用調製信號取控制載波信號的參數的過程.
其中m(t)爲調製信號,根據調製信號的不同,可分爲模擬調製和數字調製.
c(t)爲載波信號,通常有連續載波和脈衝載波兩種.連續載波又有調幅,調頻,調相三種調製方式.
sm (t)爲已調信號.根據調製前後,調製信號的頻譜是否發生線性變化,可分爲線性調製和非線性調製.
由調製信號控制載波信號的幅度.使載波信號的幅度按調製信號的規律發生變化，所以幅度調製的已調信號一般可以寫爲s m ( t ) = A m ( t ) cos ω c t
s_m(t)=Am(t)\cos\omega_ct
s m ( t ) = A m ( t ) cos ω c t S m ( ω ) = A 2 [ M ( ω − ω c ) + M ( ω + ω c ) ]
S_m(\omega)=\frac A2[M(\omega-\omega_c)+M(\omega+\omega_c)]
S m ( ω ) = 2 A [ M ( ω − ω c ) + M ( ω + ω c ) ]
在調製前後,基帶信號的頻譜 只發生了線性變化,因此幅度調製也可以稱爲線性調製.
標準調幅.將基帶信號變爲直流信號後進行幅度調製.
疊加直流分量後與載波函數相乘,所得已調信號時域表達爲s A M ( t ) = ( A 0 + m ( t ) ) cos ( ω c t )
s_{AM}(t)=(A_0+m(t))\cos(\omega_c t)
s A M ( t ) = ( A 0 + m ( t ) ) cos ( ω c t )
由於在疊加直流分量後,m(t)應爲直流信號,所以要求m a x { ∣ m ( t ) ∣ } ≤ A 0
max\left\{ |m(t)| \right\} \leq A_0
m a x { ∣ m ( t ) ∣ } ≤ A 0 S A M ( ω ) = π A 0 [ δ ( ω − ω c ) + δ ( ω + ω c ) ] + 1 2 [ M ( ω − ω c ) + M ( ω + ω c ) ]
S_{AM}(\omega)=\pi A_0[\delta(\omega-\omega_c)+\delta(\omega+\omega_c)]+\frac12 [M(\omega-\omega_c)+M(\omega+\omega_c)]
S A M ( ω ) = π A 0 [ δ ( ω − ω c ) + δ ( ω + ω c ) ] + 2 1 [ M ( ω − ω c ) + M ( ω + ω c ) ]
關於調製信號的描述
根據圖像不難看出,已調信號的帶寬BAM 爲調製信號帶寬fH 的兩倍B A M = 2 f H
B_{AM}=2f_H
B A M = 2 f H P A M = A 0 2 2 + m 2 ( t ) ˉ 2 = P c + P s
P_{AM}=\frac {A_0^2}2+\frac{\bar{m^2(t)}}2 \\
=P_c+P_s
P A M = 2 A 0 2 + 2 m 2 ( t ) ˉ = P c + P s c 爲載波功率,Ps 爲邊帶功率.
調製效率η A M = P s P A M
\eta_{AM}=\frac {P_s}{P_{AM}}
η A M = P A M P s A m cos ω m t A_m\cos\omega_mt A m cos ω m t η A M = A m 2 2 A 0 2 + A m 2
\eta_{AM}=\frac{A^2_m}{2A^2_0+A^2_m}
η A M = 2 A 0 2 + A m 2 A m 2 A 0 = A m A_0=A_m A 0 = A m
這樣調製後,效率有限,而且由於已調信號的帶寬增加,也導致傳輸頻率上的浪費.
但在解調僅需要包絡檢波即可恢復信號,解調實現非常簡單.
//鏈接包絡檢波
雙邊帶調製.與AM相比,不再疊加直流分量,此時已調信號的時域表達s D S B ( t ) = m ( t ) cos ω c t
s_{DSB}(t)=m(t)\cos \omega_c t
s D S B ( t ) = m ( t ) cos ω c t S D S B ( ω ) = 1 2 [ M ( ω − ω c ) + M ( ω + ω c ) ]
S_{DSB}(\omega)=\frac 12[M(\omega-\omega_c)+M(\omega+\omega_c)]
S D S B ( ω ) = 2 1 [ M ( ω − ω c ) + M ( ω + ω c ) ]
較AM相比,DSB由於不再疊加直流分量,使調製效率大大增加.
但相對的在解調時不能使用包絡檢波,同時調製後頻帶拓寬仍致使帶寬資源浪費.
