in question. The noise figure tells us the amount of thermal noise a device injects into a radio system. Example. Suppose a receiver has a noise figure of 2.1 dB, what is its thermal noise level in 1 Hz of bandwidth. In other words, what is No? No = -204 dBW + 2.1 dB = -201.9 dBW/Hz. Eb is the signal energy per bit. We apply this to the receive signal level (RSL). The RSL represents the total power (in dBm or dBW) entering the receiver front end, during, let's say, 1-sec duration. (energy). We want the power carried by just 1 bit. For example, if the RSL were 1 W and the signal were at 1000 bps, the energy per bit would be 1/1000 or 1 mW per bit. However, it will be more convenient here to use logarithms and decibel values (which are logarithmic). Then we define Eb as Eb = or dBW - 10 log(bit rate) (9.13) Here is an example using typical values. The RSL into a certain receiver was -89 dBW and bit rate was 2.048 Mbps. What is the value of Eb? CONCEPTS IN TRANSMISSION TRANSPORT Eb = -89 dBW - 10 log(2.048 x 106) = -89 dBW - 63.11 dB 152.11 dBW We can now develop a formula for Eb/No: Eb/No = RSLdww log(bit rate) - (-204 dBW + NF. (9.14) Simplifying, we obtain Eb/No = RSLdBW - 10 log(bit rate) + 204 dBW NFdB. (9.15) Some Notes on Eb/No and Its Use. Eb/No, for a given BER, will be different for different types of modulation (e.g., FSK, PSK, QAM, etc.). When working with Eb, we divide RSL by the bit rate, not the symbol rate nor the baud rate. There is a theoretical Eb/No and a practical Eb No. The practical is always a greater value than the theoretical, greater by the modulation implementation loss in decibels, which compensates for system imperfections. Figure 9.10 is an example of where BER is related to Eb/No. There are two curves in the figure. The first from the left is for BPSK/QPSK (binary phase shift keying/quadrature phase shift keying), and the second is for 8-ary PSK (an eight-level PSK modulation scheme). The values are for coherent detection. Coherent detection means that the receiver has a built-in phas