This is just answering one part of your question:

I don't quite understand the last part. How does he calculate X and ϕ

This is just applying the trigonometric identity
$$\sin\left(\alpha + \beta\right)=\sin\alpha\cos\beta + \cos\alpha\sin\beta$$
with \$\alpha=\omega{}t\$ and \$\beta=\phi\$.

Using this identity on the r.h.s. gives
$$X\sin\left(\omega{}t+\phi\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$
so our equation becomes
$$\frac{A}{\omega}\sin\left(\omega{}t\right)+B\cos\left(\omega{}t\right)=X\left(\sin{}\omega{}t\cos\phi+\cos\omega{}t\sin\phi\right)$$

Which we can break into two parts,
$$\frac{A}{\omega}\sin\left(\omega{}t\right)=X\cos\phi\sin{}\omega{}t$$
and
$$B\cos\left(\omega{}t\right)=X\sin\phi\cos\omega{}t$$

So,
$$\frac{A}{\omega} = X\cos\phi$$
and
$$B=X\sin\phi$$

From there you should be able to get the conclusions from your source.

The thing your confused about is what your integrating. Keep reading the book there are two things that your missing out on:

1) What your integrating:

A key thing to remember is that noise does not come in an amplitude of volts or amps, it comes in a noise power. For a resistor ( a white noise source) this noise is:

$$v_{n}^2 = \int_{0}^{\infty}S_v(f)df = 4kTR$$ the units are \$ \frac{V}{\sqrt{Hz}}\$ or \$ \frac{V^2}{Hz}\$

2) How to integrate

With noise its mostly better to think of it graphically. So this is a bandpass filter with a gain of 100. The bandpass filter is going to have a high pass, then a pass band and a low pass.

At the end of the day, if you were to have an actual noise input on Vin (like a white noise generator (a generator that has an equal amplitude of Gaussian noise on every frequency -- *on average*) or a resistor (thermal noise is the same thing as white noise), you would see band limited noise on a scope (with an FFT), the noise would start near zero, then go to 100x the amplitude as it approaches the first time constant of the band pass filter. Then stay at 100x the amplitude until the second time constant and then roll of back down near zero after the second time constant.

We can actually represent these as areas and do the integration *without integrals* and you can split them up.
I will not show you all the math and deprive you of valuable learning.

Here is what the integration is for the *passband section*:

My \$ |H(f)|^2 = 100*Vin\$ (and Vin should be in units of \$ \frac{V}{\sqrt{Hz}}\$) and the area of integration is from \$\tau_1\$ to \$ \tau_2\$ but we want this in frequency so we use the age old forumla \$ f_c = \frac{1}{2\pi\tau}\$. Another gotcha is you haven't specified what noise source you are actually trying to integrate, I'm assuming Vin. Ether way this will give you the tools to find any noise source.

If we know that the amplitude is constant H(f)^2 becomes 100*Vin (or whatever your noise source is)

$$\int_{f_{c1}}^{f_{c2}}|H(f)|^2\;df = \int_{f_{c1}}^{f_{c2}}\;df* 100*Vin = \Delta f*100*Vin $$

This means that you can solve these geometrically (remember your in log land with a 20dB rolloff) and the other two are triangles.

However if you want to double check with an integral \$ |H(f)|^2 = \frac{100*Vin}{1+RCs}\$ where \$s= j\omega\$ and (actually I'll finish this tommrow)
$$\int_{0}^{f_{c1}}|H(f)|^2\;df$$

## Best Answer

I show you how to obtain the 3dB cut-off frequency for the low pass filter \$G_1(s)\$. You can calculate the cut-off frequencies of the band pass filter \$G_2(s)\$ in a similar way, as long as you know that its maximum magnitude is attained at \$\omega=\sqrt{b}\$, as pointed out in a comment by robert bristow-johnson. The latter fact can be derived by setting the derivative of \$|G_2(j\omega)|^2\$ to zero. (Note that \$b>0\$ is always satisfied for \$G_1(s)\$ and \$G_2(s)\$ to be transfer functions of causal and stable filters.)

To compute the 3dB cut-off frequency of \$G_1(s)\$ you have to solve

$$|G_1(j\omega)|^2=\frac{|G(0)|^2}{2}=\frac{1}{2b^2}\tag{1}$$

With

$$G_1(j\omega)=\frac{1}{-\omega^2+ja\omega+b}\tag{2}$$

you get

$$|G_1(j\omega)|^2=\frac{1}{(b-\omega^2)^2+a^2\omega^2}\tag{3}$$

Plugging (3) into (1) gives

$$(b-\omega^2)^2+a^2\omega^2=2b^2\tag{4}$$

With the substitution \$x=\omega^2\$, you get the following quadratic equation

$$x^2+(a^2-2b)x-b^2=0\tag{5}$$

with the positive solution

$$x_0=b-\frac{a^2}{2}+\sqrt{\left(b-\frac{a^2}{2}\right)^2+b^2}\tag{6}$$

From (6), the value of the 3dB cut-off frequency is

$$\omega_c=\sqrt{x_0}=\sqrt{b-\frac{a^2}{2}+\sqrt{\left(b-\frac{a^2}{2}\right)^2+b^2}}\tag{7}$$