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

2-23

92 2

256log log (256) 10 10 10.24 s2 2N N μ−⇒ × × =

2.6-9 (a) Number of multiplications for standard DFT is 2 2 9(4096) 10 10 168 msN −⇒ × × = (b) Number of multiplications for FFT is

92 2

4096log log (4096) 10 10 0.246 ms2 2N N −⇒ × × =