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Exercise 3.2Z: Bessel Spectrum

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Progression of Bessel functions

Consider the complex signal

x(t)=ej⋅η⋅sin(ω0⋅t).

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  T0=2π/ω0, the Fourier series representation is:
x(t)=+∞∑n=−∞Dn⋅ej⋅n⋅ω0⋅t,
Dn=1T0⋅∫+T0/2−T0/2x(t)⋅e−j⋅n⋅ω0⋅tdt.
  • These complex Fourier coefficients can be expressed using n–th order Bessel functions of the first kind:
Jn(η)=12π⋅∫+π−πej⋅(η⋅sin(α)−n⋅α)dα.
  • These are shown on the graph in the range  0≤η≤5 . For negative values of n  one obtains:
J−n(η)=(−1)n⋅Jn(η).
  • The series representation of the Bessel functions is:
Jn(η)=∞∑k=0(−1)k⋅(η/2)n+2⋅kk!⋅(n+k)!.
  • 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:
Jn(η)=2⋅(n−1)η⋅Jn−1(η)−Jn−2(η).





Hints:


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  Dn  together with the Bessel functions of the first kind   ⇒   Jn(η).  What relationships can be seen?

All  Dn  are equal to  Jη(0).
 Dn=Jn(η) holds.
 Dn=−Jη(n) holds.

3

What are the properties of the Fourier coefficients?

All  Dn  are purely real.
All  Dn  are purely imaginary.

4

For  η=2 , the coefficients are  D0=0.224  and  D1=0.577.  From this, calculate the coefficients  D2  and  D3.

D2 = 

D3 = 

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  T0=2π/ω0 :
x(t+k⋅T0)=ej⋅η⋅sin(ω0⋅(t+k⋅T0))=ej⋅η⋅sin(ω0⋅t+k⋅2π)=ej⋅η⋅sin(ω0⋅t)=x(t).
  • This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.


(2)  The Fourier coefficients are:

Dn=1T0⋅∫+T0/2−T0/2ej⋅η⋅sin(ω0⋅t)⋅e−j⋅n⋅ω0⋅tdt.
  • Combining the two terms and after substituting   α=ω0·t , we get:
Dn=12π⋅∫+π−πej⋅(η⋅sin(α)−n⋅α)dα=Jn(η).
  • Thus, the second answer is correct.


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

Dn=12π⋅∫+π−πcos(η⋅sin(α)−n⋅α)dα+j2π⋅∫+π−πsin(η⋅sin(α)−n⋅α)dα.
  • The integrand of the first integral is an even function of  α:
I1(−α)=cos(η⋅sin(−α)+n⋅α)=cos(−η⋅sin(α)+n⋅α)=cos(η⋅sin(α)−n⋅α)=I1(α).
  • In contrast, the second integrand is an odd function:
I2(−α)=sin(η⋅sin(−α)+n⋅α)=sin(−η⋅sin(α)+n⋅α)=−sin(η⋅sin(α)−n⋅α)=−I2(α).
  • Thus, the second integral vanishes and, taking symmetry into account, we obtain:
Dn=1π⋅∫π0cos(η⋅sin(α)−n⋅α)dα.
  • Thus, the correct solution is Answer 1.


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

D2=D1−D0=0.577−0.224=0.353_,
D3=2⋅D2−D1=2⋅0.353−0.577=0.129_.


(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}.


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