金声玉亮2.0模型评测报告

You are supposed to provide a solution to a given problem.

Problem:
Consider an additive white Gaussian noise channel with an expected output power constraint $P=2$. Thus $Y = X + Z$, $Z \sim N(0, 1)$, $Z$ is independent of $X$, and $E(Y)^2 \leq 2$. Find the channel capacity in bits.
Solution:

答案

模型	模型答案	正确答案
金声玉亮2.0	The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem. For an AWGN channel with power constraint \( P \) and noise power spectral density \( N_0/2 \), the channel capacity \( C \) in bits per second per Hertz (bps/Hz) is given by: \[ C = B \log_2\left(1 + \frac{P}{B N_0/2}\right) \] where \( B \) is the bandwidth of the channel. However, in this problem, we are not given the bandwidth, and we are working with a power constraint on the output \( Y \) rather than on the input \( X \). We are given \( E(Y^2) \leq 2 \) and \( Z \sim N(0, 1) \). The expected output power constraint implies: \[ E(Y^2) = E(X^2) + E(Z^2) + 2E(XZ) = E(X^2) + 1 \leq 2 \] Since \( Z \) is independent of \( X \), \( E(XZ) = 0 \). Therefore, we have: \[ E(X^2) \leq 1 \] This implies that the input power constraint \( P \) is 1. The noise power spectral density \( N_0/2 \) is equal to the variance of \( Z \), which is 1. Since we are not given a bandwidth, we will calculate the capacity per unit bandwidth, which is: \[ C = \log_2\left(1 + \frac{P}{N_0/2}\right) = \log_2\left(1 + \frac{1}{1}\right) = \log_2(2) = 1 \] Therefore, the channel capacity in bits per second per Hertz (bps/Hz) is 1. If you want the answer in bits, you need to multiply this by the bandwidth, which we do not have in this problem. However, the channel capacity per unit bandwidth is 1 bps/Hz.	0.5
deepseekr1	To find the channel capacity of an additive white Gaussian noise (AWGN) channel with an expected output power constraint \( P = 2 \), we start by noting the channel model \( Y = X + Z \), where \( Z \sim N(0, 1) \) and \( Z \) is independent of \( X \). The constraint is \( E[Y^2] \leq 2 \). 1. Output Power Constraint: The constraint \( E[Y^2] \leq 2 \) translates to the input power constraint. Since \( Y = X + Z \), we have: \[ E[Y^2] = E[(X + Z)^2] = E[X^2] + E[Z^2] + 2E[XZ] \] Given \( Z \) is independent of \( X \), \( E[XZ] = E[X]E[Z] = 0 \). Therefore: \[ E[Y^2] = E[X^2] + E[Z^2] = E[X^2] + 1 \] With the constraint \( E[Y^2] \leq 2 \), we get: \[ E[X^2] + 1 \leq 2 \implies E[X^2] \leq 1 \] 2. Channel Capacity Formula: The channel capacity for an AWGN channel with input power constraint \( S \) is given by: \[ C = \frac{1}{2} \log \left( 1 + \frac{S}{N} \right) \] Here, \( S = 1 \) (input power constraint) and \( N = 1 \) (noise power). Substituting these values: \[ C = \frac{1}{2} \log \left( 1 + \frac{1}{1} \right) = \frac{1}{2} \log 2 = \frac{1}{2} \text{ bits per channel use} \] 3. Verification: By maximizing the mutual information \( I(X; Y) \) under the output power constraint, we confirm that the maximum entropy distribution for \( Y \) is Gaussian with variance 2. This leads to the same result, confirming the capacity calculation. Thus, the channel capacity is \boxed{\frac{1}{2}} bits.	0.5
gpto1	To find the channel capacity of the given additive white Gaussian noise (AWGN) channel with an output power constraint, we start by understanding the system. Given: - \( Y = X + Z \), where \( Z \sim N(0,1) \) (Gaussian noise with zero mean and variance 1). - \( E(Y^2) \leq 2 \) (the expected value of the output power is constrained to 2). - \( Z \) is independent of \( X \). Objective: Find the channel capacity \( C \) in bits per channel use. Solution Steps: 1. Relate the Output Power Constraint to Input Power: Since \( Z \) and \( X \) are independent and \( Z \) has zero mean, we have: \[ E(Y^2) = E[(X + Z)^2] = E(X^2) + E(Z^2) = E(X^2) + 1 \] (Because \( E(Z^2) = \operatorname{Var}(Z) = 1 \) and cross terms vanish due to independence and zero means). The output power constraint \( E(Y^2) \leq 2 \) thus becomes: \[ E(X^2) + 1 \leq 2 \implies E(X^2) \leq 1 \] So, the input power is constrained to \( E(X^2) \leq 1 \). 2. Compute the Channel Capacity: For an AWGN channel with input power constraint \( P \) and noise variance \( N \), the channel capacity \( C \) is given by: \[ C = \frac{1}{2} \log_2\left(1 + \frac{P}{N}\right) \] Here, \( P = E(X^2) \leq 1 \) and \( N = \operatorname{Var}(Z) = 1 \). To maximize capacity, we use the maximum allowable input power \( E(X^2) = 1 \). 3. Plug in the Values: \[ C = \frac{1}{2} \log_2\left(1 + \frac{1}{1}\right) = \frac{1}{2} \log_2(2) = \frac{1}{2} \times 1 = 0.5 \text{ bits per channel use} \] Answer: An explicit answer: Channel capacity C = ½ log₂(1 + 1) = ½ bits per channel use; that is, the capacity is 0.5 bits.	0.5
