Given. Two ideal inverting gates, coupling capacitor $C_1$, feedback resistor $R_1$ from $V_{out}$ to node $V_c$; gates switch exactly at $V_{DD}/2$. $R_1=10\text{k}\Omega$, $C_1=10\text{nF}$ (part d).
Given data
Quantity
Value
$R_1$
$10\text{k}\Omega$
$C_1$
$10\text{nF}$
Switching threshold
$V_{DD}/2$ (both gates)
Find. Qualitative operation; $V_c(t),V_{out}(t)$ waveforms; expression for $V_c(t)$; oscillation period $T$.
Approach. Track what happens right after $V_{out}$ flips (an instantaneous $\pm V_{DD}$ step couples through $C_1$ onto $V_c$), then let $V_c$ relax exponentially through $R_1$ toward the new $V_{out}$ level until it re-crosses $V_{DD}/2$ and triggers the next flip.
Part (a) — operation. With one input grounded, each gate is effectively an inverter of its other (signal) input. The loop is regenerative: any change in $V_{out}$ inverts through gate 1 to produce an instantaneous step at its output, which is AC-coupled through $C_1$ straight onto $V_c$ (the capacitor voltage can't jump, so the WHOLE step appears on $V_c$); $R_1$ then slowly re-charges $V_c$ toward $V_{out}$'s new level. The instant $V_c$ crosses $V_{DD}/2$, gate 2 flips $V_{out}$, which (through the global feedback to gate 1) flips gate 1's output too, delivering a fresh step onto $V_c$ and restarting the process in the opposite direction — a free-running (astable) square-wave generator.
Part (c) — expression for $V_c(t)$. Take $t=0$ at the instant $V_{out}$ has just flipped low$\to$high (so gate 1's output has just flipped high$\to$low, i.e. a $-V_{DD}$ step lands on $V_c$, which was at $V_{DD}/2$ the instant before): $V_c(0^+)=V_{DD}/2-V_{DD}=-V_{DD}/2$. $V_c$ then relaxes with $\tau=R_1C_1$ toward the new $V_{out}=V_{DD}$:
$$V_c(t)=V_{DD}+\Big(-\tfrac{V_{DD}}{2}-V_{DD}\Big)e^{-t/\tau}=\boxed{V_{DD}\Big(1-\tfrac32e^{-t/\tau}\Big)},\quad 0\le t\lt T/2$$
and by symmetry $V_c(t)=\tfrac32V_{DD}\,e^{-(t-T/2)/\tau}$ on the second half, repeating with period $T$.
Fig. Q5(b) — Vc(t) jumps by ±VDD each time Vout switches (AC-coupled through C1), then relaxes exponentially (τ=R1C1) toward the new Vout level; the next flip fires exactly when Vc re-crosses VDD/2.
Part (d) — period. Setting $V_c(T/2)=V_{DD}/2$ in the part (c) expression:
$$\tfrac12=1-\tfrac32e^{-(T/2)/\tau}\ \Rightarrow\ e^{-(T/2)/\tau}=\tfrac13\ \Rightarrow\ T/2=\tau\ln3$$
$$T=2R_1C_1\ln3$$
With $\tau=R_1C_1=(10\text{k}\Omega)(10\text{nF})=100\,\mu\text{s}$:
$$T=2(100\,\mu\text{s})\ln3=\boxed{220\,\mu\text{s}}\qquad f=1/T=\boxed{4.55\text{ kHz}}$$
