Interference Alignment and Degrees of Freedom of the K-User(9)

CADAMBE AND JAFAR: INTERFERENCE ALIGNMENT AND DEGREES OF FREEDOM OF THE-USER INTERFERENCE CHANNEL3433and . Since convex combinations are achievable by time . Tosharing between the end points, this implies that and the proof is gether with the converse, we have complete. The assumption of time varying channels is intriguing degrees of freedom will be because it is not clear if achieved with constant channels. Therefore, the validity of the Host–Madsen–Nosratinia conjecture [18] remains unknown. On the one hand the number of degrees of freedom is a discontinuous measure as evident from the point to point channel where it represents the rank of the channel matrix. Therefore the constant coef cient case may be of limited signi cance. On the other hand, the constant channel case may shed light on novel interference alignment schemes such as interference alignment in the signal “level” dimension as demonstrated for certain special cases in [22] and interference alignment through lattice codes as demonstrated for the one-sided interference channel in [21]. Next we discuss the relationship between degrees of freedom capacity characterization. and an D. The Capacity Approximation provide a capacity approximation , i.e. (21) is more accuIn general, a capacity approximation within . However, it turns rate than an approximation within out that in many cases the two are directly related. For example, it is well known that for the full rank MIMO channel with input antennas and output antennas, transmit power and i.i.d. zero mean unit variance additive white Gaussian noise may be expressed (AWGN) at each receiver, the capacity as (22) A similar relationship between the degrees of freedom and the capacity characterization also holds for the MIMO multiple access channel, the MIMO broadcast channel, and the twouser MIMO interference channel. For the MIMO, MAC, and BC, the outerbound on sum capacity obtained from full cooper. The ination among the distributed nodes is nerbound obtained from zero forcing is also so that we can write . For the two user MIMO interference channel and the two-user MIMO channel the outerbound is obtained following an extension of Carleial’s outerbound [2] which results in a MIMO MAC channel. The innerbound is obtained from zero forcing. Since of we can both of these bounds are within . However, consider similarly write user interference channel with single antennas at each the node. In this case, we have only shown (23) To claim that capacity of the user interference channel is within we need to show an innerbound. Since our achievable schemes of are based on interference alignment and zero forcing, the natural question to ask is whether an interference alignment and degrees of zero forcing based scheme can achieve exactly case to freedom. The following explanation uses the suggest that the answer is negative. symbol exConsider an achievable scheme that uses a tension of the channel. Now, consider a point that can be achieved over this extended channel using interference alignment and zero-forcing alone. If possible, let the total de. For ingrees of freedom over this extended channel be stance, . It can be argued along the is same lines as the converse part of Theorem 1 that achievable in the two-user interference channel for . ThereforeThe degrees of freedom that is accurate withinIt can be easily seen that the only point that satis es the above inequalities and achieves a total of degrees . Therefore, any scheme that achieves of freedom is degrees of freedom over the extended channel a total of . achieves the point We assume that the messages are encoded along independent streams similar to the coding scheme in the proof of Theorem 1, i.e.Now, at receiver 1, to decode an -dimensional signal using zero-forcing, the dimension of the interference has to be at most . For instance (24) Note that since column vectors and has linearly independent is full rank with probability , . Similarly, the dimension of the . Therefore, interference from transmitter 3 is also equal to the two vector spaces on the left-hand side of (24) must have full intersection, i.e., span span span span span span At receiver At receiver (25) (26) (27)represents the space spanned by the column vecwhere span tors of matrix The above equations imply that span span span spanwhere . The above equation implies that there exists at least one . Note that since all channel eigenvector of in matrices are diagonal, the set of eigenvectors of all channelAuthorized licensed use limited to: Harbin Institute of Technology. Downloaded on June 01,2010 at 01:21:44 UTC from IEEE Xplore. Restrictions apply.