Lời Giải Singals and Systems

1

CHAPTER 1

1.1 to 1.41 - part of text

1.42 (a) Periodic:

Fundamental period = 0.5s

(b) Nonperiodic

(c) Periodic

Fundamental period = 3s

(d) Periodic

Fundamental period = 2 samples

(e) Nonperiodic

(f) Periodic:

Fundamental period = 10 samples

(g) Nonperiodic

(h) Nonperiodic

(i) Periodic:

Fundamental period = 1 sample

l.43

(a) DC component =

(b) Sinusoidal component =

Amplitude =

y t( ) 3 200t π

6

+ --     cos     2

=

9 200t π

6

+ --     2

= cos

9

2

-- 400t π

3

+ --     = cos 1

9

2

--

9

2

-- 400t π

3

+ --     cos

9

2

--

2

Fundamental frequency =

1.44 The RMS value of sinusoidal x(t) is . Hence, the average power of x(t) in a 1-ohm

resistor is = A2

/2.

1.45 Let N denote the fundamental period of x[N]. which is defined by

The average power of x[n] is therefore

1.46 The energy of the raised cosine pulse is

1.47 The signal x(t) is even; its total energy is therefore

200

π

--------Hz

A ⁄ 2

( ) A ⁄ 2 2

N 2π

Ω = ------

P 1

N

---- x

2

[ ] n

n=0

N-1

= ∑

1

N

---- A2 2πn

N

--------- + φ     2

cos

n=0

N-1

= ∑

A2

N

------ 2πn

N

--------- + φ     2

cos

n=0

N-1

= ∑

E 1

4

--( ) cos( ) ωt + 1 2

dt –π ω⁄

π ω⁄

∫ =

1

2

-- ( ) ωt + 2cos( ) ωt + 1 2 ( ) cos dt

0

π ω⁄

∫ =

1

2

-- 1

2

-- ( ) 2ωt 1

2 cos + + -- 2cos( ) ωt + 1     dt

0

π ω⁄

∫ =

1

2

-- 3

2

--     π

ω

---     = = 3π ⁄ 4ω

E 2 x

2

( )t dt

0

5

∫ =

3

1.48 (a) The differentiator output is

(b) The energy of y(t) is

1.49 The output of the integrator is

for

Hence the energy of y(t) is

1.50 (a)

2 1( )2 t 2 5( ) – t 2

dt

4

5

∫ d +

0

4

∫ =

2[ ]t t=0

4 2 1

3

--( ) 5 – t 3 – t=4

5

= +

8 2

3

+ -- 26

3 = = -----

y t( )

1 for 5– < <t –4

– for 4 1 < <t 5

0 otherwise 









=

E ( ) 1 2 t ( ) –1 2

dt

4

5

∫ d + –5

–4

∫ =

= = 1 1 + 2

y t( ) A τ τd

0

t

∫ == 0 At ≤ ≤t T

E A2

t

2

dt

0

T

∫ A2

T3

3 = = ------------

-1 -0.8 0 0.8 1 t

x(5t)

1.0

-25 -20 0 20 25 t

x(0.2t)

1.0

(b)

4

1.51

1.52 (a)

0 0.1 0.5 0.9 1.0

x(10t - 5)

1.0

t

x(t)

1

-1 1 2 3

t

-1

y(t - 1)

t -1 1 2 3

-1

x(t)y(t - 1)

t

1

1

2 3

-1

-1

5

1.52 (b)

1.52 (c)

x(t - 1)

1

t

t

1

y(-t)

t

-1 1 2 3 4

-1

-2 -1 1 2 3 4

1

-1

x(t - 1)y(-t)

-2 -1 1 2 3 4

-1

-2

-1 1 2 3

-1

1 2 3 4

t

t

-2 -1

x(t + 1)y(t - 2)

t -2 -1 1 2 3 4

x(t + 1)

1

y(-t)

6

1.52 (d)

1.52 (e)

x(t)

t

t

1

y(1/2t + 1)

t

x(t - 1)y(-t)

