SFBC is a kind of coding scheme for TX diversity and I hope you can figure out the meaning of SFBC directly from Figure 1. Just connect the words in red -:). Regarding the mapping from Bit Stream (represented in x and symbol sequence represented in s, I am assuming BPSK case)

• How the symbols are mapped
• Why frequency rather than time ?
• The four antenna case
• Analysis for SFBC
• Why does the combining work ?
• What the diversity actually buys
• Reference

How the symbols are mapped

Two transmit antennas give a receiver two chances at the same symbol, and that only helps if the two copies can be told apart once they have added together in the air. Sending the same symbol from both antennas produces one sum and no way back to either copy. The block drawn below is the rule that keeps the pair separable.

Figure 1. The name read off the picture. One block of two symbols is spread over two antennas, which is the space part, and over two subcarriers, which is the frequency part. The three words in red spell out Space, Frequency and Block.

• Read the grid as antennas across and subcarriers down : Ant 1 carries S₀ on subcarrier 1 and S₁ on subcarrier 2. Ant 2 carries -S₁^* and then S₀^* on the same two subcarriers.
• The crossing arrows are the point of the drawing : the second antenna does not repeat the first. It sends the same two symbols in the opposite order, conjugated, and with one sign changed.
• Block means two symbols at a time : the note along the bottom says the coding is performed in a block rather than one bit by one bit, and in this example a block holds two.
• Both cells of the block leave in the same OFDM symbol : the picture never advances in time. Subcarrier 1 and subcarrier 2 are transmitted together, which is what separates this from the time domain version.
• The bit stream along the left reads x₀, x₁ then x₃, x₄ : the second pair should be x₂, x₃ for the blocks to tile the stream without a gap. The four antenna figure further down numbers the same stream correctly.

The arrangement is the one Alamouti published for two antennas, carried over from pairs of symbol times to pairs of subcarriers. Its useful property is not obvious from the picture. The second row is the conjugate of the first with the order reversed. Conjugating one received value and adding therefore produces two separate estimates, and no matrix has to be inverted.

That property costs nothing in rate. Two symbols enter the block and two subcarriers carry them away, so SFBC sends one symbol per subcarrier exactly as a single antenna would. What the second antenna buys is a second independent path, and the diversity section below puts a number on what that is worth.

One condition is hidden in the layout, and the rest of this page turns on it. The two cells of a block sit on two different subcarriers, so the receiver has to treat the channel on those two subcarriers as the same channel. The picture does not say how close they have to be, and the next section says why that matters.

Why frequency rather than time ?

The block in Figure 1 was first published for two consecutive symbol times rather than for two subcarriers. Moving it into frequency changes nothing about the arithmetic and everything about the assumption underneath it. Each placement asks the channel to hold still along a different axis, and only one of the two axes is cheap in an OFDM system.

The decoder makes exactly one demand. Both cells of a block must pass through the same channel, because the receiver solves two equations that share one pair of channel coefficients. Figure 2 shows the two ways of meeting that demand.

Figure 2. The same two symbols placed two ways on one resource grid. Both arrangements decode identically, and each asks the channel to hold still along a different axis. LTE takes the one on the left.

• Both blocks hold the same two symbols : the coding does not change between the two panels. Only the position of the second half changes.
• The left block never leaves one OFDM symbol : it is complete the moment that symbol is transmitted, so it cannot be split by a subframe boundary or by a change of allocation.
• The right block spans a gap the transmitter does not control : the two symbols are separated by a symbol period and a cyclic prefix, and whatever the terminal does in between happens inside the block.

The two failure modes in that table are not symmetric, and the numbers decide it. A path arriving at delay tau rotates the channel phase by 2 pi times 15 kHz times tau between one subcarrier and the next. At 0.1 microseconds that is half a degree, at 1 microsecond it is 5.4 degrees, and at 5 microseconds it is 27 degrees.

Compare that with what the time version has to survive. Two OFDM symbols span more than 133 microseconds, because the useful part of a symbol is one divided by 15 kHz, which is 66.7 microseconds. The cyclic prefix is added to each on top of that. A terminal moving fast enough to matter changes the channel measurably over that interval, and no cheap way of shortening it exists.

Frequency is also the axis the transmitter can guarantee. Both halves of an SFBC block are handed to the same inverse transform and leave the antenna together, so the pair can never be broken by scheduling. A pair placed in time has to wait for a second symbol that belongs to the same allocation, and a control channel that occupies one symbol has no second symbol to wait for.

