Interference Alignment and Degrees of Freedom of the K-User(8)

3432IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 54, NO. 8, AUGUST 2008Thus, the vectors carrying the desired signal at receiver 1 are linearly independent of the interference vectors which allows the receiver to zero force interference and obtain interference free dimensions, and therefore degrees of freedom for its message. At receiver 2 the desired signal arrives along the vectors while the interference arrives along the vectors and the vectors . As enforced by (8) the are perfectly aligned within the interference vectors . Therefore, in order to prove that interference vectors there are interference free dimensions at receiver 2 it suf ces to show that the columns of the square, -dimensional matrix (19) are linearly independent almost surely. This proof is quite similar to the proof presented above for receiver 1 and is therefore omitted to avoid repetition. Using the same arguments we can show that both receivers 2 and 3 are able to zero force the interference vectors and obtain interference free dimensions for their respective desired signals so that they each achieve degrees of freedom. Thus we established the achievability of for any . This scheme, along with the converse automatically imply thatFig. 2. Degrees of Freedom Region for the three-user interference channel.achievability as follows. Let be the degrees of freedom region of the three-user interference channel. We need to prove . We show that which along with the conthat verse proves the stated result. The points can be veri ed to lie in through trivial achievable schemes. Also, lies in (Note that Theorem 1 implies that this is the only point which achieves a total of degrees of freedom and satis es the inequalities in (20).)f Consider any as de ned by the statement of the thepoint can then be shown to lie in a convex orem. The point , J, K, L, and N. For inregion whose corner points are stance, can be expressed as a convex combination of the end points (see Fig. 2)C. The Degrees of Freedom Region for the 3 User Interference Channel Theorem 2: The degrees of freedom region of the three-user interference channel is characterized as follows:(20) Proof: The converse argument is identical to the converse argument for Theorem 1 and is therefore omitted. We showwhere the constants are de ned as shown in the table at the bottom of the page. are nonnegative for It is easily veri ed that the values of and that they add up to one. Thus, all points all in are convex combinations of achievable points. . .. . .. . ..... . .. . .. . ..... . .Authorized licensed use limited to: Harbin Institute of Technology. Downloaded on June 01,2010 at 01:21:44 UTC from IEEE Xplore. Restrictions apply.