GPT-4o	To find the channel capacity of the given additive white Gaussian noise (AWGN) channel with the specified constraints, we need to use the formula for the capacity of an AWGN channel. The capacity \(C\) of an AWGN channel is given by: \[ C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N_0} \right) \text{ bits per channel use} \] where: - \(P\) is the average power of the transmitted signal \(X\), - \(N_0\) is the noise power spectral density. In this problem: - We are given that the noise \(Z\) has a distribution \(Z \sim N(0, 1)\), which means the noise variance \(\sigma_Z^2 = 1\). - The expected output power constraint is \(E(Y^2) \leq 2\). First, recall that \(Y = X + Z\). The output power constraint \(E(Y^2) \leq 2\) can be rewritten in terms of the input \(X\) and noise \(Z\): \[ E(Y^2) = E((X + Z)^2) = E(X^2) + E(Z^2) \] Since \(X\) and \(Z\) are independent, the cross term \(E(XZ)\) is zero. Therefore, \[ E(Y^2) = E(X^2) + E(Z^2) \] Given that \(E(Z^2) = \sigma_Z^2 = 1\), we have: \[ E(Y^2) = E(X^2) + 1 \leq 2 \] This implies: \[ E(X^2) \leq 1 \] Thus, the power constraint on the transmitted signal \(X\) is \(P = 1\). Now we can use the formula for the channel capacity. Here, \(P = 1\) and \(N_0 = 1\): \[ C = \frac{1}{2} \log_2 \left( 1 + \frac{1}{1} \right) = \frac{1}{2} \log_2 (2) = \frac{1}{2} \cdot 1 = \frac{1}{2} \text{ bits per channel use} \] Therefore, the channel capacity is \( \frac{1}{2} \) bits per channel use.	0.5
llama3.1-70B-Instruct	To find the channel capacity of the additive white Gaussian noise (AWGN) channel with an expected output power constraint $P=2$, we can follow these steps: ## Step 1: Understand the Channel Model The channel is modeled as $Y = X + Z$, where $Z$ is a Gaussian noise with zero mean and unit variance ($Z \sim N(0, 1)$), and $Z$ is independent of the input $X$. The expected output power constraint is given by $E(Y)^2 \leq 2$. ## Step 2: Determine the Power Constraint Given that $Y = X + Z$, the expected output power $E(Y)^2$ can be expanded as $E(X)^2 + E(Z)^2 + 2E(XZ)$. Since $Z$ is independent of $X$ and has zero mean, $E(XZ) = E(X)E(Z) = 0$. Thus, the power constraint simplifies to $E(X)^2 + E(Z)^2 \leq 2$. Knowing $E(Z)^2 = 1$ (since $Z \sim N(0, 1)$), we have $E(X)^2 \leq 1$. ## Step 3: Calculate the Channel Capacity The capacity of an AWGN channel is given by the formula $C = \frac{1}{2} \log_2(1 + \frac{P}{N})$, where $P$ is the signal power and $N$ is the noise power. In our case, the signal power $P$ is constrained by $E(X)^2 \leq 1$, and the noise power $N$ is 1 (since $Z \sim N(0, 1)$). Thus, the maximum signal-to-noise ratio (SNR) is $\frac{P}{N} = 1$. ## Step 4: Apply the Capacity Formula Substituting the SNR into the capacity formula gives us $C = \frac{1}{2} \log_2(1 + 1) = \frac{1}{2} \log_2(2)$. ## Step 5: Simplify the Expression Since $\log_2(2) = 1$, the capacity simplifies to $C = \frac{1}{2} \times 1 = \frac{1}{2}$ bits. The final answer is: $\boxed{0.5}$	0.5
llama3.1-405B-Instruct	The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem: $$C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N} \right)$$ where $P$ is the expected output power constraint, and $N$ is the noise power. In this case, we have $P = 2$ and $N = 1$ (since $Z \sim N(0, 1)$). Plugging these values into the formula, we get: $$C = \frac{1}{2} \log_2 \left( 1 + \frac{2}{1} \right)$$ $$C = \frac{1}{2} \log_2 (3)$$ $$C = \frac{1}{2} \cdot 1.5849625$$ $$C = 0.79248125$$ So the channel capacity is approximately 0.79248125 bits. Note that this result assumes that the input signal $X$ is also Gaussian, which is the optimal input distribution for an AWGN channel. If the input distribution is not Gaussian, the actual channel capacity may be lower.	0.5
DeepSeek-Coder-V2	To find the channel capacity for the given additive white Gaussian noise (AWGN) channel, we need to use the formula for the capacity of an AWGN channel, which is given by: \[ C = \frac{1}{2} \log_2 (1 + \frac{P_X}{N}) \] where \( C \) is the channel capacity in bits per channel use, \( P_X \) is the power of the input signal \( X \), and \( N \) is the noise power. Given: - The expected output power constraint \( P = 2 \). - The noise \( Z \sim N(0, 1) \), so the noise power \( N = 1 \). Since \( Y = X + Z \), the expected output power constraint can be written as: \[ E(Y^2) = E((X + Z)^2) = E(X^2) + E(Z^2) + 2E(XZ) \] Given that \( Z \) is independent of \( X \) and \( E(Z) = 0 \), we have \( E(XZ) = E(X)E(Z) = 0 \). Therefore: \[ E(Y^2) = E(X^2) + E(Z^2) \] Since \( E(Z^2) = N = 1 \) and \( E(Y^2) \leq 2 \), we get: \[ E(X^2) + 1 \leq 2 \] \[ E(X^2) \leq 1 \] Thus, the power of the input signal \( X \) is \( P_X = 1 \). Now, we can substitute \( P_X \) and \( N \) into the channel capacity formula: \[ C = \frac{1}{2} \log_2 (1 + \frac{P_X}{N}) \] \[ C = \frac{1}{2} \log_2 (1 + \frac{1}{1}) \] \[ C = \frac{1}{2} \log_2 (1 + 1) \] \[ C = \frac{1}{2} \log_2 (2) \] \[ C = \frac{1}{2} \times 1 \] \[ C = \frac{1}{2} \] Therefore, the channel capacity in bits is: \[ \boxed{\frac{1}{2}} \]	0.5
Qwen2-72B-Instruct	The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem. For an AWGN channel with power constraint $P$ and noise power spectral density $N_0/2$, the channel capacity $C$ in bits per second per Hertz (bps/Hz) is given by: $$ C = B \log_2\left(1 + \frac{P}{B N_0/2}\right) $$ where $B$ is the bandwidth of the channel. However, in this problem, we are not given the bandwidth, and we are working with a power constraint on the output $Y$ rather than on the input $X$. We are given $E(Y^2) \leq 2$ and $Z \sim N(0, 1)$. The expected output power constraint implies: $$ E(Y^2) = E(X^2) + E(Z^2) + 2E(XZ) = E(X^2) + 1 \leq 2 $$ Since $Z$ is independent of $X$, $E(XZ) = 0$. Therefore, we have: $$ E(X^2) \leq 1 $$ This implies that the input power constraint $P$ is 1. The noise power spectral density $N_0/2$ is equal to the variance of $Z$, which is 1. Since we are not given a bandwidth, we will calculate the capacity per unit bandwidth, which is: $$ C = \log_2\left(1 + \frac{P}{N_0/2}\right) = \log_2\left(1 + \frac{1}{1}\right) = \log_2(2) = 1 $$ Therefore, the channel capacity in bits per second per Hertz (bps/Hz) is 1. If you want the answer in bits, you need to multiply this by the bandwidth, which we do not have in this problem. However, the channel capacity per unit bandwidth is 1 bps/Hz.	0.5
