数字信号处理—基于计算机的方法第5章答案

(a)
s ( t ) = 100(1 + 0.2sin ω1t + 0.5cos ω 2t ) cos ωct
2
5-4
Assume that an AM transmitter is modulated with a video testing signal given
by m ( t ) = −0.2 + 0.6sin ω1t ,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.
− min ⎡ ⎤ = 0.9 → A1 = 0.8 ⎣m ( t )⎦
另解：Amax = 2236[1 + 2 A1 ]
AMI N = 2236[1 − 1.125 A1 ] A − Amin 3.125 0.90 = max = A1 ⇒ A1 = 0.576 2 Ac 2
5
(c)
Amax = 2236[1 + 2(0.8)] = 5813.6volts A I max = max = 116.272 Amps 50 I Av = s (t ) = 2236[1 + 0.8(cos ω1t + cos 2ω1t )] ⋅ cos ωc t
⎧⎡ 1 1 1 1 ⎤ 1 ⎪ ⎢ M ( f − f c − 2 B − 2 B) + M ( f − f c − 2 B + 2 B)⎥, | f − f c − 2 B |< ⎣ ⎦ 1⎪ 1 1 1 1 ⎤ 1 = ⎨⎡ 4 ⎪⎢ M ( f + f c + B − B) + M ( f + f c + B + B)⎥, | f + f c + B |< 2 2 2 2 2 ⎦ ⎣ ⎪ f elsewhere ⎩0,
1 2 Ac 2 = 5000 50
Ac = 707 V
⎛ 5000 ⎞ 10 lg ⎜ ⎟ = 6.99 (dBK ) ⎝ 1000 ⎠
(b )
Let : m ( t ) = Am cos 2000π t
∵ 是90％调制 → Am = 0.9 ∴ 50Ω负载上通过的电压为：
f m = 1000 Hz
∴ I Av = 0 Amps 5-12 SSB signals can be generated by the phasing method shown in Fig. 5-5a, by
the filter method, of Fig. 5-5b, or by the use of Weaver’s method [Weaver, 1956], as shown in Fig. P5-12. For Weaver’s method (Fig. P5-12), where B is the bandwidth of
Ac 2 1 + max ⎡ ⎣m ( t )⎤ ⎦ 2 × 50
{
}
2
= 5000 × [1 + 0.9] = 18050 w
2
5-2
An AM transmitter is modulated with an audio testing signal given by
m ( t ) = 0.2sin ω1t + 0.5cos ω 2t , where
s ( t ) = Ac ⎡ ⎣1 + m ( t ) ⎤ ⎦ cos ω c t
s (t ) = 707[1 + 0.9 cos (2000π t )]cos[2π 850, 000t ]
(c)
s (t ) = 707 cos ω c t +
0.9(707) 0.9(707) cos[(ω c − ω m )t ] + cos[(ω c + ω m )t ] 2 2
signal is generated. (a) Evaluate the complex envelope for the AM signal in terms of A1 and ω1. (b) Determine the value of A1 for 90% modulation. (c) Find the values for the peak current and average current into the 50- Ω load for the 90% modulation case. Solution: (a)
1
(d)
1 2 1 2 1 2 ⎡ 0.92 ⎤ 2 < s (t ) > = Ac + Ac < m (t ) > = Ac ⎢1 + 2 2 2 2 ⎥ ⎣ ⎦
2
50Ω负载上的平均功率： 1 2 < s 2 ( t ) > 2 Ac ⎡ 0.92 ⎤ PAVG real = = 1+ 50 50 ⎢ 2 ⎥ ⎣ ⎦ 2 ⎡ 0.9 ⎤ = 5000 ⎢1 + = 7025w 2 ⎥ ⎣ ⎦ (e) PPEP =
5-5
A 50,000-W AM broadcast transmitter is being evaluated by means of a test. The transmitter is connected to a 50- Ω load, and
two-tone
4
m ( t ) = A1 cos ω1t + A1 cos 2ω1t , where f1=500 Hz. Assume that a perfect AM
| f |<
f
Likewise
⎫ 1 B ⎪ ⎬ 2 f elsewhere⎪ ⎭
⎡ ⎛ 1 ⎞⎤ V9 (t ) = v5 (t ) cos ⎢2π ⎜ f c + B ⎟t ⎥ 2 ⎠⎦ ⎣ ⎝ ⇒ V9 ( f ) = 1⎡ ⎛ 1 ⎞ 1 ⎞⎤ ⎛ V5 ⎜ f − f c − B ⎟ + V5 ⎜ f + f c + B ⎟⎥ ⎢ 2⎣ ⎝ 2 ⎠ 2 ⎠⎦ ⎝ 1 ⎫ B 2 ⎪ ⎪ 1 ⎬ B 2 ⎪ ⎪ ⎭
6
⎛ ⎛1 ⎞ ⎞ V1 (t ) = cos⎜ ⎜ 2π ⎜ 2 B ⎟t ⎟ ⎟ = cos(πBt ) ⎠ ⎠ ⎝ ⎝
V3 (t ) = m(t )V1 (t ) = m(t ) cos(πBt ) ↔ V3 ( f ) =
