Here's problem (Lathi): enter image description here

a. \$ x(t) = 5 sinc^2 (5 \pi t) + cos (20 \pi t) \$

Bandwidth = \$ 20 \pi \$

\$ f_s = 10 Hz, w_s = 20 \pi \$

Signal can be reconstructed from samples if \$ w_s > 2 w_{bandwidth} \$

Here's solution.

I was wondering, why is it for part B signal can be reconstructed from samples? \$ w_s \$ is equal to bandwidth of the original signal, so original signal shouldn't be reconstructed because \$w_s\$ should be greater than the bandwidth of original signal, no?

And why is that in part C signal can be reconstructed but without sine component?


1 Answer 1


The key to understanding this question comes from Shannon's Theorem #13. now called Shannon-Nyquist Sampling Theorem.

The spectrum of the sampling pulse is modeled by the Cardinal Sine function (c for cardinal) \$sinc(x)= \frac{sin(x)}{x} \$. In practical terms this is the shape of a very narrow pulse but limited by a practical number of harmonics a pulse that looks now like this.

enter image description here

Sampling a signal is a frequency mixer, that outputs sum and difference frequencies of the sampling rate and its harmonics.

Answer assumes you understand something about Shannon's sampling calculations.

a) Since sampling rate = signal spectrum f , the difference is just the DC depending on phase difference.

b) Now that the sampling rate is 2x the signal f, it is possible to obtain the fundamental f input from the mixer output and all the cross products (intermodulation) at higher f can be filtered out. The phase difference of the 2 samples/cycle determines the resulting output amplitude.

c) the sine components are shown to cancel out.

My 1st design experience using this principle out of Univ.

If you had an ultra stable ovenized , ruggedized XTAL oscillator say 10MHz and converted that square wave to (almost perfect) sawtooth and then sampled that signal with a Sample & Hold at almost the same frequency ( 0.1 ppm difference) What do you get? a 1 Hz Sawtooth signal.

Now say this is a 10MHz OCXO (ovenized ultra stable Stratum clock) being broadcast from a rocket by dividing it down to 100KHz. You are told the rocket can speed up to Mach 7 in 1 minute and is launching from Churchill MB Canada heading almost straight up into Aurora (Northern Lights) and coming down in a controlled location somewhere in the Arctic. The telemetry room sends you a feed of the subcarrier signal with your clock and you mix it in a custom box with the same OCXO and PLL to scale down the frequency to match the input signal and you create a stable Sawtooth signal from the local clock and use the incoming signal zero crossings to sample that sawtooth in a few nanoseconds so you get a tiny frequency error (1e-11) that you can adjust to get DC signal.

The rocket launches and that DC ramps up to be a sawtooth at 1 Hz then 10Hz+...then... much more then slows down and goes backwards coming down to Earth. This is what I did 40 yrs ago when WWII radar was too expensive to maintain for scientific purposes. If you added up all the sawtooths instead of returning to 0V it would the same as range ( distance the rocket has moved away from antenna) With two antenna and mixers now you sense direction. the hardest part was learning how to make it attenuate 15g of vibration and 100g shock of the launch and 50g from acceleration. I only had 1/4" of space allotted for vibe isolation. Then it had to be tested on all 3 axes and pass. The whole building shakes when the 1kW shaker table vibrates the rocket in the basement 5Hz to 3kHz. circa 1976.

  • \$\begingroup\$ what is "phase difference of the 2 samples/cycle"? How do you calculate it here? Do you just compare \$ w \$ with \$ w_s\$ ? >"all the cross products (intermodulation) "what cross products? ">the sine components are shown to cancel out" Where? I don't see it. \$\endgroup\$
    – Jack
    Dec 7, 2017 at 22:19
  • \$\begingroup\$ Phase was not included in question but affects amplitude of c) with 2 samples per sine. Ideal sampling converts all input signals to sum and difference spectrum and inverts phase of upper spectrum the sine at +j20pi cancels with -j20pi \$\endgroup\$ Dec 7, 2017 at 22:23
  • \$\begingroup\$ What is the normalized spectrum of a repetitive S(f) sample with t=0? .... S(f)= fs,2fs,3fs,4fs,5fs.... What is the output of a sampled input X(f) times S(f) = n*{(X(f)-S(f)} for n=+/- {1 ,2,3,4,5...} now compute the sum and difference of the most significant terms n=1,2 and ignore the higher than the output spectrum of interest. Time sampling is equivalent to the sum of frequency difference terms, so if the same f, the difference =0 or some DC if there was a phase included. but if sampled 2x f1 rate then 2f1-f1 =f1 and the harmonics of S(f) the sampler for n>2 are ignored for now. \$\endgroup\$ Dec 7, 2017 at 22:55
  • \$\begingroup\$ Please review Nyquist Sampling Theorem and study it. \$\endgroup\$ Dec 7, 2017 at 22:59
  • \$\begingroup\$ yeah I don't know what >"c) with 2 samples per sine" means. I know that sampling is simply shifting the original signal in this case. \$\endgroup\$
    – Jack
    Dec 7, 2017 at 23:08

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