[分享] Sensitivity
304 查看
0 回复
 楼主 | 发布于 2022-11-25 | 只看楼主
分享到:

The sensitivity is defined as the minimum signal level that a receiver can detect with“acceptable quality.”In the presence of excessive noise,the detected signal becomes unintelligible and carries little information. We define acceptable quality as sufficient signal-to-noise ratio,which itself depends on the type of modulation and the corruption(e.g.,bit error rate)that the system can tolerate. Typical required SNR levels are in the range of 6 to 25dB.

In order to calculate the sensitivity,we write

                         NF = SNRin/SNRout                             (2.146)

                           =Psig/PRS/SNRout                             (2.147)

where Psig denotes the input signal power and PRS the source resistance noise power,both per unit bandwidth. Do we express these quantities in V²/Hz or W/Hz?Since the input impedance of the receiver is typically matched to that of the antenna(Chapter 4),the antenna indeed delivers signal power and noise power to the receiver. For this reason,it is common to express both quantities in W/Hz(or dBm/Hz). It follows that

Psig = PRS *NF *SNRout                           (2.148)

Since the overall signal power is distributed across a certain bandwidth,B,the two sides of(2.148)must be integrated over the bandwidth so as to obtain the total mean squared power. Assuming a flat spectrum for the signal and the noise,we have

Psig,tot = PRS *NF *SNRout *B                         (2.149)

Equation(2.149)expresses the sensitivity as the minimum input signal that yields a given value for the output SNR. Changing the notation slightly and expressing the quantities in dB or dBm,we have(Note that in conversion to dB or dBm,we take 10log because these are power quantities.)

Psen|dBm = PRS|dBm/Hz +NF|dB + SNRmin|dB + 10logB                    (2.150)

where Psen is the sensitivity and B is expressed in Hz. Note that(2.150)does not directly depend on the gain of the system. If the receiver is matched to the antenna,then from(2.91),PRS = kT = -174dBm/Hz and

Psen = -174dBm/Hz +NF + 10logB + SNRmin                         (2.151)

Note that the sum of the first three terms is the total integrated noise of the system(sometimes called the“noise floor”)

Example 2.25

A GSM receiver requires a minimum SNR of 12dB and has a channel bandwidth of 200kHz. Awireless LAN receiver,on the other hand,specifies a minimum SNR of 23dB and has a channel bandwidth of 20MHz. Compare the sensitivities of these two systems if both have an NF of 7dB.

Solution

For the GSM receiver,Psen = -102dBm,whereas for the wireless LAN system,Psen = -71 dBm. Does this mean that the latter is inferior?No,the latter employs a much wider bandwidth and a more efficient modulation to accommodate a data rate of 54Mb/s. The GSM system handles a data rate of only 270kb/s. In other words,specifying the sensitivity of a receiver without the data rate is not meaningful.

(0 ) (0 )
回复 举报
  • 发表回复
    0/3000





    举报

    请选择举报类别

    • 广告垃圾
    • 违规内容
    • 恶意灌水
    • 重复发帖

    全部板块

    返回顶部