Aufgaben:Exercise 1.4Z: Complex Nyquist Spectrum

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Complex Nyquist spectrum

Consider a pulse  $g(t)$  with spectrum  $G(f)$  according to the sketch.  One recognizes from this representation:

• The real part of  $G(f)$  is trapezoidal with the corner frequencies  $f_{1} = 3 \, \rm kHz$  and  $f_{2} = 7 \, \rm kHz$.  In the range  $|f| < f_{1}$:

$${\rm Re}\big[G(f)\big] = A = 10^{-4} \, \rm V/Hz.$$

• The imaginary part of  $G(f)$  is always assumed to be  ${\rm Im}\big[G(f)\big] =0$   for subtasks  (1)  to  (5).  In this case  $g(t)$  is certainly a Nyquist pulse.
• From subtask  (6),  the imaginary part  ${\rm Im}[G(f)]$  in the range  $f_{1} \leq | f | \leq f_{2}$  has a triangular shape with the values  $\pm B$  at the triangle peaks.

It is necessary to check whether the pulse  $g(t)$  satisfies the first Nyquist condition even with complex spectrum:

$$g(\nu T) = \left\{ \begin{array}{c} g_0 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{1}c} {\rm{for}} \\ {\rm{for}} \\ \end{array}\begin{array}{*{20}c} \nu = 0 \hspace{0.05cm}, \\ \nu \ne 0 \hspace{0.1cm}. \\ \end{array}$$

In the course of this exercise,  reference is made to the following descriptive quantities:

• The  Nyquist frequency  gives the symmetry point of the rolloff:

$$f_{\rm Nyq}= \frac{1}{2T}= \frac{f_1 +f_2 } {2 }\hspace{0.05cm}.$$

• The  rolloff factor  is a measure of the transition steepness:

$$r = \frac{f_2 -f_1 } {f_2 +f_1 } \hspace{0.05cm}.$$

Notes:

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

• The inverse Fourier transform  $g(t)$  of a trapezoidal Nyquist spectrum with rolloff factor  $r$  can be assumed to be known:

$$g ( t )= g_0 \cdot {\rm sinc} \left ( { t}/{T}\right)\cdot {\rm sinc} \left ( { r \cdot t}/{T}\right),\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$

• A triangular low–pass spectrum  $G(f)$ limited to  $| f | < f_{0}$  and where  $G(f = 0) = B$,  after inverse Fourier transform,  leads to the time function

$$g ( t )= B \cdot f_0 \cdot {\rm sinc}^2 \left ( { f_0 \cdot t}\right)\hspace{0.05cm},\hspace{0.3cm} {\rm sinc}(x)= \sin(\pi \cdot x)/(\pi \cdot x).$$

Questions

1

For the first questions,  let  $B = 0$.  What is the Nyquist frequency?

$f_{\rm Nyq} \ = \ $ $\ \rm kHz$

2

Which rolloff factor  $r$  is present here?

$r \ = \ $

3

Calculate the maximum value  $g_{0}$  of the Nyquist pulse  $g(t)$.

$g_{0} \ = \ $ $\ \rm V$

4

Further let  $B=0$.  Calculate  $g(t)$  for the time points  $t = 100\, µ \rm s$  and  $t = 200\, µ \rm s$.

$g(t = 100\, µ \rm s) \ = \ $ $\ \rm V$

$g(t = 200\, µ \rm s) \ = \ $ $\ \rm V$

5

Calculate the pulse value at time  $t = 250\ µ \rm s$.

$g(t = 250\, µ \rm s) \ = \ $ $\ \rm V$

6

Which statements are true for  $B \neq 0$?  $G(f)$  is then complex-valued.

	The Nyquist condition is satisfied if the triangle function is between  $3 \, \rm kHz$  and  $7 \, \rm kHz$  as in the diagram.
	The Nyquist condition is fulfilled if the triangle function lies symmetrically between  $3 \, \rm kHz$  and  $5 \, \rm kHz$.
	The Nyquist condition is fulfilled if the triangle function lies symmetrically between  $4.5 \, \rm kHz$  and  $5.5 \, \rm kHz$.

7

Calculate  $g(t)$  for  $t = 250\, µ \rm s$  and  $B = A = 10^{–4} \, \rm V/Hz$   ⇒   complex spectral function.

$g(t = 250\ µ \rm s) \ = \ $ $\ \rm V$

Solution

Solution

(1)  The Nyquist frequency specifies the symmetry point of the rolloff. The following holds true:

$$f_{\rm Nyq}= \frac{f_1 +f_2 } {2 }= \frac{3\, {\rm kHz} + 7\, {\rm kHz}} {2 } \hspace{0.1cm}\underline { = 5\, {\rm kHz}} \hspace{0.05cm}.$$

(2)  The rolloff factor is also determined by the two corner frequencies $f_{1}$ and $f_{2}$:

$$r = \frac{f_2 -f_1 } {f_2 +f_1 } = \frac{7\, {\rm kHz} - 3\, {\rm kHz}} {7\, {\rm kHz} + 3\, {\rm kHz} }\hspace{0.1cm}\underline { = 0.4 }\hspace{0.05cm}.$$

