Exercise 1.5Z: Sinc-shaped Impulse Response

$\rm sinc$–shaped impulse response

The impulse response of a linear time-invariant (and non-causal) system was determined as follows (see graph):

$$h(t) = 500\hspace{0.1cm}{ {\rm s}}^{-1}\cdot{\rm sinc}\big[{t}/({ 1\hspace{0.1cm}{\rm ms}})\big] .$$

The output signals $y(t)$ should be computed if various cosine oscillations of different frequency $f_0$ are applied to the input:

$$x(t) = 4\hspace{0.05cm}{\rm V}\cdot {\rm cos}(2\pi \cdot f_0 \cdot t ) .$$

Please note:

The exercise belongs to the chapter Some Low-Pass Functions in Systems Theory.
The solution can be found in the time domain or in the frequency domain. In the sample solution you will find both approaches.
The following definite integral is given:

$$\int_{ 0 }^{ \infty } \frac{\sin(u) \cdot \cos(a \cdot u)}{u} \hspace{0.15cm}{\rm d}u = \left\{ \begin{array}{c} \pi/2 \\ \pi/4 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c}{ |a| < 1,} \\{ |a| = 1,} \\ { |a| > 1.} \\ \end{array}$$

Questions

$\Delta f \ =\ $

$\ \rm kHz$

$H(f = 0) \ =\ $

$y(t = 0) \ = \ $

$\ \rm V$

$y(t = 0) \ =\ $

$\ \rm V$

$y(t = 0) \ = \ $

$\ \rm V$

Solution

(1) A comparison with the equations on the page [[Linear_and_Time_Invariant_Systems/Some_Low-Pass_Functions_in_Systems_Theory#Ideal_low-pass_filter_–_Rectangular-in-frequency |Ideal low-pass filter]] or applying the inverse Fourier transformation shows that $H(f)$ is an ideal low-pass filter:

$$H(f) = \left\{ \begin{array}{c} \hspace{0.25cm}K \\ K/2 \\ 0 \\ \end{array} \right.\quad \quad \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} {\left| \hspace{0.005cm} f\hspace{0.05cm} \right| < \Delta f/2,} \\ {\left| \hspace{0.005cm}f\hspace{0.05cm} \right| = \Delta f/2,} \\ {\left|\hspace{0.005cm} f \hspace{0.05cm} \right| > \Delta f/2.} \\ \end{array}$$

The equidistant zero-crossings of the impulse response occur at an interval of $Δt = 1 \ \rm ms$ .
From this it follows that the equivalent bandwidth is $Δf \rm \underline{ = 1 \ \rm kHz}$.
If $K = 1$ was true, then $h(0) = Δf = 1000 \cdot \rm 1/s$ should hold.
Because of the given $h(0) = 500 \cdot{\rm 1/s} = Δf/2$ the direct signal (DC) transmission factor thus is $K = H(f = 0) \; \rm \underline{= 0.5}$.

(2) This problem is most easily solved in the spectral domain.

For the output spectrum the following holds: $Y(f) = X(f)\cdot H(f) .$
$X(f)$ consists of two Dirac functions at $± f_0$ each with weight $A_x/2 =2 \hspace{0.08cm}\rm V$.
For $f = f_0 = 1 \ {\rm kHz} > Δf/2$ , however $H(f) = 0$ holds, such that $Y(f) = 0$ and hence also $y(t) = 0$ ⇒ $\underline{y(t = 0) = 0}$.

The solution in the time domain is based on convolution:

$$y(t) = x (t) * h (t) = \int_{ - \infty }^{ + \infty } {h ( \tau )} \cdot x ( {t - \tau } ) \hspace{0.1cm}{\rm d}\tau.$$

At time $t = 0$ the following is obtained considering the symmetry of the cosine function:

$$y(t = 0 ) = \frac{A_x \cdot \Delta f}{2} \cdot \int_{ - \infty }^{ + \infty } {\rm si} ( \pi \cdot \Delta f \cdot \tau ) \cdot {\rm cos}(2\pi \cdot f_0 \cdot \tau ) \hspace{0.1cm}{\rm d}\tau.$$

With the substitution $u = π · Δf · τ$ , this can also be formulated as follows:

$$y(t = 0 ) = \frac{A_x }{\pi} \cdot \int_{ 0 }^{ \infty } \frac{\sin(u) \cdot \cos(a \cdot u)}{u} \hspace{0.15cm}{\rm d}u .$$

Here, the constant is $a = 2f_0/Δf = 2$. With this value, the given integral yields zero: $y(t = 0 ) = {A_y } = 0.$

(3) The frequency response has the value $K = 0.5$ at $f = f_0 = 100 \ \rm Hz$ according to the calculations for subtask (1) . Therefore,

$$A_y = A_x/2 = 2\ \rm V$$ is obtained.

The same result is obtained by convolution according to the above equation.
For $a = 2f_0/Δf = 0.2$ the integral is equal to $π/2$ and one obtains

$$y(t = 0 ) = {A_y } = \frac{A_x}{\pi} \cdot \frac{\pi}{2} = \frac{A_x}{2} \hspace{0.15cm}\underline{= 2\,{\rm V}}.$$

(4) The transition from the band-pass to the band-stop is exactly at $f = 0.5 \ \rm kHz$ and for this singular location the following holds:

$$H(f = f_0) = K/2.$$

Thus, the amplitude of the output signal is only half as large as calculated in subtask (3) , namely $A_y \; \underline{= 1 \, \rm V}$.
The same result is obtained with $a = 2f_0/Δf = 1$ by convolution.