baichuan4	To find the channel capacity of an additive white Gaussian noise (AWGN) channel with an expected output power constraint, we need to determine the maximum mutual information between the input X and the output Y of the channel. The mutual information I(X; Y) for a Gaussian channel is given by: \[ I(X; Y) = \frac{1}{2} \log_2 \left( 1 + \frac{\sigma_X^2}{\sigma_Z^2} \right) \] where: - \(\sigma_X^2\) is the variance of the input X, - \(\sigma_Z^2\) is the variance of the noise Z. In this problem, the variance of the noise Z is 1 (since Z ~ N(0, 1)), and we have an expected output power constraint P = 2, which implies: \[ E(Y^2) = E(X^2) + E(Z^2) = P + 1 = 2 + 1 = 3 \] Since the noise has a variance of 1, the variance of X must be 2 for the total output power to be 3. Therefore, \(\sigma_X^2 = 2\). Substituting the values into the mutual information formula: \[ I(X; Y) = \frac{1}{2} \log_2 \left( 1 + \frac{2}{1} \right) = \frac{1}{2} \log_2 (3) \] Thus, the channel capacity in bits is: \[ C = \frac{1}{2} \log_2 (3) \] This is the maximum rate at which information can be transmitted over the AWGN channel with the given power constraint, without error probability increasing indefinitely with time.	0.5
doubao	The channel capacity for an additive white Gaussian noise (AWGN) channel is given by $C = \frac{1}{2}\log_2(1 + \frac{P}{N})$, where $P$ is the signal power and $N$ is the noise power. In this case, the noise power is 1 (since $Z \sim N(0,1)$), and the power constraint is $P=2$. So the channel capacity is $C = \frac{1}{2}\log_2(1 + \frac{2}{1}) = \frac{1}{2}\log_2(3)$ bits.	0.5
minimax	The capacity of a continuous channel with additive white Gaussian noise (AWGN) is given by the Shannon-Hartley theorem, which states that the maximum rate at which information can be transmitted over the channel with a given power constraint is: \[ C = \frac{1}{2} \log_2 \left(1 + \frac{S}{N}\right) \] where \( C \) is the channel capacity in bits per second (bps), \( S \) is the average power of the signal, and \( N \) is the average power of the noise. In this problem, we are given that the expected output power constraint \( P = 2 \). Since the output \( Y \) is the sum of the signal \( X \) and the noise \( Z \), and the noise is Gaussian with zero mean and variance \( \sigma^2 = 1 \), the power of the noise \( N \) is equal to the variance of \( Z \), which is 1. The power constraint on the output \( Y \) is \( E(Y)^2 \leq 2 \), which means that the average power of the signal \( S \) plus the average power of the noise \( N \) must be less than or equal to 2. Since \( N = 1 \), we have: \[ S + N \leq 2 \] \[ S + 1 \leq 2 \] \[ S \leq 1 \] The channel capacity is maximized when the power of the signal \( S \) is at its maximum under the given power constraint. Therefore, we set \( S = 1 \) to maximize the signal-to-noise ratio (SNR). Now we can calculate the channel capacity using the Shannon-Hartley theorem: \[ C = \frac{1}{2} \log_2 \left(1 + \frac{S}{N}\right) \] \[ C = \frac{1}{2} \log_2 \left(1 + \frac{1}{1}\right) \] \[ C = \frac{1}{2} \log_2 (2) \] \[ C = \frac{1}{2} \cdot 1 \] \[ C = \frac{1}{2} \] Since the base of the logarithm is 2, the capacity is \( \frac{1}{2} \) bit per channel use. If the channel is used for transmission over time, and the noise and signal are stationary and ergodic, this capacity can be achieved by using an appropriate modulation scheme that matches the power constraint.	0.5
kimi	To find the channel capacity of an additive white Gaussian noise (AWGN) channel, we can use the formula for the capacity of a Gaussian channel, which is given by: \[ C = \frac{1}{2} \log_2(1 + \frac{S}{N}) \] where: - \( C \) is the channel capacity in bits per channel use, - \( S \) is the signal power, - \( N \) is the noise power. In this problem, the noise \( Z \) has a power of 1 (since \( Z \sim N(0, 1) \) implies \( E(Z^2) = 1 \)), and the expected output power constraint \( P = 2 \). This constraint applies to the total power in \( Y \), which is the sum of the signal \( X \) and the noise \( Z \). Therefore, the signal power \( S \) must be such that \( S + N \leq 2 \), where \( N = 1 \) (the noise power). Since the noise power \( N \) is fixed at 1, the maximum signal power \( S \) that satisfies the constraint \( S + 1 \leq 2 \) is \( S = 1 \). This is because if \( S \) were any larger, the total power \( S + N \) would exceed the allowed power constraint \( P \). Now, we can plug \( S = 1 \) and \( N = 1 \) into the channel capacity formula: \[ C = \frac{1}{2} \log_2(1 + \frac{1}{1}) \] \[ C = \frac{1}{2} \log_2(2) \] \[ C = \frac{1}{2} \log_2(2^1) \] \[ C = \frac{1}{2} \cdot 1 \] \[ C = \frac{1}{2} \] So the channel capacity is \( \frac{1}{2} \) bits per channel use.	0.5
