Applets:Frequency & Impulse Responses

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Applet Description

Real and symmetric low-passes  $H(f)$  and the corresponding impulse responses  $h(t)$  are shown,  namely

1. Gaussian low-pass,
2. rectangular low-pass,
3. triangular low-pass,
4. trapezoidal low-pass,
5. cosine-rolloff low-pass,
6. cosine-rolloff -squared low-pass.

It should be noted:

• The functions  $H(f)$  resp.  $h(t)$  are shown for up to two parameter sets in one diagram each.

• The red curves and numbers apply to the left parameter set,  the blue ones to the right parameter set.

• The abscissas  $t$  $($time$)$  and  $f$  $($frequency$)$  as well as the ordinates  $H(f)$  and  $h(t)$  are normalized in each case.

Theoretical background

Frequency response  $H(f)$  and impulse response  $h(t)$

• The  "frequency response"  $($or the  "transfer_function"$)$  $H(f)$  of a linear time-invariant transmission system gives the ratio between the output spectrum  $Y(f)$  and that of the input spectrum  $X(f)$:

$$H(f) = \frac{Y(f)}{X(f)}.$$

• If the transmission behavior at low frequencies is better than at higher frequencies,  it is called a  low-pass.

• The properties of  $H(f)$  are expressed in the time domain by the  "impulse response"  $h(t)$.   According to the  "second Fourier integral"  holds:

$$h(t)={\rm IFT} [H(f)] = \int_{-\infty}^{+\infty}H(f)\cdot {\rm e}^{+{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}f\hspace{1cm} {\rm IFT}\hspace{-0.1cm}: \rm Inverse \ Fourier transform.$$

• The inverse direction is described by the  first Fourier Integral":

$$H(f)={\rm FT} [h(t)] = \int_{-\infty}^{+\infty}h(t)\cdot {\rm e}^{-{\rm j}2\pi f t}\hspace{0.15cm} {\rm d}t\hspace{1cm} \rm FT\hspace{-0.1cm}: \ Fourier transform.$$

• In all examples we use real and even functions.  Thus:

$$h(t)=\int_{-\infty}^{+\infty}H(f)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}f \ \circ\!\!-\!\!-\! \!\!-\!\!\bullet\ \ H(f)=\int_{-\infty}^{+\infty}h(t)\cdot \cos(2\pi ft) \hspace{0.15cm} {\rm d}t .$$

• For a quadripole  $[$meaning:  $X(f)$  and  $Y(f)$  have equal units$]$:   $Y(f)$  is dimensionless. 

• Die Einheit der Impulsantwort ist  $\rm 1/s$.  Es gilt zwar $\rm 1/s = 1 \ Hz$, aber die Einheit "Hertz" ist in diesem Zusammenhang unüblich.

• The relationship between this applet and the similarly constructed applet  "Pulses and Spectra"  is based on the  "Exchange Theorem"?.

• All times are normalized to a normalization time  $T$  and all frequencies are normalized to  $1/T  \ \Rightarrow$  the numerical values of   $h(t)$  still have to be divided by  $T$.

$\text{Example:}$  If one sets a rectangle–low pass with height  $K_1 = 1$  and equivalent bandwidth  $\Delta f_1 = 1$ ,

• so the frequency response  $H_1(f)=1$  in the range  $-1 < f < 1$  and zero outside this range. 

• The impulse response  $h_1(t)$  is  $\rm sinc$–shaped with  $h_1(t= 0) = 1$  and the first zero at  $t=1$.

Now a rectangle–low-pass with  $K = 1.5$  and  $\Delta f = 2 \ \rm kHz$  is to be simulated,  where the normalization time is  $T= 1 \ \rm ms$.

• Then the first zero is at  $t=0.5\ \rm ms$  and the impulse response maximum is  $h(t= 0) = 3 \cdot 10^3 \ \rm 1/s$.

