110 年 國立臺灣聯合大學系統(清華、政治、陽明交通、中央四校聯招)研究所電機類《通訊系統》

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第 1 題15 分

  1. (Total=15%) Fig. 1 shows a series of operations that are used to separate random processes into different frequency regions. The input random process x(t) is Gaussian and white with a power spectral density function of Sx(f)=2S_x(f) = 2.

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The system functions of the various linear filters are as follows:
H1(f)={1,1000≤∣f∣≤30000,otherwiseH_1(f) = \begin{cases} 1, & 1000 \le |f| \le 3000 \\ 0, & \text{otherwise} \end{cases}
H2(f)={1,1000≤∣f∣≤20000,otherwiseH_2(f) = \begin{cases} 1, & 1000 \le |f| \le 2000 \\ 0, & \text{otherwise} \end{cases}
H3(f)={1,1500≤∣f∣≤25000,otherwiseH_3(f) = \begin{cases} 1, & 1500 \le |f| \le 2500 \\ 0, & \text{otherwise} \end{cases}
H4(f)={1,2000≤∣f∣≤30000,otherwiseH_4(f) = \begin{cases} 1, & 2000 \le |f| \le 3000 \\ 0, & \text{otherwise} \end{cases}

We developed expressions that relate the various random processes that are inputs and outputs to linear filters. For example, recall that
rzz(t)=rxx(t)∗h1(τ)∗h1(−τ)r_{zz}(t) = r_{xx}(t) * h_1(\tau) * h_1(-\tau)
rzx(t)=rxx(t)∗h1(t)r_{zx}(t) = r_{xx}(t) * h_1(t)
ry2y3(t)=rzz(t)∗h2(t)∗h3(−t)r_{y_2y_3}(t) = r_{zz}(t) * h_2(t) * h_3(-t)

Use the above properties, your knowledge of Fourier transforms, and your general knowledge of the properties of random processes to answer the following questions.

(a) (4%) Considering y1(t)y_1(t), y2(t)y_2(t), y3(t)y_3(t), and y4(t)y_4(t), the random processes that are wide-sense stationary are _____.
(b) (3%) Considering y1(t)y_1(t), y2(t)y_2(t), y3(t)y_3(t), and y4(t)y_4(t), the random processes that are white are _____.
(c) (2%) It is ____ (true/false) that the random processes y2(t)y_2(t) and y3(t)y_3(t) are statistically independent.
(d) (6%) The probability that z(t)z(t) is greater than 3 at t=20t=20 is _____.

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這一題的完整詳解

核心觀念

本題考查:

  1. 白高斯隨機過程通過 LTI 系統後的功率譜密度。
  2. 廣義平穩(WSS)與白雜訊的判斷。
  3. 高斯隨機過程中「互不相關等價於統計獨立」。
  4. 由功率譜密度求變異數,再計算高斯機率。

輸入 x(t)x(t) 為零平均白高斯隨機過程,且

Sx(f)=2S_x(f)=2

對於 LTI 系統,輸出功率譜密度為

Sy(f)=Sx(f)∣H(f)∣2S_y(f)=S_x(f)|H(f)|^2

(a) 廣義平穩性

白高斯隨機過程 x(t)x(t) 本身是 WSS。對於穩定的 LTI 濾波器,若輸入為 WSS,輸出仍為 WSS。

因此,經由 H1(f),H2(f),H3(f),H4(f)H_1(f),H_2(f),H_3(f),H_4(f) 所產生的各個隨機過程,其自相關函數只與時間差有關,平均值也不隨時間改變。

所以:

y1(t),y2(t),y3(t),y4(t) 全部皆為 WSS\boxed{y_1(t),y_2(t),y_3(t),y_4(t)\text{ 全部皆為 WSS}}

(b) 白隨機過程判斷

白隨機過程的功率譜密度必須對所有頻率皆為常數。

例如通過 H1(f)H_1(f) 後:

