Define αmax, αmin, insertion loss and gain

Attenuation (loss) is $\alpha(\omega) = -20\log_{10}|T(j\omega)|$ dB, so a large $\alpha$ means a small output.

 alpha(dB)
     |                     ______________
amin |- - - - - - - - - - |
     |                   /
     |                  /  transition
amax |~~~~~~~~~~~~~~~~ /
   0 +---------------+---+---------------> w
       passband      wp  ws   stopband

• $\alpha_{max}$ ($\alpha_p$): the maximum attenuation allowed anywhere in the passband (e.g. 0.5 dB). It sets the allowed ripple or droop.
• $\alpha_{min}$ ($\alpha_s$): the minimum attenuation required everywhere in the stopband (e.g. 40 dB).
• $\omega_p$: passband edge frequency, the last frequency at which $\alpha \le \alpha_{max}$.
• $\omega_s$: stopband edge frequency, from which $\alpha \ge \alpha_{min}$.
• Half-power frequency $\omega_{3dB}$: the frequency at which output power falls to half of its maximum, i.e. $|T(j\omega)| = |T|_{max}/\sqrt{2}$, an attenuation of 3.01 dB.
• Bandwidth: the width of the passband. For a low-pass filter it is $0$ to $\omega_p$ (or to $\omega_{3dB}$ for the half-power bandwidth); for a band-pass filter $BW = \omega_2 - \omega_1$, where $\omega_1, \omega_2$ are the lower and upper half-power frequencies.

Half-power frequency and bandwidth of a low-pass filter:

 |T|
 1.0    |------.
        |       `.
 0.707  |- - - - - -o   <- half-power point (3 dB down)
        |            `.
        |              `-.___
      0 +----------+-----------> w
        |<--- BW --->
                   w3dB

Insertion gain and insertion loss compare the load voltage with and without the filter between a source $V_1$ (resistance $R_s$) and load $R_L$.

 Without filter:  Rs                 With filter:  Rs   +--------+
  V1 o--/\/\--+-- RL  (V20)          V1 o--/\/\--| filter |-- RL (V2)
                                                 +--------+

$$V_{20} = \frac{R_L}{R_s + R_L}V_1, \qquad \text{Insertion gain} = 20\log_{10}\left|\frac{V_2}{V_{20}}\right|\ \text{dB}$$ $$\text{Insertion loss} = 20\log_{10}\left|\frac{V_{20}}{V_2}\right| = -(\text{insertion gain})\ \text{dB}$$

A passive filter normally has a positive insertion loss; an active filter can have insertion gain.