Gaussian low-pass

• Der Gauß–Tiefpass lautet mit der Höhe  $K$  und der (äquivalenten) Bandbreite  $\Delta f$:

$$H(f)=K\cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot\hspace{0.05cm}(f/\Delta f)^2}.$$

• Die äquivalente Bandbreite  $\Delta f$  ergibt sich aus dem flächengleichen Rechteck.
• Der Wert bei  $f = \Delta f/2$  ist um den Faktor  $0.456$  kleiner als der Wert bei  $f=0$.
• Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:

$$h(t)=K\cdot \Delta f \cdot {\rm e}^{-\pi(t\hspace{0.05cm}\cdot\hspace{0.05cm} \Delta f)^2} .$$

• Je kleiner  $\Delta f$  ist, um so breiter und niedriger ist die Impulsantwort   ⇒   Reziprozitätsgesetz von Bandbreite und Impulsdauer.
• Sowohl  $H(f)$  als auch  $h(t)$  sind zu keinem  $f$– bzw.  $t$–Wert exakt gleich Null.
• Für praktische Anwendungen kann der Gaußimpuls jedoch in Zeit und Frequenz als begrenzt angenommen werden. 
• Zum Beispiel ist  $h(t)$  bereits bei  $t=1.5 \cdot \Delta t$  auf weniger als  $0.1\% $  des Maximums abgefallen.

Rectangular Lowpass

• Der Rechteck–Tiefpass lautet mit der Höhe  $K$  und der (äquivalenten) Bandbreite  $\Delta f$:

$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K /2 \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f/2,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| = \Delta f/2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| > \Delta f/2.} \\ \end{array}$$

• Der  $\pm \Delta f/2$–Wert liegt mittig zwischen links- und rechtsseitigem Grenzwert.
• Für die Impulsantwort  $h(t)$  erhält man entsprechend den Gesetzmäßigkeiten der Fourierrücktransformation (2. Fourierintegral):

$$h(t)=K\cdot \Delta f \cdot {\rm si}(\pi\cdot \Delta f \cdot t) \quad \text{mit} \quad {\rm si}(x)={\sin(x)}/{x}.$$

• Der  $h(t)$–Wert bei  $t=0$  ist gleich der Rechteckfläche des Frequenzgangs.
• Die Impulsantwort besitzt Nullstellen in äquidistanten Abständen  $1/\Delta f$.
• Das Integral über die Impulsantwort  $h(t)$  ist gleich dem Frequenzgang  $H(f)$  bei der Frequenz  $f=0$, ist also gleich  $K$.

Triangular Lowpass

• Der Dreieck–Tiefpass lautet mit der Höhe  $K$  und der (äquivalenten) Bandbreite  $\Delta f$:

$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \Big(1-\frac{|f|}{\Delta f}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| \ge \Delta f.} \\ \end{array}$$

• Die absolute physikalische Bandbreite  $B$   ⇒   [nur positive Frequenzen]   ist ebenfalls gleich  $\Delta f$, ist also so groß wie beim Rechteck–Tiefpass.
• Für die Impulsantwort  $h(t)$  erhält man gemäß der Fouriertransformation:

$$h(t)=K\cdot \Delta f \cdot {\rm si}^2(\pi\cdot \Delta f \cdot t) \quad \text{mit} \quad {\rm si}(x)={\sin(x)}/{x}.$$

• $H(f)$  kann man als Faltung zweier Rechteckfunktionen  $($jeweils mit Breite  $\Delta f)$  darstellen.
• Daraus folgt:  $h(t)$  beinhaltet anstelle der  ${\rm si}$-Funktion die  ${\rm si}^2$-Funktion.
• $h(t)$  weist somit ebenfalls Nullstellen im äquidistanten Abständen  $1/\Delta f$  auf.
• Der asymptotische Abfall von  $h(t)$  erfolgt hier mit  $1/t^2$, während zum Vergleich beim Rechteck–Tiefpass  $h(t)$  mit  $1/t$  abfällt.

Trapezoidal Lowpass

Der Trapez–Tiefpass lautet mit der Höhe  $K$  und den beiden Eckfrequenzen  $f_1$  und  $f_2$:

$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \frac{f_2-|f|}{f_2-f_1} \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| \le f_1,} \\ {f_1\le \left| \hspace{0.05cm}f\hspace{0.05cm} \right| \le f_2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| \ge f_2.} \\ \end{array}$$