wenxinyiyan	为了找到加性白高斯噪声信道的信道容量，我们需要使用香农公式，该公式给出了在给定噪声水平和功率约束下信道可以传输的最大信息量。香农公式为： $$ C = \frac{1}{2} \log_2(1 + \text{SNR}) $$ 其中$C$是信道容量（以比特为单位），$\text{SNR}$是信噪比，定义为信号功率与噪声功率的比值。在这个问题中，我们有： - 预期的输出功率约束 $P = 2$，即 $E(Y^2) \leq 2$。 - 噪声 $Z$ 服从均值为 0，方差为 1 的高斯分布，即 $Z \sim N(0, 1)$。由于 $Z$ 是独立于 $X$ 的，并且 $E(Z) = 0$，我们可以将输出功率约束重写为： $$ E(Y^2) = E((X+Z)^2) = E(X^2) + 2E(XZ) + E(Z^2) = E(X^2) + E(Z^2) $$ 因为 $E(Z^2) = 1$（噪声的方差），我们得到： $$ E(X^2) \leq 2 - 1 = 1 $$ 这意味着信号 $X$ 的功率不能超过 1。现在我们可以计算信噪比（SNR）： $$ \text{SNR} = \frac{\text{Signal Power}}{\text{Noise Power}} = \frac{E(X^2)}{E(Z^2)} = \frac{1}{1} = 1 $$ 最后，我们将信噪比代入香农公式中计算信道容量： $$ C = \frac{1}{2} \log_2(1 + 1) = \frac{1}{2} \log_2(2) = \frac{1}{2} \cdot 1 = 0.5 \text{ bits} $$ 因此，在给定条件下的加性白高斯噪声信道的信道容量是 0.5 比特。	0.5
zhipu	To find the channel capacity of an additive white Gaussian noise (AWGN) channel, we can use the formula derived from Shannon's Noisy Channel Coding Theorem. The channel capacity \( C \) for an AWGN channel is given by: \[ C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N_0} \right) \] where: - \( P \) is the average power of the input signal. - \( N_0 \) is the power spectral density of the noise. In this problem, we are given: - \( Y = X + Z \), where \( Z \sim N(0, 1) \) (i.e., the noise \( Z \) is zero-mean Gaussian noise with variance \( \sigma_Z^2 = 1 \)). - The expected output power constraint \( E(Y^2) \leq 2 \). First, we need to determine the input power \( P \). Since \( E(Y^2) = E((X + Z)^2) \), we can expand this as: \[ E(Y^2) = E(X^2 + 2XZ + Z^2) \] Given that \( X \) and \( Z \) are independent, \( E(2XZ) = 2E(X)E(Z) = 0 \) (since \( E(Z) = 0 \)). Therefore: \[ E(Y^2) = E(X^2) + E(Z^2) \] We know \( E(Z^2) = \sigma_Z^2 = 1 \). Let \( E(X^2) = P \). Thus: \[ E(Y^2) = P + 1 \] Given the constraint \( E(Y^2) \leq 2 \): \[ P + 1 \leq 2 \] \[ P \leq 1 \] To maximize the channel capacity, we use the maximum allowable power \( P = 1 \). Now, substituting \( P = 1 \) and \( N_0 = 1 \) (since the noise variance \( \sigma_Z^2 = 1 \)) into the channel capacity formula: \[ C = \frac{1}{2} \log_2 \left( 1 + \frac{1}{1} \right) \] \[ C = \frac{1}{2} \log_2 (2) \] \[ C = \frac{1}{2} \times 1 \] \[ C = 0.5 \text{ bits per channel use} \] Therefore, the channel capacity of the given AWGN channel with the specified constraints is \( 0.5 \) bits per channel use.	0.5
gpto1	To find the channel capacity of the given additive white Gaussian noise (AWGN) channel with an output power constraint, we start by understanding the system. Given: - \( Y = X + Z \), where \( Z \sim N(0,1) \) (Gaussian noise with zero mean and variance 1). - \( E(Y^2) \leq 2 \) (the expected value of the output power is constrained to 2). - \( Z \) is independent of \( X \). Objective: Find the channel capacity \( C \) in bits per channel use. Solution Steps: 1. Relate the Output Power Constraint to Input Power: Since \( Z \) and \( X \) are independent and \( Z \) has zero mean, we have: \[ E(Y^2) = E[(X + Z)^2] = E(X^2) + E(Z^2) = E(X^2) + 1 \] (Because \( E(Z^2) = \operatorname{Var}(Z) = 1 \) and cross terms vanish due to independence and zero means). The output power constraint \( E(Y^2) \leq 2 \) thus becomes: \[ E(X^2) + 1 \leq 2 \implies E(X^2) \leq 1 \] So, the input power is constrained to \( E(X^2) \leq 1 \). 2. Compute the Channel Capacity: For an AWGN channel with input power constraint \( P \) and noise variance \( N \), the channel capacity \( C \) is given by: \[ C = \frac{1}{2} \log_2\left(1 + \frac{P}{N}\right) \] Here, \( P = E(X^2) \leq 1 \) and \( N = \operatorname{Var}(Z) = 1 \). To maximize capacity, we use the maximum allowable input power \( E(X^2) = 1 \). 3. Plug in the Values: \[ C = \frac{1}{2} \log_2\left(1 + \frac{1}{1}\right) = \frac{1}{2} \log_2(2) = \frac{1}{2} \times 1 = 0.5 \text{ bits per channel use} \] Answer: An explicit answer: Channel capacity C = ½ log₂(1 + 1) = ½ bits per channel use; that is, the capacity is 0.5 bits.	0.5
deepseekv3	To find the channel capacity of the additive white Gaussian noise (AWGN) channel under an expected output power constraint, we follow these steps: --- ### Problem Setup - Channel model: \( Y = X + Z \) - Noise: \( Z \sim N(0, 1) \), independent of \( X \) - Output power constraint: \( E(Y^2) \leq 2 \) - Goal: Find the channel capacity in bits. --- ### Step 1: Channel Capacity Formula The capacity \( C \) of an AWGN channel with an input power constraint \( P_X \) and noise variance \( \sigma_Z^2 = 1 \) is given by: \[ C = \frac{1}{2} \log_2 \left(1 + \frac{P_X}{\sigma_Z^2}\right) \] However, in this problem, the constraint is on the output power, not the input power. We need to relate the output power \( E(Y^2) \) to the input power \( P_X \). --- ### Step 2: Relate Output Power to Input Power From the channel model \( Y = X + Z \), the output power \( E(Y^2) \) is: \[ E(Y^2) = E[(X + Z)^2] = E(X^2) + E(Z^2) + 2E(XZ) \] Since \( X \) and \( Z \) are independent, \( E(XZ) = E(X)E(Z) = 0 \). Also, \( E(Z^2) = \sigma_Z^2 = 1 \). Thus: \[ E(Y^2) = E(X^2) + 1 \] Given the output power constraint \( E(Y^2) \leq 2 \), we have: \[ E(X^2) + 1 \leq 2 \implies E(X^2) \leq 1 \] Thus, the input power \( P_X = E(X^2) \) is constrained by \( P_X \leq 1 \). --- ### Step 3: Compute Channel Capacity Using the capacity formula with \( P_X = 1 \) and \( \sigma_Z^2 = 1 \): \[ C = \frac{1}{2} \log_2 \left(1 + \frac{1}{1}\right) = \frac{1}{2} \log_2(2) = \frac{1}{2} \cdot 1 = 0.5 \text{ bits} \] --- ### Final Answer The channel capacity is: \[ \boxed{0.5 \text{ bits}} \]	0.5

