Signal Representation/The Fourier Transform Theorems

1. Signal Representation
2. Aperiodic Signals - Impulses
3. The Fourier Transform Theorems

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Multiplication with a factor - Addition Theorem

In this section the  »Fourier Transform Theorems«  are assembled.  These can be used,  for examle,  to derive from already known transformations

$$x( t )\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X( f ),\quad x_1 ( t )\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X_1 ( f ),\quad x_2 ( t )\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,X_2 ( f )$$

new functional relationships.  Here we restrict ourselves to real time functions.

$\text{Theorem:}$  A  $\text{constant factor}$  $k$  affects the time and spectral function in the same way:

$$k \cdot x(t)\ \;\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ k \cdot X(f).$$

This relation can be used for simplification by omitting the constant  $k$  $($which can be a gain,  an attenuation or a unit factor$)$  and adding it to the result later.

The above sentence follows directly from the definition of the  »first Fourier integral«,  as well as from the  »addition theorem«,  which formulates the foundation of the  »superposition principle«.

$\text{Addition Theorem:}$  If a time function can be written as a sum of single functions, the resulting spectral function is the sum of the resulting single spectra:

$$x( t ) = x_1 ( t ) + x_2 ( t )\quad\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\quad X( f ) = X_1 (f) + X_2 ( f ).$$

Rectangular pulse,  triangular pulse and their combination

$\text{Example 1:}$  The following Fourier correspondences are known:

• The rectangular pulse:

$$x_1 ( t )\hspace{0.15cm} \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} X_1 ( f )=T \cdot {\rm sinc}(f T),$$

• the triangle pulse:

$$ x_2 ( t )\hspace{0.15cm} \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} X_2 ( f )=T /2\cdot {\rm sinc}^2(f T/2).$$

These two pulses are sketched on the right as red and blue curve respectively.

⇒   Then for the Fourier correspondences of the green drawn  $($weighted$)$  sum signal  $x(t)$  holds:

$$x(t) = {1}/{3} \cdot x_1 ( t ) + {2}/{3} \cdot x_2 ( t )\hspace{0.15cm} \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} X(f) = {1}/{3} \cdot X_1 ( f ) + {2}/{3} \cdot X_2 ( f ).$$

All theorems presented in this chapter can be found at the following  $($German language$)$  learning video with illustrated examples
        »Gesetzmäßigkeiten der Fouriertransformation«   ⇒   "Regularities to the Fourier transform".

Assignment Theorem

With the  »complex Fourier series«  for describing periodic signals,  we have found

1. that an even function always leads to real Fourier coefficients,  and
2. an odd function exclusively to imaginary Fourier coefficients. 

The Fourier transform shows similar properties.

$\text{Assignment Theorem:}$  If a real time function consists additively of an even  $($German:  "gerade"   ⇒   $\text{"g"})$  and an odd  $($German:  "ungerade"   ⇒   $\text{"u"})$  part,

$$x( t ) = x_{\rm g} ( t ) + x_{\rm u} ( t ),$$

then the following applies for its spectral function:

$$X(f) = X_{\rm R}(f) + {\rm j}\cdot X_{\rm I}(f), \hspace{0.5cm}\text{with}$$

$$ x_{\rm g} (t) \hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}X_{\rm R}(f),$$

$$x_{\rm u} (t) \hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} {\rm j} \cdot X_{\rm I} (f).$$

The real part  $X_{\rm R}(f)$  of the spectrum is then also even,  while  $X_{\rm I}(f)$  describes an odd function of frequency.

