How does SNR improve with impedance matching in a radio receiver?

Question

I've seen basically every question and answer on here relating to impedance matching, and why you need it in RF but not in audio applications, and I'm still missing something.

I initially was wondering why you need maximum power transfer and not voltage transfer, as after all, it is the voltage signal that is going to be quantised by the ADC and ultimately power is going to be V^2 and not V^2/R. So I initially had the question as this guy, who didn't get any specific answer.

But then I plotted

$$\text{SNR}(R_L) = \frac{\frac{\left(\frac{R_L10}{R_L+R_S}\right) ^2}{R_L}}{\frac{\left(\frac{R_L\left(\sqrt{4kTBR_S}+\sqrt{4kTBR_L}\right)}{R_L+R_S}\right) ^2}{R_L}}$$

Assuming the source and load have both have Johnson nyquist noise. (I know there are also active sources of noise)

And in fact

$$\text{max} (\text{SNR}) = \lim_{R_L\to0} \text{SNR}(R_L)$$

The lowest johnson nyquist noise is achieved when the load impedance or input impedance is as low as possible, so actually the best SNR in the load is when the load impedance is low?

As for reflections distorting the signal due to the mismatch, you could just multiply by the inverse of the frequency response of the reflections?

I'm missing something else that means you need maximum power transfer, and I'm not understanding.

Answer 1

I've seen basically every question and answer on here relating to impedance matching, and why you need it in RF but not in audio applications, and I'm still missing something.

Impedance matching gives you maximum power transfer and also minimizes reflections. You use impedance matching where power is scarce (often arising with an antenna, microphone or sensor), or where reflections are problematic (e.g. preserving signal bandwidth in general or waveform edges in particular). This is not an exhaustive list.

It's good that you are trying to understand noise and matching, but I think the rest of your post and answer is wrong.

Firstly,
\$\text{SNR}(R_L) = \frac{\frac{\left(\frac{R_L10}{R_L+R_S}\right) ^2}{R_L}}{\frac{\left(\frac{R_L\left(\sqrt{4kTBR_S}+\sqrt{4kTBR_L}\right)}{R_L+R_S}\right) ^2}{R_L}}\$

you seem to be adding the amplitudes of incoherent noise sources in \$\sqrt{4kTBR_S}+\sqrt{4kTBR_L}\$, the correct thing to do is to add their powers.

What you appear to be trying to show is that if you have a passive circuit (e.g. attenuator), then you get the best S/N at the output if the input is matched. This is easily proven, can I suggest that you try using power waves - this is a simple linear transformation, instead of using \$V\$ and \$I\$ you use:

\$(V+IZ_0)/\sqrt{Z_0}\$ and \$(V-IZ_0)/\sqrt{Z_0}\$

so, for example, if \$Z_0\$ is the source resistance, then the available noise power from the source is \$kT\$ per unit Hz.

When it comes to active devices, in particular LNAs, this analysis is not appropriate, particularly as the input impedance of an active device does not generally exhibit Johnson noise. There is a lot of material in the literature, including a famous paper "The theory of noisy fourpoles", then any book / paper / uni course notes on scattering parameter design of low noise amplifiers. The best S/N from a LNA is generally not when the input is matched.

Update I think your analysis is flawed. Let's start with basics

The power available from the source is \$P_{avs}=\frac{V_s^2}{4r}\$, and if you define the reflection coefficient \$\Gamma=\frac{R-Z_s}{R+Z_s}\$, then the power dissipated in the load \$P_L=P_{avs}(1-|\Gamma|^2)\$. The available noise power from the source is kT (as long as \$r>0\$), so the noise power from the source that is dissipated in the load is \$N_L=kT(1-|\Gamma|^2)\$

Any lossless matching circuit you place between the source and the load to change \$\Gamma\$ (typically trying to reduce it), changes the signal power and the noise equally, hence it doesn't change the SNR.

Now you seem to be interested in the total noise voltage across R including the noise from \$R\$ itself. This usually doesn't apply as normally \$R\$ represents an active device which doesn't normally exhibit Johnson noise from it's input impedance, and we include the effect of it's noise in the noise figure. But let's see what happens:

Working out the contribution to \$V_R\$ from \$e_n\$, let's simplify the maths by assuming \$Z = r\$, real.

\$V_R=e_n\frac{r}{r+R}\$

noting that the voltage across the physical resistor \$R\$ includes the internal noise source.

using \$1-|\Gamma|^2 = \frac{4rR}{(r+R)^2}\$

you can show \$V_R^2 = kTr(1-|\Gamma|^2)\$

Now our previous quantities are powers dissipated in \$R\$, so we could convert the previous quantities to voltage, or convert this to equivalent power. I chose to convert this to equivalent power, hence

\$N_{eq} = \frac{V_R^2}{R} = kT(1-\Gamma)^2\$

So the total equivalent noise power dissipated in \$R\$ is

\$N = kT((1-|\Gamma|^2) + (1-\Gamma)^2) = 2kT(1-\Gamma)\$

Hence \$SNR = \frac{P_{avs}}{2kT}(1+\Gamma)\$

So if your detector measures the voltage across \$R\$, you get max SNR if \$\Gamma=1\$, i.e. an open circuit. This is the actual SNR you would measure into an ideal matched load with a noise figure of 0dB.

I haven't done the analysis, but I suspect that if your detector was measuring the current into the load you would find that the SNR is max when \$\Gamma=-1\$, i.e. a short circuit.

I haven't looked at \$x\neq0\$, I can't imagine it's much different, only more complicated.