Sy1(f)=2∣H1(f)∣2={2,1000≤∣f∣≤30000,其他S_{y_1}(f) = 2|H_1(f)|^2 = \begin{cases} 2, & 1000\le |f|\le 3000\\ 0, & \text{其他} \end{cases}

其餘輸出同理,功率譜密度均只存在於特定頻帶,並非對所有頻率均為常數。因此它們都是帶限隨機過程,而非嚴格意義下的白隨機過程。

所以:

y1(t),y2(t),y3(t),y4(t) 均不是白隨機過程\boxed{y_1(t),y_2(t),y_3(t),y_4(t)\text{ 均不是白隨機過程}}

(c) y2(t)y_2(t) 與 y3(t)y_3(t) 是否統計獨立

兩個輸出之間的互功率譜密度為

Sy2y3(f)=Sz(f)H2(f)H3∗(f)S_{y_2y_3}(f) = S_z(f)H_2(f)H_3^*(f)

由於 z(t)z(t) 是經過 H1(f)H_1(f) 後的訊號,

Sz(f)=2∣H1(f)∣2S_z(f)=2|H_1(f)|^2

H2(f)H_2(f) 與 H3(f)H_3(f) 的通帶分別為:

1000≤∣f∣≤20001000\le |f|\le 2000

以及

1500≤∣f∣≤25001500\le |f|\le 2500

兩者在

1500≤∣f∣≤20001500\le |f|\le 2000

的頻帶重疊,因此

Sy2y3(f)≠0S_{y_2y_3}(f)\neq 0

所以 y2(t)y_2(t) 與 y3(t)y_3(t) 具有相關性。

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第 2 題18 分

  1. (Total=18%) A pulse communication system is being designed that will transmit one of two messages, m1m_1 and m2m_2, which occur with equal a priori probability. These messages will be signaled using the following waveforms, and we would like to select sets of transmitted signals that will either minimize the average transmitted energy or will minimize the probability of error. Assume that the channel is corrupted by an additive white Gaussian random process. Assume here that the inner product of two real-valued function of duration T=2T = 2 is defined as:
    (f(t),g(t))=∫0Tf(t)g(t)dt(f(t), g(t)) = \int_0^T f(t)g(t) dt.

Four possible signals, sa(t)s_a(t), sB(t)s_B(t), sc(t)s_c(t), and sD(t)s_D(t) are being considered, and two of them will be selected as the signal set that will indicate which message had been sent. The signals are sketched in Fig. 2.

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(a) (4%) Compute a set of orthonormal basis functions {ϕa(t),ϕB(t),ϕc(t),ϕD(t)}\{\phi_a(t), \phi_B(t), \phi_c(t), \phi_D(t)\} using the Gram-Schmidt procedure, where ϕa(t)\phi_a(t) is proportional to sa(t)s_a(t). The signals sa(t)s_a(t), sB(t)s_B(t), sc(t)s_c(t), and sD(t)s_D(t) can be expressed as a linear combination of these basis functions as _____, where the basis functions are _____.

(b) (4%) The TWO candidate signals that would result in minimum average transmitted energy, given that the two messages are equiprobable, are _____.

(c) (4%) The TWO candidate signals that would provide the lowest average probability of a bit error are _____.

(d) (6%) To increase data rate, suppose now you want to send THREE messages instead of two, which again occur with equal a priori probability. Suppose sA(t)s_A(t), sB(t)s_B(t), and sc(t)s_c(t) are chosen to represent these messages, and the power spectral density of the noise is N0/2=1N_0/2 = 1. The union bound of the average symbol error probability is _____.

