3 Validation of Transfer Functions
Lorem ipsum
3.1 Mathematical derivation of the transfer functions of the biquad
First, the individual equations of the operational amplifiers were set up:
\[ V_1 = -(V_3 + V_4) \tag{3.1}\]
\[ V_2 = -\left(\frac{1}{s} \omega_0 \cdot V_1 \right) \tag{3.2}\]
\[ V_3 = -\left(\frac{1}{s} \omega_0 \cdot V_2 \right) \tag{3.3}\]
\[ V_4 = -\left( \frac{V_2}{Q} + H_0 \cdot V_i \right) \tag{3.4}\]
3.2 Calculation of \(\frac{V_2}{V_i}\)
Insert Equation 3.1 into Equation 3.2 :
\[\begin{align} V_2 &= -\frac{\omega_0}{s} \cdot V_1 = -\frac{\omega_0}{s} \cdot (-V_3 - V_4) \nonumber \\ &= -\frac{\omega_0}{s} \cdot \left( \frac{\omega_0}{s} \cdot V_2 + \frac{1}{Q} \cdot V_2 + H_0 \cdot V_i \right) \label{eq:v2_expanded} \end{align}\]
Summarizing results in:
\[\begin{align} V_2 &= -\frac{\omega_0^2}{s^2} \cdot V_2 - \frac{\omega_0}{s Q} \cdot V_2 - \frac{\omega_0}{s} H_0 \cdot V_i \\ \Rightarrow \frac{\omega_0}{s} H_0 \cdot V_i &= -\left( \frac{\omega_0^2}{s^2} + \frac{\omega_0}{s Q} + 1 \right) V_2 \end{align}\]
Resolve to \(\frac{V_2}{V_i}\):
\[ \frac{V_2}{V_i} = \frac{\frac{\omega_0}{s} H_0}{ -\left( \frac{\omega_0^2}{s^2} + \frac{\omega_0}{s Q} + 1 \right)} = -\frac{ \frac{s}{\omega_0} H_0 }{ 1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \tag{3.5}\]
3.3 Transfer-function \(\frac{V_3}{V_i}\):
Insert Equation 3.5 into Equation 3.3 :
\[ V_3 = -\frac{\omega_0}{s} \cdot V_2 = -\frac{\omega_0}{s} \cdot \left( -\frac{\frac{s}{\omega_0} H_0 }{ 1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \cdot V_i \right) \]
\[ V_3 = \left( \frac{H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \right) V_i \]
\[ \Rightarrow \frac{V_3}{V_i} = \frac{H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \tag{3.6}\]
3.4 Transfer-function \(\frac{V_4}{V_i}\):
Insert Equation 3.5 into Equation 3.4 :
\[\begin{align} V_4 &= -\left( \frac{1}{Q} \cdot V_2 + H_0 \cdot V_i \right) = -\left( \frac{1}{Q} \cdot \left( -\frac{ \frac{s}{\omega_0} H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \cdot V_i \right) + H_0 \cdot V_i \right) \nonumber \\ &= \left( \frac{\frac{s}{\omega_0 Q} H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} - H_0 \right) V_i \\ &= \left( \frac{ \frac{s}{\omega_0 Q} - \left(1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} \right) }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \cdot H_0 \right) V_i \\ &= - \frac{ \left( 1 + \frac{s^2}{\omega_0^2} \right) H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \cdot V_i \\ \Rightarrow \frac{V_4}{V_i} &= - \frac{ \left( 1 + \frac{s^2}{\omega_0^2} \right) H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \end{align}\]
Note: In the task booklet, the following equation is given here:
\[ \frac{V_4}{V_i} = \frac{ \left( 1 + \frac{s^2}{\omega_0^2} \right) H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} } \]
This corresponds to a negation of my own result.
3.5 Calculation of the transfer function \(\frac{V_1}{V_i}\):
Based on equation Equation 3.1 : \[V_1 = -(V_3 + V_4)\]
Using the transfer-function Equation 3.6 for \(V_3\):
\[ \frac{V_3}{V_i} = \frac{H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \]
Case 1: Use of the negated form of the transfer function of \(V_4\) from my own derivation:
\[ \frac{V_4}{V_i} = - \frac{\left(1 + \frac{s^2}{\omega_0^2} \right) \cdot H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \tag{3.7}\]
Inserting Equation 3.6 and Equation 3.7 into Equation 3.1 results in:
\[ V_1 = -\left( V_3 + V_4 \right) \]
\[ V_1 = -\left( \frac{H_0}{D} + \left( -\frac{ \left( 1 + \frac{s^2}{\omega_0^2} \right) H_0 }{D} \right) \right) V_i \]
\[ V_1 = \left( \frac{ \left( \frac{s^2}{\omega_0^2} \right) H_0 }{D} \right) V_i \quad \text{where } D = 1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2} \]
\[ \Rightarrow \frac{V_1}{V_i} = \frac{ \left( \frac{s^2}{\omega_0^2} \right) H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \tag{3.8}\]
Case 2: Use the positive form of the transfer function of \(V_4\) as specified in the task booklet:
\[ \frac{V_4}{V_i} = \frac{\left(1 + \frac{s^2}{\omega_0^2} \right) \cdot H_0}{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \tag{3.9}\]
Insert into Equation 3.1 :
\[\begin{align} V_1 &= -\left( V_3 + V_4 \right) = -\left( \frac{H_0}{D} + \frac{ \left( 1 + \frac{s^2}{\omega_0^2} \right) H_0 }{D} \right) V_i \nonumber \\ &= -\left( \frac{ \left( 2 + \frac{s^2}{\omega_0^2} \right) H_0 }{D} \right) V_i \\ \Rightarrow \frac{V_1}{V_i} &= - \frac{ \left( 2 + \frac{s^2}{\omega_0^2} \right) H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \label{eq:tf_v1_falsch} \end{align}\]
3.6 Repeated calculation of the Transfer-function \(\frac{V_2}{V_i}\) via \(V_1\):
Based on the correct function for \(V_1\) Equation 3.8 :
\[ \frac{V_1}{V_i} = \frac{ \frac{s^2}{\omega_0^2} \cdot H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \]
Using the known relationship \(V_2 = -\frac{\omega_0}{s} \cdot V_1\):
\[ V_2 = -\frac{\omega_0}{s} \cdot V_1 \]
\[ V_2 = -\frac{\omega_0}{s} \cdot \frac{ \frac{s^2}{\omega_0^2} \cdot H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \cdot V_i \]
\[ V_2 = -\frac{ \frac{s}{\omega_0} \cdot H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \cdot V_i \]
\[ \Rightarrow \frac{V_2}{V_i} = - \frac{\frac{s}{\omega_0} H_0 }{1 + \frac{s}{\omega_0 Q} + \frac{s^2}{\omega_0^2}} \tag{3.10}\]
A comparison of the two transfer functions \(\frac{V_2}{V_i}\) Equation 3.5 and Equation 3.10 shows that the error is due to equation \(\frac{V_4}{V_1}\) since the derivation of \(\frac{V_1}{V_i}\) remains correct.
3.7 Conclusion
Only the use of the negated form of \(\frac{V_4}{V_i}\) from Equation 3.7 leads to a consistent result for \(\frac{V_1}{V_i}\) according to Equation 3.8 . The version from the exercise booklet (Equation 3.9) therefore appears to be incorrect.