combination of the modes
yn (t) = (c1 + c2 t)e−
√
6t
.
Solving
yn (0) = 0 = c1 and
√
ẏn (0) = 1 = − 6c1 + c2
(78)
(79)
for c1 and c2 yields
c1 = 0 and c2 = 1
so
√
yn (t) = te− 6t , and
√
√
√
ẏn (t) = − 6te− 6t + e− 6t
(80)
(81)
The impulse response is then
h(t) = b0 δ(t) + [P (D)yn (t)]u(t).
(82)
The degree of Q(D) being one greater than the degree of P (D), the value
of b0 is zero. We obtain
h(t) = [(D + 4)yn (t)]u(t)
√
√
√
= [(− 6t + 1)e− 6t + 4te− 6t ]u(t)
√
√
= [(1 + (− 6 + 4)t)e− 6t ]u(t).
(83)
(84)
(85)