Difference between revisions of "Linear and Time Invariant Systems/System Description in Time Domain"

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|Untermenü=Basics of System Theory
 
|Untermenü=Basics of System Theory
 
|Vorherige Seite=System Description in Frequency Domain
 
|Vorherige Seite=System Description in Frequency Domain
|Nächste Seite=Some Low Pass Functions in System Theory
+
|Nächste Seite=Some Low-Pass Functions in System Theory
 
}}
 
}}
  
==Impulse Response==
+
==Impulse response==
 
<br>
 
<br>
On the page&nbsp;  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|Das erste Fourierintegral]]&nbsp;  in the book &rdquo;Signal Representation&rdquo; it was explained that for any deterministic signa&nbsp; $x(t)$&nbsp; a spectral function&nbsp; $X(f)$&nbsp; can be given with the help of the Fourier transform. Often &nbsp;$X(f)$&nbsp; is referred to as the spectrum in short.  
+
On the page&nbsp;  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|The first Fourier integral]]&nbsp;  in the book&nbsp; "Signal Representation"&nbsp; it was explained that for any deterministic signal&nbsp; $x(t)$&nbsp; a spectral function&nbsp; $X(f)$&nbsp; can be given with the help of the Fourier transform.&nbsp; Often &nbsp;$X(f)$&nbsp; is referred to as the&nbsp; "spectrum"&nbsp; for short.  
  
However, all information about the spectral function is already contained in the time domain representation, even if not always immediately recognisable. The same facts apply to linear time-invariant systems.  
+
However, all information about the spectral function is already contained in the time domain representation, even if not always immediately recognisable.&nbsp; The same facts apply to linear time-invariant systems.  
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;  
 
$\text{Definition:}$&nbsp;  
The most important descriptive quantity of a linear time-invariant system in the time domain is the inverse Fourier transform of&nbsp; $H(f)$, which is called the&nbsp; '''impulse response'''&nbsp;:
+
The most important descriptive quantity of a linear time-invariant system in the time domain is the inverse Fourier transform of&nbsp; $H(f)$, which is called the&nbsp; $\text{impulse response}$&nbsp;:
 
:$$h(t) = \int_{-\infty}^{+\infty}H(f) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi ft}\hspace{0.15cm} {\rm d}f.$$}}
 
:$$h(t) = \int_{-\infty}^{+\infty}H(f) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi ft}\hspace{0.15cm} {\rm d}f.$$}}
  
  
 
The following should be noted in this regard:  
 
The following should be noted in this regard:  
*The frequency response&nbsp; $H(f)$&nbsp; and the impulse response&nbsp; $h(t)$&nbsp; are equivalent descriptive quantities that contain exactly the same information about the LTI-system.  
+
*The frequency response&nbsp; $H(f)$&nbsp; and the impulse response&nbsp; $h(t)$&nbsp; are equivalent descriptive quantities that contain exactly the same information about the LTI system.  
*If the dirac-shaped input signal&nbsp; $x(t) = δ(t)$ is used, then&nbsp; $X(f) = 1$&nbsp; is to be set and&nbsp; $Y(f) = H(f)$&nbsp; or&nbsp; $y(t) = h(t)$ are valid.  
+
*If the Dirac-shaped input signal&nbsp; $x(t) = δ(t)$ is used, then&nbsp; $X(f) = 1$&nbsp; is to be set and&nbsp; $Y(f) = H(f)$&nbsp; or&nbsp; $y(t) = h(t)$ are valid.  
*The term &rdquo;impulse response&rdquo; reflects this statement: &nbsp; $h(t)$&nbsp; is the response of the system to a (Dirac delta) function as an input signal.  
+
*The term&nbsp; "impulse response"&nbsp; reflects this statement: &nbsp; $h(t)$&nbsp; is the response of the system to a&nbsp; (Dirac delta)&nbsp; function as the input signal.  
*The above definition suggests that any impulse response must have the unit&nbsp; $\text{Hz = 1/s}$&nbsp;.  
+
*The above definition suggests that any impulse response must have the unit&nbsp; $\text{Hz = 1/s}$.  
  
  
[[File:P_ID837__LZI_T_1_2_S1_neu.png|right|frame|Rectangular Impulse Response and the associated Magnitude Spectrum|class=fit]]
+
[[File:P_ID837__LZI_T_1_2_S1_neu.png|right|frame|Rectangular impulse response and associated magnitude spectrum|class=fit]]
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 1:}$&nbsp;  
 
$\text{Example 1:}$&nbsp;  
The impulse response&nbsp; $h(t)$&nbsp; of the so-called ''slit–lowpass''&nbsp; is constant over a time interval&nbsp; $T$&nbsp; and is zero outside this time interval.  
+
The impulse response&nbsp; $h(t)$&nbsp; of the so-called&nbsp; "rectangular&ndash;in&ndash;time"&nbsp; filter is constant over a time interval&nbsp; $T$&nbsp; and is zero outside this time interval.  
 
