A CMOS Schmitt-trigger inverter has a low output level of 0 V and a high output level of 5 V. It has input thresholds of 1.6 V and 2.4 V. The input capacitance and output resistance of the Schmitt trigger are negligible. The frequency of the oscillator shown in the figure is ______ Hz. (Round off to 2 decimal places)

In the given circuit, the CMOS Schmitt-trigger inverter with feedback resistor $R=10\text{k}\Omega$ and capacitor $C=47\text{nF}$ forms an astable multivibrator (oscillator).

The parameters of the circuit are:
- Low output level: $V_{\mathrm{OL}}=0\text{V}$
- High output level: $V_{\mathrm{OH}}=5\text{V}$
- Lower threshold voltage: $V_{\mathrm{TL}}=1.6\text{V}$
- Upper threshold voltage: $V_{\mathrm{TH}}=2.4\text{V}$
- Resistor: $R=10\text{k}\Omega =10\times {10}^3\Omega$
- Capacitor: $C=47\text{nF}=47\times {10}^{-9}\text{F}$

The time constant of the RC network is:
$\tau =RC=(10\times {10}^3\Omega )\times (47\times {10}^{-9}\text{F})=4.7\times {10}^{-4}\text{s}$

Step 1: Charging Phase ($T_1$)
During the charging phase, the output of the inverter is high ($V_{\mathrm{OH}}=5\text{V}$). The capacitor voltage $V_c\left(t\right)$ charges from the initial voltage $V_{\mathrm{TL}}=1.6\text{V}$ towards $V_{\mathrm{OH}}=5\text{V}$ until it reaches the upper threshold voltage $V_{\mathrm{TH}}=2.4\text{V}$:
$V_c\left(t\right)=V_{\mathrm{final}}+(V_{\mathrm{initial}}-V_{\mathrm{final}})e^{-t/\tau }$
$2.4=5+(1.6-5)e^{-T_1/\tau }$
$2.4=5-3.4e^{-T_1/\tau }$
$3.4e^{-T_1/\tau }=2.6$
$T_1=\tau \ln \left(\frac{3.4}{2.6}\right)=\tau \ln \left(1.3077\right)\approx 0.26826\tau$

Step 2: Discharging Phase ($T_2$)
During the discharging phase, the output of the inverter is low ($V_{\mathrm{OL}}=0\text{V}$). The capacitor voltage $V_c\left(t\right)$ discharges from the initial voltage $V_{\mathrm{TH}}=2.4\text{V}$ towards $V_{\mathrm{OL}}=0\text{V}$ until it reaches the lower threshold voltage $V_{\mathrm{TL}}=1.6\text{V}$:
$1.6=0+(2.4-0)e^{-T_2/\tau }$
$1.6=2.4e^{-T_2/\tau }$
$e^{-T_2/\tau }=\frac{1.6}{2.4}=\frac{2}{3}$
$T_2=\tau \ln \left(1.5\right)\approx 0.40547\tau$

Step 3: Total Time Period ($T$) and Frequency ($f$)
The total time period $T$ of one oscillation cycle is:
$T=T_1+T_2\approx (0.26826+0.40547)\tau =0.67373\tau$
$T\approx 0.67373\times (4.7\times {10}^{-4}\text{s})\approx 3.1665\times {10}^{-4}\text{s}$

The frequency $f$ of the oscillator is:
$f=\frac{1}{T}=\frac{1}{3.1665\times {10}^{-4}}\approx 3158.12\text{Hz}$