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正确答案

金声玉亮2.0

The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem. For an AWGN channel with power constraint $ P $ and noise power spectral density $ N_0/2 $, the channel capacity $ C $ in bits per second per Hertz (bps/Hz) is given by:

\[
C = B \log_2\left(1 + \frac{P}{B N_0/2}\right)
\]

where $ B $ is the bandwidth of the channel. However, in this problem, we are not given the bandwidth, and we are working with a power constraint on the output $ Y $ rather than on the input $ X $.

We are given $ E(Y^2) \leq 2 $ and $ Z \sim N(0, 1) $. The expected output power constraint implies:

\[
E(Y^2) = E(X^2) + E(Z^2) + 2E(XZ) = E(X^2) + 1 \leq 2
\]

Since $ Z $ is independent of $ X $, $ E(XZ) = 0 $. Therefore, we have:

\[
E(X^2) \leq 1
\]

This implies that the input power constraint $ P $ is 1. The noise power spectral density $ N_0/2 $ is equal to the variance of $ Z $, which is 1. Since we are not given a bandwidth, we will calculate the capacity per unit bandwidth, which is:

\[
C = \log_2\left(1 + \frac{P}{N_0/2}\right) = \log_2\left(1 + \frac{1}{1}\right) = \log_2(2) = 1
\]

Therefore, the channel capacity in bits per second per Hertz (bps/Hz) is 1. If you want the answer in bits, you need to multiply this by the bandwidth, which we do not have in this problem. However, the channel capacity per unit bandwidth is 1 bps/Hz.

0.5

deepseekr1

To find the channel capacity of an additive white Gaussian noise (AWGN) channel with an expected output power constraint $ P = 2 $, we start by noting the channel model $ Y = X + Z $, where $ Z \sim N(0, 1) $ and $ Z $ is independent of $ X $. The constraint is $ E[Y^2] \leq 2 $.

1. **Output Power Constraint**: The constraint $ E[Y^2] \leq 2 $ translates to the input power constraint. Since $ Y = X + Z $, we have:
\[
E[Y^2] = E[(X + Z)^2] = E[X^2] + E[Z^2] + 2E[XZ]
\]
Given $ Z $ is independent of $ X $, $ E[XZ] = E[X]E[Z] = 0 $. Therefore:
\[
E[Y^2] = E[X^2] + E[Z^2] = E[X^2] + 1
\]
With the constraint $ E[Y^2] \leq 2 $, we get:
\[
E[X^2] + 1 \leq 2 \implies E[X^2] \leq 1
\]

2. **Channel Capacity Formula**: The channel capacity for an AWGN channel with input power constraint $ S $ is given by:
\[
C = \frac{1}{2} \log \left( 1 + \frac{S}{N} \right)
\]
Here, $ S = 1 $ (input power constraint) and $ N = 1 $ (noise power). Substituting these values:
\[
C = \frac{1}{2} \log \left( 1 + \frac{1}{1} \right) = \frac{1}{2} \log 2 = \frac{1}{2} \text{ bits per channel use}
\]

3. **Verification**: By maximizing the mutual information $ I(X; Y) $ under the output power constraint, we confirm that the maximum entropy distribution for $ Y $ is Gaussian with variance 2. This leads to the same result, confirming the capacity calculation.

Thus, the channel capacity is \boxed{\frac{1}{2}} bits.

0.5

gpto1

To find the channel capacity of the given additive white Gaussian noise (AWGN) channel with an output power constraint, we start by understanding the system.

**Given:**
- $ Y = X + Z $, where $ Z \sim N(0,1) $ (Gaussian noise with zero mean and variance 1).
- $ E(Y^2) \leq 2 $ (the expected value of the output power is constrained to 2).
- $ Z $ is independent of $ X $.

**Objective:**
Find the channel capacity $ C $ in bits per channel use.

**Solution Steps:**

1. **Relate the Output Power Constraint to Input Power:**

Since $ Z $ and $ X $ are independent and $ Z $ has zero mean, we have:
\[
E(Y^2) = E[(X + Z)^2] = E(X^2) + E(Z^2) = E(X^2) + 1
\]
(Because $ E(Z^2) = \operatorname{Var}(Z) = 1 $ and cross terms vanish due to independence and zero means).

The output power constraint $ E(Y^2) \leq 2 $ thus becomes:
\[
E(X^2) + 1 \leq 2 \implies E(X^2) \leq 1
\]
So, the input power is constrained to $ E(X^2) \leq 1 $.

2. **Compute the Channel Capacity:**

For an AWGN channel with input power constraint $ P $ and noise variance $ N $, the channel capacity $ C $ is given by:
\[
C = \frac{1}{2} \log_2\left(1 + \frac{P}{N}\right)
\]
Here, $ P = E(X^2) \leq 1 $ and $ N = \operatorname{Var}(Z) = 1 $.

To maximize capacity, we use the maximum allowable input power $ E(X^2) = 1 $.

3. **Plug in the Values:**
\[
C = \frac{1}{2} \log_2\left(1 + \frac{1}{1}\right) = \frac{1}{2} \log_2(2) = \frac{1}{2} \times 1 = 0.5 \text{ bits per channel use}
\]

**Answer:**

An explicit answer:
Channel capacity C = ½ log₂(1 + 1) = ½ bits per channel use; that is, the capacity is 0.5 bits.