• The assignment theorem can be easily proved by considering the theorem of  »$\text{Leonhard Euler}$«: 

$${\rm e}^{ - {\rm j}\omega _0 t} = \cos ( {\omega _0 t} ) - {\rm j}\cdot \sin ( {\omega _0 t} ).$$

• The even and odd part of each function  $x(t)$  can be calculated with the following equations:

$$x_{\rm g} (t) = {1}/{2}\big[ {x(t) + x(-t)} \big],$$

$$x_{\rm u} (t) = {1}/{2}\big[ {x(t) - x(-t)} \big].$$

Spectrum of the jump function

$\text{Example 2:}$  We consider the  »jump function«

$$x(t) = \gamma (t) = \bigg\{ \begin{array}{l} 0\quad \quad {\rm for}\;t < 0 \\ 1\quad \quad{\rm for}\; t > 0 \\ \end{array} ,$$

which can be split as follows:  

$$\gamma (t) = {1}/{2} +{1}/{2} \cdot {\rm sign}(t).$$

The  »signum function«  was used here:

$${\rm sign} (t) = \bigg\{ \begin{array}{l} -1\quad \quad {\rm for}\;t < 0, \\ +1\quad \quad{\rm for}\; t > 0. \\ \end{array} $$

Therefore the following applies:

1. The even  $($blue$)$  signal part  $x_{\rm g} (t) = {1}/{2}$  is a constant with the real spectral function  $X_{\rm R}(f) = {1}/{2} \cdot \delta(f)$.
2. The spectrum  ${\rm j} \cdot X_{\rm I}(f)$  of the odd  $($green$)$  signum function  $x_{\rm u} (t)$  was already calculated in the earlier  $\text{Example 3}$  in the section  »Fourier transform«.
3. This results for the spectrum of the  $($red$)$  sketched  jump function:

$$X(f) = X_{\rm R}(f) + {\rm j}\cdot X_{\rm I}(f)= {1}/{2} \cdot \delta (f) - {\rm j}\cdot \frac{1}{2\pi f}.$$

Similarity Theorem

The similarity theorem shows the relation between the spectral functions of two time signals of the same shape,  stretched or compressed.

$\text{Similarity Theorem:}$  If  $X(f)$  is the Fourier transform of  $x(t)$,  then with the real constant  $k$  the following relation applies:

$$x( {k \cdot t} )\hspace{0.15cm} \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} \frac{1}{\left \vert \hspace{0.05cm} k\hspace{0.05cm} \right \vert} \cdot X( {f}/{k} ).$$

$\text{Proof:}$  For positive  $k$  follows from the Fourier integral with the substitution  $\tau = k \cdot t$:

$$\int_{ - \infty }^{ + \infty } {x( {k \cdot t})} \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}ft} \hspace{0.1cm}{\rm d}t = \frac{1}{k} \cdot \int_{ - \infty }^{ + \infty } {x( \tau )} \cdot {\rm e}^{ - {\rm j}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm} f/k \hspace{0.05cm}\cdot \hspace{0.05cm}\tau } \hspace{0.1cm}{\rm d} \tau = \frac{1}{k}\cdot X( {{f}/{k}}).$$

• For negative  $k$  the integration limits would be mixed up and you get  $-1/k \cdot X(f/k)$.

• Since in the equation  $\vert \hspace{0.05cm} k \hspace{0.05cm} \vert$  is used,  the result is valid for both signs.

q.e.d.

The effects of the similarity theorem can be illustrated,  for example,  with an audio tape. 

• If such a tape is played with double speed,  this corresponds to a compression of the time signal  $(k = 2)$. 

• Thus the frequencies appear twice as high.

Two rectangles of different width

$\text{Example 3:}$  We consider two rectangles of equal height, where  $T_2 = T_1/2$  holds.

• The spectral function of  $x_1(t)$  results after the  »first Fourier Integral«  to

$$X_1 (f) = A \cdot \frac{ {1 - {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}fT_1 } } }{ { {\rm j}2\pi f} } .$$

• For this can also be written:

$$X_1 (f) = A \cdot T_1 \cdot \frac{{{\rm e}^{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}fT_1 } - {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}fT_1 } }}{{{\rm j}2\pi fT_1 }} \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}fT_1 } = A \cdot T_1 \cdot {\rm sinc}( {f T_1 } ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}fT_1 }.$$

• For the spectral function of  $x_2(t)$  follows from the similarity theorem with  $k = -2$:

$$X_2 (f) = \frac{1}{2} \cdot X_1 ( { - {f}/{2}} ) = \frac{A \cdot T_1 }{2} \cdot {\rm sinc}( { - f {T_1 }/{2} } ) \cdot {\rm e}^{ {\rm j}\pi fT_1 /2} .$$

• The function  $\text{sinc}(x) = \sin(x)/x$  is even:  $\text{sinc}(-x) = \text{sinc}(x)$.  Therefore you can omit the sign in the argument of the  $\text{sinc}$–function.

