Exercise 1.3Z: Exponentially Decreasing Impulse Response

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:

$$h(t) = {1}/{T} \cdot {\rm e}^{-t/T}.$$

Let the parameter be $T = 1 \hspace{0.05cm} \rm ms$. In the subtask (3) the 3dB–cutoff 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:

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

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.

	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.