Hi, previously we considered how to deliver digital information wirelessly. In this lecture, we'll consider how exactly a wireless digital modem realizes the information transfer in theory and principles. This picture shows some circuits in iPhone 6, and the indicated chip is the modem, modulator and demodulator. Chip used in iPhone 6 that converts digital information into radio signal which is modulation and vice versa which is the modulation. Now the processing sequence in a modem at transmitter side is introduced. First, digital information to be transferred comes at the modem, then a processing code, channel encoding is performed. It transforms incoming bit sequence into another sequence by adding more bits produced from the input bit sequence. That helps to recover erroneous bits at the receiver. Then once in every T seconds, a number of bits are mapped to the corresponding vector in a predetermined signal constellation. Here, T second means the symbol period whose inverse is proportional to the signal bandwidth. Also, a signal constellation is a set of vectors and the number of vectors in a constellation is equal to two to the power of the number of bits per symbol. In a wireless modem, two dimensional signal constellations are typically used and we can represent them in a two dimensional plot or show the component value separately. After the constellation mapping, the digital information is converted to a discrete time and continuous value to signal. Then waveform mapping maps that discrete time signal into a continuous waveform that in bit they stole information. As you know, the information bearing signal is up converted to radio frequency signal and then amplified. Finally, the radio frequency signal generates radio wave at the transmitter antenna and the radio wave propagates. Just I said before, channel encoding adds more bits. Here, the last three bits, 101 are generated from the first six bits and then added. Suppose that the channel encoder transforms k bits into n bits. Then, the number of possible candidates in the original information is 2 to the k. That means only 2 to the k elements among all possible n bit sequences are used. So if some bits are received erroneously at the receiver, it is typically not among the possible 2 to the k cases and the receiver can recover the errors by selecting the most likely one among the possible 2 to the k cases. Here, k over n is called the code of eight. Now, consider the constellation mapping. Here, the first one is called BPSK having two candidates so that it maps one bit per symbol. Also the second and third ones have four and 16 candidates so that they can map two and four bits per symbol respectively. For example, the second one is used and 01 00 11, is coming. Here, this constellation is called QPSK constellation. Then the constellation mapping produces a discrete time signal at every symbol T by selecting one vector according to the incoming 2 bits. Now consider what happens if we change constellation to the third one called 16 quad constellation. One good thing is that the number of bits delivered is doubled for ever single period that means the spectral efficiency's doubled. However, in order to keep the distance among candidates same to the previous case, more energy per bit is required. Here, as the number of bits per symbol increases the required energy per bit increases exponentially. Now back to the QPSK example. At every T, the constellation mapping produces two real values according to incoming bits and the given QPSK constellation. The two real valued discrete kind signals has the digital pulse shaping filter and then convert it to an electric current via digital to analog converting device. As I said earlier, symbol rate, the inverse of the symbol period p is proportional to the bandwidth of the information barrier signal and the symbol energies proportional to the power of the information barrier signal. The information barrier signal is up converted into radio frequency signal after appropriately amplified it generates radio wave to be propagated. At the receiver side, the propagated radio wave is received but the radio frequency signal generated from the receiver antenna is corrupted by termaloids. After down converting lowest corrupted information bearing signal is obtained. After passing a receiver filter, usually the same one use that pole shaping filter at the transmitter. A discrete time signal is obtained by sampling once in every T second. If we plot the samples as the received constellation it looks as shown. Although the original QPSK constellation has only four vectors the received symbols deviates randomly due to the noise. Then the detection is done by selecting the nearest one among the original constellation for each received symbol, because it is the most probable one. The detection convert the received the symbols into binary sequence. However, some bits may be erroneously detected. In this example, the bit in red color is erroneously detected. However, such bit error can be corrected at the channel decoder with the aid of additional bits. Here are the last three bits so that the original information is successfully delivered. Now consider when a detection fails but the channel decoder can recover it. As I've said earlier, a noisy received symbol deviates from the original constellation vector in a random manner. But statistically, the probability of larger deviation is smaller. Therefore, selecting the nearest one in the original constellation is to choose the most probable ones so that the bit error probability is minimized. However, although the probability is low, noise receive the symbols close to the boundaries are very probable to produce a bit errors which may prevent reliable information transfer. However, a channel decoder can correct such bit errors. Consider the previous example again when the erroneously detected sequence comes to the channel decoder. Then the channel decoder can find that the incoming bit sequence is not a possible candidate sequence. Then, it corrects the errors by mapping to the closest possible sequence. Although such mapping can fail either, such probability vanishes if k is large. Suppose that k is 1,000 and n is 2,000, then only one among two to the 1,000 is valid. So that valid base sequences are different in sufficiently many bit positions to each other. So such an error probability is very, very low. Can you remember the bandwidth efficiency versus the power efficiency trade of curve introduced in the previous lecture? This was a theoretical one. However, commercially used digital modems nowadays implement the theory successfully and the practically achievable tradeoff is very close to the theoretical limit. So for the questions about how to realize the theory you get the answer.