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

Gaussian low-pass,
rectangular low-pass,
triangular low-pass,
trapezoidal low-pass,
cosine-rolloff low-pass,
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.

Wir verwenden hier die die Funktion ${\rm si}(x)=\sin(x)/x)$. Der Zusammenhang mit der Funktion ${\rm sinc}(x)=\sin(\pi x)/(\pi x))$ lautet:

$${\rm sinc}(x)={\rm si}(x/\pi).$$

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.

The unit of impulse response is $\rm 1/s$. It is true that $\rm 1/s = 1 \ Hz$, but the unit "Hertz" is unusual in this context.

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 rectangular 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 $\rm si$–shaped with $h_1(t= 0) = 1$ and the first zero at $t=1$.

If a rectangular low-pass with $K = 1.5$ and $\Delta f = 2 \ \rm kHz$ should 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

The Gaussian low-pass with height $K$ and $($equivalent$)$ bandwidth $\Delta f$ reads:

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

The equivalent bandwidth $\Delta f$ is obtained from the equal-area rectangle.

The value at $f = \Delta f/2$ is smaller by a factor $\approx 0.456$ than the value at $f=0$.

For the impulse response one obtains according to the inverse Fourier transform:

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

The smaller $\Delta f$, the wider and lower is the impulse response ⇒ "Reciprocity law of bandwidth and impulse duration".

Both $H(f)$ and $h(t)$ are not exactly equal to zero at any value of $f$ resp. $t$.

However, for practical applications, the Gaussian pulse can be assumed to be limited in time and frequency.

For example, $h(t)$ has already dropped to less than $0.1\% $ of its maximum at $t=1.5 \cdot \Delta t$.

Rectangular low-pass

The rectangular low-pass with height $K$ and $($equivalent$)$ bandwidth $\Delta f$ reads:

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

The $\pm \Delta f/2$ value lies midway between the left-hand and right-hand limits.

For the impulse response $h(t)$ one obtains according to the laws of the inverse Fourier transform $($"2nd Fourier integral"$)$:

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

The $h(t)$ value at $t=0$ is equal to the square area of the frequency response.

The impulse response has zeros at equidistant intervals $1/\Delta f$.

The integral over the impulse response $h(t)$ is equal to the frequency response $H(f)$ at frequency $f=0$, thus is equal to $K$.

Triangular low-pass

The triangular low-pass with height $K$ and $($equivalent$)$ bandwidth $\Delta f$ reads:

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

The absolute physical bandwidth $B$ ⇒ [positive frequencies only] is also equal $\Delta f$, thus is as large as for the rectangular low-pass.

Korrektur: Warum sind in dieser Gleichung die Zeilenabstände rechts kleiner als lins

For the impulse response $h(t)$ one obtains according to the second Fourier transform:

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

$H(f)$ can be represented as a convolution of two rectangular functions $($each with width $\Delta f)$.

It follows: $h(t)$ contains instead of the ${\rm si}$–function the ${\rm si}^2$–function.

$h(t)$ thus also exhibits zeros at equidistant intervals $1/\Delta f$.

The asymptotic decay of $h(t)$ occurs here with $1/t^2$, while for comparison in the case of the rectangular low-pass $h(t)$ decays with $1/t$.

Trapezoidal low-pass

The trapezoidal low-pass with height $K$ and $($equivalent$)$ bandwidth $\Delta f$ reads:

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

Korrektur: Warum sind in dieser Gleichung die Zeilenabstände rechts kleiner als lins

For the equivalent bandwidth $($equal-area rectangle$)$ the following applies: $\Delta f = f_1+f_2$.

The rolloff factor $($in the frequency domain$)$ characterizes the slope:

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

The special case "$r=0$" corresponds to the rectangular low-pass and the special case "$r=1$" to the triangular low-pass.

For the impulse response, according to the inverse Fourier back transform, we obtain:

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

The asymptotic decay of $h(t)$ lies between $1/t$ $($for rectangular low-pass or $r=0)$ and $1/t^2$ $($for triangular low-pass or $r=1)$.

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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