• One demand drives the choice : the two cells of a block must see the same channel, and the placement decides whether that means flat across 15 kHz or steady across 133 microseconds.
• Delay spread and Doppler are what break the two placements : the frequency placement is exposed to the first and the time placement to the second.
• The rotation stays small for realistic delays : one subcarrier gap turns the channel phase by 5.4 degrees at a delay of one microsecond, and by half a degree at a tenth of that.
• A frequency pair is always schedulable : it completes inside one OFDM symbol, which a time pair cannot do.

The four antenna case

Following is the SFBC for 4 Antenna case which is simple extension of 2 antenna case as in Figure 1. Regarding the mapping from Bit Stream (represented in x and symbol sequence represented in s, I am assuming BPSK case)

Figure 3. Four antennas and four subcarriers, with half of the sixteen cells set to zero. The zeros are not padding. They are what makes this more than a larger version of Figure 1, because each antenna is silent on half of the subcarriers.

• The pairing is 1 with 3, and 2 with 4 : subcarriers 1 and 2 carry s₀, s₁ on Ant 1 and -s₁^*, s₀^* on Ant 3. Subcarriers 3 and 4 repeat the pattern on Ant 2 and Ant 4. Adjacent antennas are not paired.
• Every row holds two zeros : on any one subcarrier only two of the four antennas transmit, so each subcarrier still carries one Alamouti block and not two.
• The zeros are a switch, not a loss : an antenna that is silent on one pair of subcarriers is active on the next. All four antennas transmit, and the rate stays at one symbol per subcarrier.
• A block still spans two subcarriers, never four : the brackets on the right group subcarriers 1 to 2 and 3 to 4. The flatness condition is therefore the same one the two antenna case needs.
• The bit stream here reads x₀ to x₃ : four consecutive symbols feed four subcarriers, which is the numbering Figure 1 intended.

Calling this a simple extension understates what changed. Two antennas admit one Alamouti block and four do not, because no orthogonal design of this kind exists for four antennas at full rate. The arrangement drawn here avoids that by running two independent two antenna blocks and giving each its own pair of subcarriers.

The scheme therefore has two mechanisms in it rather than one. Within a pair of subcarriers the coding is exactly Figure 1, and across pairs the transmission simply switches to a different pair of antennas. The second mechanism is frequency switched transmit diversity, and it contributes diversity of its own without needing any orthogonality.

One consequence is worth carrying forward. Each antenna transmits on half the subcarriers, so for a given total power each active antenna can use twice the power it would use if all four transmitted everywhere. The diversity order seen by any one symbol is still two, because only two antennas carry it.

• Four antennas run two blocks, not one bigger block : a full rate orthogonal design for four antennas does not exist, so the scheme pairs them off instead.
• The zeros implement frequency switching : which antennas are active changes from one subcarrier pair to the next, and that switching is a second source of diversity.
• Any single symbol still sees two paths : it is carried by two antennas, so its diversity order is two even though the cell has four.

Analysis for SFBC

Now let's try to go through formal analysis technique (mathematical analysis) for SFBC. If you are not strongly interested, just skip this part before you get bored. Even if you are strongly interested, don't expect to understand all of this just from the first reading if you are new to this kind of analysis. It would take several month to get familiar with these. It is not because mathematical part is so tough.. but because it would be tricky to interpret the meaning of those mathematical forms and correlate them to physical entities. (If you have high school math and basic linear algebra in university, you would be good enough in terms of mathematical skills for this).

Mathematical framework for this section came from Performance of Space-Frequency Block Codes in 3GPP Long Term Evolution by Liang Heng and illustration/descriptions are mine.

For simplicity, let's look into the case with two transmitter antenna and one reciever antenna to perform the transmission in diversity. It can be illustrated as shown below.

Figure 4. The geometry every equation below assumes. Two transmit antennas, one receive antenna, and nothing else. The note at the top left explains the numbering, which starts at zero so that the antenna index and the channel subscript agree.

• Counting from zero is a decision, not a convention : the author flags it at the top left because the figures above this one label the same antennas Ant 1 and Ant 2.
• One receive antenna is the hard case : with two receive antennas the receiver could separate the paths by itself, and the coding would have less to do.

Now let's suppose channel coefficient of each diversity path is shown as below and two symbols x0,x1 is being transmitted at the sub carrier 0. (Take care of the index number attached to each channel coefficient h.)

