Homework # 1
Ch.2 # 21, 23, 28, 30, 32, 33, 35, 38, 40
Dr. Ray Kwok
SJSU
Fall 2013
2.21
A discrete-time signal x[n] is shown here.
Sketch and label carefully each of the following signals.
(a)
x[n-2]
shift 2 to the right
(b)
x[4-n]
flip about the 0-axis and shift 4 to the right
(c)
x[2n]
shrink by a factor of 2
(d)
x[n]u[2-n]
cut off n>2
(e)
x[n-1]δ[n-3]
x[2] when n=3, x[n]=0 for all other n.
Dr. Ray Kwok
Homework # 1
2
2.23
For each of the following systems, determine whether the system is (1) stable, (2) causal, (3) linear,
and (4) time invariant.
(a) y[ n ]. ≡ T ( x[ n]) = cos(πn) ⋅ x[ n]
Stable –x[n] is defined and cosine is finite.
Causal – y[n] only depends on present value of n.
Linear.
y1[n] ≡ T ( x1[n]) = cos(πn) ⋅ x1[n]
y2 [n] ≡ T ( x2 [n]) = cos(πn) ⋅ x2 [ n]
ay1[ n] + by2 [ n] = cos(πn) ⋅ (ax1[ n] + bx2 [n]) = T ( ax1[n] + bx2 [n])
It’s NOT time invariant.
cos(πn) = (− 1)
n
y[n] = T ( x[n]) = cos(πn) ⋅ x[n] = (− 1) x[n]
n
y[n − 1] = (− 1)
n −1
Dr. Ray Kwok
x[n − 1] ≠ (− 1) x[n − 1]
n
Homework # 1
3
2.23
(b)
T ( x[n]) = x[n 2 ]
Stable –x[n] is defined and so is x[n2].
It is NOT causal – y[n] only depends on future value of n2.
Linear.
y1[n] ≡ T ( x1[n]) = x1[n 2 ]
y2 [n] ≡ T ( x2 [n]) = x2 [ n 2 ]
ay1[n] + by2 [ n] = ax1[ n 2 ] + bx2 [n 2 ] = T (ax1[n] + bx2 [ n])
It’s NOT time invariant.
y[n] = T ( x[n]) = x[n 2 ]
y[n − no ] = x[(n − no ) ] ≠ x[n 2 − no ]
2
Dr. Ray Kwok
Homework # 1
4
2.23
∞
(c)
T ( x[ n]) = x[n]∑ δ [n − k ]
k =0
Stable –x[n] is defined and so is the delta function.
Causal – y[n] only depends on present value of n.
Linear.
∞
y1[n] ≡ T ( x1[ n]) = x1[n]∑ δ [ n − k ]
k =0
∞
y2 [n] ≡ T ( x2 [n]) = x2 [n]∑ δ [n − k ]
k =0
∞
ay1[ n] + by2 [n] = (ax1[n] + bx2 [ n])∑ δ [n − k ] = T ( ax1[n] + bx2 [ n])
k =0
It’s NOT time invariant.
∞
y[n] = T ( x[n]) = x[n]∑ δ [n − k ]
k =0
∞
∞
k =0
k =0
y[n − no ] = x[n − no ]∑ δ [n − no − k ] ≠ x[n − no ]∑ δ [ n − k ]
Dr. Ray Kwok
Homework # 1
5
2.23
(d)
T ( x[n]) =
∞
∑ x[k ]
k = n −1
NOT stable – the infinite sum can diverge easily.
NOT Causal – it sums all future k > n.
Linear.
∞
∑ x [k ]
y1[n] ≡ T ( x1[ n]) =
1
k = n −1
∞
∑ x [k ]
y2 [n] ≡ T ( x2 [n]) =
2
k = n −1
ay1[ n] + by2 [ n] =
∞
∑ (ax [k ] + bx [k ]) = T (ax [n] + bx [n])
1
2
1
2
k = n −1
It’s time invariant.
y[n] = T ( x[n]) =
∞
∑ x[k ]
k = n −1
y[n − no ] =
Dr. Ray Kwok
∞
∞
k = n − no −1
k = n −1
∞
∑ x[k ] = ∑ x[k − n ] = ∑ x[q]
o
q = n −1− no
Homework # 1
6
2.28
Four input-output pairs of a particular system S are specified in Fig.
Can system S be time-invariant? Explain.
(a)
No.
x4 = δ [ n ]
y4 = δ [n] + δ [n − 1]