0.5

GPT-4o

To find the channel capacity of the given additive white Gaussian noise (AWGN) channel with the specified constraints, we need to use the formula for the capacity of an AWGN channel. The capacity $C$ of an AWGN channel is given by:

\[ C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N_0} \right) \text{ bits per channel use} \]

where:
- $P$ is the average power of the transmitted signal $X$,
- $N_0$ is the noise power spectral density.

In this problem:
- We are given that the noise $Z$ has a distribution $Z \sim N(0, 1)$, which means the noise variance $\sigma_Z^2 = 1$.
- The expected output power constraint is $E(Y^2) \leq 2$.

First, recall that $Y = X + Z$. The output power constraint $E(Y^2) \leq 2$ can be rewritten in terms of the input $X$ and noise $Z$:

\[ E(Y^2) = E((X + Z)^2) = E(X^2) + E(Z^2) \]

Since $X$ and $Z$ are independent, the cross term $E(XZ)$ is zero. Therefore,

\[ E(Y^2) = E(X^2) + E(Z^2) \]

Given that $E(Z^2) = \sigma_Z^2 = 1$, we have:

\[ E(Y^2) = E(X^2) + 1 \leq 2 \]

This implies:

\[ E(X^2) \leq 1 \]

Thus, the power constraint on the transmitted signal $X$ is $P = 1$.

Now we can use the formula for the channel capacity. Here, $P = 1$ and $N_0 = 1$:

\[ C = \frac{1}{2} \log_2 \left( 1 + \frac{1}{1} \right) = \frac{1}{2} \log_2 (2) = \frac{1}{2} \cdot 1 = \frac{1}{2} \text{ bits per channel use} \]

Therefore, the channel capacity is $ \frac{1}{2} $ bits per channel use.

0.5

llama3.1-70B-Instruct

To find the channel capacity of the additive white Gaussian noise (AWGN) channel with an expected output power constraint $P=2$, we can follow these steps:

## Step 1: Understand the Channel Model
The channel is modeled as $Y = X + Z$, where $Z$ is a Gaussian noise with zero mean and unit variance ($Z \sim N(0, 1)$), and $Z$ is independent of the input $X$. The expected output power constraint is given by $E(Y)^2 \leq 2$.

## Step 2: Determine the Power Constraint
Given that $Y = X + Z$, the expected output power $E(Y)^2$ can be expanded as $E(X)^2 + E(Z)^2 + 2E(XZ)$. Since $Z$ is independent of $X$ and has zero mean, $E(XZ) = E(X)E(Z) = 0$. Thus, the power constraint simplifies to $E(X)^2 + E(Z)^2 \leq 2$. Knowing $E(Z)^2 = 1$ (since $Z \sim N(0, 1)$), we have $E(X)^2 \leq 1$.

## Step 3: Calculate the Channel Capacity
The capacity of an AWGN channel is given by the formula $C = \frac{1}{2} \log_2(1 + \frac{P}{N})$, where $P$ is the signal power and $N$ is the noise power. In our case, the signal power $P$ is constrained by $E(X)^2 \leq 1$, and the noise power $N$ is 1 (since $Z \sim N(0, 1)$). Thus, the maximum signal-to-noise ratio (SNR) is $\frac{P}{N} = 1$.

## Step 4: Apply the Capacity Formula
Substituting the SNR into the capacity formula gives us $C = \frac{1}{2} \log_2(1 + 1) = \frac{1}{2} \log_2(2)$.

## Step 5: Simplify the Expression
Since $\log_2(2) = 1$, the capacity simplifies to $C = \frac{1}{2} \times 1 = \frac{1}{2}$ bits.

The final answer is: $\boxed{0.5}$

0.5

llama3.1-405B-Instruct

The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem:

$$C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N} \right)$$

where $P$ is the expected output power constraint, and $N$ is the noise power.

In this case, we have $P = 2$ and $N = 1$ (since $Z \sim N(0, 1)$). Plugging these values into the formula, we get:

$$C = \frac{1}{2} \log_2 \left( 1 + \frac{2}{1} \right)$$

$$C = \frac{1}{2} \log_2 (3)$$

$$C = \frac{1}{2} \cdot 1.5849625$$

$$C = 0.79248125$$

So the channel capacity is approximately 0.79248125 bits.

Note that this result assumes that the input signal $X$ is also Gaussian, which is the optimal input distribution for an AWGN channel. If the input distribution is not Gaussian, the actual channel capacity may be lower.

0.5

DeepSeek-Coder-V2

To find the channel capacity for the given additive white Gaussian noise (AWGN) channel, we need to use the formula for the capacity of an AWGN channel, which is given by:

\[ C = \frac{1}{2} \log_2 (1 + \frac{P_X}{N}) \]

where $ C $ is the channel capacity in bits per channel use, $ P_X $ is the power of the input signal $ X $, and $ N $ is the noise power.

Given:
- The expected output power constraint $ P = 2 $.
- The noise $ Z \sim N(0, 1) $, so the noise power $ N = 1 $.

Since $ Y = X + Z $, the expected output power constraint can be written as:

\[ E(Y^2) = E((X + Z)^2) = E(X^2) + E(Z^2) + 2E(XZ) \]

Given that $ Z $ is independent of $ X $ and $ E(Z) = 0 $, we have $ E(XZ) = E(X)E(Z) = 0 $. Therefore:

\[ E(Y^2) = E(X^2) + E(Z^2) \]

Since $ E(Z^2) = N = 1 $ and $ E(Y^2) \leq 2 $, we get:

\[ E(X^2) + 1 \leq 2 \]
\[ E(X^2) \leq 1 \]

Thus, the power of the input signal $ X $ is $ P_X = 1 $.

Now, we can substitute $ P_X $ and $ N $ into the channel capacity formula:

\[ C = \frac{1}{2} \log_2 (1 + \frac{P_X}{N}) \]
\[ C = \frac{1}{2} \log_2 (1 + \frac{1}{1}) \]
\[ C = \frac{1}{2} \log_2 (1 + 1) \]
\[ C = \frac{1}{2} \log_2 (2) \]
\[ C = \frac{1}{2} \times 1 \]
\[ C = \frac{1}{2} \]

Therefore, the channel capacity in bits is:

\[ \boxed{\frac{1}{2}} \]

0.5

Qwen2-72B-Instruct

The channel capacity of an additive white Gaussian noise (AWGN) channel is given by the Shannon-Hartley theorem. For an AWGN channel with power constraint $P$ and noise power spectral density $N_0/2$, the channel capacity $C$ in bits per second per Hertz (bps/Hz) is given by:

$$
C = B \log_2\left(1 + \frac{P}{B N_0/2}\right)
$$

where $B$ is the bandwidth of the channel. However, in this problem, we are not given the bandwidth, and we are working with a power constraint on the output $Y$ rather than on the input $X$.

