Aufgaben:Exercise 1.2Z: Measurement of the Frequency Response

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Measured signal amplitudes
and phases for filter  $\rm B$

For the metrological determination of the filter frequency response, a sinusoidal input signal with an amplitude of  $2 \hspace{0.05cm} \text{V}$  and given frequency  $f_0$  is applied. The output signal  $y(t)$  or its spectrum  $Y(f)$  are then determined according to magnitude and phase.

• The magnitude spectrum at the output of filter  $\rm A$  with frequency  $f_0 = 1 \ \text{kHz}$ is:

$$|Y_{\rm A} (f)| = 1.6\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm f_0) + 0.4\hspace{0.05cm}{\rm V} \cdot {\rm \delta } (f \pm 3 f_0) .$$

• For another filter  $\rm B$  on the other hand, is always a harmonic oscillation with the (single) frequency  $f_0$. For the frequencies  $f_0$  given in the table the amplitudes  $A_y(f_0)$  and the phases  $φ_y(f_0)$  are measured. Here, the following holds:

$$Y_{\rm B} (f) = {A_y}/{2} \cdot {\rm e}^{ {\rm j} \varphi_y} \cdot {\rm \delta } (f + f_0) + {A_y}/{2} \cdot {\rm e}^{ -{\rm j} \varphi_y} \cdot {\rm \delta } (f - f_0).$$

In the task, filter  $\rm B$  should be given in the form:$$H_{\rm B}(f) = {\rm e}^{-a_{\rm B}(f)}\cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm} b_{\rm B}(f)}.$$

Here,

• $a_{\rm B}(f)$  denotes the damping curve, and
• $b_{\rm B}(f)$  the phase response.

Please note:

• The task belongs to the chapter  System Description in Frequency Domain.

Questions

1

Which of the statements are true regarding filter  $\rm A$ ?

	The following holds:   $|H(f)| = 0.8$.
	Filter  $\rm A$  does not represent an LTI–system.
	The specification of a frequency response is not possible.

2

Which of the statements are true regarding filter  $\rm B$ ?

	Filter  $\rm B$  is a low-pass filter.
	Filter  $\rm B$  is a high-pass filter.
	Filter  $\rm B$  is a band-pass filter.
	Filter  $\rm B$  is a band-stop filter.

3

Determine the damping value and the phase for filter  $\rm B$  and  $f_0 = 3 \ \text{kHz}$.

$a_{\rm B}(f_0 = \: \rm 3 \: kHz) \ = \ $  $\text{Np}$

$b_{\rm B}(f_0 = \: \rm 3 \: kHz) \ =\ $  $\text{Grad}$

4

What is the damping and phase value for for  $f_0 = 2 \ \text{kHz}$?

$a_{\rm B}(f_0 = \: \rm 2 \: kHz) \ = \ $  $\text{Np}$

$b_{\rm B}(f_0 = \: \rm 2 \: kHz) \ =\ $  $\text{Grad}$

Sample solution

Solution

(1)  Approaches 2 und 3 are correct.:

• Bei einem LTI–system gilt  $Y(f) = X(f) · H(f)$.
• Daher ist es nicht möglich, dass im Ausgangssignal ein Anteil mit  $3 f_0$  vorhanden ist, wenn ein solcher im Eingangssignal fehlt.
• Das heißt:   Es liegt hier kein LZI–System vor und dementsprechend ist auch kein Frequenzgang angebbar.

(2)  Approach 3 is correct.:

• Based on the given numerical values for  $A_y(f_0)$  filter  $\rm B$  can be assumedd to be a band-pass filter.

(3)  Mit  $A_x = 2 \text{ V}$  und  $\varphi_x = 90^\circ$  (Sinusfunktion) erhält man für  $f_0 = f_3 =3 \text{ kHz}$:

$$H_{\rm B} (f_3) = \frac{A_y}{A_x} \cdot {\rm e}^{ -{\rm j} (\varphi_x - \varphi_y)} = \frac{1\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} - 90^{\circ})} = 0.5.$$

Somit ergeben sich für  $f_0 = f_3 = 3 \text{ kHz}$  die Werte

• $a_{\rm B} (f_3)\rm \underline{\: ≈ \: 0.693 \: Np}$,
• $b_{\rm B}(f_3) \rm \underline{\: = \: 0 \: (Grad)}$.

(4)  In analoger Weise kann der Frequenzgang bei  $f_0 = f_2 =2 \text{ kHz}$  ermittelt werden:

$$H_{\rm B} ( f_2) = \frac{0.8\hspace{0.05cm}{\rm V}}{2\hspace{0.05cm}{\rm V}} \cdot {\rm e}^{ -{\rm j} (90^{\circ} - 70^{\circ})} = 0.4\cdot {\rm e}^{ -{\rm j} 20^{\circ}}.$$

Damit erhält man für  $f_0 = f_2 = 2 \ \text{ kHz}$:

• $a_{\rm B}(f_2) \rm \underline{\: ≈ \: 0.916 \: Np}$,
• $b_{\rm B}(f_2) \rm \underline{\: = \: 20°}$.

Bei  $f_0 = -f_2 =-\hspace{-0.01cm}2 \text{ kHz}$  gilt der gleiche Dämpfungswert. Die Phase hat jedoch das umgekehrte Vorzeichen. Also ist  $b_{\rm B}(–f_2) = \ –\hspace{-0.01cm}20^{\circ}.$