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.

  • \$\begingroup\$ You have the RF tag but, isn't your question about audio? You mention ADC and this is more likely to be audio? You need to be clearer about what it is you don't understand. \$\endgroup\$
    – Andy aka
    Jan 25, 2022 at 15:08
  • 3
    \$\begingroup\$ Your receiver will have its own internal noise. Anything that reduces the power of the wanted signal from outside will reduce the SNR. An impedance mismatch will reduce the power of the wanted signal. Johnson noise power is independent of impedance. Johnson voltage power reduces, as you say, at low impedance. The question you need to get your head round is why in RF, power is the key parameter, not voltage or current. \$\endgroup\$
    – Neil_UK
    Jan 25, 2022 at 15:30
  • \$\begingroup\$ @Neil_UK I mean, I get why power is important at the transmitter because the electromagnetic wave is what's transmitted physically, but I don't get the significance in the receiver circuitry. Maybe it is because the noise is independent of distributed circuit impedance effects, only loss resistance – obviously. I'll try to perform some calculations. Andy_aka ADC is used in any WiFi receiver for instance, where you take samples for digital processing, so it's about RF \$\endgroup\$ Jan 31, 2022 at 14:11

3 Answers 3


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.

\$\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

enter image description here

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:

enter image description here

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


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.

  • \$\begingroup\$ I updated my answer. Why do you add their powers? On the answer I add the voltages and then calculate the power. Is this not correct? On the lumped circuit example, calculating the powers separately and summing them still doesn't show SNR being maximum when matched. I only started to get a useful reading when I analyser it in terms of transmission lines. Anyway, you're right about LNAs having a complicated noise profile with other sources of noise and indeed 'noise matching' as opposed to impedance matching, but I don't know much about that \$\endgroup\$ Feb 2, 2022 at 23:10
  • \$\begingroup\$ See, for example, rfic.eecs.berkeley.edu/~niknejad/ee142_fa05lects/pdf/lect11.pdf p8 \$\endgroup\$
    – Tesla23
    Feb 2, 2022 at 23:41
  • \$\begingroup\$ I'm not sure what you're talking about with power waves. Can you prove that you get the best S/N when it is matched? I can prove it by varying the length of the transmission line or frequency of the signal, but I cannot prove it by varying the load resistance, I always get the same plot where S/N gets bigger the closer the load gets to 0 ohms. The proof doesn't exist online anywhere, despite all the claims, which is bizarre because I also thought it would be trivial myself \$\endgroup\$ Feb 3, 2022 at 15:58
  • \$\begingroup\$ Why $$ V_R=e_n\frac{r}{r+R}$$ and not $$V_R=e_n\frac{R}{r+R}$$ I think that's a semantic error on your part, but the maths still works out to get your final noise power if you use R and then use \$\Gamma =-\Gamma \$, and works with \$r\$ and \$\Gamma\$ like you've done \$\endgroup\$ Feb 4, 2022 at 17:03
  • 1
    \$\begingroup\$ Using practical devices, when amplifying weak signals you use devices that add noise of their own - not full Johnson noise associated with their input impedance, but some noise. So a LNA with no input (say a cooled load on the input) will give you some noise out. When you connect it to a real source, you still have this noise, plus the amplified signal plus noise from the source. Matching increases these last two, reducing the degradation to SNR from the noise added by the LNA. It's typically a bit more complicated as the LNA noise often depends on source impedance, but that's the gist. \$\endgroup\$
    – Tesla23
    Feb 4, 2022 at 20:58

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.

enter image description here

Using Rs = 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.


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

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

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:

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


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

RL = 1MΩ enter image description here $$\max(\text{SNR}) = 7.8046 \times 10^{12}= 128.92 ~\text{dB} $$

RL = 125Ω

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

RL = 1μΩ

enter image description here $$\max(\text{SNR}) = 390234 = 55.91 ~\text{dB} $$

(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.


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

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.

  • \$\begingroup\$ How is \$NF\$ max when \$R_L =R_S\$? Surely it is higher if \$R_L\$ is lower than \$R_S\$. Besides surely you want to minimise NF? You taught me one thing though: How to do inline mathjax on ee stack exchange. Weird that you have to escape the dollar. I just thought they disallowed it \$\endgroup\$ Feb 3, 2022 at 15:44
  • \$\begingroup\$ Sorry my mistake I updated it \$\endgroup\$
    – sarthak
    Feb 4, 2022 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.