As an example of a translation, consider the set of rotors from Figure 2, and suppose that

The rotors in positions 1--5 are, respectively, B, Beta, III, IV, and I.

The rotors in positions 2--5 are currently at positions A, X, L, E, respectively.

In the plugboard, the letter pair 'Y' and 'F' and the letter pair 'Z' and 'H' are both interchanged.

We input the letter 'Y', which causes the following steps:
The pawls all move. This causes rotor 5 to advance from E to F. The other two pawls are not over notches, so rotors 3 and 4 do not move.

The letter 'Y' enters the plugboard and is converted to 'F'.

The letter 'F' enters the right side of rotor 5 (an I rotor) at position 5, since 'F' is the 5th letter of the alphabet numbering from 0. Since the current setting of rotor 5 is 'F', the signal enters the rotor at position 5, but hits contact 5+5=10, or 'K'.

According to Figure 2, rotor I converts 'K' to 'N' (letter number 13). Because the setting of rotor I is 'F', however, the signal actually comes out at letter position 13−5=8 ('I').

The 'I' signal from rotor 5 now goes into the right side of rotor 4. Since rotor 4 is a IV rotor and is in the 'L' (or 11) setting, the 'I' enters at contact 8+11=19 ('T'), and is translated to 'G' (6), which comes out at position 6−11=−5=21mod26, the fifth letter from the end of the alphabet ('V').

The 'V' from rotor 4 goes now to the right side of rotor 3, a III rotor in setting 'X' (23). The signal enters the rotor at contact 21+23=44=18mod26 ('S'), is translated to 'G' (6), which exits at position 6−23=−17=9mod26 ('J').

Rotor 2 (Beta) is in position 'A', and thus translates 'J' to 'W'.

Rotor 1 (B) converts the 'W' to 'H' and bounces it back to the left side of rotor 2.

Rotor 2 (Beta in the 'A' position) converts 'H' on its left to 'X' (23) on its right.

The 'X' from rotor 2 now goes into the 23+23=46=20mod26 ('U') contact on the left side of rotor 3 (III in setting 'X'), and is converted to 'W' (22), which comes out at position 22−23=−1=25mod26 ('Z') on the right side of rotor 3.

'Z' now enters the left side of rotor 4 (IV in setting 'L') at 25+11=36=10mod26 ('K'), and is translated to 'U' (20), which comes out at position 20−11=9 ('J') on the right side of rotor 4.

The 'J' from rotor 4 enters the left side of rotor 5 (I at setting 'F') at contact 9+5=14 ('O'), is translated to 'M' (12), and comes out at position 12−5=7 ('H') on the right side of rotor 5.

Finally,the letter 'H' is converted to 'Z' by the plugboard.

Therefore, 'Y' is converted to 'Z'. Figure 3 is a visual depiction of this example. It shows the rotors after rotation; you can see the resulting setting of the last four rotors (AXLF) as the letters at the top of those rotors.