• Für die äquivalente Bandbreite  (flächengleiches Rechteck)  gilt:  $\Delta f = f_1+f_2$.
• Der Rolloff-Faktor (im Frequenzbereich) kennzeichnet die Flankensteilheit:

$$r=\frac{f_2-f_1}{f_2+f_1}.$$

• Der Sonderfall  $r=0$  entspricht dem Rechteck–Tiefpass und der Sonderfall  $r=1$  dem Dreieck–Tiefpass.
• Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:

$$h(t)=K\cdot \Delta f \cdot {\rm si}(\pi\cdot \Delta f \cdot t)\cdot {\rm si}(\pi \cdot r \cdot \Delta f \cdot t) \quad \text{mit} \quad {\rm si}(x)={\sin(x)}/{x}.$$

• Der asymptotische Abfall von  $h(t)$  liegt zwischen  $1/t$  $($für Rechteck–Tiefpass oder  $r=0)$  und  $1/t^2$  $($für Dreieck–Tiefpass oder  $r=1)$.

Cosine-rolloff Lowpass

Der Cosinus–Rolloff–Tiefpass lautet mit der Höhe  $K$  und den beiden Eckfrequenzen  $f_1$  und  $f_2$:

$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K \\ K\cdot \cos^2\Big(\frac{|f|-f_1}{f_2-f_1}\cdot {\pi}/{2}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| \le f_1,} \\ {f_1\le \left| \hspace{0.05cm}f\hspace{0.05cm} \right| \le f_2,} \\ {\left|\hspace{0.05cm} f \hspace{0.05cm} \right| \ge f_2.} \\ \end{array}$$

• Für die äquivalente Bandbreite  (flächengleiches Rechteck)  gilt:  $\Delta f = f_1+f_2$.
• Der Rolloff-Faktor (im Frequenzbereich) kennzeichnet die Flankensteilheit:

$$r=\frac{f_2-f_1}{f_2+f_1}.$$

• Der Sonderfall  $r=0$  entspricht dem Rechteck–Tiefpass und der Sonderfall  $r=1$  dem Cosinus-Quadrat-Tiefpass.
• Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:

$$h(t)=K\cdot \Delta f \cdot \frac{\cos(\pi \cdot r\cdot \Delta f \cdot t)}{1-(2\cdot r\cdot \Delta f \cdot t)^2} \cdot {\rm si}(\pi \cdot \Delta f \cdot t).$$

• Je größer der Rolloff-Faktor  $r$  ist, desto schneller nimmt  $h(t)$  asymptotisch mit  $t$  ab.

Cosine-rolloff-squared Lowpass

• Dies ist ein Sonderfall des Cosinus–Rolloff–Tiefpasses und ergibt sich aus diesem für  $r=1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}f_1=0,\ f_2= \Delta f$:

$$H(f) = \left\{ \begin{array}{l} \hspace{0.25cm}K\cdot \cos^2\Big(\frac{|f|\hspace{0.05cm}\cdot\hspace{0.05cm} \pi}{2\hspace{0.05cm}\cdot\hspace{0.05cm} \Delta f}\Big) \\ \hspace{0.25cm} 0 \\ \end{array} \right.\quad \quad \begin{array}{*{20}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.05cm} f\hspace{0.05cm} \right| < \Delta f,} \\ {\left| \hspace{0.05cm}f\hspace{0.05cm} \right| \ge \Delta f.} \\ \end{array}$$

• Für die Impulsantwort erhält man gemäß der Fourierrücktransformation:

$$h(t)=K\cdot \Delta f \cdot {\pi}/{4}\cdot \big [{\rm si}(\pi(\Delta f\cdot t +0.5))+{\rm si}(\pi(\Delta f\cdot t -0.5))\big ]\cdot {\rm si}(\pi \cdot \Delta f \cdot t).$$

• Wegen der letzten  ${\rm si}$-Funktion ist  $h(t)=0$  für alle Vielfachen von  $T=1/\Delta f$   ⇒   Die äquidistanten Nulldurchgänge des Cosinus–Rolloff–Tiefpasses bleiben erhalten.
• Aufgrund des Klammerausdrucks weist  $h(t)$  nun weitere Nulldurchgänge bei  $t=\pm1.5 T$,  $\pm2.5 T$,  $\pm3.5 T$, ...  auf.
• Für  $t=\pm T/2$  hat die Impulsanwort den Wert  $K\cdot \Delta f/2$.
• Der asymptotische Abfall von  $h(t)$  verläuft in diesem Sonderfall mit  $1/t^3$.

