Electronic – Checking whether the system is linear or non linear

The input to this system is denoted by \$x(t)\$ and its output by \$y(t)\$.

• To prove that the system is homogeneous, what you need to show is that

  if input \$x(t)\$ produces output \$y(t)\$, then the output produced by input \$\alpha\cdot x(t)\$ is \$\alpha\cdot y(t)\$.

  Here, \$\alpha\$ is an arbitrarily chosen real number, that is, the highlighted statement must hold regardless of the value of \$\alpha\$. Similarly, the requirement must be satisfied regardless of the choice of input \$x(t)\$. In other words, finding one pair of input and output signals \$x(t)\$ and \$y(t)\$ and one real number \$\alpha\$ for which the highlighted statement is true is not sufficient; it has to hold for all such choices.

• To prove that the system is additive, what you need to show is that

  if inputs \$x_1(t)\$ and \$x_2(t)\$ produce outputs \$y_1(t)\$ and \$y_2(t)\$ respectively, then the output produced by input \$x_1(t)+x_2(t)\$ is \$y_1(t)+y_2(t)\$.

  Once again, cherry-picking is not allowed; the statement has to hold for all choices of input signals and corresponding output signals.

It is not too hard to verify that your system is both homogeneous and additive.

The PAM signal \$s(t)\$ is a weighted sum of functions \$h(t)\$, where the weights are the samples of the signal \$m(t)\$:

$$s(t)=\sum_km(kT_s)h(t-kT_s)$$

This can be modeled as a multiplication of \$m(t)\$ by a comb of Dirac impulses, convolved with \$h(t)\$:

$$s(t)=\left(m(t)\sum_k\delta(t-kT_s)\right)*h(t)\tag{1}$$

From (1) it follows that the spectrum \$S(f)\$ is given by

$$S(f)=\left(M(f)*f_s\sum_k\delta(f-kf_s)\right)\cdot H(f)= f_s\sum_kM(f-kf_s)H(f)\tag{2}$$

where I've made use of the fact that convolution in one domain corresponds to multiplication in the other domain, and that a Dirac comb in one domain corresponds to a Dirac comb in the other domain (you can find this in most Fourier transform tables). \$M(f)\$ and \$H(f)\$ are of course the spectra of \$m(t)\$ and \$h(t)\$, respectively. So the spectrum \$S(f)\$ is the sum of shifted spectra \$M(f-kf_s)\$, multiplied by the spectrum \$H(f)\$. In order to sketch \$S(f)\$ you need to know \$M(f)\$ and \$H(f)\$:

$$M(f)=\frac{A_m}{2}[\delta(f-f_m)-\delta(f+f_m)]\\ H(f)=T\frac{\sin(\pi Tf)}{\pi fT}e^{-j\pi Tf}$$

For sketching \$S(f)\$ you simply ignore the phase term \$e^{-j\pi Tf}\$ of \$H(f)\$, so you just need to know that the magnitude \$|H(f)|\$ is the magnitude of a sinc function with \$H(0)=T\$ and with zeros at \$f_k=k/T\$, \$k=\pm 1,\pm 2,\ldots\$ (note that \$T\neq T_s\$!).

For (b) just remove all shifted spectra (that's what the ideal low-pass reconstruction filter does), so from (2) you're left with \$f_sM(f)H(f)\$ in the frequency range \$[0,f_s/2]\$.

For question 2 you just need to show that if \$s_1(t)\$ and \$s_2(t)\$ are the PAM signals corresponding to signals \$m_1(t)\$ and \$m_2(t)\$, respectively, then \$as_1(t)+bs_2(t)\$ is the PAM signal corresponding to the signal \$am_1(t)+bm_2(t)\$ for arbitrary constants \$a\$ and \$b\$. This is also obvious because the generation of the PAM signal only involves multiplication and convolution, so it is a linear process.