As an example as to why you are unlikely to use this, the ADI LT1028 op amp has an input impedance of about 20k, which if it exhibited Johnson noise would have a noise spectral density of about \$18nV/\sqrt{Hz}\$, but the actual input noise voltage density is around \$1nV/\sqrt{Hz}\$.

In reality, you are usually passing weak signals to some sort of low noise amplifier, if this device has constant equivalent additive input noise, then you get maximum SNR when you maximize the input signal - i.e. matching the input. Typically the input noise varies somewhat with matching which complicates things. See texts on low noise amplifier design.

Answer 2

I think this is right. If we take the lossless quarter wave transformer of my answer here, connected to a source and load that both give off Johnson Nyquist noise, we can crudely show that SNR is maximum over the load when the transformer is at a length that provides an impedance match for the given frequency.

Using R_s = 20, Z₀ = 50, T = 290K. I removed B from the noise voltage, because that's the expected noise amplitude at the specific frequency, whereas sqrt(4kTBR) is the expected RMS in the time domain of the noise, and we are using a single infinite length frequency, but I will include a noise bandwidth of 20MHz.

Using:

$$H\left(f\right)=\frac{50}{20+50}\lim_{F\to\infty}\left(\sum_{n=0}^{F}\left(\Gamma_s\Gamma_L\right)^{n}\left(1+\Gamma_L\right)\left(\cos\left(2\pi\left(1+2n\right)\left(\frac{1}{4}\frac{1}{2.4GHz}\right) f \right) + j\sin\left(2\pi\left(1+2n\right)\left(\frac{1}{4}\frac{1}{2.4GHz}\right) f \right)\right)\right) $$

$$Z_{out}(f) = 50\frac{20+50i(\tan(2\pi f (\frac{1}{9600000000})))}{50+20i(\tan(2\pi f (\frac{1}{9600000000})))} $$

The expected noise power spectral density is:

$$\mathbb{E}\{N(f)\} = \frac{\left(|H(f)|\sqrt{4kTR_s}\right) ^2} {R_L} + \frac{\left(\left|\frac{R_L}{R_L+Z_{out}(f)}\right|\sqrt{4kTR_L}\right) ^2}{R_L}$$

Power spectral density of cosine signal of 10V rms of frequency a is:

$$ S(f) = \frac{|10H(f)|^2 \frac{2\pi}{2}(\delta(f-a)+\delta(f+a))}{R_L}$$

Plotting

$$\text{SNR}(f) = \frac{\frac{1}{2\pi}\int^{f+B/2}_{f-B/2}S(x)u(x)dx}{\frac{1}{2\pi}\int_{f-B/2}^{f+B/2}\mathbb{E}\{N(x)\}dx} $$

R_L = 1MΩ $$\max(\text{SNR}) = 7.8046 \times 10^{12}= 128.92 ~\text{dB} $$

R_L = 125Ω

$$\max(\text{SNR}) = 6.7281 \times 10^{12}= 128.28 ~\text{dB} $$

R_L = 1μΩ

$$\max(\text{SNR}) = 390234 = 55.91 ~\text{dB} $$

(the plot assumes that a is at the centre of the band B, and ignores the fact that SNR is double when a is low enough for the negative frequency component to also be in the band)

So yes, highest SNR is achieved as resistance approaches infinity, except when we consider active devices like LNA, there is other non-Johnson noise which is far more substantial than the 0.64dB improvement using a higher resistance has over over the matched load, therefore the best thing to do is maximise signal power.

Answer 3

$$\text{SNR}(R_L) = \frac{\frac{\left(\frac{R_L10}{R_L+R_S}\right) ^2}{R_L}}{\frac{\left(\frac{R_L\left(\sqrt{4kTBR_S}+\sqrt{4kTBR_L}\right)}{R_L+R_S}\right) ^2}{R_L}}$$

What circuit you are talking about? I think voltage source connected to a resistive divider? This expression makes no sense.

$$\text{max} (\text{SNR}) = \lim_{R_L\to0} \text{SNR}(R_L)$$

This is also incorrect. Think about it if \$R_L\$ is zero output voltage is zero, there is no signal. How can it be maximum SNR.

The SNR at the output can easily be derived as (symbols have the usual meanings):

$$V_o = \frac{R_LV}{R_L+R_S}$$ $$V_n^2 = 4kT(R_L||R_S)\cdot\Delta f$$ $$SNR_o = \frac{R_LV^2}{R_L+R_S}\cdot\frac{1}{4kTR_S\Delta f}$$ Note that \$SNR_o = 0\$ if \$R_L = 0\$.

I've seen basically every question and answer on here relating to impedance matching, and why you need it in RF but not in audio applications, and I'm still missing something.

The type of impedance matching to be implemented depends on the application. For output stages where electromagnetic power in circuit is converted to other forms of energy, power matching has to be implemented and this is not restricted to RF case. For example, in RF communications, Power amplifier is power matched to antenna impedance. This is because the circuit energy is being converted to radiation energy. The better the matching, the more power converted to radiation. In audio applications, the electric energy is converted to sound, again better the matching, the more power converted to sound energy.
But power matching is not the only type of impedance matching. When you have a circuit noise matched, it is ensured that it adds minimum amount of noise to the input signal. In noise matching, we are not trying to maximize the SNR but the noise figure defined as the ratio of SNR at the input to that at output.

$$SNR_i = \frac{V^2}{4kTR_S\Delta f}$$ $$NF = \frac{V^2}{4kTR_S\Delta f}\cdot \frac{4kTR_S\Delta f(R_L+R_S)}{R_LV^2}$$ $$NF = \frac{R_L+R_S}{R_L}$$ Note that, \$NF\$ is minimized by maximizing \$R_L\$. So noise matching and power matching conditions can be different in general.