單邊帶調製.由於帶限信號的頻譜是對稱的,因此在理論上,僅發送半個頻譜信息就可以復原基帶信號.
這樣又進一步節約帶寬資源,但相應的在製作上要複雜許多.
在DSB的基礎上通過濾波器,保留上邊帶(USB)或下邊帶(LSB).
實際中濾波器並不具備理想濾波器這樣陡峭的邊界,濾波器截至特性越好就越貴.
殘留邊帶調製.在適當增加帶寬的條件下,優化對SSB中濾波器的要求.
VSB中不再要求嚴格的截至特性,而是需要具有互補對稱性,即相隔2 ω c 2\omega_c 2 ω c
證明如下:
根據VSB調製過程可知,未經LPF信號s(t)s ( t ) = s V S B ( t ) ⋅ 2 cos ω c t
s(t)=s_{VSB}(t)\cdot 2\cos \omega_ct
s ( t ) = s V S B ( t ) ⋅ 2 cos ω c t KaTeX parse error: No such environment: align at position 8:
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S(\omega)&=\fr… H ( ω ) H(\omega) H ( ω ) S m ( ω ) = H ( ω ) [ S V S B ( ω + ω c ) + S V S B ( ω − ω c ) ]
S_m(\omega)=H(\omega)[S_{VSB}(\omega+\omega_c)+S_{VSB}(\omega-\omega_c)]
S m ( ω ) = H ( ω ) [ S V S B ( ω + ω c ) + S V S B ( ω − ω c ) ] R ( t ) = s m ( t ) ⋅ 2 cos ω c t
R(t)=s_m(t) \cdot 2\cos \omega_c t
R ( t ) = s m ( t ) ⋅ 2 cos ω c t KaTeX parse error: No such environment: align at position 8:
\begin{̲a̲l̲i̲g̲n̲}̲
R(\omega)&=\fr… R ( ω ) = S V S B ( ω ) [ H ( ω + ω c ) + H ( ω − ω c ) ]
R(\omega)=S_{VSB}(\omega)[H(\omega+\omega_c)+H(\omega-\omega_c)]
R ( ω ) = S V S B ( ω ) [ H ( ω + ω c ) + H ( ω − ω c ) ] H ( ω + ω c ) + H ( ω − ω c ) = C o n s t a n t
H(\omega+\omega_c)+H(\omega-\omega_c) = Constant
H ( ω + ω c ) + H ( ω − ω c ) = C o n s t a n t
適用於所有線性調製.
相干解調時,爲了無失真還原基帶信號,接受端必須提供一個與接受的已調信號嚴格同步的本地載波,它與已調信號相乘後,經過LPF取出低頻分量,即可還原基帶信號
恢復載波相位對解調影響:
不考慮信道乘性噪聲干擾,則對於接受端信號爲s ( t ) = A c m ( t ) cos ( 2 π f c t + ϕ c )
s(t)= A_c m(t) \cos(2\pi f_c t +\phi_c)
s ( t ) = A c m ( t ) cos ( 2 π f c t + ϕ c ) ϕ \phi ϕ KaTeX parse error: No such environment: align at position 8:
\begin{̲a̲l̲i̲g̲n̲}̲
r(t)\cos(2\pi … y o ( t ) = A c 2 m ( t ) cos ( ϕ c − ϕ )
y_o(t)=\frac{A_c}{2}m(t)\cos(\phi_c-\phi)
y o ( t ) = 2 A c m ( t ) cos ( ϕ c − ϕ ) ϕ c \phi_c ϕ c ϕ \phi ϕ Δ ϕ \Delta \phi Δ ϕ
Δ ϕ = π 2 \Delta \phi= \frac{\pi}{2} Δ ϕ = 2 π Δ ϕ = 0 \Delta \phi= 0 Δ ϕ = 0
包絡:我們可以將任一平穩窄帶高斯隨機過程X(t)表示爲標準正態振盪的形式A ( t ) cos ( ω t + ϕ ( t ) )
A(t)\cos(\omega t + \phi(t))
A ( t ) cos ( ω t + ϕ ( t ) )
首先通過整流器,被整流信號通過LPF即可恢復原基帶信號
在理想狀態下,包絡檢波器的輸出爲y o ( t ) = g 1 + g 2 m ( t )
y_o(t)=g_1+g_2m(t)
y o ( t ) = g 1 + g 2 m ( t )
s m ( t ) s_m(t) s m ( t ) n ( t ) n(t) n ( t ) s m ( t ) s_m(t) s m ( t ) n i ( t ) n_i(t) n i ( t ) m o ( t ) m_o(t) m o ( t ) n o ( t ) n_o(t) n o ( t )