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這一題的完整詳解

核心觀念

本題考查:

  • Gram–Schmidt 正交化與正規化。
  • 訊號能量
    Ei=⟨si,si⟩E_i=\langle s_i,s_i\rangle
  • 二元 AWGN 通道中,等先驗機率下,錯誤率由兩訊號間的歐氏距離決定:
    Pe=Q(∥si−sj∥2N0)P_e=Q\left(\frac{\|s_i-s_j\|}{\sqrt{2N_0}}\right)
  • 多元訊號的聯集界(union bound)。

由圖可將四個訊號寫成

sA(t)={1,0≤t<20,其他s_A(t)= \begin{cases} 1,&0\le t<2\\ 0,&\text{其他} \end{cases} sB(t)={1,0≤t<1−1,1≤t<20,其他s_B(t)= \begin{cases} 1,&0\le t<1\\ -1,&1\le t<2\\ 0,&\text{其他} \end{cases} sC(t)=−sB(t),sD(t)=0.s_C(t)=-s_B(t),\qquad s_D(t)=0.

(a) Gram–Schmidt 正交規範化

先取與 sA(t)s_A(t) 成正比的第一個基底:

∥sA∥=∫0212dt=2\|s_A\|=\sqrt{\int_0^2 1^2dt}=\sqrt{2}

因此

ϕA(t)=sA(t)2.\phi_A(t)=\frac{s_A(t)}{\sqrt{2}}.

接著處理 sB(t)s_B(t)。其與 ϕA(t)\phi_A(t) 的內積為

⟨sB,ϕA⟩=∫02sB(t)12dt=12(1−1)=0.\langle s_B,\phi_A\rangle =\int_0^2s_B(t)\frac{1}{\sqrt{2}}dt =\frac{1}{\sqrt{2}}(1-1)=0.

所以 sB(t)s_B(t) 已與 ϕA(t)\phi_A(t) 正交。又

∥sB∥=∫0112dt+∫12(−1)2dt=2,\|s_B\|=\sqrt{\int_0^1 1^2dt+\int_1^2(-1)^2dt} =\sqrt{2},

故第二個正規化基底為

ϕB(t)=sB(t)2.\phi_B(t)=\frac{s_B(t)}{\sqrt{2}}.

由於

sC(t)=−sB(t),sD(t)=0,s_C(t)=-s_B(t),\qquad s_D(t)=0,

它們不會產生新的獨立基底。因此訊號空間實際上只有二維,不能建立四個互相正交且皆為非零的正規基底。

四個訊號在 {ϕA,ϕB}\{\phi_A,\phi_B\} 下的訊號向量為

sA(t)=2ϕA(t),s_A(t)=\sqrt{2}\phi_A(t), sB(t)=2ϕB(t),s_B(t)=\sqrt{2}\phi_B(t), sC(t)=−2ϕB(t),s_C(t)=-\sqrt{2}\phi_B(t), sD(t)=0.s_D(t)=0.

也就是

sA=(2,0),sB=(0,2),sC=(0,−2),sD=(0,0).\mathbf{s}_A=(\sqrt{2},0),\quad \mathbf{s}_B=(0,\sqrt{2}),\quad \mathbf{s}_C=(0,-\sqrt{2}),\quad \mathbf{s}_D=(0,0).

(b) 最小平均傳送能量

四個訊號的能量為

EA=EB=EC=2,ED=0.E_A=E_B=E_C=2,\qquad E_D=0.

選取兩個訊號且先驗機率相等時,平均能量為

Eˉ=Ei+Ej2.\bar E=\frac{E_i+E_j}{2}.

因此任何包含 sD(t)s_D(t) 與一個非零訊號的組合,都得到最小平均能量:

{sA,sD},{sB,sD},{sC,sD}.\{s_A,s_D\},\qquad \{s_B,s_D\},\qquad \{s_C,s_D\}.

三組的平均能量皆為

Eˉ=2+02=1.\bar E=\frac{2+0}{2}=1.

題目若要求填寫一組答案,可填:

sA(t) 與 sD(t)\boxed{s_A(t)\ \text{與}\ s_D(t)}

但須注意,{sB,sD}\{s_B,s_D\} 與 {sC,sD}\{s_C,s_D\} 也同樣正確。


(c) 最低平均錯誤機率

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第 3 題15 分

  1. (Total=15%) Consider a communication system in which the received signal y(t)y(t) is given by
    y(t)={s(t)+w(t),if "1" is sentw(t),if "0" is senty(t) = \begin{cases} s(t) + w(t), & \text{if "1" is sent} \\ w(t), & \text{if "0" is sent} \end{cases}
    where w(t)w(t) is white Gaussian noise with (zero mean) power spectral density N0/2N_0/2 and s(t)s(t) is shown below.