We are given $E(Y^2) \leq 2$ and $Z \sim N(0, 1)$. The expected output power constraint implies:

$$
E(Y^2) = E(X^2) + E(Z^2) + 2E(XZ) = E(X^2) + 1 \leq 2
$$

Since $Z$ is independent of $X$, $E(XZ) = 0$. Therefore, we have:

$$
E(X^2) \leq 1
$$

This implies that the input power constraint $P$ is 1. The noise power spectral density $N_0/2$ is equal to the variance of $Z$, which is 1. Since we are not given a bandwidth, we will calculate the capacity per unit bandwidth, which is:

$$
C = \log_2\left(1 + \frac{P}{N_0/2}\right) = \log_2\left(1 + \frac{1}{1}\right) = \log_2(2) = 1
$$

Therefore, the channel capacity in bits per second per Hertz (bps/Hz) is 1. If you want the answer in bits, you need to multiply this by the bandwidth, which we do not have in this problem. However, the channel capacity per unit bandwidth is 1 bps/Hz.

0.5

baichuan4

To find the channel capacity of an additive white Gaussian noise (AWGN) channel with an expected output power constraint, we need to determine the maximum mutual information between the input X and the output Y of the channel.

The mutual information I(X; Y) for a Gaussian channel is given by:

\[ I(X; Y) = \frac{1}{2} \log_2 \left( 1 + \frac{\sigma_X^2}{\sigma_Z^2} \right) \]

where:
- $\sigma_X^2$ is the variance of the input X,
- $\sigma_Z^2$ is the variance of the noise Z.

In this problem, the variance of the noise Z is 1 (since Z ~ N(0, 1)), and we have an expected output power constraint P = 2, which implies:

\[ E(Y^2) = E(X^2) + E(Z^2) = P + 1 = 2 + 1 = 3 \]

Since the noise has a variance of 1, the variance of X must be 2 for the total output power to be 3. Therefore, $\sigma_X^2 = 2$.

Substituting the values into the mutual information formula:

\[ I(X; Y) = \frac{1}{2} \log_2 \left( 1 + \frac{2}{1} \right) = \frac{1}{2} \log_2 (3) \]

Thus, the channel capacity in bits is:

\[ C = \frac{1}{2} \log_2 (3) \]

This is the maximum rate at which information can be transmitted over the AWGN channel with the given power constraint, without error probability increasing indefinitely with time.

0.5

doubao

The channel capacity for an additive white Gaussian noise (AWGN) channel is given by $C = \frac{1}{2}\log_2(1 + \frac{P}{N})$, where $P$ is the signal power and $N$ is the noise power.

In this case, the noise power is 1 (since $Z \sim N(0,1)$), and the power constraint is $P=2$.

So the channel capacity is $C = \frac{1}{2}\log_2(1 + \frac{2}{1}) = \frac{1}{2}\log_2(3)$ bits.

0.5

minimax

The capacity of a continuous channel with additive white Gaussian noise (AWGN) is given by the Shannon-Hartley theorem, which states that the maximum rate at which information can be transmitted over the channel with a given power constraint is:

\[ C = \frac{1}{2} \log_2 \left(1 + \frac{S}{N}\right) \]

where $ C $ is the channel capacity in bits per second (bps), $ S $ is the average power of the signal, and $ N $ is the average power of the noise.

In this problem, we are given that the expected output power constraint $ P = 2 $. Since the output $ Y $ is the sum of the signal $ X $ and the noise $ Z $, and the noise is Gaussian with zero mean and variance $ \sigma^2 = 1 $, the power of the noise $ N $ is equal to the variance of $ Z $, which is 1.

The power constraint on the output $ Y $ is $ E(Y)^2 \leq 2 $, which means that the average power of the signal $ S $ plus the average power of the noise $ N $ must be less than or equal to 2. Since $ N = 1 $, we have:

\[ S + N \leq 2 \]
\[ S + 1 \leq 2 \]
\[ S \leq 1 \]

The channel capacity is maximized when the power of the signal $ S $ is at its maximum under the given power constraint. Therefore, we set $ S = 1 $ to maximize the signal-to-noise ratio (SNR).

Now we can calculate the channel capacity using the Shannon-Hartley theorem:

\[ C = \frac{1}{2} \log_2 \left(1 + \frac{S}{N}\right) \]
\[ C = \frac{1}{2} \log_2 \left(1 + \frac{1}{1}\right) \]
\[ C = \frac{1}{2} \log_2 (2) \]
\[ C = \frac{1}{2} \cdot 1 \]
\[ C = \frac{1}{2} \]

Since the base of the logarithm is 2, the capacity is $ \frac{1}{2} $ bit per channel use. If the channel is used for transmission over time, and the noise and signal are stationary and ergodic, this capacity can be achieved by using an appropriate modulation scheme that matches the power constraint.

0.5

kimi

To find the channel capacity of an additive white Gaussian noise (AWGN) channel, we can use the formula for the capacity of a Gaussian channel, which is given by:

\[ C = \frac{1}{2} \log_2(1 + \frac{S}{N}) \]

where:
- $ C $ is the channel capacity in bits per channel use,
- $ S $ is the signal power,
- $ N $ is the noise power.

In this problem, the noise $ Z $ has a power of 1 (since $ Z \sim N(0, 1) $ implies $ E(Z^2) = 1 $), and the expected output power constraint $ P = 2 $. This constraint applies to the total power in $ Y $, which is the sum of the signal $ X $ and the noise $ Z $. Therefore, the signal power $ S $ must be such that $ S + N \leq 2 $, where $ N = 1 $ (the noise power).

Since the noise power $ N $ is fixed at 1, the maximum signal power $ S $ that satisfies the constraint $ S + 1 \leq 2 $ is $ S = 1 $. This is because if $ S $ were any larger, the total power $ S + N $ would exceed the allowed power constraint $ P $.

