Electrical – Did I correctly use the Voltage divider rule to derive the transfer function

(1) It MAY be of assistance to you to note that equation 4.29 in the above cited text COULD be called "the voltage divider rule" as it refers to R1, R2 and Vcc. ie changing what it says just slightly without changing the meaning:

• Vout = Vin x R2 / (R1 + R2)

ie R1 & R2 form a voltage divider and the above equation defines a "rule" of the result.

BUT

(2) There IS NO "voltage divider rule" as such.

Even if somebody uses that term there is still no such rule.
BECAUSE the terminology is much too general.
That's even more general than saying eg "The Ohm's law rule"
where you at least have some guide.

If you have a specific question you should explain it clearly in words and not use general terms or few words or the real requirement is liable to be missed.

Added:

Re question:

• Can you tell me how this 'rule' is derived? I'm a beginner so I didn't understand how he got the relation Vr2= (R2)(Vcc)/(R1+R2)

(1) Short answer.
Voltage across each resistor is proportional to current in it (Ohm's law).

As current in both resistors
= battery current
= the same
THEN the voltages across each resistor are proportional to their resistance value.
THIS IS THE KEY FACTOR THAT MAKES THIS WORK

Vout = Vr2 = ib x R2
Vcc = Vr1+Vr2 = ib x R1 + ib x R2 = ib x (R1 + R2)

So Vout / Vcc
= Vr2 / (Vr1 + VR2)
= ib x R2 / (ib x (R1 + R2) ) Cancel ib's Vout/Vcc= R2/(R1 + R2)
Multiply both sides by Vcc.
Vout = Vcc x R2 / (R1 + R2)
QED.

(2) Longer answer.

You MUST know Ohms law. If you don't know Ohms law and it's various re arrangements, stop reading this now, drop all lse and learn it. Wikipedia and Google know all about it N time over

... time lapse ... or no time at all as the case may be ...
So we know you know Ohm's law.

So - one version of Ohm's law says, as you know

• V = i x R

ie the voltage drop across a resistor is equal to the value of the resistor multiplied by the current flowing in it.

Now look at fig 4-29

Take this circuit in isolation.
The current from the battery flows from B+ at the top left of R1, via R1, then via R2 and back to B- and the bottom left. Look at the diagram and be SURE that you agree with the above.
Now, lets call the battery current Ib.

Call the current in R1 I_R1. It can be seen "by inspection that I_R1 = Ib.

Call the current in R2 I_R2. It can be seen "by inspection that I_R2 = Ib.

So I_R1 = IR2 = Ib.
ie the current is the same in each resistor and out of and into the battery.

Now, the voltage across R1 = VR1 is, based on Ohm's law = I_R1 x R1.
And, the voltage across R2 = VR2 is, based on Ohm's law = I_R2 x R2.

BUT I_R1 = Ib and IR2 = Ib.

So VR1 = I_R1 x R1 = Ib x R1
And VR2 = I_R2 x R2 = Ib x R2

The ratio of VR2 / VR1 = Ib x R2 / Ib x R1 = R2/R1
ie the voltages across the two resistors are proportional to their resistance values.

Look at the diagram.
Vbattery = Vcc
Vcc = the voltage across R1 + the Voltage across R2
Vcc = VR1 + VR2
Vcc = ib x R1 + ib x R2
Vcc = ib (R1 + R2)

So
To determine the ratio Vout / Vcc:

Vout / Vcc = V_R2 / Vcc
= ib x R2 / ib (R1 + R2)
but the ib's cancel so
Vout/ Vcc = R2 / (R1 + R2)
and rearranging

Vout = Vcc x R2 / (R1 + R2)

So the voltage across R2 compared to battery voltage = Vr2= (R2)(Vcc)/(R1+R2)

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\$.