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At the receiver, the received signal y(t)y(t) is passed through a filter with impulse response h(t)=s(T−t)h(t) = s(T - t). The output of the filter is sampled at time t0t_0 that provides a decision statistic YY with
Y={S+W,if "1" is sentW,if "0" is sentY = \begin{cases} S + W, & \text{if "1" is sent} \\ W, & \text{if "0" is sent} \end{cases}
where S=s(t)∗h(t)∣t=t0S = s(t) * h(t)|_{t=t_0} and W=w(t)∗h(t)∣t=t0W = w(t) * h(t)|_{t=t_0} are the filtered/sampled outputs of s(t)s(t) and w(t)w(t), respectively. The maximum likelihood (ML) decision rule is then used based on YY to decide whether "1" or "0" is sent. Assume the noise process is independent of which signal is sent.

(a) (3%) Let t0=3t_0 = 3 and T=4T=4, i.e. h(t)=s(4−t)h(t) = s(4 - t). Determine the mean and variance of the filtered and sampled noise term WW.
(b) (6%) Again, let t0=3t_0 = 3 and T=4T=4. Please find the corresponding bit error rate (BER) of the ML receiver. (Express your answer in Q-function.)
(c) (3%) Can the BER in part (b) with h(t)=s(4−t)h(t) = s(4 - t) be improved by choosing a different sampling time t0t_0? Justify your answers.
(d) (3%) Suppose now T=5T=5, i.e. h(t)=s(5−t)h(t) = s(5-t). How to determine the sampling instant t0t_0 in this case? Please provide a general rule to determine the relation between the sampling instant t0t_0 and the delay TT such that the sampled output yields the best BER.

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這一題的完整詳解

核心觀念

本題考查:

  1. 白高斯雜訊通過線性濾波器後的平均值與變異數。
  2. 匹配濾波器的取樣時刻選擇。
  3. 二元等變異數高斯訊號的 ML 判決與 BER。
  4. 自相關函數在零延遲時取得最大值。

由圖可讀得

s(t)={1,0≤t<2,−1,2≤t<4,0,其他時間.s(t)= \begin{cases} 1, & 0\le t<2,\\ -1, & 2\le t<4,\\ 0, & \text{其他時間}. \end{cases}

訊號能量為

Es=∫−∞∞s2(t) dt=∫021 dt+∫24(−1)2 dt=4.E_s=\int_{-\infty}^{\infty}s^2(t)\,dt =\int_0^2 1\,dt+\int_2^4(-1)^2\,dt=4.

(a) 雜訊項 WW 的平均值與變異數

濾波後取樣的雜訊為

W=w(t)∗h(t)∣t=t0.W=\left.w(t)*h(t)\right|_{t=t_0}.

因為 w(t)w(t) 是零平均值白高斯雜訊,經過線性濾波後仍為零平均值高斯隨機變數,因此

E[W]=0.\mathbb{E}[W]=0.

白雜訊功率頻譜密度為 N0/2N_0/2,故濾波器輸出雜訊變異數為

σW2=N02∫−∞∞h2(t) dt.\sigma_W^2=\frac{N_0}{2}\int_{-\infty}^{\infty}h^2(t)\,dt.

本題

h(t)=s(4−t),h(t)=s(4-t),

時間反轉與平移不會改變訊號能量,因此

∫−∞∞h2(t) dt=∫−∞∞s2(t) dt=Es=4.\int_{-\infty}^{\infty}h^2(t)\,dt =\int_{-\infty}^{\infty}s^2(t)\,dt =E_s=4.

所以

E[W]=0\boxed{\mathbb{E}[W]=0}

以及

Var⁡(W)=σW2=2N0.\boxed{\operatorname{Var}(W)=\sigma_W^2=2N_0}.

