Transfer function of non-inverting op-amp circuit

Question

I have this circuit:

I am asked to show that the transfer functions of the circuit is:

I know that for non-inverting op-amps:

\$ Z_{in}=\infty \$

\$ Z_{out}=0 \$

\$ H(s)=\frac{V_{out}(s)}{V_{in}(s)} = \frac{Z_{1}(s)+Z_{2}(s)}{Z_{1}(s)} = \frac{Z_{2}(s)}{Z_{1}(s)} +1 \$

But if I set:

\$ Z_1 = 1.00\mathrm{~k\Omega} \$

\$ Z_2 = 9.00\mathrm{~k\Omega} \$

I just get:

\$ H(s)=\frac{V_{out}(s)}{V_{in}(s)} = \frac{Z_{1}(s)+Z_{2}(s)}{Z_{1}(s)} = \frac{9.00\mathrm{~k\Omega} + 1.00\mathrm{~k\Omega}}{1.00\mathrm{~k\Omega}} = 10.00\mathrm{~k\Omega} \$

which is not the same as:

What am I doing wrong here?

Answer 1

You forgot to include the high-pass filter at the input. Solve for its transfer function, and then multiply it by 10, which is the gain you calculated for the op-amp transfer function.

EDIT: In one of your comments under your post you asked:

won't they be zero in this case (because its an non-inverting op-amp)? and what is then Z1 and Z2 ?

No, they won't. The only case where an impedance at that node doesn't make a difference is if it were a resistor inserted in series with the (+) input. Since it is a capacitor, you can imagine that it has an infinite at DC and becomes a short at high frequencies.

On the other hand, if there were no RC network at the (+) input of the op-amp, then you can assume its input impedance to be large at low-frequencies (~Mohms to Gohms even). However, now there is \$R_1\$ to ground. Therefore, your signal source sees an equivalent impedance of \$R_1\$ to ground approximately (it'll be \$R_1\$ in parallel with the input impedance of the Op-Amp itself), and this has an effect.

If \$R_1\$ were, on the other hand, a resistor in series with the (+) input, it'd be like adding \$R_1\$ plus the input impedance of the Op-Amp, which is already large, so now it'll be a bit larger; nothing significant changes for the input signal.

Answer 2

Well, notice that in an ideal opamp we know that:

$$\text{V}_+=\text{V}_-\tag1$$

Using the voltage divider formula, we see that:

$$\text{V}_+=\frac{\text{R}_1}{\text{R}_1+\frac{1}{\text{sC}}}\cdot\text{V}_\text{i}\tag2$$

And:

$$\text{V}_-=\frac{\text{R}_3}{\text{R}_3+\text{R}_2}\cdot\text{V}_\text{o}\tag3$$

So:

$$\frac{\text{R}_1}{\text{R}_1+\frac{1}{\text{sC}}}\cdot\text{V}_\text{i}=\frac{\text{R}_3}{\text{R}_3+\text{R}_2}\cdot\text{V}_\text{o}\tag4$$

Which gives:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}}{\text{V}_\text{i}}=\displaystyle\frac{\frac{\text{R}_1}{\text{R}_1+\frac{1}{\text{sC}}}}{\frac{\text{R}_3}{\text{R}_3+\text{R}_2}}=\frac{\text{R}_1}{\text{R}_1+\frac{1}{\text{sC}}}\cdot\frac{\text{R}_3+\text{R}_2}{\text{R}_3}=\frac{\text{sCR}_1}{1+\text{sCR}_1}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\tag5$$

So, for the amplitude we get:

\begin{equation} \begin{split} \left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\left|\frac{\text{j}\omega\text{CR}_1}{1+\text{j}\omega\text{CR}_1}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\right|\\ \\ &=\frac{\left|\text{j}\omega\text{CR}_1\right|}{\left|1+\text{j}\omega\text{CR}_1\right|}\cdot\left|1+\frac{\text{R}_2}{\text{R}_3}\right|\\ \\ &=\frac{\omega\text{CR}_1}{\left|1+\text{j}\omega\text{CR}_1\right|}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\\ \\ &=\frac{\omega\text{CR}_1}{\sqrt{1^2+\left(\omega\text{CR}_1\right)^2}}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\\ \\ &=\frac{\omega\text{CR}_1}{\sqrt{1+\left(\omega\text{CR}_1\right)^2}}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right) \end{split}\tag6 \end{equation}

And for the argument:

\begin{equation} \begin{split} \arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(\frac{\text{j}\omega\text{CR}_1}{1+\text{j}\omega\text{CR}_1}\cdot\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\right)\\ \\ &=\arg\left(\frac{\text{j}\omega\text{CR}_1}{1+\text{j}\omega\text{CR}_1}\right)+\arg\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\\ \\ &=\arg\left(\text{j}\omega\text{CR}_1\right)-\arg\left(1+\text{j}\omega\text{CR}_1\right)+\arg\left(1+\frac{\text{R}_2}{\text{R}_3}\right)\\ \\ &=\frac{\pi}{2}-\arg\left(1+\text{j}\omega\text{CR}_1\right)+0\\ \\ &=\frac{\pi}{2}-\arctan\left(\frac{\omega\text{CR}_1}{1}\right)\\ \\ &=\frac{\pi}{2}-\arctan\left(\omega\text{CR}_1\right) \end{split}\tag7 \end{equation}