A common choice for $\beta = 0.5$ (i.e., $R_1=R_2$) is not ideal for this specific formula for 50% duty cycle. A better choice for the feedback resistors to achieve $V_{Th} = \pm V_{sat}/2$ is needed.

Let's use the standard design with $R_1$ and $R_2$ for positive feedback, and $R$ and $C$ for timing, and assume an ideal Op-Amp's output swings to $\pm V_{sat}$.

The switching thresholds at the non-inverting input are $V_{TH} = +V_{sat} \frac{R_1}{R_1+R_2}$ and $V_{TL} = -V_{sat} \frac{R_1}{R_1+R_2}$.

The capacitor charges and discharges between these thresholds. For a 50% duty cycle, the charging and discharging times must be equal. This usually means the charging/discharging current paths are symmetrical.

A common simplification for 50% duty cycle when $R_1 = R_2$ (meaning $V_{TH} = \pm V_{sat}/2$) and the capacitor charges/discharges through $R$ to $\pm V_{sat}$ is:

$$ f = \frac{1}{2RC \ln(3)} \approx \frac{1}{2.2RC} $$

This approximation holds when $R_1 = R_2$ and the capacitor charges/discharges towards the full supply rails $\pm V_{sat}$.

Design Steps for 50% Duty Cycle

To achieve a 50% duty cycle, we need to ensure the capacitor charges and discharges symmetrically. A common way is to make the switching thresholds $\pm V_{sat}/2$. This means $R_1 = R_2$. And use an RC network where the capacitor effectively charges/discharges through the same resistance $R$ to $\pm V_{sat}$.

Determine Frequency: $f = 5 \text{ kHz}$. Period $T = 1/f = 1/5000 = 0.2 \text{ ms} = 200 \mu s$.
Choose Feedback Resistors ($R_1, R_2$): For symmetrical thresholds, choose $R_1 = R_2$. Let $R_1 = R_2 = 10 \text{ k}\Omega$. This sets the thresholds at $\pm V_{sat}/2$.
Choose Capacitor ($C$): Select a standard capacitor value. For audio frequencies, $1 \text{ nF}$ to $100 \text{ nF}$ is typical. Let's start with $C = 10 \text{ nF}$ ($0.01 \mu F$).
Calculate Resistor ($R$): Using the simplified formula $f = \frac{1}{2.2 RC}$ for 50% duty cycle: $$ R = \frac{1}{2.2 f C} $$ $$ R = \frac{1}{2.2 \times 5000 \text{ Hz} \times 10 \times 10^{-9} \text{ F}} $$ $$ R = \frac{1}{2.2 \times 5 \times 10^3 \times 10 \times 10^{-9}} = \frac{1}{110 \times 10^{-6}} = \frac{1}{1.1 \times 10^{-4}} $$ $$ R \approx 9090.9 \Omega $$ Choose a standard resistor value close to this, e.g., $R = 9.1 \text{ k}\Omega$.
Verify with precise formula (if applicable): If the circuit uses the configuration where $\beta = R_1 / (R_1+R_2)$ and the capacitor charges/discharges through R towards $V_{out}$, the period is $T = 2RC \ln\left(\frac{1 + \beta}{1 - \beta}\right)$. If $R_1=R_2$, then $\beta = 0.5$. $$ T = 2RC \ln\left(\frac{1 + 0.5}{1 - 0.5}\right) = 2RC \ln(3) \approx 2RC \times 1.0986 = 2.197 RC $$ So, $f = \frac{1}{2.197 RC}$. Our approximation $1/(2.2 RC)$ is very close. Using $R = 9.1 \text{ k}\Omega$ and $C = 10 \text{ nF}$: $$ f = \frac{1}{2.197 \times 9.1 \times 10^3 \times 10 \times 10^{-9}} = \frac{1}{2.197 \times 91 \times 10^{-6}} = \frac{1}{0.000200} \approx 5000 \text{ Hz} $$ This confirms our values.
Add Diodes ($D_1, D_2$) and $R_D$: To ensure symmetrical charge/discharge for 50% duty cycle, a common practice is to use two diodes and a resistor $R_D$ in series with $R$. The conceptual diagram shows this. If $R_D = R$, the effective charging/discharging resistance for the capacitor is $R$. In this case, the resistor $R$ from step 4 is the effective resistance for charging/discharging. The circuit would connect the capacitor to the inverting input, and the other end of the capacitor would be connected to $V_{out}$ through $R$. To achieve 50% duty cycle, we often use a slightly different topology or add current steering diodes. For simplicity, if we assume the equation $f = \frac{1}{2.2RC}$ holds for a standard Op-Amp astable mul