(b) ML 接收器的 BER

在 t0=3t_0=3、T=4T=4 時,

S=s(t)∗h(t)∣t=3.S=\left.s(t)*h(t)\right|_{t=3}.

由摺積定義,

S=∫−∞∞s(τ)h(3−τ) dτ.S=\int_{-\infty}^{\infty}s(\tau)h(3-\tau)\,d\tau.

又因為 h(t)=s(4−t)h(t)=s(4-t),

h(3−τ)=s(4−(3−τ))=s(1+τ).h(3-\tau)=s\bigl(4-(3-\tau)\bigr)=s(1+\tau).

因此

S=∫−∞∞s(τ)s(1+τ) dτ.S=\int_{-\infty}^{\infty}s(\tau)s(1+\tau)\,d\tau.

依照波形分段:

  • 0≤τ<10\le \tau<1:s(τ)=1s(\tau)=1、s(1+τ)=1s(1+\tau)=1,乘積為 11。
  • 1≤τ<21\le \tau<2:s(τ)=1s(\tau)=1、s(1+τ)=−1s(1+\tau)=-1,乘積為 −1-1。
  • 2≤τ<32\le \tau<3:s(τ)=−1s(\tau)=-1、s(1+τ)=−1s(1+\tau)=-1,乘積為 11。

故

S=∫011 dτ+∫12(−1) dτ+∫231 dτ=1−1+1=1.S=\int_0^1 1\,d\tau+\int_1^2(-1)\,d\tau+\int_2^3 1\,d\tau =1-1+1=1.

因此兩種假設下的決策統計量分布為

Y∣1∼N(1,2N0),Y\mid 1\sim\mathcal{N}(1,2N_0), Y∣0∼N(0,2N0).Y\mid 0\sim\mathcal{N}(0,2N_0).

由於兩種情況的變異數相同,且採用 ML 判決,判決門檻為兩個平均值的中點:

γ=S2=12.\gamma=\frac{S}{2}=\frac{1}{2}.

判決規則為

Y>12⇒判定傳送「1」,Y>\frac12 \Rightarrow \text{判定傳送「1」}, Y<12⇒判定傳送「0」.Y<\frac12 \Rightarrow \text{判定傳送「0」}.

若傳送「0」,錯誤機率為

P(e∣0)=P(W>12)=Q(1/22N0).P(e\mid 0)=P\left(W>\frac12\right) =Q\left(\frac{1/2}{\sqrt{2N_0}}\right).
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第 4 題18 分

  1. (Total=18%) Consider a binary symmetric channel with input XX, output YY, and transition probability α\alpha. More specifically, as shown in the figure below, the input and output of the channel may be "0" or "1", P(Y=0∣X=0)=P(Y=1∣X=1)=1−αP(Y=0|X=0) = P(Y=1|X=1) = 1-\alpha and P(Y=1∣X=0)=P(Y=0∣X=1)=αP(Y=1|X=0) = P(Y=0|X=1) = \alpha. The prior probability is P(X=0)=pP(X=0) = p.

🖼️【此處有附圖,請對照原卷】

(a) (4%) Find the entropy H(Y)H(Y) and the conditional entropy H(Y∣X)H(Y|X).
(b) (4%) Find α\alpha that can maximize the channel capacity.
(c) (5%) For the case of p<αp < \alpha, please give the maximum a posteriori (MAP) decision rule to decide which XX is sent based on the output YY. (Hint: In this problem, the output YY of the channel can take just two values "0" or "1". Hence, it is sufficient to find the decision result for these two output possibilities.)
(d) (5%) Suppose the prior probability is not known in advance. We manage to produce an estimated prior probability P(X=0)=qP(X=0) = q, which may or may not be equal to the true prior probability P(X=0)=pP(X=0) = p. In this case, the cross-entropy between the true and estimated prior distributions PP and P^\hat{P} is defined by H(P,P^)=−plog⁡2q−(1−p)log⁡2(1−q)H(P, \hat{P}) = -p \log_2 q - (1-p) \log_2(1-q), which can be considered as an approximated entropy of XX. Please show that the cross-entropy H(P,P^)H(P, \hat{P}) is always no less than the true entropy H(X)H(X) of XX, i.e. H(P,P^)≥H(X)H(P, \hat{P}) \ge H(X).
(Hint: You may use the Jensen's inequality: plog⁡2a+(1−p)log⁡2b≤log⁡2(pa+(1−p)b)p \log_2 a + (1-p) \log_2 b \le \log_2(pa + (1-p)b) for 0≤p≤10 \le p \le 1, a>0a > 0, and b>0b > 0.)