x3 = δ [n − 1] + ⋅ ⋅ ⋅
y3 = δ [n − 1] + ⋅ ⋅ ⋅ But it doesn’t.
or
x4 = δ [ n ]
y4 = δ [n] + δ [n − 1]
x2 = δ [ n ] + ⋅ ⋅ ⋅
y2 = δ [n] + ⋅ ⋅ ⋅
Dr. Ray Kwok
But it doesn’t.
Homework # 1
7
2.28
(b)
Can system S be linear? Explain.
No, it’s not linear.
See for example,
x1 + x2 = 2( x3 + x4 )
but
Dr. Ray Kwok
y1 + y2 ≠ 2( y3 + y4 )
Homework # 1
8
2.28
(c)
Suppose (2) & (3) are input-output pairs of a
particular system S2, and the system is
known to be LTI. What is h[n], the impulse
response of the system?
One can construct a δ[n] by
x3 [n + 1] − x3 [n] + x2 [n]
= δ [ n]
2
The output h[n] would then be
Dr. Ray Kwok
y3[n + 1] − y3[n] + y2 [n]
2
(δ [n − 1] − δ [n − 5]) + y2 [n]
h[n] =
2
2δ [n − 1] + 2δ [n − 3]
h[n] =
2
h[n] = δ [n − 1] + δ [n − 3]
h[n] =
Homework # 1
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2.28
(d)
Suppose (1) is the input-output pair of an LTI system S3.
What is the output of this system for the input in the Figure?
The input x[n] can be construct by
x[n] = x1[n] + x1[n − 2]
So the output should be
y[n] = y1[n] + y1[n − 2]
y[n] = δ [n] + δ [n − 2] + δ [n − 3] + δ [n − 4] + δ [n − 5] + δ [n − 6]
y[n] = δ [n] + u[n − 2] − u[n − 6]
Dr. Ray Kwok
Homework # 1
10
2.30
Consider the cascade connection of two LTI systems in the figure
(a)
Determine and sketch w[n] if x[n]=(-1)nu[n]. Also determine the overall output y[n].
y[n] = w[n] ∗ h2 [n] =
∞
∑ w[n − k ]h [k ]
2
k = −∞
w[n] = x[n] ∗ h1[n] =
y[n < −3] = 0
y[−3] = 1
y[−2] = 1
y[−1] = 2
y[0] = 2
y[1] = 1
y[2] = 1
y[n > 2] = 0
∞
∑ x[k ]h [n − k ]
1
k = −∞
w[n < 0] = 0
w[0] = 1
w[1] = 0
w[2] = 1
w[n ≥ 3] = 0
Dr. Ray Kwok
Homework # 1
11
2.30
(b)
Determine and sketch the overall impulse
response of the cascade system. i.e., plot
the output y[n] = h[n] (when x[n] = δ[n]).
h[ n] = h1[n] ∗ h2 [n] =
∞
∑ h [n − k ]h [k ]
1
2
k = −∞
h[n < −3] = 0
h[−3] = 1
h[−2] = 2
h[−1] = 3
h[0] = 4
h[1] = 3
h[2] = 2
h[3] = 1
h[n > 3] = 0
Dr. Ray Kwok
Homework # 1
12
2.30
(c)
Now consider the input x[n] = 2δ[n] + 4δ[n-4] – 2δ[n-12]. Sketch w[n].
w[n] = x[n] ∗ h1[n] =
∞
∑ x[k ]h [n − k ]
1
k = −∞
w[n] =
∞
∑ (2δ [k ] + 4δ [k − 4] − 2δ [k − 12])h [n − k ]
1
k = −∞
w[n] = 2h1[n] + 4h1[n − 4] − 2h1[n − 12]
Dr. Ray Kwok
Homework # 1
13
2.30
(d)
For the input in part (c), write an expression for the output y[n] in terms of the overall impulse
response h[n] as defined in part (b). Make a carefully labeled sketch of your answer.
y[ n] = x[n] ∗ h[n] =
∞
∑ x[k ]h[n − k ]
k = −∞
y[ n] = 2h[n] + 4h[n − 4] − 2h[n − 12]
Dr. Ray Kwok
Homework # 1
14
2.32
For
(a)
( )
X e jω =
(This is the Example 2.21 in the textbook.)
1
1 − ae − jω

