Exercise 1.3Z: Exponentially Decreasing Impulse Response

From LNTwww
Revision as of 19:17, 18 July 2021 by Oezer (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Decreasing impulse response

The impulse response  $h(t)$  of an LTI system, which

  • is identically zero for all times  $t < 0$,
  • changes abruptly at time  $t > 0$,  and
  • decreases for  $t > 0$  according to an exponential function:
$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T},$$ was measured.

Let the parameter be  $T = 1 \hspace{0.15cm} \rm ms$.  In the subtask  (3)   the 3dB cut-off frequency  $f_{\rm G}$  is to be determined, which is (implicitly) defined as follows:

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

Please note:

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



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

$H(f = 0) \ = \ $


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

$h(t = 0) \ = \ $

 $\rm 1/s$


Compute the 3dB cut-off frequency  $f_{\rm G}$.

$f_{\rm G} \ =\ $

 $\rm Hz$


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.


(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 cut-off 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 the coefficients one obtains:
$$f_{\rm G} = \frac{1}{2\pi T} \hspace{0.15cm}\underline{= {\rm 159 \hspace{0.1cm} Hz}}.$$

(4)  The first statement 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.