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2-1
Chapter 2 2.1-1
00
0
/ 2 2 ( )
/ 20
sinc( )0 otherwise
jj T j m n f t jn T
Ae n mAec e dt Ae m nT
φφπ φ−
−
⎧ == = − = ⎨
⎩∫
2.1-2
0 0
0
0
/ 4 / 2
0 / 40 0 0
( ) 02 2 2 2cos ( ) cos sin
2T T
n T
c v tnt nt A nc A dt A dt
T T T nπ π π
π
=
= + − =∫ ∫
n 0 1 2 3 4 5 6 7 nc 0 2 /A π 0 2 / 3A π 0 2 / 5A π 0 2 / 7A π
arg nc 0 180± ° 0 180± ° 2.1-3
0
0
/ 2
200 0 0
( ) / 2
2 2 2cos sin (cos 1)( )
T
n
c v t A
At nt A Ac A dt n nT T T n n
π π ππ π
= =
⎛ ⎞= − = − −⎜ ⎟
⎝ ⎠∫
n 0 1 2 3 4 5 6nc 0.5A 0.2A 0 0.02A 0 0.01A 0
arg nc 0 0 0 0 2.1-4
0 / 2
0 00 0
2 2cos 0T tc A
T Tπ
= =∫
2-2
( ) ( )
[ ]
00
/ 2/ 2 0 0
00 0 0 0 0 0 0
sin 2 / sin 2 /2 2 2 2cos cos4( ) / 4( ) /
/ 2 1sinc(1 ) sinc(1 )
0 otherwise2
TT
n
n t T n t Tt nt Ac A dtT T T T n T n T
A nA n n
π π π ππ ππ π π π− +⎡ ⎤
= = +⎢ ⎥− +⎣ ⎦= ±⎧
= − + + = ⎨⎩
∫
2.1-5
0
0
/ 2
00 0
( ) 02 2sin (1 cos )
T
n
c v tnt Ac j A dt j n
T T nπ π
π
= =
= − = − −∫
n 1 2 3 4 5 nc 2 /A π 0 2 / 3A π 2 / 5A π
arg nc 90− ° 90− ° 90− ° 2.1-6
0 ( ) 0c v t= =
( ) ( )
[ ]
00
/ 2/ 2 0 0
00 0 0 0 0 0 0
sin 2 / sin 2 /2 2 2 2sin sin4( ) / 4( ) /
/ 2 1sinc(1 ) sinc(1 )
0 otherwise2
TT
n
n t T n t Tt nt Ac j A dt jT T T T n T n T
jA nAj n n
π π π ππ ππ π π π− +⎡ ⎤
= − = − −⎢ ⎥− +⎣ ⎦= ±⎧
= − − − + = ⎨⎩
∫∓
2.1-7
]0 00 0
0
/ 2
0 / 20
1 ( ) ( )T Tjn t jn t
n Tc v t e dt v t e dt
Tω ω− −⎡= +⎢⎣∫ ∫
0 00 0 0 0
0
00
/ 2 / 20/ 2 0
/ 2
0
where ( ) ( / 2)
( )
T Tjn t jn jn T
T
T jn tjn
v t e dt v T e e d
e v t e dt
ω ω λ ω
ωπ
λ λ− − −
−
= +
= −
∫ ∫
∫
since 1 for even , 0 for even jnne n c nπ = =
2-3
2.1-8
Total power: / 2
2 2 2 2
0 / 2
1 1( ) ( )T T
TotalT T
P v t dt A dt A dt AT T
⎡ ⎤= = + − =⎢ ⎥
⎣ ⎦∫ ∫ ∫
Square wave has odd harmonics, and the amplitudes are given in Table T.2 as 2
nj Ac
nπ= − and the coefficients trigonometric Fourier series is 2 nc . The power in a
periodic sine wave is 2
sinewave 2mAP = then the sum of the powers of the odd harmonics are
2 2
harmonics 2 20 odd
2 16 1 1 11 ...2 2 9 25
Nn
n
c APnπ=
⎛ ⎞= = + + +⎜ ⎟
⎝ ⎠∑
2
2 2oddodd
2 2 2total odd
16 1 1 11 ...2 9 25 8 1 1 190% 0.9 1 ...
