Exercise 3.2: Laplace Transform

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Three causal time functions

Causal signals and systems are usually described by means of the Laplace transformation.  If  $x(t)$  is identical to zero for all times  $t < 0$,  then the Laplace transform is:

$$X_{\rm L}(p) = \int_{0}^{ \infty} { x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t\hspace{0.05cm}\hspace{0.05cm} .$$

In this exercise, the Laplace transforms of the causal signals shown in the graph are to be determined.  The following equations are valid in each case only for  $t \ge 0$.  For negative times, all signals are identical to zero.

  • Cosine signal with period  $T_0$:
$$x(t) = {\rm cos} (2\pi \cdot {t}/{T_0})= {\rm cos} (\omega_0 \cdot t) \hspace{0.05cm},$$
  • sine signal with period  $T_0$:
$$y(t) = {\rm sin} (2\pi \cdot {t}/{T_0})= {\rm sin} (\omega_0 \cdot t) \hspace{0.05cm},$$
  • $\sin(t)/t$–signal with equivalent zero-crossings at a distance of  $T$:
$$z(t) = {\rm si} (\pi \cdot {t}/{T})= {\rm sinc} ({t}/{T})\hspace{0.4cm}{\rm with}\hspace{0.4cm}{\rm si}(x)= {\rm sin}(x)/x ={\rm sinc}(x)/\pi \hspace{0.05cm}.$$

The following equation cannot be used to calculate the spectral function since  $z(t)$  is not energy-limited just as the signals  $x(t)$  and  $y(t)$ :

$$Z(f) = Z_{\rm L}(p)\Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm}2\pi \it f}} .$$

Rather, the fact that  $z(t) = s(t) \cdot \gamma(t)$  holds is to be considered where  $s(t)$  denotes the conventional symmetric  $\rm si$–function here:

$$s(t) = {\rm si} (\pi \cdot {t}/{T}) \quad \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad S(f).$$

$S(f)$  is a rectangular function symmetric about  $f = 0$  with height  $T$  and width  $1/T$.

The Fourier transform of the step function  $\gamma(t)$  is:

$$\gamma(t) \quad \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad {\it \Gamma}(f) = {1}/{2} \cdot \delta (f) + \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$

Please note:

  • The exercise belongs to the chapter  Laplace Transform and p-Transfer Function.
  • In the sample solution,  we use of the two comparable functions  ${\rm si}(x)$  and  ${\rm sinc}(x)$  the former.
  • The following definite integrals are given:
$$\int_{0}^{ \infty} { {\rm e}^{-p x} \cdot \cos(qx)}\hspace{0.1cm}{\rm d}x = \frac{p}{p^2 + q^2}\hspace{0.05cm} , \hspace{1.0cm}\int_{0}^{ \infty} { {\rm e}^{-p x} \cdot \sin(qx)}\hspace{0.1cm}{\rm d}x = \frac{q}{p^2 + q^2}\hspace{0.05cm} , $$
$$\int_{0}^{ \infty} { {\rm e}^{-p x} \cdot \frac{\sin(qx)}{x}}\hspace{0.1cm}{\rm d}x = {\rm arctan}\hspace{0.15cm}\frac{q}{p}\hspace{0.05cm} , \hspace{0.6cm} \int_{A}^{ B} { \frac{1}{x}}\hspace{0.1cm}{\rm d}x = {\rm ln}\hspace{0.15cm}\frac{B}{A}\hspace{0.05cm} .$$



Compute the Laplace transform  $X_{\rm L}(p)$  of the causal cosine function  $x(t)$.  What is the correct solution?

$X_{\rm L}(p) = \omega_0/(p^2 + \omega_0^2)$.
$X_{\rm L}(p) = p/(p^2 + \omega_0^2)$.
$X_{\rm L}(p) = 1/(p^2 + \omega_0^2)$.


Compute the Laplace transform  $Y_{\rm L}(p)$  of the causal sine function  $y(t)$.  What is the correct solution?

$Y_{\rm L}(p) = \omega_0/(p^2 + \omega_0^2)$.
$Y_{\rm L}(p) = p/(p^2 + \omega_0^2)$.
$Y_{\rm L}(p) = 1/(p^2 + \omega_0^2)$.


Compute the Laplace transform  $Z_{\rm L}(p)$  of the causal  $\rm si$–function  $z(t)$.  What is the correct solution?