*The associated amplitude response as the magnitude of the frequency response is&nbsp;  
 
*The associated amplitude response as the magnitude of the frequency response is&nbsp;  
:$$\vert H(f)\vert  = \vert {\rm si}(\pi fT)\vert .$$  
+
:$$\vert H(f)\vert  = \vert {\rm si}(\pi fT)\vert \hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x={\rm sinc}(x/\pi).$$  
*The area over&nbsp; $h(t)$&nbsp; is equal to&nbsp; $H(f = 0) = 1$. It follows that: &nbsp; <br>&nbsp; &nbsp; In the range&nbsp; $ 0 < t < T$&nbsp; the impulse response must be equal to&nbsp; $1/T$&nbsp;.
+
*The area over&nbsp; $h(t)$&nbsp; is equal to&nbsp; $H(f = 0) = 1$. It follows that: &nbsp; <br>&nbsp; &nbsp; In the range&nbsp; $ 0 < t < T$&nbsp; the impulse response must be constant and equal to&nbsp; $1/T$.
 
*The phase response is given by  
 
*The phase response is given by  
:$$b(f) = \left\{ \begin{array}{l} \hspace{0.25cm}\pi/T  \\  - \pi/T \\  \end{array} \right.\quad \quad\begin{array}{*{20}c}  \text{für} \\  \text{für}  \\ \end{array}\begin{array}{*{20}c}{\left \vert  \hspace{0.05cm} f\hspace{0.05cm} \right \vert  > 0,}  \\{\vert \hspace{0.05cm} f \hspace{0.05cm} \vert < 0.}  \\\end{array}$$
+
:$$b(f) = \left\{ \begin{array}{l} \hspace{0.25cm}\pi/T  \\  - \pi/T \\  \end{array} \right.\quad \quad\begin{array}{*{20}c}  \text{for} \\  \text{for}  \\ \end{array}\begin{array}{*{20}c}{\left \vert  \hspace{0.05cm} f\hspace{0.05cm} \right \vert  > 0,}  \\{\vert \hspace{0.05cm} f \hspace{0.05cm} \vert < 0.}  \\\end{array}$$
*For &nbsp; $h(t)$&nbsp; that are symmetric about&nbsp; $t = 0$&nbsp; (i.e. noncausal) &nbsp; $b(f)=0$ holds. }}
+
*With symmetrical&nbsp; $h(t)$&nbsp; around&nbsp; $t = 0$&nbsp; (i.e. non-causal) &nbsp; &rArr; &nbsp; $b(f)=0$. }}
  
==Some Laws of the Fourier Transformation==
+
==Some laws of the Fourier transform==
 
<br>
 
<br>
The&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Gesetzmäßigkeiten der Fouriertransformation]]&nbsp; have already been explained in detail in the book &rdquo;Signal Representation&rdquo;.  
+
The&nbsp; [[Signal_Representation/Fourier_Transform_Theorems|Fourier Transform Theorems]]&nbsp; have already been explained in detail in the book&nbsp; "Signal Representation".  
  
The following is a short summary, where&nbsp; $H(f)$&nbsp; describes the frequency response of an LTI-system and whose inverse Fourier transform&nbsp; $h(t)$&nbsp; is the impulse response. The laws and principles are applied more frequently in the&nbsp; [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich#Aufgaben_zum_Kapitel|Aufgaben]]&nbsp; for this chapter &rdquo;Systemtheoretische Grundlagen&rdquo;.  
+
The following is a short summary, where&nbsp; $H(f)$&nbsp; describes the frequency response of an LTI system and whose inverse Fourier transform&nbsp; $h(t)$&nbsp; is the impulse response.&nbsp; The laws and principles are applied more frequently in the&nbsp; [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Exercises_for_the_chapter|exercises]]&nbsp; for this chapter&nbsp; "System Description in Time Domain".  
  
Here, we also refer to the didactic video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]].
+
Here, we also refer to the (German language) didactic video&nbsp; [[Gesetzmäßigkeiten_der_Fouriertransformation_(Lernvideo)|Gesetzmäßigkeiten der Fouriertransformation]] &nbsp; &rArr; &nbsp; "Regularities to the Fourier transform".
  
 
In the following equations the short symbol of the Fourier transformation is used. The filled-out circle indicates the spectral domain, the white one the time domain.
 
In the following equations the short symbol of the Fourier transformation is used. The filled-out circle indicates the spectral domain, the white one the time domain.
*'''Multiplication'''&nbsp; with a constant faktor:
+
*'''Multiplication'''&nbsp; by a constant factor:
 
:$$k \cdot H(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,k \cdot h(t).$$
 
:$$k \cdot H(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,k \cdot h(t).$$
:For&nbsp; $k \lt 1$&nbsp; one deals with attenuation, while&nbsp; $k \gt 1$&nbsp; stands for amplification.
+
::For&nbsp; $k \lt 1$&nbsp; one deals with attenuation, while&nbsp; $k \gt 1$&nbsp; stands for amplification.
  