9 25
AnP
P A nπ
π
⎛ ⎞+ + +⎜ ⎟ ⎛ ⎞⎝ ⎠⇒ ⇒ = = = + + +⎜ ⎟
⎝ ⎠
3n⇒ = i.e. having the first and 3rd harmonic will represent 90% of the signal’s power and 27n = will represent 99% (98.553) of the signal’s power. 2.1-9
2 2 2 2 2 20 0 0 0 0 0 0 0
1
0
22 2 2 2
22 2 2 2 2 2
2 2 sinc 2 sinc 2 2 sinc3
1where 4
1 1 1 31 2sinc 2sinc 2sinc 0.2316 4 2 4
2 1 1 3 5 3 71 2sinc 2sinc 2sinc 2sinc 2sinc 2sinc16 4 2 4 4 2 4
nn
P c c Af Af f Af f Af f
f
Af P A
Af P
τ τ τ τ τ τ τ
τ
τ
τ
∞
=
= + = + + + +
=
⎡ ⎤> = + + + =⎢ ⎥⎣ ⎦
⎡ ⎤> = + + + + + +⎢⎣
∑
2
22 2 2
0.24
1 1 11 2sinc 2sinc 0.212 16 4 2
A
Af P Aτ
=⎥⎦
⎡ ⎤> = + + =⎢ ⎥⎣ ⎦
2-4
2.1-10
0 0
0
2
2 2/ 2 / 2
/ 2 00 0 0 0
2 2 2
2 2 2
0 02 2 2
0 even
2 oddn
41 2 4 1a) 1 13
4 4 42 2 2 0.332 so / 99.6%9 25
8 8 8b) ( ) cos cos3 cos59 25
n
T T
T
nc
n
t tP dt dtT T T T
P P P
v t t t
π
π π π
ω ω ωπ π π
−
⎧⎪= ⎨⎛ ⎞⎜ ⎟⎪⎝ ⎠⎩
⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ ′= + + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
′ = + +
∫ ∫
0t
2.1-11
( )0
0
2 2 2/ 2 2
/ 20
0 even2 odd
1 2 2 2a) 1 1 2 0.933 so / 93.3%3 5
n
T
T
nc j n
n
P dt P P PT
π
π π π−
⎧⎪= −⎨⎪⎩
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞′ ′= = = + + = =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
∫
( ) ( ) ( )
( ) ( ) ( )
0 0 0
0 0 0
4 4 4b) ( ) cos 90 cos 3 90 cos 5 903 5
4 4 4sin sin 3 sin 53 5
v t t t t
t t t
ω ω ωπ π π
ω ω ωπ π π
′ = − ° + − ° + − °
= + +
2.1-12
0
2
00 0
1/ 2 01 11/ 2 03
T
n
ntP dt cn nT T π
=⎛ ⎞ ⎧= = = ⎨⎜ ⎟ ≠⎩⎝ ⎠
∫
2-5
4 4
4 4 4 odd
2 2 1 1 1 12 21 3 5 3n
Pnπ π
∞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∑
2 2
2 2 2
1 1 1 4 1 1Thus, 1 2 3 2 3 4 6
π π⎛ ⎞+ + + = − =⎜ ⎟⎝ ⎠
2.1-13
0
2/ 2
200 0
0 even2 4 11(2 / ) odd3
T
n
ntP dt cn nT T π
⎛ ⎞ ⎧= − = = ⎨⎜ ⎟
⎩⎝ ⎠∫
2 2
2 2 2 21
1 1 1 2 1 1 1 122 2 4 4 1 2 3 3n
Pnπ π
∞
=
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + = + + + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
∑
4 4
4 4 4 4
1 1 1 1Thus, 1 3 5 2 2 3 96
π π+ + + = =
⋅
2.2-1
( )( )
( )( ) [ ]
/ 2
0( ) 2 cos cos 2
sin 2 sin 22 22 sinc( 1/ 2) sinc( 1/ 2)
22 2 2 2
tV f A ft dt
f f AA f ff f
τ
π πτ τπ πτ τ
π πτ
τ τπ π τ τ τπ π
+
+
=
⎡ ⎤−⎢ ⎥= + = − + +⎢ ⎥
−⎢ ⎥⎣ ⎦
∫
2.2-2
( )( )
( )( ) [ ]
/ 2
0
2 2
2 2
2( ) 2 sin cos 2
sin 2 sin 22 22 sinc( 1) sinc( 1)
22 2 2 2
tV f j A ft dt
f f Aj A j f ff f
τ
π πτ τπ πτ τ
π πτ
τ τπ π τ τ τπ π
+
+
= −
⎡ ⎤−⎢ ⎥= − − = − − − +⎢ ⎥
−⎢ ⎥⎣ ⎦
∫
2-6
2.2-3
2 220
2( ) 2 cos 2sin 1 1 sinc( ) 2
t AV f A A tdt A fτ τ ωτω τ τ
τ ωτ⎡ ⎤⎛ ⎞ ⎛ ⎞= − = − + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
∫
2.2-4
20
2( ) 2 sin (sin cos )( )
(sinc 2 cos 2 )
t AV f j A t dt j
Aj f ff
τ τω ωτ ωτ ωττ ωτ
τ π τπ
= − = − −
= − −
∫
2.2-5
3 303 1 10 sinc( 1 10 )
1 10t t fτ
− −−
⎛ ⎞ ⎛ ⎞Π = Π ⇔ × × ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ and we want sinc( ) 1/30x ≤ . We find
the frequency of the sinc function where the maximum value of the sidelobe is 1/30.
0
30
sin( ) =sinc( ) and max values occur at 1.5 , 2.5 , etc.
1 1 sin 9.59.5 0.335 9.5530 9.5thus 1 10 9.55 9.55 kHz.
xV f xx
x fx
f f
π π π
π τπ π
−
=
= ⇒ = ⇒ = ⇒ × ≈
× × = ⇒ >
To check, 3 3
30 3 3
sin( 9.55 10 1 10 )( 9.55 10 ) 0.0392 1/ 309.55 10 1 10
V f ππ
−
−
× × × ×= × = − ≈
× × × ×
2.2-6 We use a strategy similar to Prob. 2.2-5. 2sinc ( ) 1/ 30 sinc( ) 1/ 30 0.183x x⇒ ≤ ⇒ ≤ =
30 0 0
With 1.6 sinc( ) 0.189 and sinc( 1.65) 0.172choose 1.65 1.65 1.65 /1 10 1.650 kHz.
