Measuring a faint ripple

Question

I am working on an electronics problem where we measure some faint ripple over a huge constant. You can think of it as $100 \,\rm{V}$ DC plus a ripple of $\pm$ a couple $\rm{mV}$ over that $100\,\rm{V}$. Our signal-to-noise (SNR) calculation is $-130 \,\rm{dB}$. Is it possible to identify this faint ripple out of this huge but constant noise?

Answer 1

You can use a high pass filter for this purpose. A passive high pass filter is essentially a capacitor in series with a resistor. The input voltage is applied across the high pass filter and the output is obtained across the resistor. If the resistor has a resistance of R Ohms and the capacitor has a capacitance of C Farads, then the output as a function of the input is: $$\begin{aligned} O &= \frac{R}{R +1/(j\omega C)} * I \end{aligned}$$ where $\omega$ is the frequency of the input signal. This can be rephrased as: $$\begin{aligned} O &= \frac{j\omega CR}{j\omega CR +1} * I \end{aligned}$$ Now, the input signal may be thought of as: $$ I = I_1 + I_2$$ where $I_1$ is the DC 100V signal with frequency $\omega_1 = 0$ and $I_2$ is the ripple(a high frequency signal) with frequency $\omega_2 > 0$. Therefore the output is $$\begin{aligned} O &= \frac{j\omega CR}{j\omega CR +1} * (I_1 + I_2) \end{aligned}$$ which is: $$\begin{aligned} O &= \frac{j\omega_1 CR}{j\omega_1 CR +1} * I_1 + \frac{j\omega_2 CR}{j\omega_2 CR +1} * I_2 \end{aligned}$$ As $\omega_1 = 0$, the first term reduces to zero: $$\begin{aligned} O &= \frac{j\omega_2 CR}{j\omega_2 CR +1} * I_2 \end{aligned}$$ Thus, only input signals of a sufficiently high frequency will be passed by this filter. Input signals such as the 100V DC signal with frequency $\omega = 0$ will not be passed by the filter and the high frequency signal or the ripple can be obtained as the output of this filter (although it will be reduced in amplitude).