Exercises

• First select the number (1, ... , 6) of the exercise.  The number 0 corresponds to a "Reset":  Same setting as at the program start.
  
• A description of the exercise will be displayed.  The parameter values are adjusted.  Solution after pressing "Show solution".
  
• "Red" corresponds to the first parameter set   ⇒   $H_1(f)   \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\   h_1(t)$,  and "Blue" corresponds to the second parameter set   ⇒   $H_2(f)   \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\   h_2(t)$.
  
• Values smaller than  $0.0005$  are set to zero in the program.

(1)   Compare the  red Gaussian lowpass  $(K_1 = 1, \Delta f_1 = 1)$  to the  blue rectangular lowpass  $(K_2 = 1, \Delta f_2 = 1)$.  Questions:
        (a)  Which output signals  $y(t)$  result from the signal  $x(t) = 2 \cdot \cos (2\pi f_0 t -\varphi_0)$  with  $f_0 = 0.5$?
        (b)  What are the differences between the two lowpass filters with  $f_0 = 0.5 \pm f_\varepsilon$  and  $f_\varepsilon \ne 0, \ f_\varepsilon \to 0$?

(a)  It holds  $y(t) = A \cdot \cos (2\pi f_0 t -\varphi_0)$  with  $A = 2 \cdot H(f = f_0) \ \Rightarrow \ A_1 = 0.912, \ A_2 = 1,000$.  The phase  $\varphi_0$  remains unchanged.

(b)  For  red  $ A_1 = 0.912$  is still valid.  For  blue  it holds  $A_2 = 0$  for  $f_0 = 0.5000\text{...}001$  and  $A_2 = 2$  for  $f_0 = 0.4999\text{...}999$.

(2)   Leave the settings unchanged.  Which lowpass  $H(f)$  fulfills the first or the second Nyquist criterion?
        Here  $H(f)$  denotes the total frequency response of transmitter, channel and reception filter.

• First Nyquist criterion:  The impulse response  $h(t)$  must have equidistant zero crossings at the (normalized) times  $t = 1,\ 2$, ...
  
• The impulse response  $h(t) = {\rm si}(\pi \cdot \delta f \cdot t)$  of the rectangular lowpass filter fulfils this criterion with  $\Delta f = 1$.
  
• In contrast, the first Nyquist criterion is never fulfilled for the Gaussian lowpass and there is always impulse interference.
  
• The second Nyquist criterion is met by neither the rectangular lowpass nor the Gaussian lowpass.

(3)   Compare the  red rectangular lowpass  $(K_1 = 0.5, \Delta f_1 = 2)$  to the  blue rectangular lowpass  $(K_2 = 1, \Delta f_2 = 1)$.  Then vary  $\Delta f_1$  between  $2$  and  $0.5$.

• With  $\Delta f_1 = 2$  the zeros of  $h_1(t)$  are multiples  of  $0.5$   ⇒   $h_1(t)$   will decay twice as fast as  $h_2(t)$.
  
• With the present setting,  $h_1(t = 0) = h_2(t = 0)$ holds,  since the rectangular areas of  $H_1(f)$  and  $H_2(f)$  are equal.
  
• By decreasing  $\Delta f_1$,  the impulse response  $h_1(t)$  becomes wider and lower.  With  $\Delta f_1 = 0.5$,  $h_1(t)$ is twice as wide as  $h_2(t)$,  but simultaneously by a factor  $4$  lower.

(4)   Compare the  red trapezoidal lowpass  $(K_1 = 1, \ \Delta f_1 = 1, \ r_1 = 0.5)$  with the  blue rectangular lowpass   $(K_2 = 1, \ \Delta f_2 = 1)$.  Vary  $r_1$  between  $0$  and  $1$.

• With  $r_1 = 0.5$  the  followers/precursors of  $h_1(t)$  for the "trapezoid" are less than for the "rectangle" due to the flatter edge drop .
  
• With smaller  $r_1$  followers & precursors increase.  With  $r_1= 0$  the trapezoidal is equal to the rectangular lowpass   ⇒   $h(t)= {\rm si}(\pi \cdot t/T)$.
  
• With larger  $r_1$  followers & precursors become smaller.  With  $r_1= 1$  the trapezoidal is equal to the triangular lowpass   ⇒   $h(t)= {\rm si}^2(\pi \cdot t/T)$.