x x xx f f fτ −
= ⇒ = − = = −
⇒ ≥ ⇒ × = ⇒ = × ⇒ ≥
2-7
2.2-7
22
2
1( ) sinc 22 2
1 1 1sinc 22 2 4 2
fv t WtW W
fWt dt df dfW W W W
∞ ∞ ∞
−∞ −∞ −∞
⎛ ⎞= ↔ Π⎜ ⎟⎝ ⎠
⎛ ⎞= Π = =⎜ ⎟⎝ ⎠∫ ∫ ∫
2.2-8
( )2 2 22
2 20 0
22 arctan2 (2 )
Wbt A A A WE Ae dt E dfb b f b b
ππ π
∞ − ′= = = =+∫ ∫
50% / 22 2arctan84% 2 /
W bE WW bE b
ππππ
=′ ⎧= = ⎨ =⎩
2.2-9
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
j t
j t
v t w t dt v t W f e df dt
W f v t e dt df W f V f df
ω
ω
∞ ∞ ∞
−∞ −∞ −∞
∞ ∞ ∞− −
−∞ −∞ −∞
⎡ ⎤= ⎢ ⎥⎣ ⎦⎡ ⎤= = −⎢ ⎥⎣ ⎦
∫ ∫ ∫
∫ ∫ ∫
22( ) *( ) when ( ) is real, so ( ) ( ) *( ) ( )V f V f v t v t dt V f V f df V f df∞ ∞ ∞
−∞ −∞ −∞− = = =∫ ∫ ∫
2.2-10
2 2 2 ( )( ) ( ) ( ) ( )
Let ( ) ( ) so ( ) ( ) and ( ) ( )
Hence ( ) ( ) ( ) ( )
j ft j ft j f tw t e dt w t e dt w t e dt W f
z t w t Z f W f W f Z f
v t z t dt V f Z f df
π π π∗ ∗∞ ∞ ∞∗ − − − ∗
−∞ −∞ −∞
∗ ∗ ∗
∞ ∞
−∞ −∞
⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= = − = −
= −
∫ ∫ ∫
∫ ∫
2.2-11
1sinc so sinc
2 2( ) sinc ( ) for 2 2
t fA Af AtA A A
t fv t V f Aτ ττ τ
⎛ ⎞ ⎛ ⎞Π ↔ ↔ Π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞= ↔ = Π =⎜ ⎟⎝ ⎠
2.2-12
[ ]
[ ]
[ ]
cos sinc( 1/ 2) sinc( 1/ 2)2
( )so sinc( 1/ 2) sinc( 1/ 2) cos cos2
Let and 2 ( ) sinc(2 1/ 2) sinc(2 1/ 2)
t t BB f f
B f f f ft t B B
B A W z t AW Wt Wt
π τ τ ττ ττ π πτ τ
τ τ τ ττ
⎛ ⎞Π ↔ − + +⎜ ⎟⎝ ⎠
− −⎛ ⎞ ⎛ ⎞− + + ↔ Π = Π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= = ⇒ = − + +
2-8
2.2-13
[ ]
[ ]
[ ]
2sin sinc( 1) sinc( 1)2
2 ( ) 2so sinc( 1) sinc( 1) sin sin2
Let and 2 ( ) sinc(2 1) sinc(2 1)
t t BB j f f
B f f f fj t t B B
B jA W z t AW Wt Wt
π τ τ ττ ττ π πτ τ
τ τ τ ττ
⎛ ⎞Π ↔ − − + +⎜ ⎟⎝ ⎠
− −⎛ ⎞ ⎛ ⎞− − + + ↔ Π = − Π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= − = ⇒ = − + +
2.2-14
( )( )
( )
22 2 2 2 2 2
222
22 2 0 2 2
2
2 30 2 2
2 4 /(2 ) (2 ) (2 )
1 / 22
1 1Thus, 2 2 4
b t a t
a t
b a ae eb f a f a f
a a dfe dt dfa a f a f
dxa a aa x
π
π
π ππ π π
ππ π
π ππ
− −
∞ ∞ ∞−
−∞ −∞
∞
↔ ⇒ ↔ =+ + +
⎛ ⎞= = = ⎜ ⎟+ ⎝ ⎠ +
⎛ ⎞= =⎜ ⎟⎝ ⎠+
∫ ∫ ∫
∫
2.3-1
( ) ( ) ( ) where v( ) ( / ) sincso Z( ) ( ) ( ) 2 sinc cos 2j T j T
z t v t T v t T t A t A ff V f e V f e A f fTω ω
τ τ τ
τ τ π−
= − + + = Π ↔
= + =
2.3-2
2 2
( ) ( 2 ) 2 ( ) ( 2 ) where v( ) ( / ) sinc( ) ( ) ( ) ( ) 2 (sinc )(1 cos 4 )j T j T
z t v t T v t v t T t a t A fZ f V f e V f V f e A f fTω ω
τ τ τ
τ τ π−
= − + + + = Π ↔
= + + = +
2.3-3
2 2
( ) ( 2 ) 2 ( ) ( 2 ) where ( ) ( / ) sinc( ) ( ) 2 ( ) ( ) 2 (sinc )(cos 4 1)j T j T
z t v t T v t v t T v t a t A fZ f V f e V f V f e A f fTω ω
τ τ τ
τ τ π−
= − − + + = Π ↔
= − + = −
2-9
2.3-4
/ 2
/ 2( ) ( )2
( ) 2 sinc 2 ( ) sincj T j T
t T t Tv t A B AT T
V f AT fT e B A T fT eω ω− −
− −⎛ ⎞ ⎛ ⎞= Π + − Π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + −
2.3-5
2 2
2 2( ) ( )4 2
( ) 4 sinc 4 2( ) sinc 2j T j T
t T t Tv t A B AT T
