# Ouput SNR of an integrator

Say I have an integrator with transfer function $$\dfrac{\mathrm{F_s}}{s}$$.

This integrator has an initial condition zero and integrates from $$t=0$$ to $$t=\dfrac{1}{Fs}$$.

If I apply a constant $$V_{\mathrm{in}}$$ together with a white noise source $$e(t)$$ ~ $$N(0,\sigma^2)$$ then I would have at the input $$\mathrm{SNR}=\dfrac{{\mathrm{Vin}}^2 }{\sigma^2 }$$

At the end of integration at $$t=\dfrac{1}{Fs}$$, the output power of the signal is $$P_{\mathrm{signal}} ={V_{\mathrm{in}} }^2 {F_s^2 \left(t\right)}^2 ={V_{\mathrm{in}} }^2$$
However, the output variance of the integrated noise (Brownian motion) is
$$P_{\mathrm{noise}} =\sigma^2 F_s^2 t=\sigma^2 F_s$$ resulting in SNR at the output Fs times lower than that at the input.

This calculation does not seem consistent with simulation. Is there something wrong with my derivation?

• Hi. What is the nature of your F_s variable? What does it represent? – Richard Lyons Feb 9 '20 at 17:47
• @Lyons It represents the sampling frequency of the system, so that at the end of one integration cycle I have Vin sampled at the output. F_s is assumed to be very large. – Timothy Hannon Feb 9 '20 at 18:00
• Hi. I'm not smart enough to understand your equations. But what I would do is treat your discrete integrator (Rectangular, Trapezoidal, Simpson's,?) as a digital filter. Then measure and compare the SNRs of the integration filter's input and output sequences. – Richard Lyons Feb 9 '20 at 18:58
• @Lyons Thank you for answering. I'll look into that – Timothy Hannon Feb 11 '20 at 3:12
• Are there any difference between the regular $\mathrm{F_s}$ and slanted $F_s$? Also for $V_{in}$ and $\mathrm{V_{in}}$ – jomegaA Feb 11 '20 at 19:04

I hope you do not mind, but I am going to change terminology a little bit. Since you mentioned simulations, suppose you are sampling the Gaussian white noise, at the input to the integrator, at a constant rate of $$f_s$$ samples per second. The point spacing, $$Δt$$, is $$1/f_s$$ seconds. Suppose $$N$$ independent consecutive samples are collected. Then the $$N$$ samples span a period of $$τ_a$$ seconds and this equals $$N Δt$$. The sample mean of the $$N$$ noisy inputs to the integrator obviously provides an estimate of the known $$V_{in}$$ input voltage. The sample variance of the $$N$$ noisy inputs to the integrator provides an estimate of the known $$σ^2$$ variance of the Gaussian white noise. The power SNR at the input is $$(V_{in}/σ)^2$$. The empirical estimate would be the square of the sample mean divided by the sample variance and bigger $$N$$ values are almost surely better.

The Gaussian white noise has zero mean and $$σ^2$$ variance, so the bilateral noise power spectral density, $$η$$, is $$σ^2 Δt$$. The ideal integrator has $$τ_i$$ integration time constant. (For an ideal op amp integrator, with just an input resistor, $$R$$, and feedback capacitor, $$C$$, $$τ_i = RC$$. The sign inversion is trivial and can be ignored.) Now suppose that the integration period, i.e., integration aperture or gate duration, is $$τ_a$$ seconds, exactly as above. Assume the integrator has initial condition zero, integration starts at $$t = 0$$ and ends at $$t = τ_a$$, so $$τ_a$$ is simply the reciprocal of the OP’s $$F_s$$. N.B. My $$f_s$$ is not the OP’s $$F_s$$.

Then the DC power gain of the integrator is $$(τ_a/τ_i)^2$$ and the noise equivalent bandwidth is $$1/2τ_a$$. Now consider two cases.

Case 1: Let $$τ_i = τ_a$$. Then the signal power at the output of the integrator, at $$t = τ_a$$, is simply $$(V_{in})^2$$, because the DC power gain is unity. The noise power is $$2η$$ times unity DC power gain times $$1/2τ_a$$, which equals $$η/τ_a$$. But $$η = σ^2 Δt$$ and $$τ_a = N Δt$$. So the output power SNR is $$(V_{in})^2 N/σ^2$$. This is $$N$$ times larger than the input power SNR.

Case 2: Let $$τ_i ≠ τ_a$$. Then the signal power at the output of the integrator, at $$t = τ_a$$, is $$(V_{in} τ_a/τ_i)^2$$, because the DC power gain is $$(τ_a/τ_i)^2$$. The noise power is $$2η$$ times $$(τ_a/τ_i)^2$$ times $$1/2τ_a$$, which equals $$η τ_a /(τ_i)^2$$. But $$η = σ^2 Δt$$ and $$τ_a = N Δt$$. So the output power SNR is $$(V_{in} τ_a/τ_i)^2/[σ^2 Δt τ_a /(τ_i)^2]$$, which again equals $$(V_{in})^2 N/σ^2$$. This is again $$N$$ times larger than the input power SNR.

In Case 1, the signal power is independent of $$N$$, while the noise power is inversely proportional to $$N$$, via $$τ_a = N Δt$$. So the power SNR is proportional to $$N$$. This is simply averaging independently sampled white noise. In Case 2, signal power is proportional to $$N^2$$, while noise power is proportional to $$N$$. So the power SNR is again proportional to $$N$$.

Relationship to matched filtering

The figure below shows that, if nothing was known about the input waveform for $$t > τ_a$$, then the input could be treated as though it was a rectangular pulse starting at $$t = 0$$ and ending at $$t = τ_a$$. The integration would effectively be gated integration, which would be a simple case of matched filtering: the impulse response of the (aperture synchronized) gated integrator is a rectangular pulse of duration $$τ_a$$ and reversing it changes nothing because the gated integrator's impulse response is symmetric. The figure below shows the calculation of the matched filter power SNR. The integration time constant, $$τ_i$$, has been set equal to unity because it affects both the pulse energy, and the noise, the same way, so it divides out.

I hope this helps!

• Thank you for your detailed explanation sir. Considering the problem in frequency domain does make sense to me now, and simulation agrees with your analysis. – Timothy Hannon Feb 11 '20 at 3:09