Wireless Receiver Sensitivity


Receiver sensitivity is defined as the minimum signal power (Pi,min) at which a receiver can detect a signal while providing an adequate SNR at analog receiver output (BER at digital receiver output).

This is minimum detectable signal(MDS) power. In a multi-standard receiver, the sensitivity requirements vary depending on mode of operation, band, bandwidth, etc.,
For example,

  1. GSM standard requires a minimum sensitivity of -102dBm
  2. LTE Band-2 minimum sensitivity requirements: -102.7dBm for 1.4MHz bandwidth mode, and -92dBm for 20MHz bandwidth mode. Sensitivity requirements vary in LTE from band-to-band and across bandwidth modes.

In this post, let us go through the receiver sensitivity calculation using kTB, noise figure, and SNR of the receiver. GSM sensitivity and WLAN sensitivity calculations are explained through examples.


The noise figure of a receiver is given by

(1)   \begin{equation*} NF_{Rx} = SNR_{i}- SNR_{o} \quad\quad -----(dB) \end{equation*}

where,
SNR_{i} – SNR at the input receiver
SNR_{o} – SNR at the output of receiver

The minimum SNR required at the output of receiver is obtained from system level design calculations or simulations.

SNR at the input (SNRi) of a receiver is determined by input signal power(Pi) and noise floor(NFloor).

(2)   \begin{eqnarray*} SNR_{i}(dB) &=& P_{i}(dBm) - NFloor(dBm) \\ \end{eqnarray*}

The noise floor at the input of the receiver is mainly due to thermal noise. Available thermal noise power at temperature T_o and an equivalent noise bandwidth of B is given by P_{noise} = k T_o B. The reference temperature of 290^oK is utilized to calculate noise floor. At T_o = 290^oK, kT_o = 4 \times 10^{21} watts/Hz \approx -174 dBm/Hz. Therefore noise floor is given by

(3)   \begin{equation*} NFloor &=& -174 dBm/Hz + 10 \log(B_{Rx}) \end{equation*}

where B_{Rx} is the bandwidth of receiver in Hz.

Using Eq-(1) to Eq-(3), the noise figure of a receiver is given by

(4)   \begin{equation*} \boxed{NF_{Rx} = P_{i} + 174 ~dBm/Hz - 10 \log(B_{Rx}) - SNR_{o}} \quad\quad-----(dB) \end{equation*}

From Eq-(4), knowing the noise figure and minimum SNR requirement of the receiver, we can calculate the receiver sensitivity as

(5)   \begin{equation*} \boxed{P_{i,min} = - 174 dBm/Hz + 10 \log(B_{Rx}) + SNR_{o,min} + NF_{Rx}} \quad\quad-----(dBm) \end{equation*}

From the above equation, we can observe that sensitivity depends on receiver bandwidth, receiver noise figure and SNR(or BER) at the output of receiver. Receiver bandwidth is not same as channel bandwidth. Receiver bandwidth is mainly determined by IF filter or channel selection filter, but trade-off with lot other parameters.

For circuit designers, the design variable noise figure is of interest than the receiver sensitivity. For a communication standard given the receiver sensitivity, Eq-(4) is useful to find the noise figure of receiver. Then using Friss equation, noise figure for each block in the cascaded chain can be distributed.

Example: Sensitivity of a GSM receiver

Calculate the sensitivity of a GSM receiver whose noise figure is 10dB and minimum SNR at the output of receiver is 10dB.
Assume that input noise power is dominated by thermal noise and the bandwidth of Rx is 200kHz. Then noise floor at the input of receiver is NFloor = -174 ~dBm/Hz + 10 \log(200kHz) \approx -121~dBm
Then receiver sensitivity is given by

    \[P_{i,min} = NFloor + SNR_{omin} + NF_{Rx} = -121 + 10 + 10 = -101 ~dBm\]

 

In case of digital receivers, sensitivity is defined as minimum power level at which the receiver demodulate the received data with a specified BER, FER or PER or below that. The maximum BER dictates the minimum output SNR(SNRo,min) necessary for satisfactory reproduction or demodulation of the desired signal.

The relation between SNR and BER at the output of receiver is given as,

(6)   \begin{equation*} \boxed{SNR_o = {E_b \over N_o} + {R \over B}} \quad \quad --------(dB) \end{equation*}

R – Bit rate
B – Channel bandwidth
Eb/No – Energy per bit to noise power spectral density ratio

BER as a function of Eb/No is shown in Figure 1. Given the modulation scheme and BER required, pick the value of Eb/No from Figure 1 and compute the required output SNR using Eq-(6).

Theoretical BER over AWGN for various digital modulation techniques

Figure 1. Theoretical BER over AWGN for various digital modulation techniques
Image courtesy: Gaussian Waves

Example: Sensitivity of a WLAN receiver

Calculate the required noise figure of a WLAN receiver to detect a 64QAM signal carrying data at the rate of 54Mbps. Assume the sensitivity of the receiver as -65dBm.

IEEE standard defined the sensitivity of WLAN receiver[1]. In simple words the sensitivity of WLAN 802.11a receiver can be stated as “the minimum signal level at which the receiver detects and demodulate the signal with a PER of 10% or less than that”.

Probability of a packet received correctly(P_{PC})

(7)   \begin{equation*} P_{PC} = 1 - P_{PE} = (1- P_{BE})^{P_L} \end{equation*}

where P_{PE} is the probability of a packet received in error.
P_{BE} is the probability that a bit is received in error and P_L is the packet length.

For PER < 10%, P_{PE} < 0.1, BER \approx 3\times 10^{-5}.

From BER versus Eb/No plot shown in Figure 1, for 64QAM signal corresonding to BER of 3×10-5, Eb/No is 17dB.

Each sub-carrier codes 6bits, so 48 subcarriers code 288 bits. The duration of OFDM symbol is 4\musec. Therefore R/B = {288/4\mu sec \over 16.6MHz} \approx 6.37dB

Finally SNR at the output of receiver

(8)   \begin{equation*} SNR_o = {E_b \over N_o} + {R \over B} = 17dB + 6.37dB = 23.37dB \end{equation*}

From Eq-(4), we can compute the NF of WLAN receiver as
NF_{Rx} = -65 + 174 - 10 \log(20MHz) - 23.37 = 12.63 dB

Table below shows the required minimum sensitivity of a WLAN 802.11a receiver at different data rates[1].

Data rate (Mbps) 6 9 12 18 24 36 48 54
Rx Sensitivity (dBm) -82 -81 -79 -77 -74 -70 -66 -65

References

[1] “Receiver minimum input level sensitivity, agilent technologies.” [Online; Accessed June 2014]. Available: http://www.home.agilent.com/agilent/editorial.jspx?ckey=339504&id=339504&nid=-11143.0.00&lc=eng&cc=IN

 

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