V f AT fT e B A T fT eω ω− −
− −⎛ ⎞ ⎛ ⎞= Π + − Π⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + −
2.3-6
/ /
1Let ( ) ( ) ( ) ( / )
1Then ( ) [ ( / )] ( / ) so ( ) ( ) ( / )d dj t a j t ad d
w t v at W f V f aa
z t v a t t a w t t a Z f W f e V f a ea
ω ω− −
= ↔ =
= − = − = =
2.3-7
2 ( )( ) ( ) ( ) ( )c c cj t j t j f f tj tcv t e v t e e dt v t e dt V f fω ω πω∞ ∞ − −−
−∞ −∞⎡ ⎤ = = = −⎣ ⎦ ∫ ∫F
2.3-8
From Ex. 2.3-2, ( ) cos 2 and ( ) sinc( )2c c
t Az t A f t Z f f fτπτ⎛ ⎞= Π = ±⎜ ⎟⎝ ⎠
Using the LPF equivalent, the frequency is then 200 kHz. With
sinc( 0.1 from tables, sinc(2.6) 0.116 and sinc(2.7)=0.095
choose 2.7 2.7 200 kHz =13.5 sx
x≤ =
⇒ ≥ ⇒ = × ⇒
πτ τ μ
2.3-9 Using the same strategy as Prob. 2.3-8, we have
2sinc( 0.1 or sinc( 0.316 from tables, sinc(0.7)==0.378 and sinc(0.8)=0.234choose 0.75 0.75 200 kHz =3.75 s
x xx
π πτ τ μ
≤ ≤
⇒ ≥ ⇒ = × ⇒
2-10
2.3-10 A system is linear if proportional changes in the input give the same proportional changes in the output.
( ) 2 10 if 1 12but if we double input to 2 14y f x x x y
x x y= = + ⇒ = ⇒ =
= ⇒ =
Not linear since doubling the input did not cause the output to double. 2.3-11
2( ) 1 1With 2 4 not linear since doubling input caused output to increaseby factor of 4
y f x x x yx y
= = ⇒ = ⇒ == ⇒ = ⇒
2.3-12
2
1
22 if we double the input, orx
x
y xdx y x= ⇒ =∫
2
1
22(2 ) 2 output also doubles linearx
x
y x dx y x= ⇒ = ⇒ ⇒∫
2.3-13 10cos(20 / 5) 10cos[20 ( 1/100)] an advance of 1/100 seconds t tπ π π+ = + ⇒ 2.3-14
6 6
6 8
( ) 10cos(2 7 10 ) and ( ) 10cos(2 7 10 / 6)The second signal is delayed by /6 radians
( ) 10cos[2 7 10 ( 1.19 10 )]
c R
R
x t t x t t
x t t
π π ππ
π −
= × × = × × −
⇒ = × × − ×
Time delay is
8 8
8 8
1.19 10 s 11.9 ns. With speed of light = 3 10 m/s
minimum path delay = 1.19 10 s 3 10 m/s=3.55 m 11.9 ftdt
−
−
= × = ×
⇒ × × × ≈
Signal can also be delayed by multiples of the period of the signal or
' 7 7 87=2 / 6 1.43 10 s distance=1.43 10 s 3 10 m/s=42.86m
7 10possible path delays = (42.86 3.55) meters
dnt n
n
π π+ = = × ⇒ × × ⊗×
⇒ +
2-11
2.3-15
-9 8
6 6 8
(30 10) ns 20ns distance = 20 10 3 10 6 meters minimum path differenceGiven period =1/ 7 10 distance=1/ 7 10 3 10 42.86 meters
possible path lengths are multiples of the signal's period or 4
tΔ = − = ⇒ Δ × × × = ⇒
× ⇒ × × × =⇒ 2.86 6 metersn + 2.3-16
[ ]
( ) ( / ) cos with 2 /
( ) sinc( ) sinc( ) sinc( 1/ 2) sinc( 1/ 2)2 2 2
c c c
c c
v t A t t fA A AV f f f f f f f
τ ω ω π π ττ τ ττ τ τ τ
= Π = =
= − + + = − + +
2.3-17
[ ]
/ 2 / 2
( ) ( / ) cos( / 2) with 2 2 /
( ) sinc( ) sinc( )2 2
sinc( 1) sinc( 1)2
c c cj j
c c
v t A t t f
e eV f A f f A f f
Aj f f
π π
τ ω π ω π π τ
τ τ τ τ
τ τ τ
−
= Π − = =
= − + +
= − − − +
2.3-18
2
2 2 2 2
2( ) ( ) cos ( )1 (2 )
1 1( ) ( ) ( )2 2 1 4 ( ) 1 4 ( )
tc
c cc c
Az t v t t v t Aef
A AZ f V f f V f ff f f f
