Aufgaben:Exercise 1.4: Nyquist Criteria

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Rectangular Nyquist spectrum

Given by the sketch is the spectrum  $G(f)$  of the basic detection pulse,  where the parameter  $A$  is still to be determined.  Among other things it is to be checked whether this basic detection pulse fulfills one of the two Nyquist criteria.  These are:

• The  first Nyquist criterion  is fulfilled if for the spectral function holds:

$$\sum_{k = -\infty}^{+\infty} G(f - {k}/{T} ) = {\rm const.}$$

In this case,  the pulse  $g(t)$  has zero crossings at  $t = ν \cdot T$  for all integer values of  $ν$  except  $ν = 0$.  For the entire exercise,  $T = 0.1 \, \rm ms$  is assumed.

• If the  second Nyquist criterion  is satisfied,  $g(t)$  has zero crossings at  $\pm 1.5 T$,  $\pm 2.5 T$, etc.

Notes:

• The exercise belongs to the chapter  "Properties of Nyquist Systems".

• The two equations are assumed to be known:

$$X(f) = \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}|f| < f_0 \hspace{0.05cm}, \\ {\rm{for}}\hspace{0.15cm}|f| > f_0 \hspace{0.08cm} \\ \end{array} \hspace{0.4cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ \hspace{0.4cm} x(t) =2 \cdot A \cdot f_0 \cdot {\rm sinc}(2 f_0 T) \hspace{0.05cm},$$

$$\sin(\alpha) \cdot \cos (\beta) = \frac{1}{2} \cdot \big[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\big] \hspace{0.05cm}.$$

Questions

1

Does the given pulse  $g(t)$  satisfy the first Nyquist criterion?

	The first Nyquist criterion is fulfilled.
	The first Nyquist criterion is not fulfilled.

2

Determine the parameter  $A$  such that  $g(t = 0) = 2\, \rm V$  holds.

$A \ = \ $ $ \ \rm mV/Hz$

3

Calculate  $g(t)$  from  $G(f)$  by applying the inverse Fourier transform.  What  (normalized)  function value is obtained at  $t = T$?

$ g(t = T)/g(t = 0) \ = \ $

4

Which  (normalized)  value results for  $t = 2.5T$?

$g(t = 2.5 T)/g(t = 0)\ = \ $

5

Does the pulse  $g(t)$  satisfy the second Nyquist criterion?

	The second Nyquist criterion is fulfilled.
	The second Nyquist criterion is not fulfilled.

Solution

Solution

(1)  The following diagram shows the spectrum (the index "Per" here stands for "Periodic Continuation"):

$$G_{\rm Per}(f) = \sum_{k = -\infty}^{+\infty} G(f - \frac{k}{T} ) \hspace{0.05cm}.$$

• The indexing variable $k = 0$ indicates the original spectral function $G(f)$. This is filled in gray.
• The spectrum shifted to the right by the value $1/T = 10\, \rm kHz$ belongs to $k = 1$ and is marked in green, while $k = -1$ leads to the function highlighted in yellow.
• The red and blue areas mark the contributions of the indexing variables $k = 2$ and $k = - 2$.

Illustration of the first Nyquist criterion

It can be seen that $G_{\rm Per}(f)$ is constant. From this it follows that the first Nyquist criterion is fulfilled. Consequently, solution 1 is correct.

(2)  Due to the Fourier integrals the following relation holds:

$$g(t=0) = \int_{-\infty}^{\infty}G(f) \,{\rm d} f = A \cdot ( 2\,{\rm kHz}+6\,{\rm kHz}+2\,{\rm kHz})= A \cdot 10\,{\rm kHz}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}A = \frac{g(t=0)}{10\,{\rm kHz}} = \frac{2\,{\rm V}}{10\,{\rm kHz}} \hspace{0.1cm}\underline {= 0.2 \, {\rm mV/Hz}} \hspace{0.05cm}.$$

(3)  Let $g(t) = g_{1}(t) +g_{2}(t)$, where

• $g_{1}(t)$ contains the spectral components in the interval $\pm 3 \, \rm kHz$ and
• $g_{2}(t)$ those between $13 \, \rm kHz$ and $15 \, \rm kHz$ (and between $-13 \, \rm kHz$ and $-15 \, \rm kHz$).

