Exercise 1.1Z: VHF II Broadcasting

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Model of an FM radio transmitter

Electromagnetic waves in a frequency range from  $\rm 30$  to  $\text{300 MHz}$  – corresponding to wavelengths between one and ten meters – are referred to as  "ultra-short waves".

In common parlance, USW is also understood to mean the VHF II  ("Very High Frequency")  band, from  $\text{87.5}$  to  $\text{108 MHz}$,  which is still primarily used in Central Europe today for transmitting frequency-modulated radio programs using analog technology.

The entire frequency band is divided into several channels,  each with a channel spacing of   $\text{300 kHz}$.

The diagram displays the principle as a flow chart:

• A total of  $K$  signals  $q_k(t)$  with different carrier frequencies  $f_1$, $f_2$, ... , $f_k$, ... ,  $f_K$  are modulated and added together.
• The summed signal is then beamed from a transmitter after power amplification. 
• This outgoing signal is referred to as the transmission signal  $s(t)$.

Hints:

• This exercise belongs to the chapter   Objectives of modulation and demodulation.
• Particular reference is made to the page  Channel bundling – Frequency Division Multiplexing.

Questions

1

Which multiplexing method is used here?

	Frequency Division Multiplexing.
	Time Division Multiplexing.

2

What is the maximum number of programs that could be transmitted in the VHF II band?

$K \ = \ $

3

Which of the following statements are true?

	The VHF channels have higher bandwidth than those for long-wave, medium-wave and short-wave broadcasting.
	The range of FM radio waves is greater than that of long-wave, medium-wave and short-wave broadcasting.
	The signal arriving at the receiver  $r(t)$  is almost indistinguishable from the transmission signal  $s(t)$.
	The functional unit of a radio receiver for channel separation  ("setting the radio station")  is the  "tuner".

Solution

Solution

(1)  Answer 1 is correct.

• FDM  ("Frequency Division Multiplexing").
• The alternative multiplexing method,   TDM  ("Time Division Multiplexing")  can only be implemented with a digital communications system.

(2)  From the total bandwidth  $\text{20.5 MHz}$  and the channel bandwidth  $\text{0.3 MHz}$  we get $\underline{K = 68}$.

(3)  Answers 1 and 4 are correct:

• The channel spacing and thus the bandwidth  $\text{300 kHz}$  available for a channel is significantly larger in the VHF range than for long, medium and short wave broadcasting.  Although the frequency modulation used in FM broadcasting is characterized by better quality,  it also requires more bandwidth.
• For comparison,  in the medium wave range,  amplitude modulation and a channel spacing of  $\text{9 kHz}$  is always used  (in Europe)  
  ⇒   the source signal bandwidth is  $B_{\rm LF} = \text{9 kHz/2 = 4.5 kHz}$.
• The range of VHF is smaller than in the other frequency ranges because VHF radio waves are not reflected by the ionosphere.  Therefore,  a VHF transmission network usually consists of quite a large number of transmitters,  which are set up at small distances from each other – and mostly at an elevation   ⇒   answer 2 is incorrect.
• The signal arriving at the receiver has a much lower level than the transmission signal $s(t)$,  due to atmospheric attenuation,  which increases at least quadratically with distance  ⇒   answer 3 is incorrect.
• In a radio receiver, the tuner indeed has the task of channel separation   ⇒   answer 4 is correct.