ωπ
π π
−= = ↔+
= − + + = ++ − + +
2.3-19
/ 2 / 2
( ) ( ) cos( / 2) ( ) for 01 2
/ 2 / 2( ) ( ) ( )2 2 1 2 ( ) 1 2 ( )
/ 2 / 22 ( ) 2 ( )
tc
j j
c cc c
c c
Az t v t t v t Ae tj f
e e jA jAZ f V f f V f fj f f j f f
A Aj f f j f f
π π
ω ππ
π π
π π
−
−
= − = ≥ ↔+
−= − + + = +
+ − + +
= −− − − +
2-12
2.3-20
( )
22
( ) ( ) ( ) 2 sinc 2
sin 2 2( ) 2 (2 ) cos 2 2 sin 22 (2 )
1( ) ( ) sinc 2 cos 22
A tv t t z t z t A f
d d f AZ f A f f fdf df f f
d jAV f Z f f fj df f
ττ τ
π τ πτ π τ πτ π τπ τ π τ
τ π τπ π
⎛ ⎞= = Π ↔⎜ ⎟⎝ ⎠
⎡ ⎤ ⎡ ⎤= = −⎢ ⎥ ⎣ ⎦⎣ ⎦
−= = −−
2.3-21
2 2
22 2 2 2
2( ) ( ) ( )(2 )
1 2 8( )2 (2 ) (2 )
b t Abz t tv t v t Aeb f
d Ab j AbfZ fj df b f b f
−= = ↔+
⎡ ⎤= =⎢ ⎥− + ⎡ ⎤⎣ ⎦ +⎣ ⎦
π
ππ π π
2.3-22
( ) [ ]
2
2 3
( ) ( ) ( ) for 02
1 2( )22 2
t Az t t v t v t Ae tb j f
d A AZ fdf b j fj f b j f
π
ππ π
−= = ≥ ↔+
⎡ ⎤= =⎢ ⎥+− +⎣ ⎦
2.3-23
2 2
2 2
2 2
2 2
( ) ( / )
2 ( ) ( / )
( ) ( / )
( ) ( / )
1( ) ( )
2( ) ( ) 2
1( ) ( )2
Both results are equivalent to
bt f b
bt f b
bt f b
bt f b
v t e V f eb
d j fa v t b te edt b
d fb te V f ej df jb
bte jf e
π π
π π
π π
π π
ππ
π
− −
− −
− −
− −
= ↔ =
= − ↔
↔ =−
↔ −
2.4-1
[ ]1( ) ( ) and cos(2 ) ( ) ( )2c c cv t V f f f f f fπ δ δ⇔ ⇔ − + +
With [ ]1( )cos 2 ( ) ( ) ( )2c c cv t f t V f f f f fπ δ δ⇔ ∗ − + + We use superposition so
2-13
only a value at =
only a value at =
1 1( ) ( ) ( ) ( ) ( )2 2
and
1 1( ) ( ) ( ) ( ) ( )2 2
c
c c c
f f
c c c
f f
V f f f V f f d V f f
V f f f V f f d V f f
λ
λ
δ λ δ λ λ
δ λ δ λ λ
∞
−∞
−
∞
−∞
+
∗ − = − − = −
∗ + = − + = +
∫
∫
Therefore
[ ]1( ) cos 2 ( ) ( )2c c cv t f t V f f V f fπ∗ ⇔ − + +
2.4-2
2
0
2
0
( ) 0 0
0 22
2 2
t
y t tAtA d t
A d A t
λ λ
λ λ
= <
= = < <
= = >
∫
∫
2.4-3
2
0
2
0
2 2
3
( ) 0 0, 5
0 22
2 2 3
4 ( 3) 3 52
t
t
y t t tAtA d t
A d A t
AA d t t
λ λ
λ λ
λ λ−
= < >
= = < <
= = < <
⎡ ⎤= = − − < <⎣ ⎦
∫
∫
∫
2-14
2.4-4
2
0
1
2 2
1
( ) 0 0, 3
0 12
(2 1) 1 22
4 ( 1) 2 32
t
t
t
t
y t t tAtA d t
AA d t t
AA d t t
λ λ
λ λ
λ λ
−
−
= < >
= = < <
= = − < <
⎡ ⎤= = − − < <⎣ ⎦
∫
∫
∫
2.4-5
48
6
( ) 0 4
2 2 8 4 6
2 4 6
t
y t t
Ad At A t
Ad A t
λ
λ
= <
= = − ≤ ≤
= = >
∫∫
2.4-6
2 2
0
2 2 4
2
( ) 0 0
2 1 0 2
2 1 2
t t
t t
t
y t t
e d e t
e d e e t
λ
λ
λ
λ
− −
− −
−
= <
= = − ≤ ≤
⎡ ⎤= = − >⎣ ⎦
∫∫
2-15
2.4-7
2
10 2
1 00 1 2
2 01 2
2
( ) 0 1, 3
2(1 ) 2 1 1 0
2(1 ) 2(1 ) 2 1 0 1
2(1 ) 2(1 ) 2 1 1 2
2(1 ) 6 9 2 3
t
t
t
t
y t t t
d t t t
d d t t t
d d t t t
d t t t
λ λ
λ λ λ λ
λ λ λ λ
λ λ
−
−
−
−
= < − ≥
= + = + + − ≤ <
= + + − = − + + ≤ <
= + + − = − + + ≤ <
= − = − + ≤ <
∫∫ ∫∫ ∫∫
2.4-8
( )
0
( ) 0 0
[ /( )][ ] 0t a b t bt at
y t t
Ae Be d AB a b e e tλ λ λ− − − − −
= ≤
= = − − >∫
2.4-9
1 2
1 2
1 2
1 2 1 1 2 2
1 1 2 2
( ) ( ) sin2 2
( ) ( ) ( ) [ /( )][ ] [ /( )][ ]Let / 2, , / 2, and simplify
b t b tat j t j t
b t b tat at
j jv t Ae w t t e e B e B e
y t v w t v w t AB a b e e AB a b e eB j b j B j b j