1
1
1
1
.
+
Re X e jω = X e jω + X * e jω = 
2
2 1 − a (cos ω − j sin ω ) 1 − a(cos ω + j sin ω )
[ ( )] [ ( )
[ ( )]
Re X e jω =
(b)
( )]
1 − a cos ω
1 − a cos ω
=
2
(1 − a cos ω ) + a 2 sin 2 ω 1 − 2a cos ω + a 2

1
1 
1
1
−
Im. X e jω =
X e jω − X * e jω =
2j
2 j 1 − a (cos ω − j sin ω ) 1 − a (cos ω + j sin ω ) 
[ ( )]
[ ( )]
Im X e jω =
(c)
with -1 < a < 0, determine and sketch the following as a function of ω:
[ ( )
( )]
− a sin ω
1 − 2a cos ω + a 2



1
1


X. e jω = X e jω X * e jω = 
 1 − a(cos ω − j sin ω )  1 − a(cos ω + j sin ω ) 
1
X e jω =
1 − 2a cos ω + a 2
( )
( ) ( )
( )
(d)
( )
∠.X e
Dr. Ray Kwok
jω
[ ( )] = tan
[ ( )]
 Im X e jω
= tan 
jω
 Re X e
−1
−1
 − a sin ω 


 1 − a cos ω 
Homework # 1
15
2.33
Consider an LTI system defined by the difference equation
(a)
Determine the impulse response of this system.
y[n] = -2x[n] + 4x[n-1] – 2x[n-2]
h[n] = −2δ [n] + 4δ [n − 1] − 2δ [n − 2]
(b)
Determine the frequency response of this system. Express your answer in the form of
H (e jω ) = A(e jω )e − jωnd
where A(ejω) is a real function of ω. Explicity specify A(ejω) and the delay of nd of this system.
h[n] = −2δ [n] + 4δ [n − 1] − 2δ [n − 2]
H (e jω ) = −2 + 4e − jω − 2e − j 2ω
H (e jω ) = −2e − jω (e jω − 2 + e − jω ) = −2e − jω (2 cos ω − 2) = 4e − jω (1 − cos ω ) = 4e − jω (2 sin 2 (ω / 2))
H (e jω ) = 8e − jω sin 2 (ω / 2)
Delay nd = 1
Dr. Ray Kwok
Homework # 1
16
2.33
(c)
Sketch a plot of the magnitude |H(ejω)| and a plot of the phase
H (e jω ) = 8e − jω sin 2 (ω / 2)
H (e jω ) = 8 sin 2 (ω / 2)
∠H (e jω ) = −ω
Dr. Ray Kwok
Homework # 1
17
2.33
(d)
j 0.5πn
Suppose that the input to the system is x1[n] = 1 + e
Use the frequency response function to determine the corresponding output y1[n].
See section 2.9.6 in the textbook.
If
x[n] = e jωn
y[ n] = H (e jω )e jωn
Here
x1[ n] = 1 + e j 0.5πn = e j 0 n + e j 0.5πn
y1[n] = H (e j 0 )e j 0 n + H (e j 0.5π )e j 0.5πn
with
H (e jω ) = 8e − jω sin 2 (ω / 2)
y1[n] = 0 + 8e − jπ / 2 sin 2 (π / 4)e − jnπ / 2
y1[n] = 4e − j ( n −1)π / 2
Dr. Ray Kwok
Homework # 1
18
2.33
(
j 0.5πn
)
u[n]
Now suppose that the input to the system is x2 [n] = 1 + e
Use the defining difference equation or discrete convolution to determine the corresponding
output y2[n]. Compare y1[n] and y2[n]. They should be equal for certain values of n. Over what
range of values of n are they equal?
(d)
y 2 [ n] =
∞
∞
∑ h[k ]x [n − k ] = ∑ h[k ](1 + e
j ( n − k )π / 2
2
k = −∞
k = −∞
)u[n − k ] = ∑ h[k ](1 + e
n
j ( n − k )π / 2
)
k = −∞
(
n
y2 [n] = ∑ h[k ] 1 + e j ( n − k )π / 2
For a causal system, h[n]=0 if n<0. So, n >=0 and
)
k =0
y2 [n] = −2 + 4 − 2 − 2∑ e − jkπ / 2 + 4∑ e − jkπ / 2e jπ / 2 − 2∑ e − jkπ / 2 e jπ
n
y 2 [ n] = 4 j ∑ e
k =0
∞
(
k =0
∞
∞
y2 [n] = ∑ h[k ] + ∑ h[k ]e
k =0
∞
) ∑ h[k ](1 + e
y2 [n] = ∑ h[k ] 1 + e j ( n − k )π / 2 −
j ( n − k )π / 2
− jkπ / 2
e
jnπ / 2
)
−
∞
∑ h[k ](1 + e
j ( n − k )π / 2
)
k = n +1
y2 [n] = H (e j 0 ) + H (e jπ / 2 )e jnπ / 2 −
∞
∑ h[k ](1 + e
j ( n − k )π / 2
)
k = n +1
y2 [n] = y1[n] −
∑ h[k ](1 + e
 1 − e − jnπ / 2 