  
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*'''Displacement Theorem'''&nbsp; in the frequency and time domain:
 
*'''Displacement Theorem'''&nbsp; in the frequency and time domain:
:$$H(f - f_0) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t )\cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi f_0 t},\hspace{0.9cm}
+
:$$H(f - f_0) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t )\cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi f_0 t},$$
H(f) \cdot {\rm e}^{-{\rm j}2\pi ft_0}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t- t_0 ).$$
+
:$$H(f) \cdot {\rm e}^{-{\rm j}2\pi ft_0}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t- t_0 ).$$
:#&nbsp; A shift by&nbsp; $t_0$&nbsp; (transit time) thus leads to the multiplication by a complex exponential function in the frequency domain.  
+
:#&nbsp; A shift by&nbsp; $t_0$&nbsp; ("transit time") thus leads to multiplication by a complex exponential function in the frequency domain.  
:#&nbsp; The amplitude response&nbsp; $|H(f)|$&nbsp; is not changed by this.
+
:#&nbsp; Thereby, the amplitude response&nbsp; $|H(f)|$&nbsp; does not change.
  
  
 
*'''Differentiation Theorem'''&nbsp; in the frequency and time domain:
 
*'''Differentiation Theorem'''&nbsp; in the frequency and time domain:
:$$\frac{1}{{{\rm j}2\pi }} \cdot \frac{{{\rm d}H( f )}}{{{\rm d}f}} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,- t \cdot h( t ),\hspace{0.9cm}
+
:$$\frac{1}{{{\rm j}2\pi }} \cdot \frac{{{\rm d}H( f )}}{{{\rm d}f}} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,- t \cdot h( t ),$$
{\rm j}\cdot 2\pi f \cdot H( f ){}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \frac{{{\rm d}h( t )}}{{\rm d}t}.$$
+
:$${\rm j}\cdot 2\pi f \cdot H( f ){}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \frac{{{\rm d}h( t )}}{{\rm d}t}.$$
:A differentiating element in the LTI-system leads to a multiplication by&nbsp; ${\rm j}\cdot 2πf$&nbsp; in the frequency domain and thus among other things to a phase rotation by&nbsp; $90^{\circ}$.
+
::A differentiating element in the LTI system leads to a multiplication by&nbsp; ${\rm j}\cdot 2πf$&nbsp; in the frequency domain <br>and thus among other things to a phase rotation by&nbsp; $90^{\circ}$.
  
  
==Causal Systems==
+
==Causal systems==
 
<br>
 
<br>
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
 
$\text{Definition:}$&nbsp;  
 
$\text{Definition:}$&nbsp;  
An LTI-system is said to be&nbsp;  '''causal''' if the impulse response&nbsp;  $h(t)$&nbsp; – that is the inverse Fourier transform of the frequency response&nbsp;  $H(f)$&nbsp; – satisfies the following condition:  
+
An LTI system is said to be&nbsp;  $\text{causal}$&nbsp; if the impulse response&nbsp;  $h(t)$&nbsp; – that is the inverse Fourier transform of the frequency response&nbsp;  $H(f)$&nbsp; – satisfies the following condition:  
 
:$$h(t) = 0 \hspace{0.25cm}{\rm f\ddot{u}r}\hspace{0.25cm} t < 0.$$
 
:$$h(t) = 0 \hspace{0.25cm}{\rm f\ddot{u}r}\hspace{0.25cm} t < 0.$$
 +
 +
If this condition is not met, the system is&nbsp; '''non&ndash;causal'''&nbsp; (or "acausal").
 +
 
$\text{Please note:}$&nbsp; Any realisable system is causal. }}
 
$\text{Please note:}$&nbsp; Any realisable system is causal. }}
  
  
[[File:P_ID806__LZI_T_1_2_S3_neu.png|right|frame|Noncausal System&nbsp;  $\rm  A$&nbsp;  and causal System&nbsp;  $\rm  B$|class=fit]]
+
[[File:P_ID806__LZI_T_1_2_S3_neu.png|right|frame|Non&ndash;causal system&nbsp;  $\rm  A$&nbsp;  and causal system&nbsp;  $\rm  B$|class=fit]]
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 2:}$&nbsp;  
 
$\text{Example 2:}$&nbsp;  
The diagram illustrates the difference between the noncausal system&nbsp;  $\rm A$&nbsp;  and the causal system&nbsp;  $\rm  B$.
+
The diagram illustrates the differences between the non&ndash;causal system&nbsp;  $\rm A$&nbsp;  and the causal system&nbsp;  $\rm  B$.
*In the system&nbsp; $\rm  A$&nbsp; the effect starts earlier&nbsp; $($at &nbsp; $t =\hspace{0.05cm} –T)$&nbsp; than the cause&nbsp; $($Dirac delta function at &nbsp; $t = 0)$, which of course is not possible in practice.  
+
*In system&nbsp; $\rm  A$&nbsp; the effect starts earlier&nbsp; $($at &nbsp; $t =\hspace{0.05cm} –T)$&nbsp; than the cause&nbsp; $($Dirac delta function at&nbsp; $t = 0)$, which of course is not possible in practice.
*Almost all noncausal systems can be transformed into a feasible causal system using a runtime&nbsp; $\tau$&nbsp;.  
+
 