π ππ
π π
− −− −
− −− −
= = = − = +
= ∗ + ∗ = − − + − −= = − = − =
2.4-10
( ) ( ) ( ) let
( ) ( ) ( ) ( ) ( )
v w t v w t d t
v t w d w v t d w v t
λ λ λ μ λ
μ μ μ μ μ μ
∞
−∞
−∞ ∞
∞ −∞
∗ = − = −
= − − = − = ∗
∫∫ ∫
2.4-11
Let ( ) ( ) ( ) where ( ) ( ), ( ) ( )y t v w t d v t v t w t w tλ λ λ∞
−∞= − − = − =∫
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
y t v w t d v w t d
v w t d v w t d y t
λ λ λ λ λ λ
μ μ μ μ μ μ
∞ ∞
−∞ −∞
∞ ∞
−∞ −∞
− = − − = +
= − − − = − =
∫ ∫∫ ∫
2.4-12
Let ( ) ( ) ( ) where ( ) ( ), ( ) ( )y t v w t d v t v t w t w tλ λ λ∞
−∞= − − = − − = −∫
2-16
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
y t v w t d v w t d
v w t d v w t d y t
λ λ λ λ λ λ
μ μ μ μ μ μ
∞ ∞
−∞ −∞
∞ ∞
−∞ −∞
− = − − = − +
= − − = − =
∫ ∫∫ ∫
2.4-13
0 / 2 2 2
/ 2 0
2
/ 2
Let ( ) ( ) ( / )3( ) ( ) ( ) 0 / 24
1 3( ) / 2 3 / 22 2
t
t
t
w t v v t t
v w t d d t t
d t t
τ
τ
τ
τ
τ τ
τ λ λ τ λ λ τ τ
τ λ λ τ τ τ
+
−
−
= ∗ = Λ
∗ = + + − = − ≤ <
⎛ ⎞= − = − ≤ <⎜ ⎟⎝ ⎠
∫ ∫
∫
2 2
2
3 / 24
1 3Thus ( ) / 2 3 / 22 2
0 3 / 2
t t
v v v t t t
t
τ τ
τ τ τ
τ
⎧ − <⎪⎪⎪ ⎛ ⎞∗ ∗ = − ≤ <⎨ ⎜ ⎟
⎝ ⎠⎪⎪ ≥⎪⎩
2.4-14 { } [ ] [ ]( ) [ ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( )v t w t z t V f W f Z f V f W f Z f∗ ∗ = =F
{ }1so ( ) [ ( ) ( )] [ ( ) ( )] ( ) [ ( ) ( )] ( )v t w t z t V f W f Z f v t w t z t−∗ ∗ = = ∗ ∗F 2.4-15
1( ) ( ) 4 (2 )4 4
( ) ( ) ( ) (2 ) ( ) (1/ 2)sinc( / 2)
fV f W f f
Y f V f W f f y t t
⎛ ⎞= Π = Π⎜ ⎟⎝ ⎠
= = Π ↔ =
2-17
2.5-1
( ) cos ( ) sinc( ) sinc( )2 2
As 0 the cosine pulse ( ) gets narrower and narrower while maintaining height A.This is not the same as an impulse since the area under the cur
c c ct A Az t A t Z f f f f f
z t
τ τω τ ττ
τ
⎛ ⎞= Π = − + +⎜ ⎟⎝ ⎠
→ve is also getting smaller.
As 0 the main lobe and side lobes of the spectrum ( ) get wider and wider, however the height gets smaller and smaller. Eventually the spectrum will cover all frequenci
Z fτ →es with
almost zero energy at each frequency. Again this is different from what happens in the caseof an impulse. 2.5-2
0 0
2 20 0
2 20 0 0 0
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
d d
d d
j ft j ftv
n
j nf t j nf tv w v
n
W f v f e c nf f nf e
c nf e f nf c nf c nf e
π π
π π
δ
δ
− −
− −
⎡ ⎤= = −⎢ ⎥⎣ ⎦⎡ ⎤= − ⇒ =⎣ ⎦
∑
∑
2.5-3
[ ]0 0 0 0 0
0 0 0
( ) 2 ( ) 2 ( ) ( ) 2 ( ) ( )
( ) 2 ( )
v vn n
w v
W f j fV f j f c nf f nf j nf c nf f nf
c nf j nf c nf
π π δ π δ
π
⎡ ⎤= = − = −⎢ ⎥⎣ ⎦⇒ =
∑ ∑
2.5-4
[ ]
{ }
{ }
0 0 0 0 0 0 0
0 0 0 0
0 0 0
0 0 0
1 1( ) ( ) ( ) ( ) ( ) ( )2 2
1 [( ) ] ( ) [( ) ] ( )2
1 [( ) ] [( ) ] ( )2
1so ( ) [( ) ] [( ) ]2
v vn n
v vk k
v vn
w v v
W f V f mf c nf f kf mf c nf f kf mf
c k m f f kf c k m f f kf
c n m f c n m f f nf
c nf c n m f c n m f
δ δ
δ δ
δ
⎡ ⎤= − = − − + − +⎢ ⎥⎣ ⎦
⎡ ⎤= − − + + −⎢ ⎥⎣ ⎦
= − + + −
= − + +
∑ ∑
∑ ∑
∑
2-18
2.5-5
( )
4
4 0
