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 one $X(f)$ refers to it 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 Fourier retransform of $H(f)$, which is called the 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 LZI system.
- If one uses the dirac-shaped input signal $x(t) = δ(t)$, then $X(f) = 1$ is to be set and $Y(f) = H(f)$ respectively $y(t) = h(t)$ is valid.
- The term „impulse response” reflects this statement: $h(t)$ is the response of the system to a (Dirac) impulse at the input.
- The above definition suggests that any impulse response must have the unit $\text{Hz = 1/s}$ .
. $\text{Example 1:}$ The impulse response $h(t)$ of the so-called slit low-pass 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
- $$\vert H(f)\vert = \vert {\rm si}(\pi fT)\vert .$$
- 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 equal $1/T$ . - The phase response is given by
- $$b(f) = \left\{ \begin{array}{l} \hspace{0.25cm}\pi/T \ - \pi/T \end{array} \right.\quad \quad\begin{array}{*{20}c} \text{for} \text{for} \end{array}\begin{array}{*{20}c}{\left \vert \hspace{0.05cm} f\hspace{0.05cm} \right \vert{ 0.} \vert \hspace{0.05cm} f \hspace{0.05cm} \vert < 0,} \hspace{0.05cm} \hspace{0.05cm} \vert < 0.} *If symmetric $h(t)$ um $t = 0$ (i.e. acausal) would $b(f)=0$. <div style="clear:both;"> </div> </div> =='"`UNIQ--h-1--QINU`"'Some laws of the Fourier transform== <br> The [[Signal_Representation/Fourier_Transform_Laws|Laws of the Fourier Transform]] have already been explained in detail in the book „Signal Representation”. Here now follows a short summary, where $H(f)$ describes the frequency response of an LZI system and whose Fourier retransform $h(t)$ is the impulse response. These laws are applied more frequently in the [[Linear_and_Time_Invariant_Systems/System_Description_in_the_Time_Domain#Tasks_to_Chapter|Tasks]] to this chapter „System-Theoretical Basics”. We also refer here to the learning video [[Laws_of_the_Fourier_transform_(learning video)|Laws of the Fourier transform]]. In the following equations the short symbol of the Fourier transform is used. The filled circle indicates the spectral domain, the white one the time domain. *'''Multiplication''' with a constant factor: :$$k \cdot H(f)\bullet\!\!\!\!\!\!\!\circ\,k \cdot h(t).$$ :At $k \lt 1$ one speaks of attenuation, while $k \gt 1$ stands for amplification. *''Similarity theorem'': :$$H({f}/{k})\bullet\!\!-\!\!-\!\!-\!\!\circ\,|k| \cdot h(k\cdot t).$$ :# This states: A compression $(k < 1)$ of the frequency response leads to a broader and lower impulse response. :# Stretching $(k > 1)$ of $H(f)$ makes $h(t)$ narrower and higher. *'''Displacement theorem''' in the frequency domain and in the time domain: :$$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) \cdot {\rm e}^{-{\rm j}2\pi ft_0}\bullet\!\!-\!\!-\!\!-\!\!\circ\, h( t- t_0 ).$$ :# A shift by $t_0$ (transit time) thus leads in the frequency domain to the multiplication by a complex exponential function. :# The amplitude response $|H(f)|$ is not changed by this. *''Differentiation theorem''' in the frequency domain and in the time domain: :$$\frac{1}{{\rm j}2\pi }} \cdot \frac{{\rm d}H( f )}}{{\rm d}f}} \bullet\!\!\!\!\!\!\!\circ\,- t \cdot h( t ),\hspace{0.9cm} {\rm j}\cdot 2\pi f \cdot H( f ){}\bullet\!\!\!-\!\!\!-\!\!-\!\circ\, \frac{{\rm d}h( t )}}{\rm d}t}.$$ :A differentiating element in the LZI system leads in the frequency domain to a multiplication by ${\rm j}\cdot 2πf$ and thus among other things to a phase rotation by $90^{\circ}$. =='"`UNIQ--h-2--QINU`"'Causal systems== <br> <div class="bluebox"> $\text{Definition:}$ An LZI system is said to be '''causal'''' if the impulse response $h(t)$ - that is, the Fourier retransform of the frequency response $H(f)$ - satisfies the following condition: :$$h(t) = 0 \hspace{0.25cm}{\rm f\ddot{u}r}\hspace{0.25cm} t < 0.$$ $\text{Please note:}$ Any realisable system is causal. <div style="clear:both;"> </div> </div> [[File:P_ID806__LZI_T_1_2_S3_new.png|right|frame|Acausal system $\rm A$ and causal system $\rm B$|class=fit]] {{GreyBox|TEXT=. $\text{example 2:}$ The diagram illustrates the difference between the acausal system $\rm A$ and the causal system $\rm B$. *In the system $\rm A$ the effect starts earlier $($at $t =\hspace{0.05cm} -T)$ than the cause $($Dirac function at $t = 0)$, which of course is not possible in practice. *Almost all acausal systems can be transformed into a feasible causal system using a runtime $\tau$ . *For example, with $\tau = T$: :$$h_{\rm B}(t) = h_{\rm A}(t - T).