Electronic – the filter and the transfer function that shown below

The problem has come from the substitution of s for 2*PI*f.

s is a substitution for j*w

thus the equation actually is: 1/(1+j)
If you resolve this back to magnitude: 1+j = sqrt(1^2 + 1^2) = sqrt(2)

resulting in: 1/sqrt(2) = 0.707 as you expect

So take a low-pass filter: RC

^{simulate this circuit – Schematic created using CircuitLab}

$$ Vout = Vin*\frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} $$ $$ Vout = Vin*\frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{j\omega RC + 1} $$ $$ \frac{Vout}{Vin} = \frac{1}{\sqrt{(\omega RC)^{2} + 1^{2}}} $$ From a magnitude point of view.

Your expression for \$H(s)\$ is correct:

$$H(s)=\frac{RCs}{1+RCs}$$

where \$\tau=RC\$ is the time constant of the high pass filter. The most important features of this transfer function are the location of the pole (\$s_{\infty}=-1/RC\$) and the location of the zero (\$s_0=0\$). Note that the location of the pole and the zero is of course the same for the transfer function in your lecture notes. However, the gain factor in your notes is wrong. The consequence of this is that for \$s\rightarrow\infty\$ (i.e. for very high frequencies) your transfer function converges to unity gain, whereas the one in your notes converges to \$1/RC\$.