Aufgaben:Exercise 3.2: From the Spectrum to the Signal

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Spektraldarstellung der Sprungfunktion

Given the spectral function

$$X(f) = \frac{{2\,{\rm V}}}{ { {\rm j}\pi f}}.$$

The associated time function  $x(t)$  can be determined with the help of  The Second Fourier Integral :

$$x(t) = \int_{ - \infty }^{ + \infty } {X(f)} \cdot {\rm e}^{{\rm j}2\pi ft} {\rm d} f = x_{\rm R} (t) + {\rm j} \cdot x_{\rm I} (t),$$

where holds for the real part and the imaginary part, respectively:

$$x_{\rm R} (t) = 2\,{\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{{\sin ( {2\pi ft} )}}{ {\pi f}}}\hspace{0.1cm} {\rm d}f, $$

$$x_{\rm I} (t) = -2\, {\rm V} \cdot \int_{ - \infty }^{ + \infty } {\frac{ {\cos ( {2\pi ft} )}}{ {\pi f}}} \hspace{0.1cm}{\rm d}f.$$

Hints:

• This exercise belongs to the chapter  Fourier Transform and Its Inverse.

• If necessary, use the following information for the solution:

$$x( {t = 0}) = \int_{ - \infty }^{ + \infty } {X( f )}\hspace{0.1cm} {\rm d}f,\hspace{0.5cm} X( {f = 0} ) = \int_{ - \infty }^{ + \infty } {x( t)}\hspace{0.1cm} {\rm d}t ,\hspace{0.5cm}\int_0^\infty {\frac{{\sin ( {ax} )}}{x}}\hspace{0.1cm} {\rm d}x = {\rm sign} ( a ) \cdot{\pi }/{2}. $$

Questions

1

Which of the following statements are true for the time signa  $x(t)$ ?

	$x(t)$  is a complex function.
	$x(t)$  is purely real.
	$x(t)$  is purely imaginary.

2

Calculate the signal curve  $x(t)$  in the entire definition area. Which values occur at the times  $t = 1\, \text{ms}$  and  $t = -\hspace{-0.05cm}1\, \text{ ms}$  auf?

$x(t=+1\, \text{ms}) \ = \ $ $\ \text{V}$

$x(t=-1 \text{ms})\hspace{0.2cm} = \ $ $\ \text{V}$

3

What is the signal value at time  $t = 0$?

$x(t=0) \ = \ $ $\ \text{V}$

4

What is the spectral value at the frequency  $f = 0$?

$X(f=0) \ = \ $ $\ \text{V/Hz}$

Solution

Solution

(1)  Correct is the proposed solution 2   ⇒  $x(t)$  is purely real::

• For the imaginary signal component  ⇒   $x_{\rm I}(t)$  the integrand is an odd function (gerader Zähler, ungerader Nenner).
• Thus the integral from  $-\infty$  bis  $+\infty$  is zero.
• In contrast, for the real component  $x_{\rm R}(t)$   ⇒   even integrand (odd numerator, odd denominator) yields a non-zero value.

(2)  With the abbreviation  $a = 2\pi t$  can be written for the time signal:

$$x(t) = x_{\rm R} \left( t \right) = \frac{{4\,{\rm V}}}{\pi }\int_0^\infty {\frac{{\sin( {af} )}}{f}}\hspace{0.1cm} {\rm d}f.$$

This leads to the result using the given definite integral:

$$x(t) = \frac{{4\,{\rm V}}}{\pi } \cdot \frac{\pi }{2} \cdot {\mathop{\rm sign}\nolimits} ( t ) = 2\;{\rm V} \cdot {\mathop{\rm sign}\nolimits} ( t ).$$

• For  $t > 0$    $x(t) = +2\,\text{V}$ .
• Correspondingly,  $x(t) = -\hspace{-0.1cm}2\,\text{V}$  applies for  $t < 0$.
• The signal  $x(t)$  thus describes a step function from  $-\hspace{-0.05cm}2\,\text{V}$ auf $+2\,\text{V}$.

(3)  At  $t = 0$    $x(t)$  has a jump point. The right-hand limit value for  $t \rightarrow 0$  lautet  $x_+ = +2\,\text{V}$.

• If one approaches the jump point of negative times as close as desired, one obtains  $x_– = -\hspace{-0.05cm}2\,\text{V}$. The following then applies to the actual signal value at  $t = 0$ :

$$x( {t = 0} ) = {1}/{2}\cdot ( x_{+} + x_{-} ) \hspace{0.15 cm}\underline{= 0}.$$

• The same result is obtained by considering the relation

$$x( t = 0) = \int_{ - \infty }^{ + \infty } {X( f)}\hspace{0.1cm} {\rm d}f = 0.$$

(4)  The spectral value at  $f = 0$  is equal to the integral from  $-\infty$  to  $+\infty$  over the time function  $x(t)$:

$$X( f = 0) = \int_{ - \infty }^{ + \infty } {x( t)}\hspace{0.1cm} {\rm d}t \hspace{0.15 cm}\underline{= 0}.$$

Here is a second solution:

• The right-hand limit for  $f → 0$  is  $X_+ = -\text{j} \cdot \infty$, the left-hand limit  $X_- = \text{j} \cdot \infty$.
• So the relationship also applies with regard to the spectral value at   $f = 0$ :

$$X( {f = 0}) = {1}/{2}\cdot \left( {X_{ +} + X_{-} } \right) = 0.$$