Aufgaben:Exercise 1.6: Rectangular-in-Time Low-Pass Filter

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Rectangular impulse response, non-causal and causal

We consider below the constellation shown in the graph:

• The frequency response  $H(f) = H_1(f) · H_2(f)$  in the lower branch is determined by the impulse responses of its two subcomponents.
• Here,  $h_1(t)$  is constantly equal to  $k$  in the reange from  $-1\ \rm ms$  to  $+1\ \rm ms$  and zero outside.
• At the range limits, half the value is valid in each case.
• The time variable drawn in the figure is thus  $Δt = 2 \ \rm ms$.

The impulse response of the second system function  $H_2(f)$  is:

$$h_2(t) = \delta(t - \tau).$$

The frequency response between the signals  $x(t)$  and  $z(t)$  is of high-pass character and generally:

$$H_{\rm HP}(f) = 1 - H_1(f) \cdot {\rm e}^{-{\rm j\hspace{0.05cm}\cdot \hspace{0.05cm}2 \pi}f \tau}.$$

• For the subtasks  (1)  to  (4)  the following holds:   $τ = 0$   ⇒   $H(f) = H_1(f)$.
• However, using  $τ = 0$  this can also be formulated as follows  $(Δt = 2 \ \rm ms)$:

$$H_{\rm HP}(f) = 1 - {\rm si}( \pi \cdot {\rm \Delta}t \cdot f).$$

• With no effect on the solution of the problem, note that this equation is not applicable for  $τ ≠ 0$  because of:

$$|H_{\rm HP}(f)|\hspace{0.09cm} \ne \hspace{0.09cm}1 - |H_1(f)| .$$

Please note:

• The exercise belongs to the chapter  Some Low-Pass Functions in Systems Theory.
• Reference is made especially to the page  Slit low-pass filter.

Questions

1

Calculate the height  $k$  of the impulse response  $h_1(t)$  on the side condition of  $H_1(f = 0) = 1$.

$k \ =\ $ $\ \rm 1/s$

2

Let the input signal  $x(t)$  be a rectangle symmetric about  $t = 0$  of duration $T = 2 \ \rm ms$  and height $1 \, \rm V$. Let  $τ = 0$ hold.
Which statements are true?

	$y(t)$  is rechteckförmig.
	$y(t)$  is dreieckförmig.
	$y(t)$  is trapezförmig.
	Der Maximalwert von  $y(t)$  beträgt  $ 1\hspace{0.05cm} \rm V$.

3

Which statements are true, if  $x(t)$  has a rectangle width of  $T = 1 \ \rm ms$ ?

	$y(t)$  ist rectangular.
	$y(t)$  ist triangular.
	$y(t)$  ist trapezoidal.
	The maximum value of  $y(t)$  is  $1\hspace{0.05cm} \rm V$.

4

The following still holds:   $τ = 0$. Compute the output signal  $z(t)$, if  $x(t)$  jumps from zero to  $1\hspace{0.05cm} \rm V$  at time  $t = 0$ .
Which statements are true?

	$z(t)$  is an even function of time.
	$z(t)$  has a jump discontinuity at  $t = 0$ .
	At time  $t = 0$ ,   $z(t) = 0$ holds.
	For  $t > 1 \ \rm ms$ ,   $z(t) = 0$ is true.

5

What is the curve shape of  $z(t)$  in response to the step-shaped input signal  $x(t)$, if the runtime is  $τ =1 \hspace{0.05cm} \rm ms$ ?
What signal value occurs at $t =1 \hspace{0.05cm} \rm ms$ ?

$z(t = 1 \rm \ ms) =\ $ $\ \rm V$

Solution

Solution

(1)  The condition  $H(f = 0) = 1$  means that the area of the impulse response is equal to  $1$ .   From this it follows that:

$$k = {1}/{\Delta t} \hspace{0.15cm}\underline{= 500\hspace{0.1cm}{ 1/{\rm s}}} .$$

(2)  Approaches 2 and 4 are correct:

• The output signal  $y(t)$  is obtained as the convolution product of  $x(t)$  and  $h(t)$.
• Convolution of two rectangles of equal width results in a triangle with its maximum at  $t = 0$:

$$y(t = 0 ) = 1\hspace{0.05cm}{\rm V}\cdot \int_{ - 1\,{\rm ms} }^{ 1\,{\rm ms} } {k \hspace{0.1cm}}{\rm d}\tau = 1\hspace{0.05cm}{\rm V}\cdot \int_{ - 1\,{\rm ms} }^{ 1\,{\rm ms} } {\frac{1}{2\,{\rm ms}} \hspace{0.1cm}}{\rm d}\tau= 1\hspace{0.05cm}{\rm V}.$$

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(3)  Approach 3 is correct:

• Convolution of two rectangles of different widths results in the trapezoidal output signal as shown in the sketch.
• The maximum value occurs in the constant range of $-0.5 \hspace{0.05cm} \rm ms$  to  $+0.5 \hspace{0.05cm} \rm ms$  and is

$$y(t = 0 ) = 1\hspace{0.05cm}{\rm V} \cdot \frac{1}{2\,{\rm ms}} \hspace{0.05cm}\cdot 1\,{\rm ms} = 0.5\hspace{0.05cm}{\rm V}.$$

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(4)  Richtig sind die Lösungsvorschläge 2, 3 und 4:

• Die Impulsantwort des Gesamtsystems lautet:   $h_{\rm HP}(t) = \delta(t) - h(t).$  Beide Anteile sind in der Skizze dargestellt.
• Durch Integration über  $h_{\rm HP}(t)$  und Multiplikation mit  $1 \hspace{0.05cm} \rm V$  kommt man zum gesuchten Signal  $z(t)$.
  In der unteren Skizze sind dargestellt:

1. das Integral über  $δ(t)$  blau,
2. die Funktion  $-σ(t)$  rot,  und
3. das gesamte Signal  $z(t)$  grün.

• $z(t)$  ist eine ungerade Funktion in  $t$  mit einer Sprungstelle bei  $t = 0$:   Der Signalwert bei  $t = 0$  liegt genau in der Mitte zwischen dem links– und dem rechteckseitigem Grenzwert und ist somit Null.
• Für  $t > 1 \hspace{0.05cm} \rm ms$  gilt ebenfalls  $z(t) = 0$, da das Gesamtsystem eine Hochpass-Charakteristik aufweist.

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(5)  Die untere Grafik zeigt die resultierende Impulsantwort  $h_{\rm HP}(t)$  und die Sprungantwort  $σ_{\rm HP}(t)$.

• Diese springt bei  $t = 0$  auf  $1$  und klingt bis zum Zeitpunkt  $t = 2 \hspace{0.05cm} \rm ms$  auf den Endwert "Null" ab.
• Zum Zeitpunkt  $t = 1\ \rm ms$  ergibt sich  $σ_{\rm HP}(t) = 0.5$.

• Das Signal  $z(t)$  ist formgleich mit der Sprungantwort  $σ_{\rm HP}(t)$, ist jedoch noch mit  $1 \hspace{0.05cm} \rm V$  zu multiplizieren.
• Der gesuchte Signalwert zur Zeit  $t_1 = 1 \hspace{0.05cm} \rm ms$  ergibt sich also zu  $z(t_1) \; \rm \underline{ = \ 0.5 \: {\rm V}}$.