Pulse Generator - I (Art of Electronics)
Consider \$V_{in}\$ to be 0V and for that to have been the case for some considerable time. \$C_1 \$ will have 5V on its LHS (left hand side) because it has been charged by \$ R_2\$ and approximately 0.6V on its RHS (right hand side) because \$ Q_2 \$ is on and the voltage is limited by its base, \$R_3\$ is providing the base current.

We now take \$V_{in}\$ high to say 5V. This turns \$Q_1\$ on so the voltage on the LHS of \$C_1\$ falls to 0V, because the voltage across a capacitor can't change instantly the voltage on the RHS of \$C_1\$ falls to about -4.4V. Since the base of \$Q_2\$ is negative there is \$ 5-(-4.4) = 9.4\$ volts across \$R_3\$ so it must have \$940\mu\text{A}\$ flowing through it from top to bottom as drawn. \$Q_2\$ base-emitter junction is reverse biased (there is no base current) so the it must be flowing through \$C_1\$ into the collector of \$Q_1\$ and out of its emitter.  This has the effect of discharging \$C_1\$ and starting to charge it in the other direction.

This turns \$Q_2\$ off and \$V_{out}\$ goes immediately to 5V.

\$R_3\$ now starts to charge the RHS of \$C_1\$ towards 5V, it will never get there because once it reaches about 0.6V \$Q_2\$ starts to turn on and \$V_{out}\$ begins to fall.  The falling edge on \$V_{out}\$ is softer than the rising edge because transistors are current driven and because some of the current in \$R_3\$ is used to charge the capacitor as \$Q_2\$ starts to conduct.

We now take \$V_{in}\$ to 0V again and the LHS of \$C_1\$ charges via \$R_2\$ to 5V once it gets there we are ready to generate our next pulse.