Aufgaben:Exercise 3.2Z: Bessel Spectrum

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Progression of Bessel functions

Consider the complex signal

$$x(t) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t) }\hspace{0.05cm}.$$

For example, the equivalent low-pass signal at the output of an angle modulator (PM, FM) can be represented in this form if appropriate normalizations are made.

• When  $T_0 = 2π/ω_0$, the Fourier series representation is:

$$x(t) = \sum_{n = - \infty}^{+\infty}D_n \cdot{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t }\hspace{0.05cm},$$

$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}x(t) \cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

• These complex Fourier coefficients can be expressed using $n$–th order Bessel functions of the first kind:

$${\rm J}_n (\eta) = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$

• These are shown on the graph in the range  $0 ≤ η ≤ 5$ . For negative values of $n$  one obtains:

$${\rm J}_{-n} (\eta) = (-1)^n \cdot {\rm J}_{n} (\eta)\hspace{0.05cm}.$$

• The series representation of the Bessel functions is:

$${\rm J}_n (\eta) = \sum\limits_{k=0}^{\infty}\frac{(-1)^k \cdot (\eta/2)^{n \hspace{0.05cm} + \hspace{0.05cm} 2 \hspace{0.02cm}\cdot \hspace{0.05cm}k}}{k! \cdot (n+k)!} \hspace{0.05cm}.$$

• If the function values for  $n = 0$  and  $n = 1$  are known, the Bessel functions for  $n ≥ 2$  can be determined from them by iteration:

$${\rm J}_n (\eta) = \frac{2 \cdot (n-1)}{\eta} \cdot {\rm J}_{n-1} (\eta) - {\rm J}_{n-2} (\eta) \hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  Phase Modulation.
• Particular reference is made to the page  Equivalent low-pass signal in phase modulation.
• The values of the Bessel functions can be found in collections of formulae in table form.
• You can also use the interactive applet   Bessel functions of the first kind  to solve this task.

Questions

1

What are the properties of the signal  $x(t)$?

	$x(t)$  is imaginary for all times  $t$ .
	$x(t)$  is periodic.
	The spectral function $X(f)$  is obtained via the Fourier integral.

2

Write the Fourier coefficients  $D_n$  together with the Bessel functions of the first kind   ⇒   ${\rm J}_n(η)$.  What relationships can be seen?

	All  $D_n$  are equal to  ${\rm J}_η(0)$.
	$D_n = {\rm J}_n(η)$ holds.
	$D_n = -{\rm J}_η(n)$ holds.

3

What are the properties of the Fourier coefficients?

	All  $D_n$  are purely real.
	All  $D_n$  are purely imaginary.

4

For  $η = 2$ , the coefficients are  $D_0 = 0.224$  and  $D_1 = 0.577$.  From this, calculate the coefficients  $D_2$  and  $D_3$.

$D_2 \ = \ $

$D_3 \ = \ $

5

What are the Fourier coefficients  $D_{-2}$  and  $D_{-3}$ ?

$D_{-2} \ = \ $

$D_{-3} \ = \ $

Solution

Solution

(1)  Only the second answer is correct:

• $x(t)$  is a complex signal that only becomes real in exceptional cases, for example at time   $t = 0$.
• A purely imaginary value (at certain times) can only result when $η ≥ π/2$    ⇒   Answer 1 is incorrect.
• For example, when  $T_0 = 2π/ω_0$ :

$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$

• This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.

(2)  The Fourier coefficients are:

$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t) }\cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$

• Combining the two terms and after substituting   $α = ω_0 · t$ , we get:

$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm} = {\rm J}_n (\eta) .$$

• Thus, the second answer is correct.

(3)  Using Euler's theorem, the Fourier coefficients can be represented as follows:

$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha + \frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$

• The integrand of the first integral is an even function of  $\alpha$:

$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}= {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$

• In contrast, the second integrand is an odd function:

$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}= -{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$

• Thus, the second integral vanishes and, taking symmetry into account, we obtain:

$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$

• Thus, the correct solution is Answer 1.

(4)  According to the formula for iterative calculation, when  $η = 2$:

$$ D_2 = D_1 - D_0 = 0.577 - 0.224 \hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$

$$D_3 = 2 \cdot D_2 - D_1 = 2 \cdot 0.353 - 0.577 \hspace{0.15cm}\underline {= 0.129} \hspace{0.05cm}.$$

(5)  Due to the given symmetry relation, it further holds that:

$$ D_{–2} = D_2\hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$

$$D_{–3} = -D_3 \hspace{0.15cm}\underline {= -0.129} \hspace{0.05cm}.$$