Exercise 1.1: Simple Filter Functions

Considered two-port networks

A filter with the frequency response

$$H_{\rm LP}(f) = \frac{1}{1+ {\rm j}\cdot f/f_0}$$

is described as a low-pass filter of first order.  Out of it, a first order high-pass filter can be designed according to the following rule:

$$H_{\rm HP}(f) = 1- H_{\rm LP}(f) .$$

In both cases  $f_0$  represents the so-called  $\text{3 dB}$–cutoff frequency.

The figure shows two two-port networks  $\rm A$  and  $\rm B$.  The task is to clarify which of the two networks has a low-pass characteristic and which has a high-pass characteristic.

The components of circuit  $\rm A$  are given as follows:

$$R = 50 \,\, {\rm \Omega}, \hspace{0.2cm} C = 637 \,\, {\rm nF} .$$

The inductivity  $L$  of circuit  $\rm B$  is to be computed in subtask  (6) .

• The task belongs to the chapter  System Description in Frequency Domain.
• For the subtask  (4)  cosine-shaped input signals are assumed. The frequency  $f_x$  is variable, the power is  $P_x = 10\,{\rm mW}.$

Questions

1

Compute the frequency response  $H_{\rm A}(f)$  of the two-port network  $\rm A$  and check the following statements.

 Two-port network  $\rm A$  is a low-pass filter. Two-port network  $\rm A$  is a high-pass filter.

2

Compute the reference frequency  $f_0$  using the components  $R$  and  $C$.

 $f_0 \ = \$  $\text{kHz}$

3

Compute the amplitude response  $|H_{\rm A}(f)|$. What are the numerical values for  $f = f_0$  and  $f = 2f_0$?

 $|H_{\rm A}(f = f_0)|\ = \$ $|H_{\rm A}(f = 2f_0)|\ = \$

4

What is the power  $P_y$  of the output signal  $y(t)$ if a cosine signal of frequency  $f_x = 5\,{\rm kHz}$  or  $f_x = 10\,{\rm kHz}$  is applied to the input?

 $P_y(f_x = 5 \ \rm kHz)\ = \$  $\text{mW}$ $P_y(f_x = 10 \ \rm kHz)\ = \$  $\text{mW}$

5

Compute the amplitude response  $|H_{\rm B}(f)|$  of the two-port network  $\rm B$  with the components  $R$  and  $L$  using the reference frequency  $f_0 = R/(2πL)$.
What are the values for  $f = 0$,  $f = f_0$, $f = 2f_0$  and for  $f → ∞$?

 $|H_{\rm B}(f = 0)|\ = \$ $|H_{\rm B}(f = f_0)|\ = \$ $|H_{\rm B}(f = 2f_0)|\ = \$ $|H_{\rm B}(f → ∞)|\ = \$

6

What inductivity  $L$  results in the reference frequency  $f_0 = 5 \,\text{kHz}$?

 $L\ = \$  $\text{mH}$

Sample solution

Solution

(1)  Approach 1 is correct:

• The complex resistance of the capacitance  $C$  is equal to  $1/({\rm j}ωC)$, where  $ω = 2πf$  represents the so-called angular frequency.
• The frequency response can be calculated according to the voltage divider principle:
$$H_{\rm A}(f) = \frac{Y_{\rm A}(f)}{X_{\rm A}(f)} = \frac{1/({\rm j}\omega C)}{R+1/({\rm j}\omega C)}=\frac{1}{1+{\rm j \cdot 2\pi}\cdot f \cdot R\cdot C}.$$
• Because of  $H_{\rm A}(f = 0) = 1$  this cannot be a high-pass filter; rather it is a low-pass filter.
• At low frequencies, the reactance of the capacitance is very large and  $y_{\rm A}(t) ≈ x_{\rm A}(t)$ applies.
• In contrast to this, at very high frequencies, the capacitor acts like a short circuit and   $y_{\rm A}(t) ≈ 0$  holds.

(2)  By comparing the coefficients between  $H_{\rm LP}(f)$  given in the statement of the task and  $H_{\rm A}(f)$  according to subtask  (1)  the following is obtained:

$$f_0 = \frac{1}{2\pi \cdot R \cdot C} = \frac{1}{2\pi \cdot{\rm 50\hspace{0.05cm} \Omega}\cdot {\rm 637 \cdot 10^{-9}\hspace{0.05cm} s/\Omega}}\hspace{0.15cm}\underline{\approx 5 \, {\rm kHz}}.$$

(3)  The amplitude response is:

$$|H_{\rm A}(f)| = \frac{1}{\sqrt{1+ (f/f_0)^2}}.$$
• For  $f = f_0$  the numerical value  $1/\sqrt{2}\hspace{0.1cm} \underline{≈ 0.707}$  is obtained, and
• for  $f = 2f_0$  approximately the value  $1/\sqrt{5}\hspace{0.1cm} \underline{≈ 0.447}$.

(4)  The output power can be calculated using the following equation:

$$P_y = P_x \cdot |H_{\rm A}(f = f_x)|^2.$$
• For  $f_x = f_0$  ⇒   $P_y = P_x/2 \hspace{0.1cm} \underline{ = 5\hspace{0.1cm} {\rm mW}}$, so this results in half of the power at the output.
• In logarithmic representation, this relationship is:
$$10 \cdot {\rm lg}\hspace{0.2cm} \frac{P_x(f_0)}{P_y(f_0)} = 3\,{\rm dB}.$$
Accordingly, for  $f_0$  the term  "3dB cut-off frequency"  is also common.
• For  $f_x = 2f_0$,  on the other hand, a smaller value is obtained:  $P_y = P_x/5 \hspace{0.1cm}\underline{= 2\hspace{0.1cm} {\rm mW}}$.

(5)  Analogous to subtask  (1)  the following holds:

$$H_{\rm B}(f) = \frac{Y_{\rm B}(f)}{X_{\rm B}(f)} = \frac{{\rm j}\omega L}{R+{\rm j}\omega L}=\frac{{\rm j2\pi}\cdot f \cdot L/R}{1+{\rm j2\pi}\cdot f \cdot L/R}.$$
• Using the reference frequency  $f_0 = R/(2πL)$  this can also be written as in the following:
$$H_{\rm B}(f) = \frac{{\rm j}\cdot f/f_0}{1+{\rm j}\cdot f/f_0}\hspace{0.5cm}\Rightarrow \hspace{0.5cm}|H_{\rm B}(f)| = \frac{|f/f_0|}{\sqrt{1+ (f/f_0)^2}}.$$
• Hence, these numerical values are obtained:
$$|H_{\rm B}(f = 0)| \hspace{0.15cm}\underline{= 0}, \hspace{0.5cm} |H_{\rm B}( f_0)| \hspace{0.15cm}\underline{=0.707}, \hspace{0.5cm}|H_{\rm B}(2f_0)| \hspace{0.15cm}\underline{= 0.894}, \hspace{0.5cm}|H_{\rm B}(f \rightarrow \infty)|\hspace{0.15cm}\underline{ = 1}.$$
• Therefore, the circuit  $\rm B$  is a high-pass filter.

(6)  According to the above definition of the reference frequency it follows that:

$$L = \frac{R}{2\pi \cdot f_0} = \frac{{\rm 50\hspace{0.05cm} \Omega}}{2\pi \cdot{\rm 5000 \hspace{0.05cm} Hz}}= {\rm 1.59 \cdot 10^{-3}\hspace{0.05cm} \Omega s}\hspace{0.15cm}\underline{= {\rm 1.59 \hspace{0.1cm} mH}} .$$