# 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 $$\ m(t) = A_{m} cos(2\pi f_{m}t) \$$is PAM modulated to produce the signal $$\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.

• What have you done so far? Where do you get stuck? May 8, 2014 at 12:12
• PAM = S(f) = fs * ΣM(f-kfs) * H(f) May 8, 2014 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)? May 8, 2014 at 13:05
• I understand that m(t) -> Sampler -> mδ(t) -> PAM generator h(t) -> PAM May 8, 2014 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.