Commit d22cac29 authored by ignacio's avatar ignacio
Browse files

Added md solutions of homework 2 and 3

parent 0c7c7a8e
# Exercise Solutions for Lecture 1
## 1a)
First we derive the expression for the reflectance R(x). The general form of a sinusoidal is
$$
R(x) = A + B \cos(kx)
$$
In this case we know the period is $\frac1\lambda$ so $k=\frac{4\pi}{\lambda}$.
Applying the conditions $R_{max}=0.5$, $R_{min}=0$ we get $A=B=\frac14$, so
$$
R(x) = \frac14 (1+\cos(\frac{4\pi}{\lambda}x))
$$
Therefore
$$
\left|\frac{dR}{dx}\right| = \frac\pi\lambda \sin(\frac{4\pi}{\lambda}x)
$$
The maximum slope is $\left|\frac{dR}{dx}\right|_{max}=\frac\pi\lambda$, when the sine is equal to 1, or equivalently when
$$
\frac{4\pi}{\lambda}x_n = n\frac\pi2 \ , \ \forall n\in \mathrm{Z} \\
x_n = n \frac\lambda8
$$
At $x_1$ the reflectance is $R_0 = R(x_1) = \frac14$.
## 1b)
The responsivity $R_{xp}$ is defined as
$$
R_{xp} = \frac{dP}{dx}
$$
We can write the output power $P$ as
$$
P = P_{in} R(x)
$$
Using the results from section 1a) we can approximate the reflectance $R(x)$ by doing a a Taylor expansion around the point of maximum slope $x_1$
$$
R(x) \approx R(x_1) + \frac{dR}{dx}(x_1) \ x = \frac14 + \frac\pi\lambda x
$$
We can then write
$$
P \approx P_{in} \left( \frac14 + \frac\pi\lambda x \right)
$$
so
$$
R_{xp} =\frac{dP}{dx}= \frac{\pi P_{in}}{\lambda}
$$
## 1c)
In the lecture we saw that the Power Spectral Density of the reflected laser beam is given by
$$
S_{pp} = P \hbar \omega_0
$$
From b) , we know that at the point of maximum slope $P=P_{in}R_0$ so
$$
S_{pp} =P_{in}R_0 \hbar \omega_0
$$
Then we have that
$$
S_{xx} = \frac{S_{pp}}{R_{xp}^2} = \frac{P_{in}R_0 \hbar \omega_0 }{P_{in}^2 \frac{\pi^2}{\lambda^2}}
$$
using $\omega_0 = \frac{2\pi c}{\lambda}$
$$
S_{xx} = 2\frac{R_0 \hbar c\lambda}{P_{in}\pi} \approx 1,8 \cdot 10^{-29} \text{$m^2$/Hz}
$$
so the sensitivity is $\sqrt{S_{xx}}= 4,24\cdot 10^{-14} \ m/\sqrt{Hz}$
Is this reasonable? An Atomic Force Microscope, which is not shot noise limited, can have sensitivities of 2 $\text{fm}/\sqrt{\text{Hz}}$. The Laser Interferometer Gravitational-Wave Observatory (LIGO), which is shot noise limited, has a sensitivity of $10^{-19}\text{m}/\sqrt{\text{Hz}}$. Ligo uses a 2 Watt laser but uses tricks in designing their interferometer differently to get better sensitivity.
## 1d)
Measuring for a time $T_m$ would give us an error
$$
\sigma_x = \sqrt{\frac{S_{xx}}{T_m}}
$$
If we want to detect a displacement equal to the diameter of a proton ($D_p \approx 10^{-15}m$), the error has to at least equal to that displacement. We would therefore have to measure for a time
$$
T_m = \frac{S_{xx}}{\sigma_x^2} = 50,7 \text{ ms}
$$
# Homework 3
## Exercise 1
### a)
$$
S_{xx}^{th}(\omega) = \frac{k_B T \gamma }{m\omega_0^2} \frac{1}{(\omega-\omega_0)^2 + (\gamma/2)^2}
$$
Peak value at $\omega = \omega_0$
$$
S_{xx}^{th}(\omega_0) = \frac{4k_B T }{m\omega_0^2\gamma} = \frac{4k_B TQ }{m\omega_0^3} = 3,38 \cdot 10^{-26} \ m^2/Hz
$$
where we used $Q = \frac{\omega_0}{\gamma}$. The sensitivity is then
$$
\sqrt{S_{xx}^{th}(\omega_0)} = 183 \ fm/\sqrt{Hz}
$$
![](figures/2a_thermal_sensitivity.png)
#
### b)
If the signal to noise ratio of our spectral noise peak $S_{xx}^{th}$ is $3,38\cdot 10^{-26} \ m^2/Hz$ then our detector noise level should be $3,38\cdot 10^{-28} \ m^2/Hz$ (Note: strange concept: the signal to noise ratio of detecting noise...)