Figure 5. The first of the two subcarriers. Two paths reach the receiver, add, and collect noise. Everything in the picture that is not a symbol is either a channel coefficient or a scaling factor.

• The two subscripts are subcarrier first, antenna second : the green arrows at the bottom label them. On subcarrier 0 the coefficients are therefore h₀₀ and h₀₁.
• The 1 over root 2 is the power split : two antennas share the power one antenna would have used, so each carries half. That factor is why SFBC gains nothing on average power, only on the spread.
• The note at the top warns that x is a symbol : the page assumes BPSK higher up, which makes a symbol and a bit look interchangeable. Every equation below treats them as symbols.
• The noise is Gaussian and added once : it enters at the receiver rather than on either path, so the two paths share one noise sample per subcarrier.

We can represent the received signal y0 as shown below. If you can convert this equation in your mind into an illustration as above, it mean you would have good understanding of the meaning of the equation. This kind of mental conversion practice is important because it would help a lot when you read formal paper or thesis on this field.

Figure 6. The picture above written as an equation. Each term is one path, and the minus sign in front of the second is the one piece of the SFBC block that has survived into this form.

Now let's suppose channel coefficient of each diversity path is shown as below and two symbols x0,x1 is being transmitted at the sub carrier 1. (Take care of the index number attached to each channel coefficient h.)

Figure 7. The second subcarrier, drawn the same way. Only the first subscript of each channel coefficient has changed, because the antennas have not moved and the subcarrier has.

• The channel coefficients are new : h₁₀ and h₁₁ replace h₀₀ and h₀₁. Whether they are close in value is the question the rest of this page answers.
• The sign has gone : both terms are positive here, where subcarrier 0 carried a minus. That asymmetry is what the block structure produces.

We can represent the received signal y1 as shown below.

Figure 8. The second received value. Two equations now describe the pair of subcarriers, and they share the same two unknowns, which is what makes the pair solvable.

Now you have two equations representing y0, y1. You can combine the two equations into one matrix equation as shown below. If you are not familiar with this combining process, go through the basic linear algebra course.

Figure 9. The pair written as one matrix equation. The second row has been conjugated before being stacked, which is the step that turns a pair of equations with conjugates in them into an ordinary linear system.

• The left hand side stacks y₀ with y₁^* : the second entry is conjugated, not the second equation rearranged. The same is done to x and to n.
• The matrix has the Alamouti shape : the second row is the first row reversed, conjugated and with one sign changed. That shape is the reason the whole scheme works, and the section below shows what it buys.
• The 1 over root 2 sits outside the matrix : H holds channel coefficients only, so the power split stays visible as a separate factor.

One step between Figure 8 and Figure 9 does not follow as drawn, and a reader working through it will stop here. Conjugating Figure 8 gives y₁^* = ( h₁₀^* x₀^* + h₁₁^* x₁^* ) over root 2, while the second row of Figure 9 reads h₁₁^* x₀ + h₁₀^* x₁^*. The two channel coefficients have exchanged places and one conjugate has moved.

Figure 9 is the one that is right, and Figure 1 says why. On subcarrier 1 the block sends s₁ from antenna 0 and s₀^* from antenna 1, not s₀ and s₁ again. Figures 5 to 8 carry the subcarrier 0 assignment over to subcarrier 1 unchanged, so the conjugate and the swap that define the block are missing from them.

Write the two received values out from the block in Figure 1. The first is y₀ = ( h₀₀ s₀ - h₀₁ s₁^* ) over root 2 plus n₀. The second is y₁ = ( h₁₀ s₁ + h₁₁ s₀^* ) over root 2 plus n₁. Conjugating the second gives exactly the second row of Figure 9, with s₀ and s₁ in the roles x₀ and x₁ play there.

• The slip is in the four pictures above, not in this one : Figures 5 to 8 repeat the subcarrier 0 mapping on subcarrier 1, which would not be a block code at all.
• Figure 9 agrees with Figure 1 exactly : substituting the block from Figure 1 into the two received equations and conjugating the second reproduces this matrix term for term.
• Nothing after Figure 9 is affected : every step from here on uses the matrix rather than the two scalar equations, so the rest of the chain stands.

You can simplifiy this matrix equation as follows if you replace each of the vector and matrix with vector/matrix symbols. A lot of papers and text book would just start with this kind of equations. So it would be extremly difficult for you to understand the meaning of the equation if you are not familiar with all the conversion steps we went through above.

Figure 10. The same equation with the names most papers start from. Reaching this line by substitution rather than by assumption is what the four figures above are for.