*For example, with&nbsp; $\tau = T$ the following holds:  
+
*Almost all non&ndash;causal systems can be transformed into a feasible causal system using a runtime&nbsp; $\tau$.
:$$h_{\rm B}(t) = h_{\rm A}(t - T).$$}}
+
 +
*For example, with&nbsp; $\tau = T$ the following holds: &nbsp; $h_{\rm B}(t) = h_{\rm A}(t - T).$
 +
 
  
 +
All statements made so far apply for causal as well as non&ndash;causal systems.
  
*All statements made so far apply for causal as well as noncausal systems.
+
*However, for the description of causal systems some specific properties can be used as explained in the third main chapter &rdquo;Description of Causal Realisable Systems&rdquo;&nbsp;  [[Lineare_zeitinvariante_Systeme|of this book]]&nbsp;.
+
However, for the description of causal systems some specific properties can be used as explained in the third main chapter "Description of Causal Realisable Systems"&nbsp; of&nbsp;  [[Lineare_zeitinvariante_Systeme|this book]].}}
  
  
 
{{BlaueBox|TEXT=
 
{{BlaueBox|TEXT=
In this first and the following second main chapter we mainly consider noncausal systems, since their mathematical description is usually simpler.  
+
In this first and the following second Main Chapter we mainly consider non&ndash;causal systems since their mathematical description is usually simpler.  
*So, the frequency response&nbsp; $H_{\rm A}(f)$&nbsp; real,  
+
*So in this example, the frequency response&nbsp; $H_{\rm A}(f)$&nbsp; is real,  
 
*while for&nbsp; $H_{\rm B}(f)$&nbsp; the additional term&nbsp; ${\rm e}^{–{\rm j2π}f\hspace{0.05cm}T}$&nbsp; has to be considered. }}
 
*while for&nbsp; $H_{\rm B}(f)$&nbsp; the additional term&nbsp; ${\rm e}^{–{\rm j2π}f\hspace{0.05cm}T}$&nbsp; has to be considered. }}
  
  
  
==Computation of the Output Signal==
+
==Computation of the output signal==
 
<br>
 
<br>
We consider the following problem: &nbsp; Let the input signal&nbsp; $x(t)$&nbsp; and the frequency response&nbsp; $H(f)$ be known. The output signal&nbsp; $y(t)$ is to be determinded.  
+
We consider the following problem: &nbsp; Let the input signal&nbsp; $x(t)$&nbsp; and the frequency response&nbsp; $H(f)$ be known.&nbsp; The output signal&nbsp; $y(t)$ is to be determinded.  
  
[[File:EN_LZI_T_1_2_S4.png|right|frame|To Determine the Output Quantities of an LTI-System|class=fit]]
+
[[File:EN_LZI_T_1_2_S4.png|right|frame|To determine the output quantities of an LTI system|class=fit]]
  
If the solution is to be determined in the frequency domain, first the spectrum&nbsp; $X(f)$&nbsp; must be determined from the given input signal $x(t)$ via&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|Fouriertransformation]]&nbsp; and multiplied by the frequency response&nbsp; $H(f)$&nbsp;. By a&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_zweite_Fourierintegral|Fourierrücktransformation]]  of the product the signal &nbsp; $y(t)$ is obtained.
+
If the solution is to be determined in the frequency domain, first the spectrum&nbsp; $X(f)$&nbsp; must be determined from the given input signal $x(t)$ via&nbsp; [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral|Fourier transform]]&nbsp; and multiplied by the frequency response&nbsp; $H(f)$.&nbsp; By the&nbsp; [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_second_Fourier_integral|inverse Fourier transform]]&nbsp; of the product the signal &nbsp; $y(t)$ is obtained.
  
 
Here is a summary of the entire solution process:  
 
Here is a summary of the entire solution process:  
Line 113: Line 119:
  
  
The same result is obtained after the computation in the time domain by first determining the impulse response&nbsp; $h(t)$&nbsp; from the frequency response&nbsp; $H(f)$&nbsp; by means of inverse Forier transformation and then applying the convolution operation:
+
The same result is obtained after the computation in the time domain by first determining the impulse response&nbsp; $h(t)$&nbsp; from the frequency response&nbsp; $H(f)$&nbsp; by means of the&nbsp; "inverse Forier transform"&nbsp; and then applying the convolution operation:
 
:$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau  )}  \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
 
:$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau  )}  \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
 
*The results are identical for both approaches.  
 
