Exercise 2.4: Frequency and Phase Offset

Model of a synchronous demodulator

Consider the source signal $q(t) = A_{\rm 1} \cdot \cos(2 \pi f_{\rm 1} t ) +A_{\rm 2} \cdot \sin(2 \pi f_{\rm 2} t )$ with the signal parameters

$$ A_1 = 2\,{\rm V}, \hspace{0.15cm}f_1 = 2\,{\rm kHz} \hspace{0.05cm},$$

$$A_2 = 1\,{\rm V}, \hspace{0.15cm}f_2 = 5\,{\rm kHz}\hspace{0.05cm}.$$

This signal is DSB amplitude-modulated.

Thus, the modulated signal $s(t)$ has spectral components at $±45$ kHz, $±48$ kHz, $±52$ kHz and $±55$ kHz.
It is also known that the transmitter-side carrier $z(t)$ is sinusoidal $(ϕ_{\rm T} = -90^\circ)$.

The demodulation to be performed with the circuit sketched here, which is defined by the following parameters
("E" ⇒ "empfägerseitig" ⇒ "receiver-side"):

Amplitude $A_{\rm E}$ (no unit),
frequency $f_{\rm E}$,
phase $ϕ_{\rm E}$.

The $H_{\rm E}(f)$ block represents an ideal, rectangular low-pass filter, which is suitably dimensioned.

Hints:

This exercise belongs to the chapter Synchronous Demodulation.
Particular reference is made to the pages Influence of a frequency offset and Influence of a phase offset.

Take the following trigonometric transformations into account:

$$\cos(\alpha)\cdot \cos(\beta) = {1}/{2} \cdot \left[ \cos(\alpha-\beta) + \cos(\alpha+\beta)\right] \hspace{0.05cm},$$

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

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

Questions

	The demodulator would work better for DSB-AM with carrier.
	The carrier would unnecessarily increase the transmit power.
	The correct dimensioning of the low-pass $H_{\rm E}(f)$ is essential.
	One could also use an envelope demodulator.
	Envelope demodulation is only applicable for $m \le 1$ .

$A_{\rm E} \ = \ $

$f_{\rm E} \ \hspace{0.05cm} = \ $

$\ \text{kHz}$

$\phi_{\rm E} \ = \ $

$\ \text{deg}$

$v(t = 0)\ = \ $

$\ \text{V}$

$v(t = 0)\ = \ $

$\ \text{V}$

	It holds that $v(t) = q(t) · \cos(2π · Δ\hspace{-0.05cm}f_{\rm T} · t).$
	$v(t)$ contains a spectral component at $2$ kHz.
	$v(t)$ contains a spectral component at $4$ kHz.
	$v(t)$ contains a spectral component at $6$ kHz.

Solution

(1) Answers 2, 3 and 5 are correct:

Envelope demodulation is not applicable for DSB-AM without carrier and a modulation depth of $m > 1$.
The performance of the synchronous demodulator is not increased by the additional carrier component, but only leads to an unnecessary increase in the transmit power to be applied.
The third statement is also correct. The solution to Exercise 2.4Z shows the effects of omitting or incorrectly dimensioning $H_{\rm E} (f)$.

(2) As the name "synchronous demodulator" already implies, the signals $z(t)$ and $z_{\rm E} (t)$ must be synchronous in frequency and phase:

$$f_{\rm E} = f_{\rm T} \hspace{0.15cm}\underline {= 50\,{\rm kHz}}, \hspace{0.15cm}\phi_{\rm E} = \phi_{\rm T} \hspace{0.15cm}\underline {= - 90^{\circ}} \hspace{0.05cm}.$$

The carrier frequency $f_{\rm T} $ at the transmitter can be determined from the transmission spectrum $S(f)$. In the case of perfect synchronisation:

$$v(t) = {A_{\rm E}}/{2} \cdot q(t) + {A_{\rm E}}/{2} \cdot q(t)\cdot \cos(2 \cdot \omega_{\rm T} \cdot t ) \hspace{0.05cm}.$$

The second term is removed by the low-pass filter. Thus, with $A_{\rm E}\hspace{0.15cm}\underline{ = 2}$, $v(t) = q(t)$ holds.

(3) In the theory section, it was shown that in general for DSB-AM and synchronous demodulation:

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

Even insufficient phase synchronisation does not lead to distortions, only to a frequency-independent attenuation.
With $ϕ_{\rm T} =-90^\circ$ and $ϕ_{\rm E} = -120^\circ$ ⇒ $Δϕ_{\rm T} = -30^\circ$:

$$ v(t) = \cos(30^{\circ}) \cdot q(t)= 0.866 \cdot q(t) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t= 0) = 0.866 \cdot A_1 \hspace{0.15cm}\underline {= 1.732\,{\rm V}}\hspace{0.05cm}.$$

(4) Now the phase difference is $Δϕ_{\rm T} = 90^\circ$ and we get $v(t) \equiv 0$.

It is pointless to discuss whether this is still a distortion-free system.
The result $v(t) \equiv 0$ is due to the fact that cosine and sine are orthogonal functions.
This principle is made use of, for example, in what is known as quadrature amplitude modulation..

(5) The equation for the signal after multiplication is:

$$b(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t - 90^{\circ}) \cdot 2 \cdot \cos(\omega_{\rm E} \cdot t - 90^{\circ})= 2 \cdot q(t) \cdot \sin(\omega_{\rm T} \cdot t ) \cdot \sin(\omega_{\rm E} \cdot t )\hspace{0.05cm}.$$

This result can also be rewritten using the trigonometric transformation

$$\sin(\alpha)\cdot \sin(\beta) = {1}/{2} \cdot \left[ \cos(\alpha-\beta) - \cos(\alpha+\beta)\right]$$

as follows:

$$ b(t) = q(t) \cdot \cos((\omega_{\rm T} - \omega_{\rm E}) \cdot t ) + q(t) \cdot \cos((\omega_{\rm T} + \omega_{\rm E}) \cdot t ) \hspace{0.05cm}.$$

The second term lies in the vicinity of $2f_{\rm T}$ for $f_{\rm E} = f_{\rm T}$ and is removed by the low-pass.
With the frequency difference $Δ\hspace{-0.05cm}f_{\rm T} = f_{\rm E} - f_{\rm T}= 1$ kHz, this leaves:

$$ v(t) = q(t) \cdot \cos(2 \pi \cdot \Delta \hspace{-0.05cm}f_{\rm T} \cdot t) \hspace{0.05cm}.$$

The first statement is correct. This states that now the signal $v(t)$ becomes quieter and louder again after demodulation according to a cosine function (a "beat").
The cosine component of $q(t)$ with frequency $f_1 = 2\text{ kHz}$ now becomes two components (each of half the amplitude) at $1\text{ kHz}$ and $3\text{ kHz}$.
Similarly, the sink signal does not include a component at $f_2 = 5\text{ kHz}$, only components at $4\text{ kHz}$ and at $6\text{ kHz}$:

$$1\,{\rm V} \cdot \sin(2 \pi \cdot 5\,{\rm kHz} \cdot t)\cdot \cos(2 \pi \cdot 1\,{\rm kHz} \cdot t) = 0.5\,{\rm V} \cdot \sin(2 \pi \cdot 4\,{\rm kHz} \cdot t) + 0.5\,{\rm V} \cdot \sin(2 \pi \cdot 6\,{\rm kHz} \cdot t)\hspace{0.05cm}.$$

Answers 1, 3 and 4 are correct.