-1

-3 -2 -1 1 2 3

-1

1

1 2 4 6

-1.0

-3 -2 -1 1 2 3

6 -5 -4 -3 -2 -1

1 2 3

-4 -3 -2 -1

1

-1

x(t)

t

-4 -3 -2 -1

1 2 3

t

y(2 - t)

t

1 2 3

-1

-1

x(t)y(2 - t)

7

1.52 (f)

1.52 (g)

-2 -1 1 2

t

1

-1

x(t)

x(2t)y(1/2t + 1)

+1

-1

-0.5

-1

t

1 2

1 1 2 3

t

-1.0

y(t/2 + 1)

-3 -2 -1

1.0

-5

-6

-7 -6 -5 -4 -3 -2

1

-1

x(4 - t)

y(t)

-2 -1 1 2 4

t

t

-3 -2 -1 1 2 3

t

x(4 - t)y(t) = 0

8

1.53 We may represent x(t) as the superposition of 4 rectangular pulses as follows:

To generate g1(t) from the prescribed g(t), we let

where a and b are to be determined. The width of pulse g(t) is 2, whereas the width of

pulse g1(t) is 4. We therefore need to expand g(t) by a factor of 2, which, in turn, requires

that we choose

The mid-point of g(t) is at t = 0, whereas the mid-point of g1(t) is at t = 2. Hence, we must

choose b to satisfy the condition

or

Hence,

Proceeding in a similar manner, we find that

Accordingly, we may express the staircase signal x(t) in terms of the rectangular pulse g(t)

as follows:

1

1 2 3 4

t

g1(t)

1

11 2 3 4

t

g2(t)

1

1 2 3 4

t

g3(t)

1

1 2 3 4

t

g4(t)

0

g1( )t = g at b ( ) –

a

1

2 = --

a( ) 2 – 0 b =

b 2a 2 1

2

--     == = 1

g1( )t g

1

2

--t – 1     =

g2( )t g

2

3

--t 5

3 – --     =

g3( )t = g t( ) – 3

g4( )t = g( ) 2t – 7

9

1.54 (a)

(b)

(c)

(d)

(e)

x t( ) g

1

2

--t – 1     g

2

3

--t 5

3 – --     = + ++ g t( ) – 3 g( ) 2t – 7

0 1 2

t

x(t) = u(t) - u(t - 2)

0 1 2

-1 3

-2

-1

t

x(t) = u(t + 1) - 2u(t) + u(t - 1)

t

x(t) = -u(t + 3) + 2u(t +1) -2u(t - 1) + u(t - 3)

-3 1 2 3

-1

0

t

x(t) = r(t + 1) - r(t) + r(t - 2)

-2 -1 0 1 2 3

1

t

x(t) = r(t + 2) - r(t + 1) - r(t - 1)+ r(t - 2)

1

-3 -2 -1 0 1 2

10

1.55 We may generate x(t) as the superposition of 3 rectangular pulses as follows:

All three pulses, g1(t), g2(t), and g3(t), are symmetrically positioned around the origin:

1. g1(t) is exactly the same as g(t).

2. g2(t) is an expanded version of g(t) by a factor of 3.

3. g3(t) is an expanded version of g(t) by a factor of 4.

Hence, it follows that

That is,

1.56 (a)

(b)

-4 -2 0 2 4

1

g1(t)

t

-4 -2 0 2 4

1

g2(t)

t

-4 - 2 0 2 4

1

g3(t)

t

g1( )t = g t( )

g2( )t g

1

3

--t     =

g3( )t g

1

4

--t     =

x t( ) g t( ) g

1

3

--t     g

1

4

--t     = + +

o o 2

-1 0 1

n

x[2n]

o

o

o

-1 0 1

o n

2

1

x[3n - 1]

11

1.56 (c)

(d)

(e)

