Objective 3.8 — Signal-to-Noise Ratio#
LO# |
Description |
|---|---|
3.8 |
I can calculate signal-to-noise ratio (SNR) and determine the effects on wireless communications. |
By the end of this reading you should be able to:
Explain where noise comes from and what sets the noise floor.
Compute noise power from system temperature and bandwidth.
Compute SNR for a communications link and for a radar link.
Work backward from a required SNR to a minimum detectable signal.
Signal-to-Noise Ratio (SNR)#
Imagine yourself trying to have a conversation with someone in a very noisy room. We can use the signal-to-noise ratio (SNR) to describe how well the other person can hear you:
If the other person can understand you, then the SNR is acceptable. If she can’t hear you, the SNR is too low. In this case, you have two options: either increase the signal (talk louder) or decrease the noise (tell everyone else to be quiet). Either option will effectively increase the SNR. In electrical engineering, we describe SNR in terms of power:
In the above example, the noise came from everybody else talking. But, in wireless transmissions, there are several sources of noise: jamming, ambient noise from other signals, and thermal noise. We will discuss jamming in the lesson on Electronic Warfare, and for our purposes, we can assume the ambient noise is small compared to other sources. Thermal noise is caused by the thermal agitation of charge carriers.
Recall that temperature is actually a measurement of the energy of particles in a system. Inside a wireless system, this inherent particle movement manifests as noise, and can interfere destructively as noise. The equation to calculate the power of thermal noise is:
In this equation,
k is the Boltzmann constant, \(1.38 \times 10^{-23}\) J/K.
\(T_{sys}\) is the effective noise temperature of the system in kelvins (K).
BW is the bandwidth of the receiver in Hz. It is set by the receiver’s filter, which is matched to the bandwidth of the signal the receiver is built for (Objectives 2.1 and 3.1). A simple pulse radar’s receiver is matched to its pulse, so \(BW \approx 1/\tau\) (Objective 3.7).
With k in J/K, the temperature in K, and the bandwidth in Hz, the noise power comes out in watts. The powers in this objective are tiny, so they are usually written in femtowatts: 1 fW = \(10^{-15}\) W.
Since we know how to calculate the noise power, the signal-to-noise ratio is therefore:
Every receiver needs some minimum SNR to do its job, written \(SNR_{\min}\) (problems often call it the required SNR). A link closes, or a radar detects its target, when \(SNR \geq SNR_{\min}\). Setting the two equal and solving for the received power gives the receiver’s sensitivity, or minimum detectable signal, \(P_{R,\min} = SNR_{\min} \cdot k\left( T_{sys} \right)BW\), so comparing \(P_{R}\) with \(P_{R,\min}\) is the same test. This is where the minimum received powers quoted in Objectives 3.5 and 3.6 came from.
An SNR below 1 means only that the signal is below the noise as it is received. \(SNR_{\min}\) is a property of the receiver’s processing: a receiver that integrates many pulses, or de-spreads a spread-spectrum signal, can work with an \(SNR_{\min}\) far below 1, while one without that processing gain needs an \(SNR_{\min}\) above 1. When a jammer is present (Objective 3.9), its received power adds to the noise and the same \(SNR_{\min}\) still decides: \(P_{R}/(P_{noise} + P_{R,jam}) \geq SNR_{\min}\).
Key Concept
\(P_{noise} = k\,T_{sys}\,BW\), so the minimum detectable signal is just \(P_{R,\min} = SNR_{\min} \times k\,T_{sys}\,BW\), where \(SNR_{\min}\) is the minimum SNR the receiver needs (its required SNR). It is a property of your receiver and its bandwidth, not of the transmitter — which is where the sensitivity figures in Objectives 3.5 and 3.6 came from.
Example Problem 1#
A 500 mm-radius antenna receives 166.0 fW of power. If the received signal has a bandwidth of 10 MHz and the system has an effective temperature of 310 K, what is the signal-to-noise ratio?
Understand: In this situation, we already know how much power arrives at the receiver, but we want to know how the inherent thermal noise of the system affects system performance. In other words, SNR measures how hard we have to work to process the signal.
Identify Key Information:
Knowns: We know the received power, the radius of the antenna (which is irrelevant, since we already know the received power), the bandwidth of the signal, and the effective temperature of the system.
Unknowns: Signal-to-noise ratio (SNR).
