Commit c67dab1a authored by Gary Steele's avatar Gary Steele
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added some figures, but i'm gonna move the solutions out of this repo anyway

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# Exercise Solutions for Lecture 1
## 1a
First, we need to establish the upper and lower frequency limits of the noise spectrum we need for our measurement. The lower frequency limit is based on the measurement time of 1 hour:
$$
\begin{align}
f_1 = \frac{1}{1 \text{hour}} = \frac{1}{3600 \text{s}} = 0.277 \text{mHz}
\end{align}
$$
The upper frequency limit is set by how long we average each point (the sampling rate):
$$
\begin{align}
f_2 = \frac{100}{1 \text{hour}} = \frac{100}{3600 \text{s}} = 27.7 \text{mHz}
\end{align}
$$
Now we can compute the integral
$$
\begin{align}
\sigma_v^2 &= \int_{f_1}^{f_2} \frac{A^2}{f} df \\
&= A^2 \log( f_2 / f_1 )\\
&= A^2 \log(100)\\
\sigma_v &= 2.14 \text{mV}
\end{align}
$$
![](figures/1_f.PNG)
An interesting feature of pure 1/f noise for this type of "DC" measurement is that the signal to noise ratio is independent of the measurement time, and depends only on the number of data points; measuring more slowly doesn't help. Measuring for a longer time scales both $f_1$ and $f_2$ to lower frequencies. This decreases the bandwidth, and for white noise, would indeed decrease $\sigma_v$. However, with 1/f noise, the averaging window moves to lower frequencies where $S_{vv}$ is larger. For 1/f noise, these two effects exactly cancel (the blue and green bordered areas in the above image have the same area, for the same number of data points).
A bit more information on understanding the frequency bounds:
There is a total measurement time, $T_m$, and a single data point averaging time $T_{avg}$, where if we gather $n$ data points, then $T_m = n T_{avg}$. The lower limit of the integral says what is the lowest frequency we should care about–if some component of the signal/noise “changes” once every $T_m$ seconds, then it is just barely resolvable in our data. If the total measurement time is very long, then $1/T_m$ is very small and we can probably set the bound to 0 – but we are assuming that the extra contribution is insignificant compared to the rest of the integral, which depends on the shape of the spectrum and the upper bound itself.
On the other hand, what is the fastest frequency we should care about? It would be component of the signal that “changes” as quickly as possible in our data, which means it has to change in every sample we take, so its frequency goes like $1/T_{avg}$, so this is the upper bound. It is true that the underlying signal has higher frequencies, but we cannot reliably sample those because of an effect called aliasing – this is why, for example, we low-pass filtered the output of the amplifier with a cutoff frequency of our sampling rate (in reality, the Nyquist limit says that if we want to accurately sample a component of frequency $f$, we need to sample at a rate $2f$, though we are being rough with our estimates so this factor of 2 has been dropped.
## 1b
For the "lock-in" measurement, the situation is different. The filter of the lock-in reduces the bandwidth of your measurement by changing the width $\Delta f$ of its filter while keeping the center frequency $f_0$ fixed. Since we will probably be averaging for a while, we can assume $\Delta f \ll f_0$ (e.g 100 seconds total measuring time needs $\Delta f = 1$ Hz $ \ll 50$ KHz). In this case, we can approximate
$$
\begin{align}
\sigma_v^2 = \int_{f_1}^{f_2} S_{vv} (f) df \approx S_{vv} (f_0) \Delta f
\end{align}
$$
with $\Delta f = \frac{100}{T_m}$, we get
$$
\begin{align}
\sigma_v^2 &= \frac{100 S_{vv}(f_0)}{T_m} \rightarrow T_m = \frac{100A^2}{f_0 \sigma_v^2}\\
T_m &= \frac{100 (1 \text{mV})^2}{5 \cdot 10^4 \text{Hz} \big( \frac{2.14\text{mV}}{100} \big)^2} = 20 \cdot (1/2.14)^2 \text{s}\\
&= 4 \text{s}
\end{align}
$$
Using a lock-in amplifier, we can achieve 100 times better noise with 900 times less averaging time (compare 40ms to 36s per data point). Furthermore, the noise level $\sigma_v$ in the case of "DC" measurements increases as we add more points to our measurement, while for the lock-in amplifier, $\sigma_v$ is *independent* of the number of data points we take!
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