Unformatted text preview: 1 EE113 Homework 8 Problem 12.5 Determine the unilateral z transforms of each of the following sequences and indicate their regions
of convergence:
(a) x(n) = (n − 1)u(n + 1).
(b) x(n) = (1 + n2 αn−2 )u(n + 31).
(c) x(n) = αn with α > 0.
(d) The impulse response sequence of the relaxed causal system y (n) − 3 y (n − 1) + 1 y (n − 2) = x(n − 1).
4
8
Solution: Note that the unilateral z transform of x(n) is the bilateral z transform of x(n)u(n).
(a) The equivalent sequence is x′ (n) = x(n)u(n) = (n − 1)u(n) = nu(n) − u(n). The unilateral z transform of
x(n) then is
X + (z ) = z
z
−
,
2
(z − 1)
z−1 z  > 1 (b) The equivalent sequence is x′ (n) = x(n)u(n) = (1 + n2 αn−2 )u(n) = u(n) + α−2 n2 αn u(n). The unilateral
z transform of x(n) then is
X + (z ) = z
z+α
−
,
z − 1 α(z − α)3 {z  > 1} ∩ {z  > α} (c) The equivalent sequence is x′ (n) = x(n)u(n) = αn u(n) = αn u(n) with α > 0. The unilateral z transform
of x(n) then is
X + (z ) = z
,
z−α z  > α (d) Because this system is causal, its impulse response sequence h(n) = h(n)u(n) is a casual sequence. The
unilateral z transform of h(n) then is the same as its transfer function H (z ):
H (z ) = November 21, 2011 Y (z )
8z
=2
,
X (z )
8z − 6z + 1 z  > 1
2 DRAFT 2 Problem 12.21 Consider a causal system that is described by the difference equation
y (n) = 5
1
y (n − 1) − y (n − 2) + x(n − 2),
6
6 y (−2) = 0, y (−1) = 1 Use the unilateral z transform to determine its complete response to the sequence
1
4 x(n) = (n − 1) n−2 u(n − 1) Solution: Because
x(n) = (n − 1) 1
4 n−2 u(n − 1) = (n − 2) n−2 1
4 u(n − 2) if we introduce an equivalent input sequence xo (n) as
1
4 xo (n) = n n u(n) and modify the difference equation as
y (n) = 5
1
y (n − 1) − y (n − 2) + x(n − 4),
6
6 y (−2) = 0, y (−1) = 1 then we should have the same output sequence. Let us ﬁnd out the unilateral z transform of the equivalent input
sequence xo (n):
+
Xo (z ) = z 1
4z
2,
−1
4 z  > 1
4 Take the unilateral z transform on both sides of the equivalent difference equation. We can get
Y + (z ) = 5 −1 +
1 −2 +
+
z Y (z ) + y (−1) −
z Y (z ) + y (−1)z −1 + y (−2) + z −4 Xo (z )
6
6 Substituting the initial conditions leads to
Y + (z ) = 5 −1 +
1 −2 +
+
z Y (z ) + 1 −
z Y (z ) + z −1 + z −4 Xo (z )
6
6 Solving for Y + (z ) yields
Y + (z ) = 1− 5
1 −1
6 − 6z
5 −1
+ 1 z −2
6z
6 + 1− 1
z −4
4z
1 −2 ·
1
+ 6z
z−4 5 −1
6z 2, z  > 1
2 Resorting to the partial fractions, we can recover the sequence y (n) as
y (n) = 3 · November 21, 2011 1
2 n−1 −2· 1
3 n−1 u(n) + 96 1
2 n−1 −9· 1
3 n−1 + 2 · (n + 3) · 1
4 n−1 u(n − 4) DRAFT 3 Problem 12.25 Consider the constantcoefﬁcient difference equation
1
1
y (n) − y (n − 1) − y (n − 2) = x(n − 1)
6
6
with initial conditions y (−2) = 0 and y (−1) = 6. Use the z transform technique to ﬁnd the answers to parts (a)(c):
(a) The zeroinput response.
(b) The zerostate response to x(n) = u(n).
(c) The complete response using the unilateral z transform.
(d) Check that your answer to part (c) is the sum of the answers to parts (a) and (b).
(e) Now determine the answers to parts (a)(c) by using the timedomain techniques you learned earlier for
solving constantcoefﬁcient difference equations by working with the modes of the system. Compare your
answers to those obtained by using the z transform technique.
(f) Find the response of the system to x(n) = ( 1 )n u(n − 2) in three different ways.