*The results are identical for both approaches.  
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{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
 
$\text{Example 3:}$&nbsp;  
 
$\text{Example 3:}$&nbsp;  
At the input of a slit-lowpass with rectangular impulse response of width&nbsp; $T$&nbsp; (see [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich#Impulsantwort|$\text{Example 1}$]])&nbsp;  a rectangular pulse &nbsp; $x(t)$&nbsp; of duration&nbsp; $2T$&nbsp; is applied.  
+
At the input of a filter with rectangular impulse response&nbsp; $h(t)$&nbsp; of width&nbsp; $T$&nbsp; (see&nbsp; [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich#Impulsantwort|$\text{Example 1}$]])&nbsp;  a rectangular pulse &nbsp; $x(t)$&nbsp; of duration&nbsp; $2T$&nbsp; is applied.  
[[File:P_ID812__LZI_T_1_2_S4b_neu.png|right|frame|Trapezoidal Output Signal since&nbsp; $x(t)$&nbsp; and&nbsp; $h(t)$&nbsp; are rectangular|class=fit]]
+
[[File:P_ID812__LZI_T_1_2_S4b_neu.png|right|frame|Trapezoidal output signal since&nbsp; $x(t)$&nbsp; and&nbsp; $h(t)$&nbsp; are rectangular|class=fit]]
  
 
In this case, direct computation in the time domain is more convenient: &nbsp;  
 
In this case, direct computation in the time domain is more convenient: &nbsp;  
 
*The convolution of two rectangles &nbsp; $x(t)$&nbsp; and&nbsp; $h(t)$&nbsp; of different widhts results in a trapezoidal output signal&nbsp; $y(t)$.
 
*The convolution of two rectangles &nbsp; $x(t)$&nbsp; and&nbsp; $h(t)$&nbsp; of different widhts results in a trapezoidal output signal&nbsp; $y(t)$.
  
*The low-pass property of the filter can be seen from the finite edge steepness of &nbsp; $y(t)$.  
+
*The low-pass property of the filter can be seen from the finite edge steepness of&nbsp; $y(t)$.  
  
*The pulse height&nbsp; $3\text{ V}$&nbsp; is preserved in this example because of&nbsp;  
+
*The pulse height&nbsp; $(3\text{ V)}$&nbsp; is preserved in this example because of&nbsp;  
 
:$$H(f = 0) = 1/T · T = 1.$$ }}
 
:$$H(f = 0) = 1/T · T = 1.$$ }}
  
  
==Step Response==
+
==Step response==
 
<br>
 
<br>
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
Line 144: Line 150:
  
  
Die Berechnung im Frequenzbereich wäre hier etwas umständlich, denn man müsste dann folgende Gleichung anwenden:  
+
The computation in the frequency domain would be a bit laborious here because the following equation would have to be applied:  
 
:$${\rm \sigma}(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, X(f ) \cdot H(f) =\left({1}/{2}\cdot \delta(f) + \frac{1}{{\rm j}\cdot 2\pi f} \right) \cdot H(f).$$
 
:$${\rm \sigma}(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, X(f ) \cdot H(f) =\left({1}/{2}\cdot \delta(f) + \frac{1}{{\rm j}\cdot 2\pi f} \right) \cdot H(f).$$
  
Die Berechnung im Zeitbereich führt dagegen direkt zum Ergebnis:
+
In contrast to this, the computation in the time domain leads directly to the result:
 
:$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
 
:$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
  
Bei kausalen Systemen gilt&nbsp; $h(\tau) = 0$&nbsp;  für&nbsp; $\tau \lt 0$, so dass die untere Integrationsgrenze in obiger Gleichung zu&nbsp; $\tau = 0$&nbsp; gesetzt werden kann.  
+
For causal sytems&nbsp; $h(\tau) = 0$&nbsp;  holds for&nbsp; $\tau \lt 0$, such that the lower limit of integration in the above equation can be set to&nbsp; $\tau = 0$&nbsp;.  
  
 
{{BlaueBox|TEXT=   
 
{{BlaueBox|TEXT=   
$\text{Beweis:}$&nbsp;  
+
$\text{Proof:}$&nbsp;  
Das genannte Ergebnis ist auch aus folgendem Grunde einsichtig:  
+
The above result is also insightful for the following reason:  
*Die Sprungfunktion&nbsp; $\gamma(t)$&nbsp; hängt mit der Diracfunktion&nbsp; $\delta(t)$&nbsp; wie folgt zusammen:  
+
*The step function&nbsp; $\gamma(t)$&nbsp; is related to the Dirac delta function&nbsp; $\delta(t)$&nbsp; as follows:  
 
:$${\rm \gamma}(t) = \int_{ - \infty }^{ t } {\delta ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
 
:$${\rm \gamma}(t) = \int_{ - \infty }^{ t } {\delta ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
*Da wir Linearität vorausgesetzt haben und die Integration eine lineare Operation darstellt, gilt auch für das Ausgangssignal der entsprechende Zusammenhang:  
+
*Since we have assumed linearity and integration is a linear operation the corresponding relationship also applies to the output signal:  
 
