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$\DeclareMathOperator{\sinc}{sinc}$ Given 3 signals

$x_1(t) = \sinc(t) , x_2(t) = 10\cos(2\pi f_2t)$ and $x_3(t) = 5\cos(2\pi f_1t)$,

where $f_2 = 100\text{ kHz}$ and $f_1 = 1\text{ kHz}$.

I must create the general form of the simple amplitude modulation AM ($S(t)= A_c\left[1+k_am(t)\right]\cos(2\pi f_ct)$) , the DSB-SC modulation ( $S(t)=A_cm(t)\cos(2\pi f_ct)$ ), the PM phase modulation and the FM frequency modulation, but I must use the sum of the 2 signals to create the information signal ( $m(t)$ )and the third must be the carrier.

What I can't understand is which one must be the carrier in every single case, the one with the highest frequency in our example $x_2(t)$ or the one with lowest?

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  • $\begingroup$ I tried to fix your question, because it was really unclear what was formula symbols with indices and what products of formula symbols. Please check whether it makes sense! Especially, you define $f_1$ and $f_2$, but there's no $f_1$, and $f_3$ remains undefined. $\endgroup$ Aug 26, 2023 at 20:03
  • $\begingroup$ Thanks a lot , there is no f3 that was a mistake the question remains the same : which one should i use as a carrier ? $\endgroup$
    – Thanos
    Aug 26, 2023 at 20:41
  • $\begingroup$ I'd go with the one called "$f_c$", $c$ as in $c$arrier! $\endgroup$ Aug 26, 2023 at 21:27

1 Answer 1

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The $\mathrm{sinc}$ function has the following Fourier Transform also shown below:

Sinc-rect

Here in your question the amplitudes do not matter much.

Consider the DSB-SC case:

We obviously cannot choose the $\mathrm{sinc}$ as our carrier since it does not have any modulating properties. We want something that will take our baseband (not necessarily, more appropriately lowpass) signal and put it at a higher available frequency band such that it can be easily transmitted without any error/loss of data.

To do this we could perhaps use $x_3(t) = 5\cos(2\pi f_1 t)$

This signal has the Fourier Transform as below:

$$5\mathrm{cos\:\omega_{0} t\overset{FT}{\leftrightarrow}\frac 52[\delta(f -f_{0})+\delta(f+f_{0})]} \quad \quad \tag{1}$$

where $\omega_0 =2\pi f_0$. Now if our $m(t) = \mathrm{sinc}(t) + 10\cos(2\pi f_2 t)$

then our resulting $S(t) = A_c(\mathrm{sinc}(t) + 10\cos(2\pi f_2 t))\cdot 5\cos(2\pi f_1 t)$ which can be written as:

$$ S(t) = \underbrace{25A_c(\mathrm{sinc}(t)\cos(2\pi f_1 t))}_{p_1(t)} + \underbrace{50A_c(\cos(2\pi f_2 t) \cos(2\pi f_1 t))}_{p_2(t)}$$

Remember that multiplication in time-domain is convolution in Fourier domain and that convolution with a shifted impulse merely shifts our signal corresponding to the shift of the impulse. Keeping these two things in mind, consider $p_1(t)$. We have a $\mathrm{rect}(\omega)$ (rectangle in the figure above) convolved with the two shifted impulses from $(1)$ where $f_0 = f_1 =1\texttt{kHz}$. This convolution will essentially place the $\mathrm{rect}$ at $\pm f_1$ in the frequency domain which looks something like this:

P_1(f)

As you can see we have successfully placed a part of our message signal at a higher frequency. However, the task is not done. We still have $p_2(t)$ to analyze for which you can utilize the following identity:

$$ \cos(x)\cdot \cos(y) = 0.5 \cos(x + y) + 0.5\cos(x - y)$$

and then you can use $(1)$ to find the FT. If your analysis leads to overlapping between the Fourier Transforms of the different parts of your message signal then you can try a different carrier signal. This is how you are supposed to perform this analysis.

As per DSP.SE's rules it is discouraged to outright answer HW questions but I hope I have provided you with enough intuition that you can solve the rest of the exercise on your own.

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