Assumptions: Thermal noise is the only appreciable source of noise in the system. Although the effective noise temperature is 310 K, which is equivalent to 98.3° F, this does not mean that the temperature of the antenna is necessarily 98.3°F. Instead, it means that all the energy in the system is equivalent to thermal noise at a temperature of 98.3°F.
Plan: We simply employ the SNR equation to calculate the SNR.
Solve: The SNR equation gives us,
Answer: For the given system, the signal-to-noise ratio is 3.88. Whether that is enough depends on the \(SNR_{\min}\) of the receiver: one that needs 3.5, like the receiver in Example Problem 2, would work.
Example Problem 2#
You are designing a communication system to receive a signal with a bandwidth of 12 MHz. In order to process the signal, your receiver requires an SNR of 3.5 (\(SNR_{\min} = 3.5\)). Assuming the effective noise temperature will not be higher than 320 K, what is the minimum received power required by your system?
Understand: In earlier examples, we knew the minimum received power. Now, however, we need to calculate it.
Identify Key Information:
Knowns: We know the required SNR, the maximum system temperature, and the bandwidth of the system.
Unknowns: The minimum received power required to process the signal.
Assumptions: Thermal noise is the only appreciable source of noise in the system.
Plan: We can rearrange the SNR equation to solve for the received power, \(P_R\). Since we know the maximum system temperature and the minimum required SNR, this gives us the worst case scenario, which is the minimum required receive power.
Solve: First, we need to solve the SNR equation for \(P_R\), with the SNR set to its minimum value:
Answer: The minimum power required by the system is 185.5 fW.
Example Problem 3#
Consider a radar that transmits 1 kW of power at a frequency of 300 MHz using an antenna with a gain of 150 toward an aircraft with a radar cross section (RCS, the \(\sigma\) of Objective 3.6) of 100 \(m^2\), which is 86 km away. How much power will the radar receive and what will be the signal to noise ratio (SNR) at the radar’s antenna? Solving for the received power, using the radar equation, gives:
Now for many ultra-sensitive receivers, 20.73 fW is an acceptable power. However, what if the radar is still transmitting its 1 kW of power at the same time? If we treat this transmitted power as noise, we can calculate a signal to noise ratio of:
This number is unacceptably small – no receiver will be able to distinguish the returned signal from the “noise” of the transmitter. There are two distinct ways to solve this problem: use separate antennas for transmitting and receiving (this is called bistatic) or send pulses of energy, instead of a continuous stream, with a monostatic radar (aka pulse-Doppler radar!).
Type-along
A datalink receiver has \(T_{sys} = 290\) K and needs \(SNR_{\min} = 10\).
With \(BW = 5\) MHz, find the noise power and the minimum detectable signal \(P_{R,\min}\).
The signal arrives at 150 fW. Find the SNR. Does the link close?
The link is redesigned for a lower data rate, dropping \(BW\) to \(500\) kHz. What is the new minimum detectable signal, and does 150 fW close the link now?
By what factor does the maximum communication range improve?
Why does a radar gain much less from the same change?
Solution
1. \(P_{noise} = (1.38\times10^{-23})(290)(5\times10^6) = 20.0\) fW, so \(P_{R,\min} = 10 \times 20.0 = 200\) fW.
2. \(SNR = 150/20.0 = 7.50\), which is below \(SNR_{\min} = 10\), so the link does not close. The same verdict comes from \(150\) fW \(< 200\) fW.
3. Bandwidth drops 10×, so \(P_{noise} = 2.00\) fW and \(P_{R,\min} = 20.0\) fW. Now \(SNR = 150/2.00 = 75.0 \ge 10\), so the link closes.
4. For comms \(P_R \propto 1/R^2\), so a 10× sensitivity gain improves range by \(\sqrt{10} = 3.16\times\).
5. For radar \(P_R \propto 1/R^4\), so the same 10× buys only \(10^{1/4} = 1.78\times\). The fourth-power law punishes every improvement equally.
Wrap-Up#
Noise is thermal and unavoidable: \(P_{noise} = k T_{sys} BW\). What decides whether a link works is SNR, not received power in isolation: the link closes when \(SNR \ge SNR_{\min}\), so the ‘minimum detectable signal’ is \(SNR_{\min}\) times the noise floor. Narrower bandwidth means less noise, a lever the transmitter does not have.
Objective 3.9 asks what happens when someone raises your noise floor on purpose.