Exercise 1.3Z: Exponentially Decreasing Impulse Response

From LNTwww
Revision as of 12:51, 1 July 2021 by Oezer (talk | contribs)

Dropping impulse response

Gemessen wurde die Impulsantwort  $h(t)$  eines LZI–Systems,

  • die für alle Zeiten  $t < 0$  identisch Null ist,
  • sich zur Zeit  $t > 0$  sprungartig verändert, und
  • für  $t > 0$  entsprechend einer Exponentialfunktion abfällt:
$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}.$$

Der Parameter sei  $T = 1 \hspace{0.05cm} \rm ms$. In der Teilaufgabe  (3) ist nach der  3dB–Grenzfrequenz  $f_{\rm G}$  gefragt, die wie folgt (implizit) definiert ist:

$$|H(f = f_{\rm G})| = {1}/{\sqrt{2}} \cdot|H(f = 0)| .$$





Please note:

$$\int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x = {\pi}/{2} .$$



Questions

1

Compute the frequency response  $H(f)$. What value is obtained for  $f = 0$?

$H(f = 0) \ = \ $

2

What is the value of the impulse response at time  $t = 0$?

$h(t = 0) \ = \ $

 $\rm 1/s$

3

Compute the 3dB–cutoff frequency  $f_{\rm G}$.

$f_{\rm G} \ =\ $

 $\rm Hz$

4

Which of the following statements are true?

The considered system is causal.
The considered system has high-pass filter characteristics.
If a cosine signal of frequency  $f_{\rm G}$  is applied to the system input, the output signal is also cosine-shaped.


Sample solution

(1)  The frequency response  $H(f)$  is the Fourier transform of  $h(t)$:

$$H(f) = \int_{-\infty}^{+\infty}h(t) \cdot {\rm e}^{\hspace{0.05cm}{-\rm j}2\pi ft}\hspace{0.15cm} {\rm d}t = \frac{1}{T} \cdot \int_{0}^{+\infty} {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\hspace{0.15cm} {\rm d}t.$$
  • Integration leads to the result:
$$H(f) = \left[ \frac{-1/T}{{\rm j}2\pi f+{1}/{T}} \cdot {\rm e}^{\hspace{0.05cm}{-(\rm j}2\pi f+ {1}/{T}) t}\right]_{0}^{\infty}= \frac{1}{1+{\rm j} \cdot 2\pi fT}.$$
  • At frequency  $f = 0$  the frequency response has the value  $H(f = 0) \; \underline{= 1}$.


(2)  This frequency response can also be written with real and imaginary parts as follows:

$$H(f) = \frac{1}{1+(2\pi fT)^2} -{\rm j} \cdot \frac{2\pi fT}{1+(2\pi fT)^2}.$$
  • The impulse response at time  $t = 0$  is equal to the integral over  $H(f)$.
  • Since the imaginary part is odd only the real part has to be integrated over.
  • Using the symmetry property one obtains:
$$h(t=0)=2 \cdot \int_{ 0 }^{ \infty } \frac{1}{1+(2\pi fT)^2} \hspace{0.1cm}{\rm d}f = \frac{1}{\pi T} \cdot \int_{ 0 }^{ \infty } \frac{1}{1+x^2} \hspace{0.1cm}{\rm d}x .$$
  • Using the given definite integral with the result  $π/2$  the following is obtained:
$$h(t=0)= \frac{1}{2 T} \hspace{0.15cm}\underline{= {\rm 500\cdot 1/s}}.$$
  • The result shows that the impulse response at  $t = 0$  is equal to the mean value of the left-hand– and right-hand limits.


(3)  The amplitude response in this task or in general with the 3dB-cutoff frequency  $f_{\rm G}$ is:

$$|H(f)| = \frac{1}{\sqrt{1+(2\pi fT)^2}} = \frac{1}{\sqrt{1+(f/f_{\rm G})^2}}.$$
  • By comparing coefficients one obtains:
$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$


(4)  Richtig ist der erste Lösungsvorschlag}.:

  • Wegen  $h(t) = 0$   für  $t < 0$  ist das System tatsächlich kausal. Es handelt sich um einen Tiefpass erster Ordnung.
  • Dagegen müsste ein Hochpass folgende Bedingung erfüllen:
$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$
  • $H(f)$  ist eine komplexe Funktion. Der Phasengang lautet  (siehe  Aufgabe 1.1Z):
$$b(f) = \arctan {f}/{f_{\rm G}}.$$
  • Für die Frequenz  $f = f_{\rm G}$  erhält man  $b(f = f_{\rm G}) = π/4 = 45^\circ$.
  • Liegt am Eingang ein Cosinussignal der Frequenz  $f = f_{\rm G}$  an, so ergibt sich für das Ausgangssignal:
$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$
  • Dieses Signal ist zwar eine harmonische Schwingung, aber kein Cosinussignal.