This would require a sensitivity
$$
\sqrt{S_{xx}^{det}} = \frac{S^{th}_{xx}}{10} = 18 \ \text{fm}
$$
As we saw om the lecture 2 homework, this should not be too hard for a laser interferometer, for examples.
#
## Exercise 2
### a)
![](figures/2a_circuit.png)
In parallel, impedances add in inverse
$$
\frac1Z = \frac1{Z_R} + \frac1{Z_L} + \frac1{Z_C} = \frac1R + i\omega C + \frac1{i\omega L}
$$
Working out the above expression a bit we get
$$
\begin{aligned}
Z &= \frac1{ \frac1R + i\left( \omega C - \frac1{\omega L}\right)} \frac{\frac1R - i(\omega L - \frac1{\omega C})}{\frac1R - i(\omega L - \frac1{\omega C})} \\
&= \frac{\frac1R - i(\omega L - \frac1{\omega C})}{\frac1{R^2} + (\omega C- \frac1{\omega L})^2}
\end{aligned}
$$
$$
\text{Re}(Z) = \frac{\frac1R}{\frac1{R^2} + (\omega C- \frac1{\omega L})^2}
$$
Using $\Gamma = \frac1{RC}$ and $\omega_0^2=\frac1{LC} $:
$$
\begin{aligned}
\text{Re}(Z) &= \frac{\frac1R}{\frac1{R^2} + \frac{C^2}{\omega^2}(\omega ^2- \frac1{C L})^2} \\
&=\frac{\omega^2}{RC^2} \frac{\frac1R}{\frac{\omega^2}{R^2C^2} + (\omega ^2- \frac1{C L})^2} \\
&= R\frac{\omega^2}{R^2C^2} \frac1{\omega^2\Gamma^2 + (\omega ^2- \omega_0^2)} \\
&= R \frac{\omega^2 \Gamma^2}{\omega^2 \Gamma^2+ (\omega ^2- \omega_0^2)}
\end{aligned}
$$
(ok, a big pain but now in a reasonable form...). The PSD of the cavity thermal noise is then
$$
S_{vv}(\omega) = 4 k_B T R \ \frac{ \omega^2 \Gamma^2}{\omega^2 \Gamma^2+ (\omega ^2- \omega_0^2)}
$$
### 2b)
At resonance:
$$
S_{vv}(\omega_0) = 4 k_B T R
$$
Note: this is the PSD that would be produced by the (equivalent) resistor in parallel with my LC circuit that I need to produce the quoted quality factor.
In fact , a nice way to interpret the noise out of the cavity is that it corresponds to the Johnson noise of the resistor filtered by the LC cavity. Another way of seeing it is the following: at $\omega_0$, $\frac1{Z_C}+\frac1{Z_L} = 0$ and so $Z(\omega_0)=R$.
How big is R?
$$
R = \frac{Q}{\omega_0 C} = \frac{Q}{C}\sqrt{LC} = Q \sqrt{\frac{L}{C}} = 3.16 \ k\Omega
$$
This gives:
$$
S_{vv} = 4 \cdot (1.38 \cdot 10^{-23} \ \frac{J}{K}) \cdot (300 \ K) \cdot (3.16 \cdot 10^{3} \Omega) = 5.23 \cdot 10^{-17} \ V^2/Hz
$$
(recall that $1\ \Omega = 1 \ V^2 \cdot J ^{-1} \cdot s^{-1}$). Then
$$
\sqrt{S_{vv}} = 7.23 \ nV/ \sqrt{Hz}
$$
This should be easily measurable by my amplifier with a sensitivity of $84 \ pV/\sqrt{Hz}$.
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