1⎡ 1 1 ⎤ M ( f − B) + M ( f + B)⎥ ⎢ 2⎣ 2 2 ⎦ 1 2 1 1 ⎤ ⎡ j ⎢− M ( f − B ) + M ( f + B )⎥ 2 2 ⎦ ⎣ ⎫ 1 B ⎪ 2 ⎬ elsewhere⎪ ⎭
3
Solution:
(a)
m ( t ) = −0.2 + 0.6sin ω1t f m = f1 = 3.57 MHz ; Ac = 100
s ( t ) = 100 (1 + m ( t ) ) cos ω c t
= 100 ( 0.8 + 0.6sin ω1t ) cos ω c t
(b)
%pos. mod. = %neg. mod. =
பைடு நூலகம்5-1
An AM broadcast transmitter is tested by feeding the RF output into a 50- Ω
(dummy) load. Tone modulation is applied. The carrier frequency is 850 kHz and the FCC licensed power output is 5,000 W. The sinusoidal tone of 1,000 Hz is set for 90% modulation. (a) Evaluate the FCC power in dBk (dB above 1 kW) units. (b) Write an equation for the voltage that appears across the 50- Ω load, giving numerical values for all constants. (c) Sketch the spectrum of this voltage as it would appear on a calibrated spectrum analyzer. (d) What is the average power that is being dissipated in the dummy load? (e) What is the peak envelope power? Solution： (a) FCC power:
=0
for ω c >> ω1 Amax = 96.238 Amps 50 s (t ) = 2236[1 + 0.576(cos ω1t + cos 2ω1t )] ⋅ cos ω ct I max =
=0
另解：Amax = 2236[1 + 2(0.576)] = 4811.9volts
for ω c >> ω1
to find [m(t)]min :
x(θ) = cosθ +cos2θ
dx(θ ) = − sin θ − 2sin 2θ = 0 dθ
− sin θ = 4sin θ cosθ
θ = 104.5° x(104.5°) = −1.125
o min ⎡ ⎣m ( t )⎤ ⎦ = A1 x (104.5 ) = −1.125 A1
f1 = 500 Hz , f 2 = 500 2 Hz ,and Ac = 100 .
Assume that the AM signal is fed into a 50Ω load. (a) Sketch the AM waveform. (b) What is the modulation percentage? (c) Evaluate and sketch the spectrum of the AM waveform. Solution:
1 1 1 1 1 B |< B ⇒ − B < f − f c − B < B 2 2 2 2 2 1 1 1 1 ⇒ fc + B − B < f < fc + B + B ⇒ fc < f < fc + B 2 2 2 2 1 1 Likemise | f c + f c + B | < B ⇒ − f c − B < f < − f c 2 2 Aside :| f − f c − Thus,
Amax − Ac 140 − 100 = = 40% Ac 100 Ac − Amin 100 − 20 = = 80% Ac 100
(c)
f >0 S ( f ) = 40δ ( f − fc ) − j15 ⎡ ⎣δ ( f − f c − f m ) − δ ( f − fc + f m ) ⎤ ⎦
V4 (t ) = m(t )V2 (t ) = m(t ) sin(πBt ) ↔ V4 ( f ) =
1 ⎧ ⎫ 1 1 ⎤ V ( f ), | f | < B ⎪ ⎧1 ⎡ 4 ⎪ ⎪ ⎪ ⎪ ⎢ M ( f − B) + M ( f + B)⎥, 2 V5 ( f ) = ⎨ 2 2 ⎦ ⎬ = ⎨2 ⎣ ⎪ ⎪ ⎪ 0 , ⎩ f elsewhere⎪ ⎪ ⎩0, ⎭ ⎧1 ⎡ 1 1 ⎤ ⎪ ⎢− M ( f − B) + M ( f + B)⎥, V6 ( f ) = ⎨ 2 ⎣ 2 2 ⎦ ⎪ 0, ⎩ | f |<
m(t),
(a) Find a mathematical expression that describes the waveform out of each block on the block diagram. (b) Show that s(t) is an SSB signal.
Figure P5-12 Weaver’s method for generating SSB. Solution:
Ac2 50,000 = ⇒ Ac = 2236V 2(50) g (t ) = AC [1 + m(t )] = 2236[1 + A1 (cos ω 1t + cos 2ω 1t )]
(b)
m ( t ) = A1 cos ω1t + A1 cos 2ω1t = A1[cos ω1t + cos 2ω1t ]