# Differential encoder output of PSK modulation input

I am a bit confused to the answer given for the following question.

If the input to a binary differential PSK modulation system is 01100101000111 tabulate The differential coder output

I know that for the differential coder output: 0 is encoded by the same pulse used to encode the previous data bit and 1 is encoded by the negative(opposite) of the pulse used to encode the previous data bit.

I am confused because Online and what I calculated does not match the answer that my professor gave.

My answer: For 01100101000111 input --> the output 001000110000101

Professors answer: --> 1 -1 -1 -1 1 -1 -1 1 1 -1 1 -1 -1 -1 -1

I found that this is only true if I first invert the input then apply the method, (in other words For 01100101000111 input becomes 10011010111000 )

Im so confused now, because I do not think that is how he obtained his final answer.

• FWIW, it doesn't look right to me either. Besides nobody uses DPSK becuase it is not bandwidth efficient. But conceptually it is the same as Differential Bi-phase or Manchester codes ( 4 types) one of which became called Bi-phase Mark (toggle on 1) en.wikipedia.org/wiki/Manchester_code This then migrated to QPSK etc for analog and RLL encoding for digital and magnetic encoding. – Sunnyskyguy EE75 Mar 17 '17 at 1:57

You seem to have a binary system. This is how I would solve it. First, assume a "dummy" inital input and output of 0; then, if the current input is different from the previous input, output a 1; otherwise output a 0. It may be easier to place the data vertically:

input output
----- ------
0
0     0
1     1
1     0
0     1
0     0
1     1
0     1
1     1
0     1
0     0
0     0
1     1
1     0
1     0

The sequence in your professor's answer looks wrong given the context. However, the reason they may be using 1 and -1 is that, in BPSK, a bit '1' is mapped to a positive amplitude V, and a bit '0' to a negative amplitude -V. The professor seems to have incorporated this step into their solution.