# pulse amplitude modulation

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) = Am cos(2*pi*fm*t) is PAM modulated to produce the signal s(t). Assume that Am = 5V, fm = 3Hz. the sampling period Ts = 0.1 sec and the PAM pulses are of duration T = 0.02 sec.

*PAM -> Pulse Amplitude Modulation.

(a) Derive an expression for the spectrum of s(t) and plot this over the frequency range ±2fs, where fs = 1/Ts.

(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.

• What have you done so far? Where do you get stuck? – Andy aka May 8 '14 at 12:12
• PAM = S(f) = fs * ΣM(f-kfs) * H(f) – Junta May 8 '14 at 13:04
• 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)? – Junta May 8 '14 at 13:05
• I understand that m(t) -> Sampler -> mδ(t) -> PAM generator h(t) -> PAM – Junta May 8 '14 at 13:11

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.