Aufgaben:Exercise 2.12: Non-coherent Demodulation

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ASK Demodulation
(non-coherent)

Consider an amplitude modulated signal:

$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

Reaching the receiver based on the channel propagation time,  the signal is

$$ r(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \hspace{0.05cm}.$$

The arrangement shown here allows perfect demodulation – that is:  $v(t) = q(t)$ – without knowledge of the phase  $Δϕ_T$,  but only if the source signal  $q(t)$  satisfies certain conditions.

The two receiver-side carrier signals are:

$$ z_{\rm 1, \hspace{0.08cm}E}(t) = 2 \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm},$$

$$ z_{\rm 2, \hspace{0.08cm}E}(t) = -2 \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

$\rm LP_1$  and  $\rm LP_2$  denote two ideal  (rectangular)  low-pass filters,  each with cutoff frequency equal to the carrier frequency  $f_{\rm T}$.

We consider as (digital) source signals:

1. the unipolar square wave signal  $q_1(t)$  with dimensionless amplitude values  $0$  and  $3$,
2. the bipolar square wave signal  $q_2(t)$  with the dimensionless amplitude values  $±3$.

With respect to  $s(t)$,  these two signals result in

1. an  ASK signal, 
2. a  BPSK signal.

The nonlinear function  $v = g(b)$  is to be determined in this exercise.

Hints:

• This exercise belongs to the chapter  Further AM Variants.
• Particular reference is made to the section  Incoherent (non-coherent) Demodulation.

• The following trigonometric transformations are given:

$$ \cos(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)+ \cos(\alpha + \beta) \big],$$

$$ \sin(\alpha) \cdot \sin(\beta) = 1/2 \cdot \big[ \cos(\alpha - \beta)- \cos(\alpha + \beta) \big],$$

$$ \sin(\alpha) \cdot \cos(\beta) = 1/2 \cdot \big[ \sin(\alpha - \beta)+ \sin(\alpha + \beta) \big] \hspace{0.05cm}.$$

Questions

1

What are the signals  $b_1(t)$  and  $b_2(t)$  in both branches – after multiplier and low-pass respectively?  Which statements apply?

	$b_1(t) = q(t) · \cos(Δϕ_{\rm T})$.
	$b_2(t) = q(t) · \cos(Δϕ_{\rm T})$.
	$b_1(t) = q(t) · \sin(Δϕ_{\rm T})$.
	$b_2(t) = q(t) · \sin(Δϕ_{\rm T})$.
	$b_1(t) = b_2(t) = q(t)$.

2

What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the unipolar source signal  $q_1(t)$  is applied to the input?

$b_{\rm min} \ = \ $

$b_{\rm max} \ = \ $

3

How should the characteristic curve  $v = g(b)$  be chosen,  so that  $v(t) = q(t)$  holds?

	$v=g(b) = b^2$.
	$v=g(b) = \sqrt{b}$.
	$v=g(b) = \arctan(b).$

4

What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the bipolar source signal  $q_2(t)$  is applied to the input?

$b_{\rm min} \ = \ $

$b_{\rm max} \ = \ $

Solution

Solution

(1)  Applying the trigonometric transformations given on the exercise page and taking into account the two low-pass filters  (the components around twice the carrier frequency are removed),  we obtain:

$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$

$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$

• Thus,  the first and fourth answers  are correct.

(2)  The sum of the squares of the two partial signals gives:

$$ b(t) = b_1^2(t) + b_2^2(t)= q^2(t) \cdot \left( \cos^2(\Delta \phi_{\rm T})+ \sin^2(\Delta \phi_{\rm T})\right) = q^2(t)\hspace{0.05cm}.$$

The possible amplitude values are thus:

$$b_{\rm min}\hspace{0.15cm}\underline{ = 0},$$

$$ b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$

(3)  The second answer is correct:

$$v=g(b) = \sqrt{b} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t) = \sqrt{ q^2(t) } = q(t)\hspace{0.05cm}.$$

(4)  The result  $b(t) = q^2(t)$ – see subtask (2)  – leads here to the result:

$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$

$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$

This shows that the demodulator considered here only functions,

• if at all times   $q(t) ≥ 0$   or   $q(t) ≤ 0$   holds,
• and this is known at the receiver.