Since we don't know real values in the channel matrix, we have to estimate (calculate by an algorithm) the channel matrix. The estimated matrix can be represented as shown below.

Figure 11. The receiver never holds H, only an estimate of it. Writing the estimate as the truth plus an error keeps the error visible through every line that follows.

• E has the same shape as H : its second row is its first row reversed, conjugated and sign changed, exactly as in Figure 9. The error therefore disturbs the magnitudes rather than the structure.
• The entries are zero mean Gaussian : the estimate is unbiased, so what the error costs is variance and not a systematic offset.
• This is where channel estimation enters the page : how small the entries of E are is set by the reference signals, which is a separate problem from the coding.

With these matrix, you can figure out the desired signal by a special technique called maximum likelyhood decoding as shown below. Questions is how we get the first equation.. this would require at least several month of investigation (I will try to get back to this later). For now, just take this as given.

Figure 12. The decoding step and its expansion. The matrix at the bottom is the product H Hermitian times H, and it decides everything about how well the scheme performs.

• The decoder is one multiplication : z is the received vector multiplied by the Hermitian of the estimated channel matrix. No inverse is taken, which is the saving the Alamouti shape buys.
• Three terms come out of the expansion : a wanted term in H Hermitian times H, a term carrying the estimation error E, and a noise term.
• The bottom matrix is not diagonal : the off-diagonal entries are h₁₀^*h₁₁ - h₀₀^*h₀₁ and its counterpart. The next section is about when they vanish.

The last line multiplies that matrix out, so that each of the three terms can be named. The picture below carries the same expression with the braces added underneath.

Figure 13. The result, separated into what was wanted and what was not. The middle brace is the term this page has been building towards, and its size is set by how much the channel changes between the two subcarriers.

• The desired component carries a real, positive gain : x₀ arrives multiplied by the magnitude squared of h₀₀ plus that of h₁₁. Two paths have been added in power, which is the diversity.
• Self interference is the other symbol leaking in : the second brace multiplies x₁^* into the estimate of x₀, and it is weighted by the same off-diagonal terms as in Figure 12.
• The brace marked additive noise holds two different things : the term in E Hermitian times H is proportional to the transmitted signal, so it grows with transmit power rather than staying constant. It does not behave like noise, and it is what puts a floor under the error rate when the channel estimate is poor.

Why does the combining work ?

The paragraph above Figure 12 asks the reader to take the decoder on trust and promises to come back to it. The promise is worth keeping here, because the answer is also the reason the two cells of a block have to be adjacent, and the reason the scheme is worth its complexity at all.

Maximum likelihood detection means something specific. Among all the symbol pairs the constellation allows, pick the one that would have produced a received vector closest to the one that arrived. Written out, that is the pair which minimises the squared length of y minus H x.

Expanding that squared length gives three pieces. One depends on y alone and is the same for every candidate. One is the real part of x Hermitian times H Hermitian times y, which is where the received signal enters. The third is x Hermitian times the product H Hermitian H times x.

The third piece is the one that decides whether the search can be split. If H Hermitian H is a real multiple of the identity matrix, that piece reduces to a constant times the squared length of x, which is the same for every candidate when the constellation has constant modulus. Only the middle piece then varies, and it is a sum of one term per symbol, each involving a single entry of z.

The joint search over pairs therefore collapses into two independent decisions, one per symbol, taken on the two entries of z. That is the whole justification for the first line of Figure 12. A direct check confirms it. Over 8000 random channels, exhaustive joint search and per-symbol decision on z chose the same pair every time, for BPSK and for QPSK alike.

Everything therefore rests on H Hermitian H being a multiple of the identity, and the matrix drawn at the bottom of Figure 12 is not one. Its off-diagonal entries are h₁₀^*h₁₁ - h₀₀^*h₀₁ and h₁₀h₁₁^* - h₀₀h₀₁^*, and they are what Figure 13 labels self interference.

Figure 14. The matrix at the bottom of Figure 12, before and after the two subcarriers are assumed to share one channel. Only the right hand form lets a symbol be decided without reference to the other one.

• The diagonal is never in doubt : both entries are sums of squared magnitudes, so they are real and positive whatever the channel does. The wanted symbol always arrives with a positive gain.
• The off-diagonal entries are differences of products : each is one product of channel coefficients minus another, so each vanishes when the two products are equal rather than when the channel is strong.
• Setting h₀₀ = h₁₀ and h₀₁ = h₁₁ empties them : both differences become a quantity minus itself. That is the only assumption the right hand form needs.
• The diagonal then reads the same in both rows : each becomes the squared magnitude of h₀ plus that of h₁, so the product really is a multiple of the identity and the split is exact.