$$} *For causal systems, all the statements made so far apply in the same way as for acausal systems. *For the description of causal systems, however, some specific properties can be used, as will be explained in the third main chapter „Description of Causal Realisable Systems” [[Linear_Time_invariant_Systems|of this book]] . {{BlueBox|TEXT= In this first and the following second main chapter we mainly consider acausal systems, since their mathematical description is usually simpler. *So the frequency response $H_{\rm A}(f)$ is real, *while for $H_{\rm B}(f)$ the additional term ${\rm e}^{-{\rm j2π}f\hspace{0.05cm}T}$ has to be considered. }} =='"`UNIQ--h-3--QINU`"'Calculation of the output signal== <br> We consider the following problem: Let the input signal $x(t)$ and the frequency response $H(f)$ be known. We are looking for the output signal $y(t)$. [[File:EN_LZI_T_1_2_S4.png|right|frame|To determine the output quantities of an LZI system|class=fit]] If the solution is to be in the frequency domain, the spectrum $X(f)$ must first be determined from the given input signal $x(t)$ by [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_FourierIntegral|Fourier Transform]] and multiplied by the frequency response $H(f)$ . By [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_second_Fourier_integral|Fourier_back_transform]] of the product one then arrives at the signal $y(t)$. Here is a summary of the entire calculation process: :$${\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}.$$ The same result is obtained after the calculation in the time domain by first calculating the impulse response $h(t)$ from the frequency response $H(f)$ by means of Fourier back transformation and then applying the convolution operation: :$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {x ( \tau )} \cdot h ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$ *The results are identical for both approaches. *Purposefully, one should choose the procedure that leads to the goal with less computational effort. <div class="greybox"> $\text{Example 3:}$ At the input of a slit lowpass with rectangular impulse response of width $T$ (see [[Linear_and_Time_Invariant_Systems/System_Description_in_TimeDomain#ImpulseResponse|$\text{Example 1}$]]) a rectangular impulse $x(t)$ of duration $2T$ is applied. [[File:P_ID812__LZI_T_1_2_S4b_new.png|right|frame|Trapezoidal output pulse, since $x(t)$ and $h(t)$ are rectangular|class=fit]] In this case, direct computation in the time domain is more convenient: *Folding two rectangles of different widths $x(t)$ and $h(t)$ leads to the trapezoidal output pulse $y(t)$. *The low-pass property of the filter can be seen from the finite slope of $y(t)$. *The pulse height $3\text{ V}$ is preserved in this example, because of :$$H(f = 0) = 1/T - T = 1.$ <div style="clear:both;"> </div> </div> =='"`UNIQ--h-4--QINU`"'Step response== <br> {{BlueBox|TEXT= $\text{Definitions:}$ An input function often used in practice $x(t)$ to measure $H(f)$ is the '''step response'''' :'"`UNIQ-MathJax15-QINU`"' The '''step response''' $\sigma(t)$ is the response of the system when the step function $\gamma(t)$ is applied to the input: :'"`UNIQ-MathJax16-QINU`"'} The calculation in the frequency domain would be a bit awkward here, because one would then have to apply the following equation: :'"`UNIQ-MathJax17-QINU`"' The calculation in the time domain, on the other hand, leads directly to the result: :'"`UNIQ-MathJax18-QINU`"' For causal systems $h(\tau) = 0$ holds for $\tau \lt 0$, so the lower limit of integration in the above equation can be set to $\tau = 0$ . <div class="bluebox"> $\text{Proof:}$ The above result is also insightful for the following reason: *The jump function $\gamma(t)$ is related to the Dirac function $\delta(t)$ as follows: :'"`UNIQ-MathJax19-QINU`"' *Since we have assumed linearity and integration is a linear operation, the corresponding relationship also applies to the output signal: :'"`UNIQ-MathJax20-QINU`"' <div align="right">q.e.d.</div> <div style="clear:both;"> </div> </div> [[File:EN_LZI_T_1_2_S5.png|right|frame|calculation of step response for rectangular impulse response|class=fit]] <div class="greybox"> $\text{Example 4:}$ The graph illustrates the situation for the rectangular–impulse response $h(\tau)$. *The abscissa has been renamed to $\tau$ . *Drawn in blue is the step function $\gamma(\tau)$. *By mirroring and shifting one obtains $\gamma(t - \tau)$ ⇒ violet dashed curve. *The red shaded area thus gives the step response $\sigma(\tau)$ at time $\tau = t$
Aufgaben zum Kapitel
Aufgabe 1.3: Gemessene Sprungantwort
Aufgabe 1.3Z: Exponentiell abfallende Impulsantwort
Aufgabe 1.4: Zum Tiefpass 2. Ordnung
Aufgabe 1.4Z: Alles rechteckförmig