4 2
2
( ) ( ) ( 2 )
1 1 1 1( ) ( ) ( )2 2 2 2
But ( ) ( ), so
( ) 1 2 sinc 22
Agrees with ( ) 2 sinc 22
j f
j ft j
j f j f
j f
v t Au t Au t
V f A f f ej f j f
f e e fAV f e A f e
j ftv t A f e
π τ
π
π τ π τ
π τ
τ
δ δπ π
δ δ
τ τπ
τ τ ττ
−
− −
− −
−
= − −
⎧ ⎫⎡ ⎤= + − +⎨ ⎬⎢ ⎥
⎣ ⎦⎩ ⎭=
= − =
−⎛ ⎞= Π ↔⎜ ⎟⎝ ⎠
2.5-6
( )
2 2
2 2 0
2 2
( ) ( ) ( )
1 1 1 1( ) ( ) ( ) ( )2 2 2 2
But ( ) ( ) ( ), so
1( ) ( ) ( ) 2 sinc 22
Agrees with ( )
j f j f
j f j f j
j f j f
v t A Au t Au t
V f A f f e f ej f j f
f e f e e f
V f A f e e A f A fj f
v t A
π τ π τ
π τ π τ
π τ π τ
τ τ
δ δ δπ π
δ δ δ
δ δ τ τπ
−
−
−
= − + + −
⎧ ⎫⎡ ⎤ ⎡ ⎤= − + − +⎨ ⎬⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎩ ⎭= =
⎡ ⎤= − − = −⎢ ⎥
⎣ ⎦= ( / 2 ) ( ) 2 sinc 2A t A f A fτ δ τ τ− Π ↔ −
2.5-7
( )
2 2
2 2 0
2 2
( ) ( ) ( )
1 1 1 1( ) ( ) ( ) ( )2 2 2 2
But ( ) ( ) ( ) ( ), so
( ) cos 22
If 0, ( ) sgn ( )
j fT j fT
j fT j fT j
j fT j fT
v t A Au t T Au t T
V f A f f e f ej f j f
f e f e e f fA AV f e e fT
j f j fAT v t A t V f
j
π π
π π
π π
δ δ δπ π
δ δ δ δ
ππ π
π
−
−
−
= − + − −
⎧ ⎫⎡ ⎤ ⎡ ⎤= − + − +⎨ ⎬⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎩ ⎭= = =
− −= + =
−→ = − ↔ = , which agrees with Eq. (17)
f
2-19
2.5-8
( ) sinc and (0) 1, sosinc 1( ) ( )
2 21 1If 0, ( ) ( ) and ( ) ( ), which agrees with Eq. (18)
2 2
V f f VfW f f
j f
w t u t W f fj f
εε δ
π
ε δπ
= =
= +
→ = = +
2.5-9
1/( ) and (0) 1, so1/ 2
1/ 1( ) ( )( 2 )(1/ 2 ) 2
1 1If 0, ( ) ( ) and ( ) ( ), which agrees with Eq. (18)2 2
V f Vj f
W f fj f j f
w t u t W f fj f
εε π
ε δπ ε π
ε δπ
= =+
= ++
→ = = +
2.5-10
( ) [ ]( )
( ) / ( ) ( )
so Z( ) ( sinc ) 2 sinc cos 2j T j T
z t A t t T t T
f A f e e A f fTω ω
τ δ δ
τ τ τ τ π−
= Π ∗ − + +
= + =
2.5-11
( ) [ ]( )2 2
( ) / ( 2 ) 2 ( ) ( 2 )
so ( ) ( sinc ) 2 2 sinc (1 cos 4 )j T j T
z t A t t T t t T
Z f A f e e A f fTω ω
τ δ δ δ
τ τ τ τ π−
= Π ∗ − + + +
= + + = +
2.5-12
( ) [ ]( )2 2
( ) / ( 2 ) 2 ( ) ( 2 )
so ( ) ( sinc ) 2 2 sinc (cos 4 1)j T j T
z t A t t T t t T
Z f A f e e A f fTω ω
τ δ δ δ
τ τ τ τ π−
= Π ∗ − − + +
= − + = −
2-20
2.5-13
n 0 1 2 3 4 5 6 7 8sin( ) ( 0.5 )t t nπ δ − 0 1 0 1 0 1 0 1 0
( )v t 0 1 1 2 2 3 3 4 4 2.5-14
n -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 cos(2 ) ( 0.1 )t t nπ δ − 1 0.81 0.31 -0.31 -0.81 -1 -0.81 -0.31 0.31 0.81 1
( )v t 1 1.81 2.12 1.81 1 0 -0.81 -1.12 -0.81 0 1
( ) for 1,10v t n = 1.81 2.12 1.81 1 0 -0.81 -1.12 -0.81 0 1
2.6-1 Recall for a rectangular pulse train, with 0 0amplitude, frequency and period , and A f T
0 0
22
0 00 0
1 1( ) ( )ntj
j nf t Tnc v t e dt v t e dt
T T
ππ
∞ ∞ −−= =∫ ∫
0
With sampling ( ) ( ) and with ( ) ( ). With samples
s s
s
v t v k t T t v t v kTN T NT
⇒ Δ = Δ ⇒ ⇒⇒ =
Putting this back into the above integral, we have
0
2 2
,0 0
1 1( ) ( )snkT nkj jT N
n ks s
c v k e dt v k e dtNT NT
π π∞ ∞− −= =∫ ∫ .