:$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
 
:$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau  )}  \hspace{0.1cm}{\rm d}\tau.$$
 
<div align="right">q.e.d.</div>}}  
 
<div align="right">q.e.d.</div>}}  
  
  
[[File:EN_LZI_T_1_2_S5.png|right|frame|Berechnung der Sprungantwort bei rechteckförmiger Impulsantwort|class=fit]]
+
[[File:EN_LZI_T_1_2_S5.png|right|frame|Computation of the step response for rectangular impulse response|class=fit]]
 
{{GraueBox|TEXT=   
 
{{GraueBox|TEXT=   
$\text{Beispiel 4:}$&nbsp;  
+
$\text{Example 4:}$&nbsp;  
Die Grafik verdeutlicht den Sachverhalt für die Rechteck&ndash;Impulsantwort&nbsp; $h(\tau)$.
+
The graph illustrates the facts for the rectangular impulse response&nbsp; $h(\tau)$.
* Die Abszisse wurde in&nbsp; $\tau$&nbsp; umbenannt.  
+
*The abscissa has been renamed to&nbsp; $\tau$&nbsp;.  
*Blau eingezeichnet ist die Sprungfunktion&nbsp; $\gamma(\tau)$.
+
*The step function&nbsp; $\gamma(\tau)$ is drawn in blue.
*Durch Spiegelung und Verschiebung erhält man&nbsp; $\gamma(t - \tau)$ &nbsp; &rArr; &nbsp; violett gestrichelte Kurve.  
+
*&nbsp; $\gamma(t - \tau)$ &nbsp; is obtained by mirroring and shifting &rArr; &nbsp; curve dashed in violet.  
*Die rot hinterlegte Fläche gibt somit die Sprungantwort&nbsp; $\sigma(\tau)$&nbsp; zum Zeitpunkt&nbsp; $\tau = t$&nbsp; an.}}
+
*The red shaded area thus gives the step response&nbsp; $\sigma(\tau)$&nbsp; at time&nbsp; $\tau = t$&nbsp;.}}
  
  
==Exercises for the Chapter==
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==Exercises for the chapter==
 
<br>
 
<br>
[[Aufgaben:1.3 Gemessene Sprungantwort|Aufgabe 1.3: Gemessene Sprungantwort]]
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[[Aufgaben:Exercise_1.3:_Measured_Step_Response|Exercise 1.3: Measured Step Response]]
  
[[Aufgaben:Aufgabe_1.3Z:_Exponentiell_abfallende_Impulsantwort|Aufgabe 1.3Z: Exponentiell abfallende Impulsantwort]]
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[[Aufgaben:Exercise_1.3Z:_Exponentially_Decreasing_Impulse_Response|Exercise 1.3Z: Exponentially Decreasing Impulse Response]]
  
[[Aufgaben:1.4 Zum Tiefpass 2. Ordnung|Aufgabe 1.4: Zum Tiefpass 2. Ordnung]]
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[[Aufgaben:Exercise_1.4:_Low-Pass_Filter_of_2nd_Order|Exercise 1.4: Low-Pass Filter of 2nd Order]]
  
[[Aufgaben:1.4Z Alles rechteckförmig|Aufgabe 1.4Z: Alles rechteckförmig]]
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[[Aufgaben:Exercise_1.4Z:_Everything_Rectangular|Exercise 1.4Z: Everything Rectangular]]
  
 
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Revision as of 18:16, 18 July 2021

Impulse response


On the page  The first Fourier integral  in the book  "Signal Representation"  it was explained that for any deterministic signal  $x(t)$  a spectral function  $X(f)$  can be given with the help of the Fourier transform.  Often  $X(f)$  is referred to as the  "spectrum"  for short.

However, all information about the spectral function is already contained in the time domain representation, even if not always immediately recognisable.  The same facts apply to linear time-invariant systems.

$\text{Definition:}$  The most important descriptive quantity of a linear time-invariant system in the time domain is the inverse Fourier transform of  $H(f)$, which is called the  $\text{impulse response}$ :

$$h(t) = \int_{-\infty}^{+\infty}H(f) \cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi ft}\hspace{0.15cm} {\rm d}f.$$


The following should be noted in this regard:

  • The frequency response  $H(f)$  and the impulse response  $h(t)$  are equivalent descriptive quantities that contain exactly the same information about the LTI system.
  • If the Dirac-shaped input signal  $x(t) = δ(t)$ is used, then  $X(f) = 1$  is to be set and  $Y(f) = H(f)$  or  $y(t) = h(t)$ are valid.
  • The term  "impulse response"  reflects this statement:   $h(t)$  is the response of the system to a  (Dirac delta)  function as the input signal.
  • The above definition suggests that any impulse response must have the unit  $\text{Hz = 1/s}$.


Rectangular impulse response and associated magnitude spectrum

$\text{Example 1:}$  The impulse response  $h(t)$  of the so-called  "rectangular–in–time"  filter is constant over a time interval  $T$  and is zero outside this time interval.