= 4 j 
− jπ / 2 
1
−
e


k = n +1
k =0
∞
− jkπ / 2
j ( n − k )π / 2
h[n] = −2δ [n] + 4δ [n − 1] − 2δ [n − 2]
h[n > 2] = 0
∴ y2 [ n] = y1[n]
)
for n>2
k = n +1
Dr. Ray Kwok
Homework # 1
19
2.35
An LTI system has the input and impulse response given by the following plots:
(a)
Use discrete convolution to determine the output of the system y[n] = x[n]*h[n] for the above input.
Give your answer as a carefully labeled sketch of y[n] over a range sufficient to define it completely.
y[n] = x[n] ∗ h[ n] =
∞
∑ x[k ]h[n − k ]
k = −∞
y[n<0]=0
y[0] = 1
y[1] = 0
y[2] = 1
y[3] = 0
y[4] = 5
y[5] = 0
y[6] = 1
y[7] = 0
y[8] = 1
y[n>8] = 0
Dr. Ray Kwok
Homework # 1
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2.35
b)
The deterministic autocorrelation of a signal x[n] is defined in Eq.(2.188) as cxx[n] = x[n]*x[-n].
The system defined by figures above is a matched filter for the input shown. Noting that h[n] =
x[-(n-4)], express the output in part (a) in terms of cxx[n].
chh [l] ≡
∞
∑ h[k ]h[l + k ]
k = −∞
h[n] = x[−(n − 4)]
y[n] = x[n] ∗ h[n] =
y[n] =
∞
∞
k = −∞
k = −∞
∑ x[k ]h[n − k ] = ∑ x[k ]x[−((n − k ) − 4)]
∞
∑ x[k ]x[(4 − n) + k ] = c
xx
[ 4 − n]
k = −∞
Dr. Ray Kwok
Homework # 1
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2.35
c)
Determine the output of the system whose impulse
response is h[n] when the input is
x[n] = u[n+2]. Sketch your answer.
y[n] = x[ n] ∗ h[n] =
∞
∞
k = −∞
k = −∞
∑ x[k ]h[n − k ] = ∑ u[k + 2]h[n − k ]
⋅⋅⋅
Dr. Ray Kwok
Homework # 1
y[n<-2]=0
y[-2] = 1
y[-1] = 2
y[0] = 3
y[1] = 2
y[2] = 3
y[n>2] = 3
22
2.38
Consider the cascade of two LTI systems in the figure
The impulse responses of the two systems are:
h1[n] = u[n − 5]
1, 0 ≤ n ≤ 4
h2 [n] = 
0, otherwise
(a)
Make a sketch showing both h2[k] and h1[n-k] (for some arbitrary n<0) as function of k.
Dr. Ray Kwok
Homework # 1
23
2.38
b)
Determine h[n] = h1[k]*h2[n] , the impulse response of the overall system. Give your answer as
an equation (or set of equations) that define h[n] for − ∞ < n < ∞
or as a carefully labeled plot of h[n] over a range sufficient to define it completely.
h[n] = h1[n] ∗ h2 [n] =
∞
∑ h [k ]h [n − k ]
1
2
k = −∞
y[n<5]=0
y[5] = 1
y[6] = 2
y[7] = 3
y[8] = 4
y[9] = 5
y[n>9] = 5
Dr. Ray Kwok
Homework # 1
24
2.40
Determine which of the following signals is periodic. If a signal is periodic, determine its period.
x[n] = e j (2πn / 5 )
Periodic.
(b)
x[n] = sin(πn / 19)
Periodic.
(c)
x[n] = ne jπn
Not periodic because of the “n” in front of the exponential.
(c)
x[n] = e jn
Not periodic.
(a)
x[n + N ] = e j 2π ( n + N ) / 5 = x[n]
2πN
= 2π
5
N =5
π

x[n + N ] = sin (n + N )  = x[n]
 19

πN
= 2π
19
N = 38
x[n + N ] = e j ( n + N ) = x[ n]
N = 2π
Dr. Ray Kwok
Homework # 1
which is not an integer.
25