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這一題的完整詳解

核心觀念

本題考二元對稱通道的熵、通道容量、最大後驗機率(MAP)判決,以及交叉熵與真實熵的關係。

定義二元熵函數

h2(x)=−xlog⁡2x−(1−x)log⁡2(1−x),h_2(x)=-x\log_2 x-(1-x)\log_2(1-x),

其中 0log⁡200\log_2 0 以極限值 00 計算。

解題方法

(a) 求 H(Y)H(Y) 與 H(Y∣X)H(Y|X)

由全機率公式,

P(Y=0)=P(Y=0∣X=0)P(X=0)+P(Y=0∣X=1)P(X=1)=p(1−α)+(1−p)α.P(Y=0)=P(Y=0|X=0)P(X=0)+P(Y=0|X=1)P(X=1) =p(1-\alpha)+(1-p)\alpha.

整理得

P(Y=0)=α+p(1−2α).P(Y=0)=\alpha+p(1-2\alpha).

因此

H(Y)=h2 ⁣(α+p(1−2α)).H(Y)=h_2\!\left(\alpha+p(1-2\alpha)\right).

當 XX 固定時,輸出以機率 α\alpha 翻轉、以機率 1−α1-\alpha 不翻轉,因此不論 X=0X=0 或 X=1X=1,條件熵都是 h2(α)h_2(\alpha)。所以

H(Y∣X)=p h2(α)+(1−p)h2(α)=h2(α).H(Y|X)=p\,h_2(\alpha)+(1-p)h_2(\alpha)=h_2(\alpha).

(b) 求可使通道容量最大的 α\alpha

通道互資訊為

I(X;Y)=H(Y)−H(Y∣X).I(X;Y)=H(Y)-H(Y|X).

對固定的 α\alpha,調整輸入機率可使 H(Y)H(Y) 最大為 11 bit;例如令 p=12p=\frac12,輸出也均勻。因此通道容量為

C=1−h2(α).C=1-h_2(\alpha).

二元熵 h2(α)h_2(\alpha) 在 α=12\alpha=\frac12 時最大,在 α=0\alpha=0 或 α=1\alpha=1 時最小。因此若 α\alpha 的範圍是 0≤α≤10\leq\alpha\leq1,容量最大時

α=0或α=1,\alpha=0 \quad\text{或}\quad \alpha=1,

且此時 C=1C=1 bit。α=0\alpha=0 表示通道不翻轉;α=1\alpha=1 表示通道必定翻轉,兩者都能無誤判地還原輸入。

若課程將交越機率限定在 0≤α≤120\leq\alpha\leq\frac12,則最大容量發生在 α=0\alpha=0。

(c) 在 p<αp<\alpha 時求 MAP 判決規則

MAP 判決是比較兩個輸入在已知輸出下的聯合機率。收到 Y=0Y=0 時,比較

P(X=0,Y=0)=p(1−α),P(X=1,Y=0)=(1−p)α.P(X=0,Y=0)=p(1-\alpha),\qquad P(X=1,Y=0)=(1-p)\alpha.

選 X=0X=0 的條件為

p(1−α)≥(1−p)α  ⟺  p≥α.p(1-\alpha)\geq(1-p)\alpha \iff p\geq\alpha.

