Aufgaben:Exercise 2.12: Non-coherent Demodulation

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Non-coherent
ASK Demodulation

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:

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

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

$\rm TP_1$  and  $\rm TP_2$  denote two ideal (rectangular) lowpass filters, each with cutoff frequency equal to the carrier frequency  $f_{\rm T}$ .

As (digital) source signals we consider:

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

With respect to  $s(t)$ , these two signals result in an  ASK signal  and a  BPSK signal, respectively.

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 page  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 lowpass 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 lowpass 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)  – here leads 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 function when for all times   $q(t) ≥ 0$  or  $q(t) ≤ 0$  holds, and this is known at the receiver.