Exercise 3.2Z: Bessel Spectrum

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

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

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

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

$D_2 \ = \ $

$D_3 \ = \ $

$D_{-2} \ = \ $

$D_{-3} \ = \ $

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:

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

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