Im bit struggling with PAM question. I've been trying to solve this for few hours but I am stuck at the stage where I need to sample \$m(t)\$. Could you please help me with detailed answers and workings. Thank you :D

1st Question

The sinusoidal signal \$ m(t) = A_{m} cos(2\pi f_{m}t) \$is PAM modulated to produce the signal \$s(t)\$. Assume that \$A_{m} = 5V\$, \$f_{m} = 3Hz\$. the sampling period \$T_{s} = 0.1s\$ and the PAM pulses are of duration \$T = 0.02s\$.

*PAM -> Pulse Amplitude Modulation.

(a) Derive an expression for the spectrum of \$s(t)\$ and plot this over the frequency range \$±2f_{s}\$, where \$f_{s} = \frac{1}{T} s\$.

(b) Assuming an ideal reconstruction filter, plot the spectrum of the filter output, \$g(t)\$. Compare this spectrum with the output that would occur if there was no aperture effect.

2nd Question

Prove that the PAM generator, \$h(t)\$, is a linear process, where \$h(t)\$ is square wave between 0 and T with amplitude of 1.

  • 1
    \$\begingroup\$ What have you done so far? Where do you get stuck? \$\endgroup\$
    – Andy aka
    Commented May 8, 2014 at 12:12
  • \$\begingroup\$ PAM = S(f) = fs * ΣM(f-kfs) * H(f) \$\endgroup\$
    – Junta
    Commented May 8, 2014 at 13:04
  • \$\begingroup\$ Im stuck at stage where I get m(f-kfs) I got 5cos(6 * pi * n * Ts) but I am not sure if it is correct. and to get M(f-kfs), I should use FT to m(f-kfs)? \$\endgroup\$
    – Junta
    Commented May 8, 2014 at 13:05
  • \$\begingroup\$ I understand that m(t) -> Sampler -> mδ(t) -> PAM generator h(t) -> PAM \$\endgroup\$
    – Junta
    Commented May 8, 2014 at 13:11

1 Answer 1


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


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


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.


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