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数字信号处理—基于计算机的方法第5章答案 DOC

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数​ 字​ 信​ 号​ 处​ 理​ —​ 基​ 于​ 计​ 算​ 机​ 的​ 方​ 法​ 第​ 5​ 章​ 答​ 案 (8人评价)

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数​ 字​ 信​ 号​ 处​ 理​ —​ —​ 基​ 于​ 计​ 算​ 机​ 的​ 方​ 法​ ​ 第​ ​ 三​ 版​ 清​ ​ 华​ 大​ 学​ 出

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giter 贡献于2011-09-16

5-1

评价文档：

An AM broadcast transmitter is tested by feeding the RF output into a 50-W

(dummy) load. Tone modulation is applied. The carrier frequency is 850 kHz and

一亿文档冲亿季 文档冲亿季，好礼乐...

the FCC licensed power output is 5,000 W. The sinusoidal tone of 1,000 Hz is set

Excel使用技巧大全(超全..

for 90% modulation. (a) Evaluate the FCC power in dBk (dB above 1 kW) units.

Power Point的使用技巧 CAD技巧120个绝对实用 Word2003使用技巧大全

(b) Write an equation for the voltage that appears across the 50- W load, giving

数字信号处理—基于计算.

numerical values for all constants.

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(c) Sketch the spectrum of this voltage as it would appear on a calibrated

数字信号处理—基于计 24页

spectrum analyzer.

数字信号处理—基于计 33页

(d) What is the average power that is being dissipated in the dummy load?

数字信号处理—基于计 5页

(e) What is the peak envelope power? Solution： 分享到：

(a)

1 2

数字信号处理-基于计算 579页

数字信号处理—基于计

FCC power:

1 /24

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Ac

2

数字信号处理—基于计

Ac = 707 V

= 5000

33页

数字信号处理—基于计

50

5页

æ 5000 ö 10 lg ç ÷ = 6.99 è 1000 ø

(b )

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Let :

数字信号处理—基于计

( dBK )

23页

数字信号处理—基于计 5页

m ( t ) = Am c o s 2 0p 0t 0

fm =

1H 0z 0 0

Q 是 9 0％ 调 制 ® Am = 0 . 9 \ 50 W 负 载 上 通 过 的 电 压 为 ：

诉中心；如要提出功能问题或意 议，请点击此处进行反馈。

s ( t ) = 707[1 + 0.9 cos (2000p t )] cos[2p 850, 000 t ]

s ( t ) = 707 cos w c t +

0 . 9 ( 707 ) 2

cos[( w c - w m ) t ] +

44页 如要投诉违规内容，请到

s t = A é1 + m t ù cos w t c c ë ( ) ( )û

(c)

数字信号处理 DSP_Ch

0 . 9 ( 707 ) 2

cos[( w c + w m ) t ]

1

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(d) 2

< s (t ) > =

1

2

Ac +

2

1

2

2

A c < m (t ) > =

2

1

2

Ac

2

2

0.9 ù é 1+ 2 úû ëê

50 W 负 载 上 的 平 均 功 率 ：

PA V G real =

< s

2

(t ) >

1 =

2

Ac

2

50 50 2 0.9 ù é = 5000 ê1 + ú = 7025 w 2 û ë

(e)

5-2

PPEP =

Ac

2

2

0.9 ù é ê1 + ú 2 û ë

{1 + max éë m ( t ) ùû} 2 ´ 50

2

2

= 5000 ´ [1 + 0.9 ] = 18050 w

An AM transmitter is modulated with an audio testing signal given by

m ( t ) = 0.2 sin w 1t + 0.5 cos w 2 t , where

f1 = 500 H z

, f 2 = 500 2 H z ,and Ac = 100 .

Assume that the AM signal is fed into a 50 W load. (a) Sketch the AM waveform. (b) What is the modulation percentage? (c) Evaluate and sketch the spectrum of the AM waveform. Solution:

(a)

s ( t ) = 100(1 + 0.2 sin w 1t + 0.5 cos w 2 t ) cos w c t

2

5-4 Assume that an AM transmitter is modulated with a video testing signal given by m ( t ) = - 0.2 + 0.6 sin w 1t ,where f1 =3.57MHz. Let Ac=100. (a) Sketch the AM waveform. (b) What are the percentages of positive and negative modulation? (c) Evaluate and sketch the spectrum of the AM waveform about fc. 3

Solution:

(a)

m ( t ) = - 0.2 + 0.6 sin w 1t f m = f 1 = 3.57 M H z

Ac = 100

;

s ( t ) = 100 ( 1 + m ( t ) ) cos w c t = 100 ( 0.8 + 0.6 sin w 1t ) cos w c t

(b )

% pos. m o d . = % neg. m o d . =

(c)

Ac Ac - Am in Ac

= =

140 - 100 100 100 - 20 100

= 40%

= 80%

f >0 S

5-5

Am ax - Ac

(f)=

40 d

(

f - f c ) - j15 éë d

(

f - fc - fm ) - d

(

f - f c + f m ) ùû

A 50,000-W AM broadcast transmitter is being evaluated by means of a

two-tone

test.

The

transmitter

is

connected

4

to

a

50- W

load,

and

m t = A cos

t + A cos 2

( )is generated. signal

w

1

1

1

t 1 w , where f1 =500 Hz. Assume that a perfect AM

(a) Evaluate the complex envelope for the AM signal in terms of A1 and w1 . (b) Determine the value of A1 for 90% modulation. (c) Find the values for the peak current and average current into the 50- W load for the 90% modulation case. Solution: (a) 2

50 , 000 =

Ac

2 ( 50 )

Þ A c = 2236 V

g ( t ) = A C [1 + m ( t )] = 2236 [1 + A1 (cos w 1 t + cos 2w 1 t )]

(b)

m ( t ) = A1 c os w 1t + A1 cos 2w 1t = A1 [ cos w 1t + c os 2w 1t ]

to find [m(t)]min :

x(q) = cosq +cos2q

d x (q ) dq

= - sin q - 2 sin 2q = 0

- sin q = 4 sin q cos q q = 104 . 5 °

x (104 . 5 ° ) = - 1 . 125

min éë m ( t ) ùû = A1 x (104.5

o

) = - 1.12 5 A

1

- min éë m ( t ) ùû = 0.9 ® A1 = 0 .8 另 解 ： Am ax = 2236[1 + 2 A1 ] AM I N = 2236[1 - 1.125 A1 ] A 3.125 - Am in 0.90 = m ax A1 Þ A1 = 0.576 = 2 Ac 2 5