With
2 2
,0 0
we can replace the integral by a summation and giving
1 1 1( ) ( )
s snk nkN Nj j
N Nn k s n
k ks s
dt t T dt T
c v k e T c v k eNT NT N
π π− −
= =
≈ Δ = →
= ⇒ =∑ ∑
With 2
' ' '
0If ( ) were a rectangular pulse then the [ ( )] ( ) ( )
knN jN
kx k DFT x k X k x k e
π−
=
= =∑
Thus ( )nc NX k= 2.6-2 If output is average of present and past 3 inputs then
1 1 1 1( ) ( ) ( 1) ( 2) ( 3)4 4 4 4
y k x k x k x k x k= + − + − + −
2-21
Using superposition of the input with a set of delayed impulses, we get
1 1 1 1( ) ( ) ( 1) ( 2) ( 3)4 4 4 4
h k k k k kδ δ δ δ= + − + − + −
2 21 7 3
8 4
0 0 0( ) ( ) ( ) ( )
kn kn knN j j jN
k k kH n h k e h k e h k e
π π π− − − −
= = =
= = =∑ ∑ ∑
1 1 1 1[cos( / 4) sin( / 4)] [cos(2 / 4) sin(2 / 4)] [cos(3 / 4) sin(3 / 4)]4 4 4 41 1 [cos( / 4) cos(2 / 4) cos(3 / 4)] [sin( / 4) sin(2 / 4) sin(3 / 4)]4 4 4
n j n n j n n j n
jn n n n n n
π π π π π π
π π π π π π
= + − + − + −
= + + + − + +
(0) 1, (1) [0.25 0.604], (2) 0.0, (3) [0.25 0.104]H H j H H j= = − = = − 2.6-3 If the outputs are the sum of the weighted inputs, then
8 4 3 1( ) ( ) ( 1) ( 2) ( 3)16 16 16 16
y k x k x k x k x k= + − + − + −
Using superposition of the input with a set of delayed impulses, we get
8 4 3 1( ) ( ) ( 1) ( 2) ( 3)16 16 16 16
h k k k k kδ δ δ δ= + − + − + − 2 21 7 3
8 4
0 0 0( ) ( ) ( ) ( )
kn kn knN j j jN
k k kH n h k e h k e h k e
π π π− − − −
= = =
= = =∑ ∑ ∑
1 1 3 1[cos( / 4) sin( / 4)] [cos(2 / 4) sin(2 / 4)] [cos(3 / 4) sin(3 / 4)]2 4 16 161 1 3 1 1 3 1cos( / 4) cos(2 / 4) cos(3 / 4) sin( / 4) sin(2 / 4) sin(3 / 4)2 4 16 16 4 16 16
n j n n j n n j n
n n n j n n n
π π π π π π
π π π π π π
= + − + − + −
⎡ ⎤ ⎡ ⎤= + + + − + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦(0) 1, (1) [0.63 0.41], (2) 0.31 0.19, (3) [0.37 0.033]H H j H j H j= = − = − = −
2.6-4 (a) ( ) [6,0,0,4,0,4,0,0] and [0,0, 1,0,0,1,0,0]R IX n X= = −
2 /
0 0
1 1 2 2( ) ( ) ( )cos ( )sinN N
j nk NR I
n n
nk nkx k X n e X n jX nN N N N
π π π= =
⎡ ⎤= = +⎢ ⎥⎣ ⎦∑ ∑
Because of symmetry
2-22
/ 2 1
1
(0) 2 2 2( ) ( ) cos ( )sin
(0) 6 2 = 4cos sin8 8 8
0.75 4cos 0.75 sin 0.5
N
R In
X nk nkx n X n jX nN N N N
X n nj
n j n
π π
π π
π π
−
=
= + +
+ +
= + −
∑
(b) with /160 160
( ) 0.75 4cos(0.75 160 ) sin(0.5 160 )( ) 0.75 4cos120 sin 80
st kT k k tx t t j t
x t t j tπ ππ π
→ = ⇒ = ⇒= + × − ×
⇒ = + −
(c) DC value = 0.75 (d) 1 160 / 8 20 Hznf f= = = 2.6-5
1 12 /
0 1( ) [3,1,1,0], ( ) ( ) (0) ( )[cos 2 / sin 2 / ]
N Nj kn N
k kx k X n x k e x x k kn N j nk Nπ π π
− −−
= =
= = = + −∑ ∑
3
14 ( ) (0) ( )[cos 2 / 4 sin 2 / 4]
kN X n x x k kn j nkπ π
=
= ⇒ = + −∑
(0) 5, (1) 2 , (2) 3, (3) 2X X j X X j= = − = = + 2.6-6
1 1
66
20 MHz 40 MHz
We want 0.01 MHz and
10.01 10 40001/ 40 10
s
j j
W fnf f f f
NT
NN
+
= ⇒ ≥
Δ = − = =
⇒ × = ⇒ ≥× ×
To reduce computation speed, we want to use the FFT 4096N⇒ = 2.6-7
1 2 1 2 1 14 and 7 make by padding ( ) with zeros so that (2 1)13
N N N N x k N NN= = ⇒ = ≥ −
⇒ =
2.6-8 (a) Number of multiplications for standard DFT is 2 2 9(256) 10 10 655.4 sN μ−⇒ × × = (b) Number of multiplications for FFT is