• With  $T_2 = T_1/2$  one gets:

$$X_2 (f) = A \cdot T_2 \cdot {\rm sinc}( {fT_2 } ) \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}T_2 } .$$

Reciprocity Theorem of time duration and bandwidth

This law follows directly from the  »similarity theorem«:   The wider a pulse is in its extension,  the narrower and higher is the corresponding spectrum and vice versa.

To be able to make quantitative statements,  we define two parameters for energy-limited signals.  Both quantities are shown in the diagram in  $\text{Example 4}$  for a Gaussian pulse and its likewise Gaussian spectrum.

$\text{Definition:}$  The  »equivalent pulse duration«  is derived from the time course.  It is equal to the width of an area–equal rectangle with same height as  $x(t)$:

$$\Delta t = \frac{1}{{x( {t = 0} )}} \cdot \int_{ - \infty }^{ + \infty } {x( t)} \hspace{0.1cm}{\rm d}t.$$

$\text{Definition:}$  The  »equivalent bandwidth«   is defined in the frequency domain.  It gives the width of the area–equal rectangle with same height as spectrum  $X(f)$:

$$\Delta f = \frac{1}{{X( {f = 0} )}}\cdot \int_{ - \infty }^{ + \infty } {X( f )} \hspace{0.1cm} {\rm d}f.$$

$\text{Reciprocity Theorem:}$  The product of the equivalent pulse duration and the equivalent bandwidth is always the same  $1$:

$$\Delta t \cdot \Delta f = 1.$$

$\text{Proof:}$  Based on the two Fourier integrals,  for  $f = 0$  resp.  $t = 0$:

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

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

If you take this result into account in the above definitions, you get

$$\Delta t = \frac{{X( {f = 0} )}}{{x( {t = 0} )}}, \hspace{0.5cm}\Delta f = \frac{{x( {t = 0} )}}{{X( {f = 0} )}}.$$

From this   $\Delta t \cdot \Delta f = 1$   follows directly.

q.e.d.

Note that  $\Delta f$  is defined over the actual spectrum  $X(f)$  and not over  $|X(f)|$.

• For real functions the integration over the even function part is sufficient,  since the integral over the odd part is always zero due to the  »assignment theorem«.

• For odd time functions and thus purely imaginary spectra,  the two definitions of  $\Delta t$  and  $\Delta f$  fail.

$\text{Example 4:}$  The graph illustrates the equivalent pulse duration  $\Delta t$  and the equivalent bandwidth  $\Delta f$  exemplary for the Gaussian pulse. 

Gaussian example for the reciprocity theorem

Furthermore,  it is valid:

• Widening the Gaussian pulse by the factor  $3$  will reduce the equivalent bandwidth by the same factor.

• If the pulse amplitude  $x(t = 0)$  is not changed,  the integral area above  $X(f)$  remains constant.

• This means that  $X(f=0)$  is simultaneously increased by the factor  $3$ .

Duality Theorem

This regularity is particularly useful for obtaining new Fourier correspondences.

$\text{Duality Theorem:}$  If  $X(f)$  is the Fourier transform of  $x(t)$,  then:

$$X^{\star}(t)\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}x^{\star}( f ).$$

If we restrict ourselves to real time functions,  the signs for  »conjugated complex»  can be omitted on both sides of the Fourier correspondence.