題設 p<αp<\alpha,故收到 Y=0Y=0 時判定 X=1X=1。

收到 Y=1Y=1 時,比較

P(X=0,Y=1)=pα,P(X=1,Y=1)=(1−p)(1−α).P(X=0,Y=1)=p\alpha,\qquad P(X=1,Y=1)=(1-p)(1-\alpha).

選 X=0X=0 的條件為

pα≥(1−p)(1−α)  ⟺  p≥1−α.p\alpha\geq(1-p)(1-\alpha) \iff p\geq1-\alpha.

因此完整判決規則為:

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第 5 題34 分

  1. (Total=34%) The figure below depicts a scenario where two base stations transmit simultaneously the same Quadrature Amplitude Modulation (QAM) signal to a car that is in the middle of two base stations and moves toward Base Station 2. The signal is expressed as
    s(t)=Acp(t)cos⁡(2πf0t)−Asp(t)sin⁡(2πf0t)s(t) = A_c p(t) \cos(2\pi f_0 t) - A_s p(t) \sin(2\pi f_0 t)
    in which f0f_0 is the carrier frequency, p(t)p(t) is a rectangular pulse with the width of TT, and the amplitudes AcA_c and AsA_s can take on values from {±1,±3}\{\pm 1, \pm 3\}.

🖼️【此處有附圖,請對照原卷】

(a) (5%) Sketch the signal space diagram (constellation diagram) of this QAM signal.
(b) (10%) Sketch the block diagram of the coherent receiver of a typical QAM system. Furthermore, derive what happens if the local oscillation used by the "coherent" receiver is incorrect (say, the frequency is f0+Δff_0 + \Delta f).
(c) (5%) Let the symbol error rate of QAM be approximated as 4Q(γ)4Q(\sqrt{\gamma}), where QQ represents the Q function and γ\gamma is the SNR. To improve the symbol error rate from 10−210^{-2} to 10−310^{-3}, roughly how much shall the SNR increase?
(d) (7%) Assume that the channel gain and propagation delay from two base stations to the car are the same. Derive the mathematical expression of the received signal, ignoring the speed of the car.
(e) (7%) Now consider the car's speed vv and the Doppler effect (the carrier frequency is modified with f=cf0f = c f_0, where cc is the speed of light, and choosing the sign according to the moving direction). Derive the mathematical expression of the received signal.

🖼️ 本題含圖表,以下為原卷對應頁面:
原卷第 8 頁

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這一題的完整詳解

核心觀念

本題綜合考查:

  1. 正交振幅調變(QAM)的訊號空間與星座圖。
  2. 同調接收器的正交解調,以及載波頻率誤差造成的 I/Q 互相干擾。
  3. QQ 函數與訊雜比(SNR)的關係。
  4. 多個基地台訊號的相干疊加。
  5. 都卜勒效應對載波頻率與接收訊號的影響。

令

s(t)=Acp(t)cos⁡(2πf0t)−Asp(t)sin⁡(2πf0t)s(t)=A_c p(t)\cos(2\pi f_0t)-A_s p(t)\sin(2\pi f_0t)

其中

Ac,As∈{±1,±3}.A_c,A_s\in\{\pm1,\pm3\}.

(a) QAM 星座圖

選擇正交基底

ϕc(t)=2Tp(t)cos⁡(2πf0t)\phi_c(t)=\sqrt{\frac{2}{T}}p(t)\cos(2\pi f_0t)

以及

ϕs(t)=−2Tp(t)sin⁡(2πf0t).\phi_s(t)=-\sqrt{\frac{2}{T}}p(t)\sin(2\pi f_0t).

則訊號可表示為

s(t)=T2Acϕc(t)+T2Asϕs(t).s(t)=\sqrt{\frac{T}{2}}A_c\phi_c(t)+\sqrt{\frac{T}{2}}A_s\phi_s(t).

因此星座點的座標比例為

(Ac,As)∈{±1,±3}×{±1,±3}.(A_c,A_s)\in\{\pm1,\pm3\}\times\{\pm1,\pm3\}.