  • The associated amplitude response as the magnitude of the frequency response is 
$$\vert H(f)\vert = \vert {\rm si}(\pi fT)\vert \hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x={\rm sinc}(x/\pi).$$
  • The area over  $h(t)$  is equal to  $H(f = 0) = 1$. It follows that:  
        In the range  $ 0 < t < T$  the impulse response must be constant and equal to  $1/T$.
  • The phase response is given by
$$b(f) = \left\{ \begin{array}{l} \hspace{0.25cm}\pi/T \\ - \pi/T \\ \end{array} \right.\quad \quad\begin{array}{*{20}c} \text{for} \\ \text{for} \\ \end{array}\begin{array}{*{20}c}{\left \vert \hspace{0.05cm} f\hspace{0.05cm} \right \vert > 0,} \\{\vert \hspace{0.05cm} f \hspace{0.05cm} \vert < 0.} \\\end{array}$$
  • With symmetrical  $h(t)$  around  $t = 0$  (i.e. non-causal)   ⇒   $b(f)=0$.

Some laws of the Fourier transform


The  Fourier Transform Theorems  have already been explained in detail in the book  "Signal Representation".

The following is a short summary, where  $H(f)$  describes the frequency response of an LTI system and whose inverse Fourier transform  $h(t)$  is the impulse response.  The laws and principles are applied more frequently in the  exercises  for this chapter  "System Description in Time Domain".

Here, we also refer to the (German language) didactic video  Gesetzmäßigkeiten der Fouriertransformation   ⇒   "Regularities to the Fourier transform".

In the following equations the short symbol of the Fourier transformation is used. The filled-out circle indicates the spectral domain, the white one the time domain.

  • Multiplication  by a constant factor:
$$k \cdot H(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,k \cdot h(t).$$
For  $k \lt 1$  one deals with attenuation, while  $k \gt 1$  stands for amplification.


  • Similarity Theorem:
$$H({f}/{k})\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,|k| \cdot h(k\cdot t).$$
  1.   This implies:   Compression   $(k < 1)$  of the frequency response results in a wider and lower impulse response.
  2.   Dilation  $(k > 1)$  of  $H(f)$  makes  $h(t)$  narrower and higher.


  • Displacement Theorem  in the frequency and time domain:
$$H(f - f_0) \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t )\cdot {\rm e}^{\hspace{0.05cm}{\rm j}2\pi f_0 t},$$
$$H(f) \cdot {\rm e}^{-{\rm j}2\pi ft_0}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, h( t- t_0 ).$$
  1.   A shift by  $t_0$  ("transit time") thus leads to multiplication by a complex exponential function in the frequency domain.
  2.   Thereby, the amplitude response  $|H(f)|$  does not change.


  • Differentiation Theorem  in the frequency and time domain:
$$\frac{1}{{{\rm j}2\pi }} \cdot \frac{{{\rm d}H( f )}}{{{\rm d}f}} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\,- t \cdot h( t ),$$
$${\rm j}\cdot 2\pi f \cdot H( f ){}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \frac{{{\rm d}h( t )}}{{\rm d}t}.$$
A differentiating element in the LTI system leads to a multiplication by  ${\rm j}\cdot 2πf$  in the frequency domain
and thus among other things to a phase rotation by  $90^{\circ}$.


Causal systems


$\text{Definition:}$  An LTI system is said to be  $\text{causal}$  if the impulse response  $h(t)$  – that is the inverse Fourier transform of the frequency response  $H(f)$  – satisfies the following condition:

$$h(t) = 0 \hspace{0.25cm}{\rm f\ddot{u}r}\hspace{0.25cm} t < 0.$$

If this condition is not met, the system is  non–causal  (or "acausal").

$\text{Please note:}$  Any realisable system is causal.


Non–causal system  $\rm A$  and causal system  $\rm B$

$\text{Example 2:}$  The diagram illustrates the differences between the non–causal system  $\rm A$  and the causal system  $\rm B$.

  • In system  $\rm A$  the effect starts earlier  $($at   $t =\hspace{0.05cm} –T)$  than the cause  $($Dirac delta function at  $t = 0)$, which of course is not possible in practice.
  • Almost all non–causal systems can be transformed into a feasible causal system using a runtime  $\tau$.
  • For example, with  $\tau = T$ the following holds:   $h_{\rm B}(t) = h_{\rm A}(t - T).$


All statements made so far apply for causal as well as non–causal systems.


However, for the description of causal systems some specific properties can be used as explained in the third main chapter "Description of Causal Realisable Systems"  of  this book.


In this first and the following second Main Chapter we mainly consider non–causal systems since their mathematical description is usually simpler.

  • So in this example, the frequency response  $H_{\rm A}(f)$  is real,
  • while for  $H_{\rm B}(f)$  the additional term  ${\rm e}^{–{\rm j2π}f\hspace{0.05cm}T}$  has to be considered.