$\text{Proof:}$  The  »first Fourier integral«  is after successive renaming  $t \to u$,  $f \to t$:

$$X( f ) = \int_{ - \infty }^{ + \infty } {x( u )} \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}u} \hspace{0.1cm}{\rm d}u, \hspace{1cm} X(t ) = \int_{ - \infty }^{ + \infty } {x( u )} \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}t\hspace{0.05cm}\cdot \hspace{0.05cm}u}\hspace{0.1cm} {\rm d}u.$$

• If you change the sign in the exponent,  you have to replace  $X(t)$  by  $X^*(t)$  and  $x(u)$  by  $x^*(u)$ :

$$X^{\star}(t ) = \int_{ - \infty }^{ + \infty } {x^{\star}( u )} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi\hspace{0.05cm}\cdot \hspace{0.05cm} t\hspace{0.05cm}\cdot \hspace{0.05cm}u}\hspace{0.1cm} {\rm d}u.$$

• With the further renaming  $u \to f$  one gets to the »second Fourier integral«:

$$X^{\star}(t ) = \int_{ - \infty }^{ + \infty } {x^{\star}( f )} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm} {\rm d}f.\hspace{7.9cm}$$

q.e.d.

$\text{Above:  Rectangular-in-time}$   ⇒ $\text{sinc-in-frequency};$
$\text{below;  sinc-in-time}$   ⇒ $\text{rectangular-in-frequency}$

$\text{Example 5:}$ 

The figure on the right shows an application of the duality theorem,  namely the functional relations between

• a signal  $x_1(t)$  with rectangular time function,  and

• a signal  $x_2(t)$  with rectangular spectral function.

$\text{Another Example:}$ 

• The spectrum  $X(f) = \delta(f)$  of the DC signal  $x(t) = 1$  is assumed to be known.

• According to the  »duality theorem«,  the spectral function of the Dirac delta is therefore:

$$ x(t) = \delta(t)\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm} X(f)= 1.$$

Shifting Theorem

We now consider

• a shift of the time function,  e.g. caused by a delay;

• or a frequency shift,  as it occurs for example with  »analog double-sideband amplitude modulation«.

$\text{Shifting Theorem:}$  If  $X(f)$  is the Fourier transform of  $x(t)$,  the following correspondences also apply:

$$\mathbf{(1)}\hspace{0.5cm}x( {t - t_0 } )\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}X( f ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm} f\hspace{0.05cm}\cdot \hspace{0.05cm}t_0 },$$

$$\mathbf{(2)}\hspace{0.5cm}x( t ) \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f_0 \hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}X( {f - f_0 } ).$$

Here,  $t_0$  and  $f_0$  are any time or frequency values.

$\text{Proof of Equation (1):}$  The  »first Fourier integral«  for signal   $x_{\rm V}(t) = x(t-t_0)$  shifted to the right by  $t_0$  is defined with the substitution  $\tau = t - t_0$:

$$X_{\rm V} ( f ) = \int_{ - \infty }^{ + \infty } {x( {t - t_0 } ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm}{\rm d}t} = \int_{ - \infty }^{ + \infty } {x( \tau ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}( {\tau + t_0 } )}\hspace{0.1cm} {\rm d}\tau .}$$

• The term independent from the integration variable  $\tau$  can be dragged in front of the integral. 

• With the renaming  $\tau \to t$  one then obtains

$$X_{\rm V}( f ) = {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t_0 } \cdot \int_{ - \infty }^{ + \infty } {x( t )} \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm} {\rm d}t = {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t_0 } \cdot X( f).$$

q.e.d.

Shifting theorem example

$\text{Example 6:}$  As already mentioned, the symmetrical rectangular pulse  $x_1(t)$  has the spectrum

$$X_1 ( f ) = A \cdot T \cdot {\rm si}( {\pi fT} )= A \cdot T \cdot {\rm sinc}( {fT} )$$

$$\hspace{0.9cm} \text{with} \hspace{0.5cm} {\rm si}(x)= \sin(x)/x,\hspace{0.5cm}{\rm sinc}(x)= \sin(\pi x)/(\pi x)={\rm si}(x/\pi).$$

• The rectangular pulse  $x_2(t)$  displayed below is shifted to the right with respect to  $x_1(t)$  by  $T/2$:  

$$x_2(t) = x_1(t-T/2).$$

• Thus its spectrum is:

$$X_2( f ) = A \cdot T \cdot {\rm sinc}( {fT} ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}T} .$$

• This spectral function can also be written as follows with the  »theorem of Euler«  and some trigonometric conversions:

$$X_2( f ) = \frac{A }{2\pi f} \cdot \sin ( 2\pi fT) + {\rm j}\cdot \frac{A }{2\pi f} \cdot \big[ {\cos ( {2\pi fT} ) - 1} \big] .$$

• The same result can be obtained with the  »assignment theorem«;  
  ⇒   The real part of the spectrum belongs to the even signal part  $x_{\rm g}(t)$,  the imaginary part to the odd signal part  $x_{\rm u}(t)$.