3
Solution:
(a) The zeroinput response has to be solved by the general solutions. The general solution for the homogeneous
equation is
n 1
2 yh (n) = C1 + C2 − 1
3 n Substituting the initial conditions y (−2) = 0 and y (−1) = 6 leads to
C1 = 9
,
5 C2 = − 4
5 Hence, the zeroinput response for n ≥ 0 is
yzi = 9
5 n 1
2 − 4
5 − n 1
3 u(n) (b) The zerostate response can be solved by using the z transform technique. The transfer function of the system
is
H (z ) = 1− z −1
z
=2 1
,
1
z − 6z − 1
− 6 z −1
6 z  > 1 −1
6z 1
2 The z transform of the input sequence x(n) = u(n) is
X (z ) = z
,
z−1 z  > 1 The z transform of the zerostate response can be expressed as
Yzs (z ) = z
z2 − 1 z −
6 1
6 · z
z2
=
z−1
(z − 1) z − 1
2 z+ 1
3 = 3
2 z−1 + 3
1
−5
+ 10 1 ,
z−1
z+ 3
2 z  > 1 which leads to
yzs (n) = November 21, 2011 33
−
25 1
2 n−1 + 1
10 − 1
3 n−1 u(n − 1) DRAFT 4 (c) Taking the unilateral z transform on both sides of the difference equation yields
Y + (z ) − 1 −1 +
1 −2 +
z Y (z ) + y (−1) −
z Y (z ) + y (−1)z −1 + y (−2) = z −1 X + (z )
6
6 Substituting the initial conditions yields
Y + (z ) − 1 −1 +
1 −2 +
z Y (z ) + 6 −
z Y (z ) + 6z −1 = z −1 X + (z )
6
6 Solving for Y + (z ) yields
Y + (z ) =
=
= 1 + z −1
z −1
X + (z )
1 −2 +
1 −1
1−
− 6z
1 − 6 z − 1 z −2
6
1 −1
6z
2 z2 z +z
− 1z −
6 9
5z
1
2 z− 1+
6
4
−5z
z+1
3 + z
− 1z
6
3
2 z2
+ z−1 · z
z−1 − 1
6 + 1
−3
5
10
,
1+
z−2
z+1
3 z  > 1 which leads to
y (n) = 9
5 1
2 n − 4
5 − n 1
3 33
−
25 u(n) + 1
2 n−1 + 1
10 − 1
3 n−1 u(n − 1) (d) Compare the two results straightforwardly.
(e) We only need to ﬁnd out the zerostate response in the time domain. The impulse response sequence can be
solved from the general solution by plugging the corresponding initial conditions h(0) = 0 and h(1) = 1,
which results in
n 1
2 6
5 h(n) = 6
5 − − n 1
3 u(n − 1) Hence, the zerostate response can be found by
6
5 1
2 n 6
5 yzs (n) =
= k=1 =
= 1
2 6
5 n − 33
−
25 − k 1
2 −1
2
1− 6
5
− 6
5 n+1 − 1
2 1
2 1
3 n u(n − 1) ⋆ u(n)
− 1
3 k u(n − 1) 1
1
−3 − −3
1+ 1
3 n−1 1
10 − 33
−
25 1
2 + 1
3 n+1 u(n − 1)
n−1 u(n − 1) Therefore, by part (a), the complete response is
y (n) = 9
5 1
2 n − 4
5 − 1
3 n u(n) + n−1 + 1
10 − 1
3 n−1 u(n − 1) which coincides with the result obtained in part (c).
(f) One can use the z transform, or use the timedomain convolution, or use the unilateral z transform to ﬁnd
1
out the zerostate or complete response to x(n) = ( 3 )n u(n − 2). Here we only give the ﬁrst method. The November 21, 2011 DRAFT 5 z transform of the input sequence is
X (z ) = z −2
z
·
1,
9 z−3 1
3 z  > The z transform of the zerostate response can be expressed as
Yzs (z ) = z
z2 − 1
6z − 1
6 · z −2
z
·
9
z− 1
3 = 1
9 z− z+ 1
2 1
3 1
3 n−1 − 1
2 n−1 = z− 1
3 4
5 z− 1
2 + 1
5 z+ 1
3 + −1
,
z−1
3 z  > 1
2 which leads to
4
5 yzs (n) = 1
2 1
3 n−1 n + 1
5 1
3 − n−1 u(n − 1) Then, the complete response is
y (n) = November 21, 2011 9
5 1
2 n − 4
5 − u(n) + 4
5 + 1
5 − 1
3 n−1 − 1
3 n−1 u(n − 1) DRAFT 6 Problem 13.9 Find the DTFTs of the following sequences:
(a) x(n) =
(b) x(n) =
(c) x(n) = 1 n−1
u(n + 1).