(f)

o

o o o o

o o o o

o

n -4 -3 -2 -1 0

1

1

2 3 4 5

-1

y[1 - n]

o

o o o

o o o o

o

n

-3 -2 -1

1

1

2 3 4 5

-1

y[2 - 2n]

o o o o o

o

o

o

o

o

o

o

o o o

o

4

3

2

1

-7 -6 -5 -4 -3

-2 -1 0 1 2 3 4 5 6 7 8

n

x[n - 2] + y[n + 2]

o o o o

o o

o o o

o

o o o

-5 -4 -3 -2 2 3

-1 4 5 6 7

1

-1

n

x[2n] + y[n - 4]

12

1.56 (g)

(h)

(i)

(j)

o o n

o

o

o

o

o

o

o

o -5 -4 -3 -2 -1 1

1

2

3

x[n + 2]y[n - 2]

o o o

o

o

o o

o

o

o

3

2

1

-3 -2 -1 1 2 3 4 5 6 7 8

n

x[3 - n]y[-n]

o

o

o

o

o o

o

o

o o o

o

-5 -4 -3 -2 -1

3

2

1

-1

-2

-3

1 2 3 4 5 6

n

x[-n] y[-n]

o o o

o

o

o o

o

o

o o o

3

2

1

-1

-2

-3

-6 -5 -4 -3

-2 -1 1 2 3 4 5 6

n

x[n]y[-2-n]

o

13

1.56 (k)

1.57 (a) Periodic

Fundamental period = 15 samples

(b) Periodic

Fundamental period = 30 samples

(c) Nonperiodic

(d) Periodic

Fundamental period = 2 samples

(e) Nonperiodic

(f) Nonperiodic

(g) Periodic

Fundamental period = 2π seconds

(h) Nonperiodic

(i) Periodic

Fundamental period = 15 samples

1.58 The fundamental period of the sinusoidal signal x[n] is N = 10. Hence the angular

frequency of x[n] is

m: integer

The smallest value of is attained with m = 1. Hence,

radians/cycle

o o o o o

o

o

o

o

o

o

o o o

3

2

1

-1

-2

-3

1 2 3 4 5 6

-8 -7 -6 -5 -4 -3

-2 -1

x[n + 2]y[6-n]

n

Ω 2πm

N = ----------

Ω

Ω 2π

10------ π

5 = = --

14

1.59 The amplitude of complex signal x(t) is defined by

1.60 Real part of x(t) is

Imaginary part of x(t) is

1.61 We are given

The waveform of x(t) is as follows

xR

2

( )t xI

2 + ( )t A2 ( ) ωt + φ A2 ( ) ωt + φ 2 + sin

2 = cos

A () ω ωt + φ ( ) t + φ 2 + sin

2 = cos

= A

Re{ } x t( ) Aeαt = cos( ) ωt

Im{ } x t( ) Aeαt = sin( ) ωt

x t( )

t

∆

--- for ∆

2

--- t

∆

2 – ≤ ≤ ---

1 for t

∆

2 ≥ ---

2 for t ∆

2 < –--- 

















=

x(t)

1

1

2

1

2

-∆/2

∆/2

t

-

0

15

The output of a differentiator in response to x(t) has the corresponding waveform:

y(t) consists of the following components:

1. Rectangular pulse of duration ∆ and amplitude 1/∆ centred on the origin; the area

under this pulse is unity.

2. An impulse of strength 1/2 at t = ∆/2.

3. An impulse of strength -1/2 at t = -∆/2.

As the duration ∆ is permitted to approach zero, the impulses (1/2)δ(t-∆/2) and

-(1/2)δ(t+∆/2) coincide and therefore cancel each other. At the same time, the rectangular

pulse of unit area (i.e., component 1) approaches a unit impulse at t = 0. We may thus state

that in the limit:

1.62 We are given a triangular pulse of total duration ∆ and unit area, which is symmetrical

about the origin:

-∆/2 ∆/2

1/∆

y(t)

1

2 δ(t - ) 1

2

1

2 δ(t + ) ∆

2

t 0

y t( )

∆ → 0

lim

∆ → 0

lim d

dt = ---- x t( )

= δ( )t

x(t)

2/∆

slope = -4/∆2

area = 1

slope = 4/∆2

-∆/2 0 ∆/2

t