Ps:BPF的作用爲濾除帶外噪聲
解調器輸出有用信號的平均功率與輸出噪聲平均功率之比S o N o = m o 2 ( t ) ‾ n o 2 ( t ) ‾
\frac{S_o}{N_o}=\frac{\overline{m_o^2(t)}}{\overline{n_o^2(t)}}
N o S o = n o 2 ( t ) m o 2 ( t )
對於不同形式的s m ( t ) s_m(t) s m ( t ) 單邊 功率譜密度爲n 0 n_0 n 0 N i = n 0 B
N_i= n_0 B
N i = n 0 B
輸出信噪比與輸入信噪比的比值G = S o / N o S i / N i
G=\frac{S_o / N_o}{S_i / N_i}
G = S i / N i S o / N o
輸出信號平均功率:1 4 m 2 ( t ) ‾ \frac 14 \overline{m^2(t)} 4 1 m 2 ( t ) 1 4 n 0 B \frac 14 n_0B 4 1 n 0 B S o N o = m 2 ( t ) ‾ n 0 B
\frac {S_o}{N_o}=\frac{\overline{m^2(t)}}{n_0B}
N o S o = n 0 B m 2 ( t ) S i N i = 1 2 m 2 ( t ) ‾ n 0 B
\frac{S_i}{N_i}=\frac{\frac 12 \overline{m^2(t)}}{n_0B}
N i S i = n 0 B 2 1 m 2 ( t ) G = 2 G=2 G = 2
具體計算流程:
設調製信號爲m ( t ) c o s 2 π f c t m(t)cos 2\pi f_ct m ( t ) c o s 2 π f c t m o ( t ) = 1 2 m ( t )
m_o (t) =\frac 12 m(t)
m o ( t ) = 2 1 m ( t ) S o = 1 4 m ( t ) ‾ 2 S_o = \frac 14 \overline{m(t)}^2 S o = 4 1 m ( t ) 2
再考慮噪聲通過相干解調後的情況:n i ( t ) = n c ( t ) cos 2 π f c t − n s ( t ) sin 2 π f c t
n_i(t)=n_c(t)\cos 2\pi f_ct -n_s(t)\sin 2\pi f_c t
n i ( t ) = n c ( t ) cos 2 π f c t − n s ( t ) sin 2 π f c t 1 2 n c ( t ) + 1 2 n c ( t ) cos 2 π ⋅ 2 f c t − 1 2 n s ( t ) sin 2 π ⋅ 2 f c t
\frac 12n_c(t) + \frac 12n_c(t)\cos 2\pi\cdot2f_ct-\frac 12 n_s(t)\sin2\pi\cdot 2f_c t
2 1 n c ( t ) + 2 1 n c ( t ) cos 2 π ⋅ 2 f c t − 2 1 n s ( t ) sin 2 π ⋅ 2 f c t n o ( t ) = 1 2 n c ( t ) n_o(t)=\frac 12n_c(t) n o ( t ) = 2 1 n c ( t ) N o = 1 4 n c ( t ) ‾ 2 = 1 4 n 0 B
N_o=\frac 14 \overline{n_c(t)}^2=\frac 14n_0B
N o = 4 1 n c ( t ) 2 = 4 1 n 0 B S o N o = m ( t ) ‾ 2 n 0 B
\frac{S_o}{N_o}=\frac{\overline{m(t)}^2}{n_0B}
N o S o = n 0 B m ( t ) 2 S i N i = 1 2 m ( t ) ‾ 2 n 0 B
\frac {S_i}{N_i}=\frac {\frac 12 \overline{m(t)}^2}{n_0B}
N i S i = n 0 B 2 1 m ( t ) 2
輸出信號平均功率:1 4 m 2 ( t ) ‾ \frac 14 \overline{m^2(t)} 4 1 m 2 ( t ) 1 4 n 0 B \frac 14 n_0B 4 1 n 0 B S o N o = m 2 ( t ) ‾ 4 n 0 B
\frac{S_o}{N_o}=\frac{\overline{m^2(t)}}{4n_0B}
N o S o = 4 n 0 B m 2 ( t ) S i N i = m 2 ( t ) ‾ 4 n 0 B
\frac{S_i}{N_i}=\frac{\overline{m^2(t)}}{4n_0B}
N i S i = 4 n 0 B m 2 ( t ) G = 1 G=1 G = 1
具體計算流程:
不能單單從制度增益比較兩種調製方式,考慮到二者帶寬,輸入信號功率不同,這樣的比較是不合理的.
