4.1. Find the expected number of bit errors

### Ideas for Solving the Problem
1. **Bit Error Probability for Coherent BPSK:** The probability of bit error ($P_b$) for a coherent Binary Phase Shift Keying (BPSK) receiver is given by $P_b = Q\left(\sqrt{\frac{2E_b}{N_0}}\right)$, where $E_b$ is the energy per bit and $N_0$ is the single-sided noise power spectral density. The Q-function is defined as $Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^{\infty} e^{-t^2/2} dt$.
2. **Energy per Bit:** The energy per bit ($E_b$) can be calculated as $E_b = P \cdot T_b$, where $P$ is the signal power and $T_b$ is the bit duration.
3. **Bit Duration:** The bit duration ($T_b$) is the inverse of the data rate ($R_b$), i.e., $T_b = \frac{1}{R_b}$.
4. **Expected Number of Bit Errors:** The expected number of bit errors in a given time period is the product of the bit error probability, the data rate, and the time period.

### Calculation Steps
#### Step 1: Calculate the bit duration ($T_b$).
The data rate is $R_b = 5000$ bits/s. Therefore, the bit duration is:
$T_b = \frac{1}{R_b} = \frac{1}{5000} = 0.0002$ s.

#### Step 2: Calculate the signal power ($P$).
The signal amplitude is $A = 1 \text{ mV} = 1 \times 10^{-3} \text{ V}$. Since the signal power is normalized to a 1-Ω resistive load, the power is:
$P = \frac{A^2}{2} = \frac{(1 \times 10^{-3})^2}{2} = \frac{1 \times 10^{-6}}{2} = 0.5 \times 10^{-6} \text{ W}$.

#### Step 3: Calculate the energy per bit ($E_b$).
$E_b = P \cdot T_b = (0.5 \times 10^{-6} \text{ W}) \cdot (0.0002 \text{ s}) = 1 \times 10^{-10} \text{ J}$.

#### Step 4: Calculate the argument of the Q-function.
$\sqrt{\frac{2E_b}{N_0}} = \sqrt{\frac{2 \times 1 \times 10^{-10}}{1 \times 10^{-11}}} = \sqrt{\frac{2 \times 10^{-10}}{10^{-11}}} = \sqrt{20} \approx 4.472$.

#### Step 5: Calculate the bit error probability ($P_b$).
$P_b = Q\left(\sqrt{\frac{2E_b}{N_0}}\right) = Q(4.472)$. Using a Q-function table or calculator, we find that $Q(4.472) \approx 3.87 \times 10^{-6}$.

#### Step 6: Calculate the number of bits transmitted in one day.
One day has 24 hours, each hour has 60 minutes, and each minute has 60 seconds. So, one day has $24 \times 60 \times 60 = 86400$ seconds.
The number of bits transmitted in one day is $N = R_b \times \text{time} = 5000 \text{ bits/s} \times 86400 \text{ s} = 432 \times 10^6$ bits.

#### Step 7: Calculate the expected number of bit errors in one day.
The expected number of bit errors is $N_{\text{errors}} = N \times P_b = (432 \times 10^6) \times (3.87 \times 10^{-6}) = 432 \times 3.87 = 1671.84$.

### Final Answer
The expected number of bit errors made in one day is approximately 1672.

### Highlights
– The Q-function is crucial for calculating the bit error probability in BPSK systems.
– The energy per bit is a key parameter that depends on both the signal power and the bit duration.
– The expected number of errors is directly proportional to the bit error probability and the number of transmitted bits.
– Be careful with units and conversions (e.g., mV to V, seconds to days).