Computation of the output signal


We consider the following problem:   Let the input signal  $x(t)$  and the frequency response  $H(f)$ be known.  The output signal  $y(t)$ is to be determinded.

To determine the output quantities of an LTI system

If the solution is to be determined in the frequency domain, first the spectrum  $X(f)$  must be determined from the given input signal $x(t)$ via  Fourier transform  and multiplied by the frequency response  $H(f)$.  By the  inverse Fourier transform  of the product the signal   $y(t)$ is obtained.

Here is a summary of the entire solution process:

$${\rm 1.\,\, step\hspace{-0.1cm} :}\hspace{0.5cm} X(f)\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, x( t )\hspace{1.55cm}{\rm input\:spectrum},$$
$${\rm 2.\,\, step\hspace{-0.1cm}:}\hspace{0.5cm}Y(f)= X(f) \cdot H(f) \hspace{0.82cm}{\rm output\:spectrum},$$
$${\rm 3.\,\, step\hspace{-0.1cm}:}\hspace{0.5cm} y(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, Y(f )\hspace{1.55cm}{\rm output\:signal}.$$


The same result is obtained after the computation in the time domain by first determining the impulse response  $h(t)$  from the frequency response  $H(f)$  by means of the  "inverse Forier transform"  and then applying the convolution operation:

$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau )} \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$
  • The results are identical for both approaches.
  • Purposefully, the procedure of solution with less computational effort should be chosen.


$\text{Example 3:}$  At the input of a filter with rectangular impulse response  $h(t)$  of width  $T$  (see  $\text{Example 1}$)  a rectangular pulse   $x(t)$  of duration  $2T$  is applied.

Trapezoidal output signal since  $x(t)$  and  $h(t)$  are rectangular

In this case, direct computation in the time domain is more convenient:  

  • The convolution of two rectangles   $x(t)$  and  $h(t)$  of different widhts results in a trapezoidal output signal  $y(t)$.
  • The low-pass property of the filter can be seen from the finite edge steepness of  $y(t)$.
  • The pulse height  $(3\text{ V)}$  is preserved in this example because of 
$$H(f = 0) = 1/T · T = 1.$$


Step response


$\text{Definition:}$  In practice, one of the often used input functions   $x(t)$  in order to measure  $H(f)$  is the  step function

$${\rm \gamma}(t) = \left\{ \begin{array}{l} \hspace{0.25cm}0 \\ 0.5 \\ \hspace{0.25cm} 1 \\ \end{array} \right.\quad \quad\begin{array}{*{20}c} \text{for} \\ \text{for}\\ \text{for} \\ \end{array}\begin{array}{*{20}c}{\vert \hspace{0.05cm} t\hspace{0.05cm} \vert < 0,} \\ {\vert \hspace{0.05cm}t\hspace{0.05cm} \vert = 0,} \\ {\vert \hspace{0.05cm} t \hspace{0.05cm} \vert > 0.} \\ \end{array}$$

The  step response  $\sigma(t)$  is the response of the system if the step function  $\gamma(t)$  is applied to the input:

$$x(t) = {\rm \gamma}(t)\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y(t) = {\rm \sigma}(t).$$


The computation in the frequency domain would be a bit laborious here because the following equation would have to be applied:

$${\rm \sigma}(t)\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, X(f ) \cdot H(f) =\left({1}/{2}\cdot \delta(f) + \frac{1}{{\rm j}\cdot 2\pi f} \right) \cdot H(f).$$

In contrast to this, the computation in the time domain leads directly to the result:

$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$

For causal sytems  $h(\tau) = 0$  holds for  $\tau \lt 0$, such that the lower limit of integration in the above equation can be set to  $\tau = 0$ .

$\text{Proof:}$  The above result is also insightful for the following reason:

  • The step function  $\gamma(t)$  is related to the Dirac delta function  $\delta(t)$  as follows:
$${\rm \gamma}(t) = \int_{ - \infty }^{ t } {\delta ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$
  • Since we have assumed linearity and integration is a linear operation the corresponding relationship also applies to the output signal:
$${\rm \sigma}(t) = \int_{ - \infty }^{ t } {h ( \tau )} \hspace{0.1cm}{\rm d}\tau.$$
q.e.d.


Computation of the step response for rectangular impulse response

$\text{Example 4:}$  The graph illustrates the facts for the rectangular impulse response  $h(\tau)$.

  • The abscissa has been renamed to  $\tau$ .
  • The step function  $\gamma(\tau)$ is drawn in blue.
  •   $\gamma(t - \tau)$   is obtained by mirroring and shifting ⇒   curve dashed in violet.
  • The red shaded area thus gives the step response  $\sigma(\tau)$  at time  $\tau = t$ .


Exercises for the chapter


Exercise 1.3: Measured Step Response

Exercise 1.3Z: Exponentially Decreasing Impulse Response

Exercise 1.4: Low-Pass Filter of 2nd Order

Exercise 1.4Z: Everything Rectangular