2
2n−1
1
u(n − 1).
3
1 n−2
−4
u(−n − 1). In each case, ﬁnd expressions for the magnitude and phase of the DTFT.
Solution:
(a) Notice that
1
2 x(n) = 4 · n+1 u(n + 1) Thus we can get
X (ejω ) = 4ejω
=
1 − 1 e−jω
2 4 · exp(jω )
5
4 − cos(ω ) · exp j arctan sin(ω )
2−cos(ω ) where the amplitude and the phase are
4 X (ejω ) = 5
4 ∠X (ejω ) = ω − arctan , − cos(ω ) sin(ω )
2 − cos(ω ) (b) Notice that
x(n) = 1
·
3 1
9 n−1 u(n − 1) Thus we can get
X (ejω ) = 1
3 1
3 · e−jω
=
1 − 1 e−jω
9 82
81 − 2
9 · exp(−jω ) · cos(ω ) · exp j arctan sin(ω )
9−cos(ω ) where the amplitude and the phase are
1
3 X (ejω ) = 82
81 − 2
9 , ∠X (ejω ) = −ω − arctan · cos(ω ) sin(ω )
9 − cos(ω ) (c) Notice that
x(n) = −64 · 4−n−1 u(−n − 1)
Thus we can get
X (ejω ) = −64 · ejω
=
1 − 4 ejω −64 · exp(jω )
17 − 8 · cos(ω ) · exp −j arctan sin(ω )
1
4 −cos(ω ) where the amplitude and the phase are
X (ejω ) = November 21, 2011 64
17 − 8 · cos(ω ) , ∠X (ejω ) = π + ω + arctan 1
4 sin(ω )
− cos(ω ) DRAFT 7 Problem 13.13 Find the DTFTs of the following sequences:
(a) x(n) =
(b) x(n) = 1 n−3
2
1 n−1
4 (c) x(n) = cos 1n
3 u(n) + π
3n u(n) + cos π
4n u(n − 1).
. + δ (n − 1) + δ (n + 1). In each case, ﬁnd expressions for the magnitude and phase of the DTFT.
Solution:
(a) Notice that
n 1
2 x(n) = 8 · u(n) + 1
·
3 n−1 1
3 u(n − 1) Thus we can get
X (ejω ) = 1 −jω
e
8
+ 3 1 −jω =
1 −jω
1 − 2e
1 − 3e
1
3 + 10
9 − 2
3 8
5
4 − cos(ω ) · exp j arctan sin(ω )
2−cos(ω ) · exp(−jω ) cos(ω ) · exp j arctan sin(ω )
3−cos(ω ) Let
a1 − arctan θ1 8
5
4 , 1
3 a2 10
9 − cos(ω ) sin(ω )
2 − cos(ω ) , θ2 − 2
3 cos(ω ) −ω − arctan sin(ω )
3 − cos(ω ) Then, X (ejω ) can be rewritten as
X (ejω ) = a1 ejθ1 + a2 ejθ2
= [a1 cos(θ1 ) + a2 cos(θ2 )] + j [a1 sin(θ1 ) + a2 sin(θ2 )]
a2 + 2a1 a2 cos(θ1 − θ2 ) + a2 · e
1
2 = j arctan a1 sin(θ1 )+a2 sin(θ2 )
a1 cos(θ1 )+a2 cos(θ2 ) Hence, the amplitude and the phase are
X (ejω ) = a2 + 2a1 a2 cos(θ1 − θ2 ) + a2
1
2 ∠X (ejω ) = arctan a1 sin(θ1 ) + a2 sin(θ2 )
a1 cos(θ1 ) + a2 cos(θ2 ) (b) Notice that
x(n) = 4 · 1
4 n u(n) + cos π
n
3 Thus we can get
X (ejω ) =
= November 21, 2011 4
1− 1 −jω
4e +π δ ω− π
π
+δ ω+
3
3 4
17
16 − 1
2 · cos(ω ) · exp j arctan sin(ω )
4−cos(ω ) +π δ ω− π
π
+δ ω+
3
3 DRAFT 8 Let
a= 4
17
16 − 1
2 , θ = − arctan · cos(ω ) sin(ω )
4 − cos(ω ) The amplitude is X (ejω ) = a, and the phase is jω ∠X (e ) = (c) We can get
X (ejω ) = π δ ω − a2 + 2πa cos(θ) + π 2 , otherwise arctan a sin(θ )
a cos(θ )+π θ, π
π
+δ ω+
4
4 ω = ±π
3 , ω = ±π
3
otherwise + e−jω + ejω = π δ ω − π
π
+δ ω+
4
4 + 2 cos (ω ) The amplitude is
X (ejω ) = 2 cos (ω ) + π δ ω − π
π
+δ ω+
4
4 and the phase is
∠X (ejω ) = 0 November 21, 2011 DRAFT 9 Problem 13.21 Find the DTFTs of the following sequences:
(a) x(n) =
(b) x(n) = sin( π n)
3
.