$Z_{\rm L}(p)$  has a rectangular shape.
$Z_{\rm L}(p) = \arctan (1/p)$.
$Z_{\rm L}(p) = T/\pi \cdot \arctan (\pi/(pT))$.


Compute the real part of the spectrum  $Z(f)$.  Which statements are true?

${\rm Re}\big[Z(f)\big]$  has a rectangular shape.
${\rm Re}\big[Z(f)\big]$  is proportional to $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$


Compute the imaginary part of  $Z(f)$.  Which statements are true?

${\rm Im}\big[Z(f)\big]$  has a rectangular shape.
${\rm Im}\big[Z(f)\big]$  is proportional to  $\ln\; \big|(f \cdot T -0.5)/(f \cdot T +0.5)\big|.$


(1)  Suggested solution 2  is correct:

  • According to the Laplace definition, the following holds with the given equations:
$$X_{\rm L}(p) = \int_{0}^{ \infty} { x(t) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \int\limits_{0}^{ \infty} { {\rm cos} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{p}{p^2 + \omega_0^2} \hspace{0.05cm} .$$
  • Suggestion 3 is ruled out since  $X_{\rm L}(p)$  must have the unit "second" (integral over time) while  $p$  and  $\omega_0$  each have the unit "1/s".

(2)  Suggested solution 1  is correct:

  • Here, the following holds using the same approach as in the subtask  (1):
$$Y_{\rm L}(p) = \int_{0}^{ \infty} { {\rm sin} (\omega_0 \cdot T) \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{\omega_0}{p^2 + \omega_0^2} \hspace{0.05cm} .$$

(3)  Suggested solution 3  is correct:

  • The  $p$–transfer function of the causal  $\rm si$–function is as follows considering the integral given above:
$$Z_{\rm L}(p) = \int_{0}^{ \infty} { \frac{\sin(\pi \cdot t/T)}{\pi \cdot t/T} \hspace{0.05cm}\cdot \hspace{0.05cm} {\rm e}^{-p t}}\hspace{0.1cm}{\rm d}t = \frac{T}{\pi} \cdot {\rm arctan} \; \frac{\pi}{p\cdot T} \hspace{0.05cm} .$$
  • Suggestion 1 only applies to the Fourier transform of the non-causal  $\rm si$–function.
  • Since here the argument of the  $\rm arctan$  function is dimensional, suggestion 2 cannot be true for this reason alone.

(4)  Suggested solution 1  is correct:

  • The following arises as a result from  $z(t) = s(t) \cdot \gamma(t)$  with the convolution theorem:
$$Z(f) = S(f) \star {\it \Gamma}(f) = {1}/{2} \cdot S(f) \star \delta (f) + S(f) \star \frac{1}{{\rm j} \cdot 2\pi f}\hspace{0.05cm}.$$
  • Since  $S(f)$  is real, the real part of  $Z(f)$  is obtained as the first term of this equation:
$${\rm Re}[ Z(f)] = {1}/{2} \cdot S(f) \star \delta (f) = {1}/{2} \cdot S(f) \hspace{0.05cm}.$$
  • The real part of  $Z(f)$  thus has the same rectangular shape as  $S(f)$,  but it is only half as high:
$${\rm Re}\{ Z(f)\}= \left\{ \begin{array}{c} T/2 \\ 0 \end{array} \right. \begin{array}{c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \end{array} \begin{array}{*{20}c} { |f|< 1/(2T)\hspace{0.05cm},} \\ { |f|> 1/(2T)\hspace{0.05cm},} \end{array} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 1}}.$$

(5)  Suggested solution 2  is correct:

  • With the result of the last subtask,  it follows for the imaginary part:
$${\rm Im}\{ Z(f)\} = S(f) \star \frac{(-1)}{{\rm j} \cdot 2\pi f} \hspace{0.05cm}.$$
  • This convolution integral yields the following for sufficiently large frequencies  $f \ge 1/(2T)$:
$${\rm Im}\{ Z(f)\} = -T \cdot \int_{f- 1/(2T)}^{ f+ 1/(2T)} { \frac{1}{2\pi x}}\hspace{0.1cm}{\rm d}x = \frac{T}{2\pi } \cdot {\rm ln}\hspace{0.15cm}\left |\frac{f- 1/(2T)}{f+ 1/(2T)}\right | \hspace{0.05cm} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\underline{{\rm suggestion \hspace{0.15cm} 2}}.$$