(5)   Compare the  trapezoidal lowpass  $(K_1 = 1, \ \Delta f_1 = 1, \ r_1 = 0.5)$  to the  cosine-rolloff lowpass  $(K_2 = 1, \ \Delta f_2 = 1, \ r_2 = 0.5)$.
        Vary  $r_2$  between  $0$  and  $1$.  Interpret the impulse response for  $r_2 = 0.75$.  Which lowpass satisfies the first Nyquist criterion?

• With  $r_1 = r_2= 0.5$  the edge drop of  $H_2(f)$  is steeper by the frequency  $f = 0.5$  than the edge drop of  $H_1(f)$.
  
• With the same rolloff  $r= 0.5$  the impulse response  $h_2(t)$  for  $t > 1$  has larger magnitudes than  $h_1(t)$.
  
• With  $r_1 = 0.5$  and  $r_2 = 0.75$  $H_1(f) \approx H_2(f)$  holds and therefore also  $h_1(t) \approx h_2(t)$. 
  
• $H_1(f)$  and  $H_2(f)$  both fulfill the first Nyquist criterion:  Both functions are point-symmetrical around the "Nyquist point".
  
• Because of  $\Delta f = 1$  both  $h_1(t)$  and  $h_2(t)$  have zero crossings at  $\pm 1$,  $\pm 2$   ⇒   in each case maximum vertical eye opening.

(6)   Compare the  cosine-square lowpass  $(K_1 = 1, \ \ \Delta f_1 = 1)$  with the  cosine-rolloff lowpass  $(K_2 = 1, \ \ \Delta f_2 = 1,\ r_2 = 0.5)$.
        Vary  $r_2$  between  $0$  and  $1$.  Interpret the results.  Which low-pass satisfies the second Nyquist criterion?

• $H_1(f)$  is a special case of the cosine-rolloff lowpass with rolloff  $r_2 =1$.  The first Nyquist criterion is also fulfilled with  $r_2 \ne 1$.
  
• According to the second Nyquist criterion  $h(t)$  must also have zeros at  $t=\pm 1.5$,  $\pm 2.5$,  $\pm 3.5$, ...  $($ but not, however, at  $t = \pm 0.5)$.
  
• For the cosine-square lowpass,  $h_1(t=\pm 0.5) = 0.5$  and it therefore holds  $h_1(t=\pm 1) = h_1(t=\pm 1.5) = h_1(t=\pm 2)= h_1(t=\pm 2.5) = \text{...} =0$.
  
• Only the cosine-square lowpass fulfils the first and second Nyquist criteria simultaneously:  Maximum vertical and horizontal eye opening.

Applet Manual

    (A)     Theme (changeable graphical user interface design)

• Dark:   dark background  (recommended by the authors)
• Bright:   white background  (recommended for beamers and printouts)
• Deuteranopia:   for users with pronounced green visual impairment
• Protanopia:   for users with pronounced red visual impairment

    (B)     Preselection for frequency response  $H_1(f)$  (red curve)

    (C)     Parameter definition for  $H_1(f)$ 

    (D)     Numeric output for  $H_1(f_*)$  and  $h_1(t_*)$

    (E)     Preselection for frequency response  $H_2(f)$   (blue curve)

    (F)     Parameter definition for  $H_2(f)$ 

    (G)     Numeric output for  $H_2(f_*)$  and  $h_2(t_*)$

    (H)     Setting the frequency  $f_*$  for the numeric output

    (I)      Setting the time  $t_*$  for the numeric output

    (J)     Graphic field for the frequency domain

    (K)     Graphic field for the time domain

    (L)     Selection of the exercise according to the numbers

    (M)     Task description and questions

    (N)     Show and hide sample solution

About the authors

This interactive calculation tool was designed and implemented at the  Institute for Communications Engineering  at the  Technical University of Munich.

• The first version was created in 2005 by Ji Li as part of her diploma thesis with “FlashMX – Actionscript” (Supervisor: Günter Söder).
• In 2017 the program was redesigned by David Jobst (Ingenieurspraxis_Math, Supervisor: Tasnád Kernetzky ) via "HTML5".
• Last revision and English version 2020 by  Carolin Mirschina  in the context of a working student activity.  Translation using DEEPL.com.

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