Aufgaben:Exercise 1.3Z: Exponentially Decreasing Impulse Response

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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:

• The task belongs to the chapter  System Description in Time Domain.
• The following definite integral is given:

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

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

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)  The first approach is correct:

• Due to  $h(t) = 0$   for  $t < 0$  the system is indeed causal. It is a low-pass filter of first order.
• In contrast, a high-pass filter would have to satisfy the following condition:

$$H(f = 0) = \int_{-\infty}^{+\infty}h(t) \hspace{0.15cm}{\rm d}t = 0.$$

• $H(f)$  is a complex function. The phase response is  (see  Exercise 1.1Z):

$$b(f) = \arctan {f}/{f_{\rm G}}.$$

• For the frequency  $f = f_{\rm G}$  one obtains  $b(f = f_{\rm G}) = π/4 = 45^\circ$.
• If a cosine signal of frequency  $f = f_{\rm G}$  is applied ot the input, the output signal is given by:

$$y(t) = K \cdot \cos( 2 \pi f_{\rm G} t - 45^{\circ}).$$

• This signal is a harmonic oscillation but not a cosine signal.