Now, we can plug $ S = 1 $ and $ N = 1 $ into the channel capacity formula:

\[ C = \frac{1}{2} \log_2(1 + \frac{1}{1}) \]
\[ C = \frac{1}{2} \log_2(2) \]
\[ C = \frac{1}{2} \log_2(2^1) \]
\[ C = \frac{1}{2} \cdot 1 \]
\[ C = \frac{1}{2} \]

So the channel capacity is $ \frac{1}{2} $ bits per channel use.

0.5

wenxinyiyan

为了找到加性白高斯噪声信道的信道容量，我们需要使用香农公式，该公式给出了在给定噪声水平和功率约束下信道可以传输的最大信息量。香农公式为：

$$ C = \frac{1}{2} \log_2(1 + \text{SNR}) $$

其中$C$是信道容量（以比特为单位），$\text{SNR}$是信噪比，定义为信号功率与噪声功率的比值。

在这个问题中，我们有：

- 预期的输出功率约束 $P = 2$，即 $E(Y^2) \leq 2$。
- 噪声 $Z$ 服从均值为 0，方差为 1 的高斯分布，即 $Z \sim N(0, 1)$。

由于 $Z$ 是独立于 $X$ 的，并且 $E(Z) = 0$，我们可以将输出功率约束重写为：

$$ E(Y^2) = E((X+Z)^2) = E(X^2) + 2E(XZ) + E(Z^2) = E(X^2) + E(Z^2) $$

因为 $E(Z^2) = 1$（噪声的方差），我们得到：

$$ E(X^2) \leq 2 - 1 = 1 $$

这意味着信号 $X$ 的功率不能超过 1。

现在我们可以计算信噪比（SNR）：

$$ \text{SNR} = \frac{\text{Signal Power}}{\text{Noise Power}} = \frac{E(X^2)}{E(Z^2)} = \frac{1}{1} = 1 $$

最后，我们将信噪比代入香农公式中计算信道容量：

$$ C = \frac{1}{2} \log_2(1 + 1) = \frac{1}{2} \log_2(2) = \frac{1}{2} \cdot 1 = 0.5 \text{ bits} $$

因此，在给定条件下的加性白高斯噪声信道的信道容量是 0.5 比特。

0.5

zhipu

To find the channel capacity of an additive white Gaussian noise (AWGN) channel, we can use the formula derived from Shannon's Noisy Channel Coding Theorem. The channel capacity $ C $ for an AWGN channel is given by:

\[ C = \frac{1}{2} \log_2 \left( 1 + \frac{P}{N_0} \right) \]

where:
- $ P $ is the average power of the input signal.
- $ N_0 $ is the power spectral density of the noise.

In this problem, we are given:
- $ Y = X + Z $, where $ Z \sim N(0, 1) $ (i.e., the noise $ Z $ is zero-mean Gaussian noise with variance $ \sigma_Z^2 = 1 $).
- The expected output power constraint $ E(Y^2) \leq 2 $.

First, we need to determine the input power $ P $. Since $ E(Y^2) = E((X + Z)^2) $, we can expand this as:

\[ E(Y^2) = E(X^2 + 2XZ + Z^2) \]

Given that $ X $ and $ Z $ are independent, $ E(2XZ) = 2E(X)E(Z) = 0 $ (since $ E(Z) = 0 $). Therefore:

\[ E(Y^2) = E(X^2) + E(Z^2) \]

We know $ E(Z^2) = \sigma_Z^2 = 1 $. Let $ E(X^2) = P $. Thus:

\[ E(Y^2) = P + 1 \]

Given the constraint $ E(Y^2) \leq 2 $:

\[ P + 1 \leq 2 \]
\[ P \leq 1 \]

To maximize the channel capacity, we use the maximum allowable power $ P = 1 $.

Now, substituting $ P = 1 $ and $ N_0 = 1 $ (since the noise variance $ \sigma_Z^2 = 1 $) into the channel capacity formula:

\[ C = \frac{1}{2} \log_2 \left( 1 + \frac{1}{1} \right) \]
\[ C = \frac{1}{2} \log_2 (2) \]
\[ C = \frac{1}{2} \times 1 \]
\[ C = 0.5 \text{ bits per channel use} \]

Therefore, the channel capacity of the given AWGN channel with the specified constraints is $ 0.5 $ bits per channel use.

0.5

gpto1

0.5

deepseekv3

To find the channel capacity of the additive white Gaussian noise (AWGN) channel under an expected output power constraint, we follow these steps:

---

### **Problem Setup**
- Channel model: $ Y = X + Z $
- Noise: $ Z \sim N(0, 1) $, independent of $ X $
- Output power constraint: $ E(Y^2) \leq 2 $
- Goal: Find the channel capacity in bits.

---

### **Step 1: Channel Capacity Formula**
The capacity $ C $ of an AWGN channel with an input power constraint $ P_X $ and noise variance $ \sigma_Z^2 = 1 $ is given by:

\[
C = \frac{1}{2} \log_2 \left(1 + \frac{P_X}{\sigma_Z^2}\right)
\]

However, in this problem, the constraint is on the **output power**, not the input power. We need to relate the output power $ E(Y^2) $ to the input power $ P_X $.

---

### **Step 2: Relate Output Power to Input Power**
From the channel model $ Y = X + Z $, the output power $ E(Y^2) $ is:

\[
E(Y^2) = E[(X + Z)^2] = E(X^2) + E(Z^2) + 2E(XZ)
\]

Since $ X $ and $ Z $ are independent, $ E(XZ) = E(X)E(Z) = 0 $. Also, $ E(Z^2) = \sigma_Z^2 = 1 $. Thus:

\[
E(Y^2) = E(X^2) + 1
\]

Given the output power constraint $ E(Y^2) \leq 2 $, we have:

\[
E(X^2) + 1 \leq 2 \implies E(X^2) \leq 1
\]

Thus, the input power $ P_X = E(X^2) $ is constrained by $ P_X \leq 1 $.

---

### **Step 3: Compute Channel Capacity**
Using the capacity formula with $ P_X = 1 $ and $ \sigma_Z^2 = 1 $:

\[
C = \frac{1}{2} \log_2 \left(1 + \frac{1}{1}\right) = \frac{1}{2} \log_2(2) = \frac{1}{2} \cdot 1 = 0.5 \text{ bits}
\]

---

### **Final Answer**
The channel capacity is:

\[
\boxed{0.5 \text{ bits}}
\]

0.5