Differentiation Theorem

This theorem shows, how the differentiation of a function  $x(t)$  resp.  $X(f)$  affects the corresponding Fourier transform;  it is also applicable several times.  A simple example for the application of this theorem is the relation between current  $i(t)$  and voltage  $u(t)$  of a capacitance  $C$  according to the equation   $i(t) = C \cdot \text{d}u(t)/\text{d}t$.

$\text{Differentiation Theorem:}$  If  $X(f)$ is the Fourier transform of  $x(t)$, the following two correspondences are also valid:

$$\mathbf{(1)}\hspace{0.5cm}\frac{{{\rm d}x( t )}}{{{\rm d}t}}\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f \cdot X( f ),$$

$$\mathbf{(2)}\hspace{0.5cm}- t \cdot x( t )\hspace{0.15cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\hspace{0.15cm}\frac{1}{{{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi }} \cdot \frac{{{\rm d}X( f )}}{{{\rm d}f}}.$$

$\text{Proof of Equation (1):}$ This equation results from differentiation of the  $\text{second Fourier integral}$:

$$y(t) = \frac{\text{d}x(t)}{\text{d}t} = \frac{\text{d} }{\text{d}t}\int_{ - \infty }^{ + \infty } X( f ) \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\hspace{0.03cm}\pi\hspace{0.05cm}\cdot \hspace{0.05cm} f\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm} {\rm d}f = \int_{ - \infty }^{ + \infty } {X( f )} \cdot {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi f \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\pi \hspace{0.05cm}\cdot \hspace{0.05cm} f\hspace{0.05cm}\cdot \hspace{0.05cm}t} {\rm d}f.$$

• At the same time is:

$$y( t ) = \int_{ - \infty }^{ + \infty } {Y( f )} \cdot {\rm e}^{ {\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}2\hspace{0.03cm}\pi \hspace{0.05cm}\cdot \hspace{0.05cm}f\hspace{0.05cm}\cdot \hspace{0.05cm}t}\hspace{0.1cm} {\rm d}f.$$

• By comparing the integrands, the variation  $\mathbf{(1)}$  of the differentiation theorem is obtained.
• To derive the second variant one proceeds from the $\text{first Fourier integral}$  in an analogous manner.
• The negative exponent in the first Fourier integral leads to the minus sign in the time function.
  q.e.d.

Correlation between  "Jump"  and  "Dirac delta"

$\text{Example 7:}$  The spectra of the signals  $x_1(t)$  and  $x_2(t)$  were already calculated in the previous examples:

$$X_1( f ) = \frac{1 }{ {\rm j\cdot \pi} f}, $$

$$X_2( f ) = 2 = {\rm const.}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} X_2(f) = X_1(f) \cdot {\rm j}\cdot 2\pi f.$$

• From the differentiation theorem it follows that  $x_2(t)$  is equal to the time-derivative of  $x_1(t)$ .
• This is actually correct:  For  $t \neq 0$  ⇒   $x_1(t)$  is constant, i.e. the derivative is zero.
• For  $t=0$  the gradient is infinitely large, which is also expressed in the equation  $x_2(t) = 2 \cdot \delta(t)$.
• The impulse weight  $2$  of the Dirac delta function considers that the jump within  $x_1(t)$  at  $t = 0$  has the height  $2$.