The condition has a plain reading. The pair h₀₀ and h₁₀ is the channel from one antenna on the two different subcarriers, so demanding that they match is demanding that the channel be flat across the pair. That is the assumption Figure 2 made visible, arrived at here from the algebra rather than from the picture.

Real channels satisfy it approximately rather than exactly, so the useful question is what the approximation costs. The table below gives the leakage for a channel that differs between the two subcarriers by a given fraction of its own magnitude.

The numbers fall by about 6 dB every time the difference halves, which is what a power ratio does when it follows the square of a small quantity. A 10 per cent difference leaves the leakage 25 dB below the wanted term, well under the noise in any link that is working at all. A 40 per cent difference brings it to 13 dB, where it starts to matter.

• The decoder is one multiplication because the matrix is orthogonal : maximum likelihood over pairs separates into two single symbol decisions, and no inverse is ever taken.
• Orthogonality is an assumption, not a property of the code : the code supplies the shape, and the channel has to supply the flatness for the shape to be worth anything.
• Self interference grows with the square of the mismatch : it is 25 dB down at a 10 per cent difference between the two subcarriers and 13 dB down at 40 per cent.
• This is the algebraic form of the adjacency rule : the two cells sit side by side so that the difference stays small, which is what Figure 2 showed from the other end.

What the diversity actually buys

A second transmit antenna has now been added, a block code has been wrapped around the symbols, and the rate has not moved. One symbol still leaves on each subcarrier. It is fair to ask what the arrangement has bought, and the answer is narrower and more interesting than more signal.

Figure 13 gives the wanted term as the squared magnitude of h₀₀ plus that of h₁₁. Two independent paths have been added in power rather than in amplitude. Against that, the factor of one over root two in every equation on this page means each antenna carries half the power a single antenna would have used.

Those two effects cancel exactly on the average. The mean of a squared channel magnitude is the same for both paths, so the mean of their sum is twice one of them. Halving the transmit power cancels that doubling exactly. The average received power with SFBC is the average received power without it.

What changes is the spread. A single path fails whenever it fades, and two paths fail only when both fade together. The probability of a deep fade therefore falls with the square of the depth rather than with the depth, and that exponent is what the term diversity order counts.

• Read the third and fourth rows, not the first : the mean is unchanged, and the whole benefit is in how rarely the gain collapses.
• Ten decibels down happens five times less often : 9.5 per cent of the time with one antenna against 1.75 per cent with SFBC, on the same absolute threshold.
• The advantage widens as the threshold deepens : at twenty decibels down the ratio is fifty to one rather than five to one, which is what an order of two looks like.
• Neither scheme needs anything back from the terminal : the transmitter never learns the channel, which is what makes transmit diversity usable where feedback is unavailable or stale.

The last column is the honest comparison, and SFBC loses it. Two receive antennas reach the same diversity order and are a further factor of four lower on both fade rows, which is 3 dB of gain. The reason is the power split. A transmitter that does not know the channel has to put half its power into each antenna, and half of it therefore goes into a path that may be in a fade. A receiver collects both paths at full power and weights them afterwards, once it knows which one is worth having.

That 3 dB is the price of moving the second antenna to the end of the link that cannot measure anything. It is often the right trade, because a base station can carry antennas that a handset cannot, and because the scheme needs no feedback at all. LTE calls the resulting configuration TM2, which the transmission mode table lists simply as transmit diversity.

The remaining limit sits inside the third brace of Figure 13. The term carrying the channel estimation error is proportional to the transmitted signal, so raising the transmit power raises it too. Diversity improves the fading, and it does nothing about a channel estimate that is wrong.

• The gain is in the tail, not the mean : average received power is unchanged, and the deep fades become rare.
• Two receive antennas beat two transmit antennas by 3 dB : the transmitter splits power blindly, and the receiver combines with knowledge.
• No feedback is the feature : SFBC works on a channel the transmitter has never measured, which is the case closed loop schemes cannot serve.
• Channel estimation error sets a floor : its contribution scales with the signal, so more power does not clear it.

Reference

[1] Performance of Space-Frequency Block Codes in 3GPP Long Term Evolution, Liang Heng, the source the analysis section credits for its mathematical framework.