共有 4×4=164\times4=16 個星座點:

                 A_s
                  ↑
          ●       ●       ●       ●
        (−3,3)  (−1,3)  (1,3)   (3,3)

          ●       ●       ●       ●
        (−3,1)  (−1,1)  (1,1)   (3,1)

          ●       ●       ●       ●
        (−3,−1) (−1,−1) (1,−1) (3,−1)

          ●       ●       ●       ●
        (−3,−3) (−1,−3) (1,−3) (3,−3)
                  └──────────────→ A_c

若直接以 sin⁡(2πf0t)\sin(2\pi f_0t) 作為第二基底,因原式為 −Assin⁡(2πf0t)-A_s\sin(2\pi f_0t),第二座標會上下顛倒;採用上述負號基底後,即可直接標成 (Ac,As)(A_c,A_s)。

每一個星座點攜帶

log⁡216=4\log_2 16=4

個位元。


(b) 同調 QAM 接收器與載波頻率誤差

同調接收器架構

典型 QAM 同調接收器如下:

                 ┌──× 2cos(2πf0t) ──低通濾波器── I(t) ──取樣/判決── Ac
r(t) ────────────┤
                 │
                 └──× −2sin(2πf0t) ─低通濾波器── Q(t) ──取樣/判決── As
                              ↑
                         本地振盪器

接收訊號分別與同相載波及正交載波相乘,再經低通濾波器去除高頻項。

若本地載波完全正確,則

r(t)=Acp(t)cos⁡(ω0t)−Asp(t)sin⁡(ω0t),r(t)=A_cp(t)\cos(\omega_0t)-A_sp(t)\sin(\omega_0t),

其中 ω0=2πf0\omega_0=2\pi f_0。使用 2cos⁡(ω0t)2\cos(\omega_0t) 與 −2sin⁡(ω0t)-2\sin(\omega_0t) 解調後,得到

I(t)=Acp(t),I(t)=A_cp(t), Q(t)=Asp(t).Q(t)=A_sp(t).

本地振盪器有頻率誤差

若本地振盪器頻率為

fLO=f0+Δf,f_{\mathrm{LO}}=f_0+\Delta f,

定義

Δω=ω0−ωLO=−2πΔf.\Delta\omega=\omega_0-\omega_{\mathrm{LO}} =-2\pi\Delta f.

經過正交解調與低通濾波後:

I′(t)=Acp(t)cos⁡(Δωt)−Asp(t)sin⁡(Δωt),I'(t)=A_cp(t)\cos(\Delta\omega t) -A_sp(t)\sin(\Delta\omega t), Q′(t)=Acp(t)sin⁡(Δωt)+Asp(t)cos⁡(Δωt).Q'(t)=A_cp(t)\sin(\Delta\omega t) +A_sp(t)\cos(\Delta\omega t).

寫成矩陣形式:

[I′(t)Q′(t)]=[cos⁡(Δωt)−sin⁡(Δωt)sin⁡(Δωt)cos⁡(Δωt)][Acp(t)Asp(t)].\begin{bmatrix} I'(t)\\ Q'(t) \end{bmatrix} = \begin{bmatrix} \cos(\Delta\omega t)&-\sin(\Delta\omega t)\\ \sin(\Delta\omega t)&\cos(\Delta\omega t) \end{bmatrix} \begin{bmatrix} A_cp(t)\\ A_sp(t) \end{bmatrix}.

這表示原本的星座圖會隨時間旋轉,且 II 與 QQ 兩路彼此混合,造成符號判決錯誤。若 Δf\Delta f 持續存在,星座點會持續旋轉;若頻率誤差很小但存在相位追蹤,則仍須靠載波恢復電路補償。


(c) SNR 增加量

題目給定符號錯誤率近似為

Ps≈4Q(γ).P_s\approx4Q(\sqrt{\gamma}).

原本

Ps=10−2,P_s=10^{-2},

所以

Q(γ1)=10−24=2.5×10−3.Q(\sqrt{\gamma_1})=\frac{10^{-2}}{4}=2.5\times10^{-3}.

查 QQ 函數表:

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