(3)  For a pulse with real low–pass spectrum, the maximum is always at $t = 0$ and it holds:

$$g_0 = g(t=0) = \int_{-\infty}^{+\infty}G(f) \,{\rm d} f = A \cdot 2 f_{\rm Nyq} = 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 5 \cdot10^{3} \,{\rm Hz}\hspace{0.1cm}\underline { = 1\,{\rm V}}\hspace{0.05cm}.$$

(4)  For the Nyquist pulse, the equidistant zero crossings occur at the interval $T = 1/(2f_{\rm Nyq}) = 100 \, \rm µ s$. From this one obtains directly:

$$g(t= 100\,{\rm µ s}) = \ \hspace{0.1cm}\underline { g(T) = 0,}$$

$$g(t= 200\,{\rm µ s}) = \ \hspace{0.1cm}\underline {g(2T) = 0} \hspace{0.05cm}.$$

This result also follows from the given equation with $r = 0.4$:

$$g ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot 0.4 \cdot t}/{T}\right) \hspace{0.05cm}.$$

Responsible for satisfying the first Nyquist condition is the first term.

(5)  According to the equation given in (4):

$$g(t= 250\,{\rm µ s})= g_0 \cdot {\rm si} ( {2.5 \cdot \pi })\cdot {\rm si} ( \pi )\hspace{0.1cm}\underline { = 0} \hspace{0.05cm}.$$

This zero is due to the second term and does not lie in the Nyquist time grid $\nu T$

(6)  For the following derivation, let $g(t)= g_{\rm R}(t) + g_{\rm I}(t) \hspace{0.05cm},$ where $g_{\rm R}(t)$ is the real part and $g_{\rm I}(t)$ is the imaginary part of $G(f)$.

• The first part is calculated exactly as in point (4):

$$g_{\rm R} ( t )= g_0 \cdot {\rm si} \left ( {\pi \cdot t}/{T}\right)\cdot {\rm si} \left ( {\pi \cdot 0.4 \cdot t}/{T}\right) \hspace{0.05cm}.$$

• To satisfy the first Nyquist criterion, the following must hold for the imaginary part with $1/T = 10 \, \rm kHz$:

$$\sum_{k = -\infty}^{+\infty} {\rm Im}\left[G \left ( f - {k}/{T} \right)\right]= 0 \hspace{0.05cm}.$$

• With the given corner frequencies $f_{1} = 3 \, \rm kHz$ and $f_{2} = 7 \ \rm kHz$, the two triangles are around $\pm 5\, \rm kHz$, so the above equation is satisfied.
• The same is true for $f_{1} = 4.5\, \rm kHz$ and $f_{2} = 5.5 \, \rm kHz$.
• In contrast, the triangle peaks with $f_{1} = 3\, \rm kHz$ and $f_{2} = 5 \, \rm kHz$ are at $\pm 4 \ \rm kHz$.
• In this case the triangle functions do not cancel due to the periodic continuation and the Nyquist condition is not fulfilled.

Therefore the correct solutions are 1 and 3.

(7)  With the result  $g_{\rm R}(2.5T) = 0$  from (3), it follows  $g(2.5T) = g_{\rm I}(2.5T)$, where  $g_{\rm I}(t)$  is the Fourier inverse transformation of ${\rm j}\cdot \ G_{\rm I}(f)$. It holds:

$${\rm j} \cdot G_{\rm I}(f) = {\rm j} \cdot\left[ \delta(f + f_{\rm Nyq}) - \delta(f - f_{\rm Nyq})\right] \star D(f) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} g_{\rm I}(t) = 2 \cdot {\rm sin} ( 2 \pi\cdot f_{\rm Nyq} \cdot t )\cdot d(t)\hspace{0.05cm}.$$

• The sine function enforces the required zero crossings at multiples of $T = 100 \, \rm µ s$.
• $D(f)$ is a triangular function around $f = 0$ with $D(f = 0) = B$ and one-sided width $f_{0}= f_{2} – f_{\rm Nyq} = f_{\rm Nyq} – f_{1} = 2 \, \rm kHz$.
• For the associated time function we can thus write according to the specification:

$$g_{\rm I}(t ) = 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2 \pi\cdot f_{\rm Nyq} \cdot t)\cdot {\rm si}^2(\pi\cdot f_{\rm 0} \cdot t) \hspace{0.05cm}.$$

• In particular, for time $t = 250 \, \rm µ s$ (green square):

$$g(t = 2.5 T) = g_{\rm I}(t = 2.5 T) = \ 2 \cdot B \cdot f_0 \cdot{\rm sin} ( 2.5 \pi )\cdot {\rm si}^2(\frac{\pi}{2}) = \ \frac{8}{\pi^2} \cdot 10^{-4 }\,\frac{\rm V}{\rm Hz}\cdot 2 \cdot 10^{3} \,{\rm Hz}\hspace{0.1cm}\underline {= 0.162\,{\rm V}} \hspace{0.05cm}.$$

Asymmetric Nyquist pulse

The diagram shows the change of the time function due to the imaginary part (green time course):

• The result is now an asymmetric function curve $g(t)$, shown in blue.
• However, the zero crossings of $g_{\rm R}(t)$ at the distance $T$ remain.