若在相同條件下比較.二者抗噪能力基本近似
已知輸入信號KaTeX parse error: No such environment: align at position 8:
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s_m(t)+n_i(t) … E ( t ) = [ A 0 + m ( t ) + n c ( t ) ] 2 + n s 2 ( t ) E(t) = \sqrt{[A_0+m(t)+n_c(t)]^2+n_s^2(t)} E ( t ) = [ A 0 + m ( t ) + n c ( t ) ] 2 + n s 2 ( t ) S i = A 0 2 2 + m 2 ( t ) ‾ 2 S_i = \frac{A_0^2}{2} + \frac{\overline{m^2(t)}}{2} S i = 2 A 0 2 + 2 m 2 ( t ) N i = n 0 B N_i = n_0 B N i = n 0 B
對於輸出的包絡信號,現考慮輸入信噪比的兩種情況:
大信噪比[ A 0 + m ( t ) ] ≫ n c 2 ( t ) + n s 2 ( t )
[A_0+m(t)] \gg \sqrt{n_c^2(t)+n_s^2(t)}
[ A 0 + m ( t ) ] ≫ n c 2 ( t ) + n s 2 ( t )
計算:
此時輸出信號功率:S o = m 2 ( t ) ‾ S_o=\overline{m^2(t)} S o = m 2 ( t )
輸出噪聲功率:N o = n 0 B N_o=n_0B N o = n 0 B
輸出信噪比S o N o = m 2 ( t ) ‾ n 0 B
\frac{S_o}{N_o}=\frac{\overline{m^2(t)}}{n_0B}
N o S o = n 0 B m 2 ( t ) G A M = 2 m 2 ( t ) ‾ A 0 2 + m 2 ( t )
G_{AM}=\frac{2\overline{m^2(t)}}{A_0^2+m^2(t)}
G A M = A 0 2 + m 2 ( t ) 2 m 2 ( t ) A 0 = ∣ m ( t ) ∣ m a x A_0= |m(t)|_{max} A 0 = ∣ m ( t ) ∣ m a x 2 3 \frac23 3 2
小信噪比[ A 0 + m ( t ) ] ≪ n c 2 ( t ) + n s 2 ( t )
[A_0+m(t)] \ll \sqrt{n_c^2(t)+n_s^2(t)}
[ A 0 + m ( t ) ] ≪ n c 2 ( t ) + n s 2 ( t ) E ( t ) = R ( t ) + [ A 0 + m ( t ) ] cos θ ( t )
E(t)=R(t)+[A_0+m(t)]\cos \theta(t)
E ( t ) = R ( t ) + [ A 0 + m ( t ) ] cos θ ( t )
輸出噪聲功率:N o = n 0 B N_o=n_0B N o = n 0 B
輸出信噪比S o N o = m 2 ( t ) ‾ n 0 B
\frac{S_o}{N_o}=\frac{\overline{m^2(t)}}{n_0B}
N o S o = n 0 B m 2 ( t ) G A M = 2 m 2 ( t ) ‾ A 0 2 + m 2 ( t )
G_{AM}=\frac{2\overline{m^2(t)}}{A_0^2+m^2(t)}
G A M = A 0 2 + m 2 ( t ) 2 m 2 ( t ) A 0 = ∣ m ( t ) ∣ m a x A_0= |m(t)|_{max} A 0 = ∣ m ( t ) ∣ m a x 2 3 \frac23 3 2
小信噪比[ A 0 + m ( t ) ] ≪ n c 2 ( t ) + n s 2 ( t )
[A_0+m(t)] \ll \sqrt{n_c^2(t)+n_s^2(t)}
[ A 0 + m ( t ) ] ≪ n c 2 ( t ) + n s 2 ( t ) E ( t ) = R ( t ) + [ A 0 + m ( t ) ] cos θ ( t )
E(t)=R(t)+[A_0+m(t)]\cos \theta(t)
E ( t ) = R ( t ) + [ A 0 + m ( t ) ] cos θ ( t )
當輸入信噪比小於一定程度時,輸出信噪比急劇下降,這一現象被稱爲門限效應 ,開始出現門限效應的輸入信噪比值稱爲門限值.
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