White Noise Noise Unit 1000 Hz 0 kHz to 3 kHz -4.8 dBm/kHz dBrnc 90.0 dBrnc 88.0 dBrnc 88.4 dBrnc dBrn 3 kHz FLAT 90.0 dBrn 88.8 dBrn 90.3 dBrn dBrn 15 kHz FLAT 90.0 dBrn 90.0 dBrn 97.3 dBrn Psophometric voltage 870 mV 582 mV 604 mV (600 S2) 1.26 X 10 pWp 5.62 X 108 pWp 6.03 X 108 pWp 91.0 dBp 87.5 dBp 87.8 dBp Source: Based on Table 4.2, p. 60, Ref. 1. LEARNING DECIBELS AND THEIR APPLICATIONS Example 2. We measure noise in the voice channel at 37 dBrnc. What is the equivalent noise in pWp? 37 dBrnC = 10(10g X x 10-9) - + 90 dB, -52.5 = 10(log X x 10-9), -5.25 = log X x 10-9, = antilog(-5.25) = = 5623 x 10-9, X = 5623 pWp. Carry out the following exercises. The answers follow. 1. -83 dBmp =? pWp 2. 47,000 pWp =? dBmp 3. -47 dBm =? dBmp 4. 33 dBrnC =? dBmp 5. 20,000 pWp =? dBrnC 6. 50,000 pWp =? dBm 7. 2000 pW =? pWp 8. 4000 pWp =? dBrnC Answers: 1: 5 pWp. 2: -43.28 dBmp. 3: -49.5 dBmp. 4: 2238 pWp = - -56.5 dBmp = -54 dBm. 5: 42.5 dBrnC. 6: -43 dBmp = -40.5 dBm. 7: 1120 pWp. 8: 35.5 dBrnC. C.8 INSERTION LOSS AND INSERTION GAIN When dealing with the broad field of telecommunication engineering, we will often encounter the terms insertion loss and insertion gain. These terms give us important information about a two-port network in place in a circuit. Two-port just means we have an input (port) and an output (port). A major characteristic of this device is that it will present a loss in the circuit or it will present a gain. Losses and gains are expressed in dB. In the following we show a simple circuit terminated in its characteristic impedance, Zo. Zo=600s2 We now insert into this same circuit a two-port network as follows: Z Network First for the case of insertion loss: Let us suppose the device is an attenuator, a length of waveguide, a mixer with loss, or any other lossy device. Suppose we are delivering power P2 to the load ZL with the network in place and power Po with the network removed. The ratio expressed in dB of Po to P2 is called the insertion loss of the network: Insertion lossdb = 10 log(po/p2). If 