Bandwidth of low pass filter
In your case $$G(s) = \frac{a}{s+b} = \left(\frac{a}{b}\right)\left( \frac {1}{1+\frac {s}{b}}\right)$$

The output will be -3dB compared to the passband when \$s = jb\$  (equivalent to \$\frac{b}{2\pi}\$ Hz) so the bandwidth is \$b\$ radians/second and the passband gain is (a/b).