Integration Theorem

Integration is a linear operation just like differentiation. This results in the following theorem:

$\text{Integration Theorem:}$  If  $X(f)$ is the Fourier transform (spectral function) of  $x(t)$, then the following Fourier correspondences also apply:

$$\mathbf{(1)}\hspace{0.5cm}\int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )} \hspace{0.1cm}{\rm d}\tau \ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ X( f )\left( {\frac{1}{{{\rm j}\cdot 2\pi f}} + \frac{1}{2}\cdot \delta ( f )} \right),$$

$$\mathbf{(2)}\hspace{0.5cm}x( t )\left( { - \frac{1}{{{\rm j}\cdot 2\pi t}} + \frac{1}{2}\cdot \delta ( t )} \right)\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \ \ \int_{ - \infty }^f {X( \nu ) \hspace{0.1cm}{\rm d}\nu .}$$

$\text{Illustration – no exact proof:}$ 

The integration theorem represents exactly the inversion of the  $\text{differentiation theorem}$.  If one applies the differentiation theorem to the equation  $\mathbf{(1)}$  one obtains

$$\frac{ {\rm d} }{ {\rm d}t} \int_{ - \infty }^{\hspace{0.05cm}t} {x( \tau )} \hspace{0.1cm}{\rm d}\tau\ \ \circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\,\ \ X( f )\cdot \left( {\frac{1}{ { {\rm j}\cdot 2\pi f} } + \frac{1}{2}\cdot \delta ( f )} \right) \cdot {\rm j}\cdot 2\pi f.$$

This example shows the validity of the integration theorem:

• The differentiation according to the upper limit on the left side yields exactly the integrand  $x(t)$.
• The right side of the correspondence correctly results in  $X(f)$, since the Dirac delta function is hidden with  $f=0$  because of the multiplication with  $\text{j}\cdot 2\pi f$.

Notes:   All theorems shown in this chapter – such as the integration and differentiation theorem – will be elucidated with examples in the (German language) learning video
     "Gesetzmäßigkeiten der Fouriertransformation"   ⇒   "Regularities to the Fourier transform".

Correlation between  "Rectangle"  and  "Ramp"

$\text{Example 8:}$  The sketched signals  $x_1(t)$  and  $x_2(t)$  are related as follows

$$x_2( t ) = \frac{1}{T}\cdot \int_{ - \infty }^{\hspace{0.05cm}t} {x_1 } ( \tau )\hspace{0.1cm}{\rm d}\tau .$$

Due to the integration theorem the following relation between the spectra applies:

$$X_2 ( f ) = \frac{1}{T}\cdot X_1 ( f ) \cdot \left( {\frac{1}{{{\rm j}\cdot 2\pi f}} + \frac{1}{2}\cdot \delta ( f )} \right).$$

With the spectral function

$$X_1 ( f ) = A \cdot T \cdot {\rm sinc}( {fT} ) \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm}\pi fT}$$

one gets

$$X_2 ( f ) = \frac{ {A } }{2}\cdot \delta ( f ) + \frac{ {A \cdot T} }{ {2{\rm j} } } \cdot \frac{ {\sin( {\pi fT}) } }{ {\left( {\pi fT} \right)^2 } } \cdot {\rm e}^{ - {\rm j}\hspace{0.05cm}\cdot\hspace{0.05cm}\pi fT},$$

or after trigonometric transformations:

$$X_2 ( f ) = \frac{ {A} }{2}\cdot \delta ( f ) + \frac{ {A \cdot T} }{ { ( {2\pi fT} )^2 } }\cdot \big[ {\cos( {2\pi fT} ) - 1 - {\rm j}\cdot \sin ( {2\pi ft} )} \big].$$

It should be noted here:

• The Dirac delta function at  $f = 0$  with weight  $A/2$  considers the DC component of the ramp function  $x_2(t)$.
• This also means:   The DC component of the ramp function is exactly the same as the DC component of the jump function.
• The missing triangle with the corner point coordinates  $(0, 0)$, $(T, A)$  and  $(0, A)$  does not change the DC component.
• This triangular area has no effect compared to the infinite remaining area (going to infinity).

Exercises for the chapter

Exercise 3.4: Trapezoidal Spectrum and Pulse

Exercise 3.4Z: Trapezoid, Rectangle and Triangle

Exercise 3.5: Differentiation of a Triangular Pulse

Exercise 3.5Z: Integration of Dirac Functions

Exercise 3.6: Even/Odd Time Signal

Exercise 3.6Z: Complex Exponential Function