n
π
sin( 3 (n−1))
.
n−1 Solution: See example 13.7 Lowpass DTFT on the textbook.
(a) We can rewrite x(n) as
x(n) =
and identify ωc = π
3. π sin π n
3
·
π
3
n
3 Then, we can get
X (ejω ) = π · rect where we deﬁne the rectangular function rect(·) as 1 , ω
rect ωc
0 , ω
π
3 ω  < ω c
ω c ≤ ω  ≤ π (b) Notice that this is a timeshifted version of part (a). We can immediately get
X (ejω ) = π · rect November 21, 2011 ω
π
3 · e−jω DRAFT 10 Problem 13.23 Invert the following DTFTs:
(a) X (ejω ) = cos( π ω ).
2
(b) X (ejω ) =
(c) X (ejω ) = 1
.
1− 1 e−jω
2
π
cos( 3 ω ) · sin( π ω ).
3 (d) X (ejω ) = cos2 ( π ω ).
6
(e) X (ejω ) = sin( π ω ) · cos2 ( π ω ).
3
4
Solution:
(a) Recall the Euler formula. Then, mathematically, we can get
x(n) = 1
π
π
δ n+
+δ n−
2
2
2 However, from the signal processing point of view, where we interpret the delay
apparently, the noninteger number π
2 π
2 as multiples of samples, is not a legal one. In other words, in general we shall not see this kind of DTFT in signal processing area.
(b) We can get
x(n) = 1
2 n u(n) (c) We notice that
X (ejω ) = 2π
ω
3 1
sin
2 which leads to
x(n) = j
2π
δ n+
4
3 −δ n− 2π
3 Again, this is not a legal signal.
(d) We notice that
X (ejω ) = cos π
3ω +1 2 which leads to
x(n) = 1
π
π
δ n+
+ 2δ (n) + δ n −
4
3
3 Again, this is not a legal signal.
(e) We notice that
X (ejω ) = sin cos
π
ω·
3 π
2ω 2 +1 = 1
π
1
sin
ω+
sin
2
3
4 5π
π
ω − sin
ω
6
6 which leads to
x(n) = j
π
π
δ n+
−δ n−
2
3
3 + j
5π
δ n+
4
6 −δ n 5π
6 −δ n+ π
π
+δ n−
6
6 Again, this is not a legal signal. November 21, 2011 DRAFT 11 Problem 13.35 The DTFTs of two sequences {x(n), y (n)} are shown in Fig. 13.18. Determine the sequences.
Find also their energies. Solution: Notice that X (ejω ) can be expressed as
ω X (ejω ) = 2 · rect π
4 ω − rect π
8 Therefore, the sequence x(n) is
x(n) = 1
π
1
π
sinc
n − sinc
n
2
4
8
8 where the sinc(·) function is deﬁned as
sin(x)
x sinc(x) =
The energy of the signal is
Ex = 1
2π π X (ejω )2 dω =
−π 5
8 jω Similarly, we can express Y (e ) as
Y (ejω ) = 2 · rect ω − 2 · tri π
4 ω
π
4 where we deﬁne the triangular function as
1− ω 
ωc The sequence corresponding to the triangular function tri ω
ωc tri ω
ωc x(n) = November 21, 2011 rect ω
ωc is ωc
ωc
· sinc2
n
2π
2 DRAFT 12 Therefore, the sequence y (n) is
y (n) = 1
π
1
π
sinc
n − sinc2
n
2
4
8
8 The energy of the signal is
Ey = November 21, 2011 1
2π π Y (ejω )2 dω =
−π 1
3 DRAFT ...
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 Cos, Recurrence relation, Fibonacci number, Rectangular function

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