With the Fourier correspondence given, the two components are:

$$g_1(t) \ = \ A \cdot 6\,{\rm kHz} \cdot {\rm si}(\pi \cdot 6\,{\rm kHz} \cdot t) \hspace{0.05cm},$$

$$g_2(t) \ = \ A \cdot 2\,{\rm kHz} \cdot{\rm si}(\pi \cdot 2\,{\rm kHz} \cdot t) \cdot 2 \cdot {\rm cos}(2 \pi \cdot 14\,{\rm kHz} \cdot t) \hspace{0.05cm}.$$

The second equation follows from the relation:

$$G_2(f) = \left[ \delta(f + 14\,{\rm kHz}) + \delta(f - 14\,{\rm kHz})\right] \star \left\{ \begin{array}{c} A \\ 0 \\\end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}}\hspace{0.15cm}|f| < 1\,{\rm kHz} \hspace{0.05cm}, \\ {\rm{for}}\hspace{0.15cm}|f| > 1\,{\rm kHz} \hspace{0.05cm}. \\ \end{array}$$

The bottom diagram shows the numerically determined time history $g(t)$. For the time $t = T = 0.1\, \rm ms$ (yellow square) we get:

$$g_2(t = T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.2 \cdot \pi )\cdot \cos (2.8 \cdot \pi) = \frac{ A \cdot 4\,{\rm kHz}}{0.2 \cdot \pi}\cdot {\rm sin}(0.2 \cdot \pi )\cdot\cos (0.8 \cdot \pi) = \frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot [{\rm sin}(-0.6 \cdot \pi)+ {\rm sin}(\pi)] $$

$$\Rightarrow \hspace{0.3cm} g_2(t = T ) = -\frac{ A \cdot 10\,{\rm kHz}}{ \pi}\cdot {\rm sin}(0.6 \cdot \pi).$$

For the first component $g_1(t)$, at time $t = T$:

$$g_1(t = T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(0.6 \cdot \pi )= \frac{ A \cdot 6\,{\rm kHz}}{0.6 \cdot \pi}\cdot {\rm sin}(0.6 \cdot \pi )= - g_2(t = T )$$

$$\Rightarrow \hspace{0.3cm} g(t = T ) = g_1(t = T ) + g_2(t = T )\hspace{0.1cm}\underline {= 0 } \hspace{0.05cm}.$$

This result is not surprising because of the Nyquist property.

Higher frequency Nyquist pulse

(4)  For $t = 2.5 T$ (green square), the following partial results are obtained:

$$g_1(t = 2.5 T ) = A \cdot 6\,{\rm kHz} \cdot {\rm si}(1.5 \cdot \pi )= \frac{ A \cdot 6\,{\rm kHz}}{1.5 \cdot \pi}\cdot {\rm sin}(1.5 \cdot \pi )= - \frac{ A \cdot 4\,{\rm kHz}}{ \pi}\hspace{0.05cm},$$

$$g_2(t = 2.5 T ) = 2A \cdot 2\,{\rm kHz} \cdot {\rm si}(0.5 \cdot \pi )\cdot \cos (7 \cdot \pi)=- \frac{ A \cdot 8\,{\rm kHz}}{ \pi} $$

$$\Rightarrow \hspace{0.3cm} g(t = 2.5 T ) = g_1(t = 2.5 T ) +g_2(t = 2.5 T ) = - \frac{ A \cdot 12\,{\rm kHz}}{ \pi} \hspace{0.05cm}.$$

Considering $g(t = 0) = A \cdot 10 \ \rm kHz$, we get:

$$\frac{g(t = 2.5 T )}{g(t = 0)} = -\frac{ 1.2}{ \pi} \hspace{0.1cm}\underline {= -0.382 } \hspace{0.05cm}.$$

(5)  The second Nyquist criterion states that the Nyquist pulse $g(t)$ has zero crossings at $\pm 1.5T, \pm 2.5T, \pm 3.5T, ...$

• According to the result from (4) this condition is not fulfilled here.
• Therefore solution 2 is correct.