Report of Numerical Acquisition
Xavier and Sigismond 2022.10.16
Part of preparation
1. In the area of audio sampling and video sampling, if the sampling frequency is not appropriate, maybe we can see the backflow of the river.
2. We can clearly see that T t o t = T e × N T_{tot} =T_{e} \times N Ttot=Te×N
3. u ( t ) = E 0 c o s ( 2 π f 0 t ) , e ( t ) = E 1 c o s ( 2 π f e t ) u(t)=E_0cos(2\pi f_0t), e(t)=E_1cos(2\pi f_et) u(t)=E0cos(2πf0t),e(t)=E1cos(2πfet), so u s ( t ) = k u ( t ) × e ( t ) = k E 1 E 0 ( c o s ( 2 π ( f 0 + f e ) t ) 2 + c o s ( 2 π ( f 0 − f e ) t ) 2 ) u_s(t)=ku(t) \times e(t)=kE_1E_0(\frac{cos(2\pi(f_0+f_e)t)}{2}+\frac{cos(2\pi(f_0-f_e)t)}{2}) us(t)=ku(t)×e(t)=kE1E0(2cos(2π(f0+fe)t)+2cos(2π(f0−fe)t))
4.
5. When f e > 2 f 0 f_e>2f_0 fe>2f0 we can get the frequencies f 0 + f e f_0+f_e f0+fe and f 0 − f e f_0-f_e f0−fe.
6. f e = 3 f 0 2 , s o u s ( t ) = k E 1 E 0 ( c o s ( 2 π × 5 f 0 2 t ) 2 + c o s ( 2 π f 0 2 t ) 2 ) f_e=\frac{3f_0}{2}, so\,u_s(t)=kE_1E_0(\frac{cos(2\pi \times \frac{5f_0}{2} t)}{2}+\frac{cos(2\pi \frac{f_0}{2} t)}{2}) fe=23f0,sous(t)=kE1E0(2cos(2π×25f0t)+2cos(2π2f0t)) ,we can see the frequency f 0 2 \frac{f_0}{2} 2f0 in the spectrum, which means that there is a “parasitic” peak appearing between 0 and f 0 f_0 f0
7. If we want the “parasitic” peak is outside the interval [ 0 ; f 0 ] [0;f_0] [0;f0], we must have
f ( x ) = { ∣ f 0 + f e 2 ∣ > f 0 ∣ f 0 − f e 2 ∣ > f 0 f(x)=\left \{ \begin{aligned} \vert \frac{f_0+f_e}{2}\vert &>& f_0 \\ \vert \frac{f_0-f_e}{2}\vert &>& f_0 \\ \end{aligned} \right. f(x)=⎩ ⎨ ⎧∣2f0+fe∣∣2f0−fe∣>>f0f0
So we have f e > 2 f 0 f_e > 2f_0 fe>2f0
8. Maybe this setup is called follower ,with a follower we can make the voltage of the positive-going input is equal to the voltage of the voltage of the sortie, it can improve the clarity of the signal.
Part of experiment and the data processing
Part I of the experiment
9. We took some pictures when we chose the different samping frequency, and the parameter is shown on the screen.
When we choose f e = 10 k H z f_e=10kHz fe=10kHz ,we have the
f e = 2 k H z f_e=2kHz fe=2kHz , we have a list of images likes theses
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f e = 1 k H z f_e=1kHz fe=1kHz, we have a list of images likes theses:
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f e = 400 H z f_e=400Hz fe=400Hz, we have a list of image likes theses:
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f e = 300 H z f_e=300Hz fe=300Hz, we can have an image likes this:
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11. - We can see that the more we increase the sampling frequency f e f_e fe, the more similar the signal u e b ( t ) u_{eb}(t) ueb(t)we get is like the sinusoidal signal.
- And we can see that if the samping frequency f e < 2 f 0 f_e<2f_0 fe<2f0, for this setup, f e < 400 H z f_e<400Hz fe<400Hz, it’s diificult for the u e b ( t ) u_{eb}(t) ueb(t) to reflect the tendency of the input signal (In this setup, the input signal is u ( t ) = 3 s i n ( 200 × 2 π t ) u(t)=3sin(200 \times 2\pi t) u(t)=3sin(200×2πt)), which means that the signal is distorted.
Applications issues
1. The sampling frequency should be chosen 8kHz for telephony;
2. T = 1 2 k H z × 10000 = 5 s T=\frac{1}{2kHz}\times 10000=5s T=2kHz1×10000=5s, so the maximum signal length that can be recorded is 5s.
3.The period of the electrical signal T e = 1 50 H z = 0.02 s T_e=\frac{1}{50Hz}=0.02s Te=50Hz1=0.02s,and we know that a neon lamp emits two flashes per period of the electrical signal, which means that it flashes 60 s 0.02 s × 2 = 6000 \frac{60s}{0.02s}\times2=6000 0.02s60s×2=6000 in a minute, so it’s not satisfied for the Shannon criterion, so it’s suitable for use a neon lamps to light the machine if we want to record the movement of the machine.
4. Because normally the human can hear the voice at frequencies ranging from 20Hz to 20kHz, to satisfy the Shannon criterion, we must choose f e > 40 k H z f_e>40kHz fe>40kHz ,but sometimes it’s better to give a little bigger f e f_e fe to suit some special demand.
Part II of the experiment
14. The recording obtained is many values which are near to 2, what is different from I might expect is that all of the values are less than, instead of some are bigger than 2 and some are less than 2.
15. We remove the repetitive value in the Excel table exported by the sortware and sort the data in descending order,and this is the processed Excel table :
| Amplitude - Voltage |
| 1.99673 |
| 1.997056 |
| 1.997382 |
| 1.997708 |
| 1.998034 |
| 1.99836 |
| 1.998686 |
| 1.999012 |
we can find that the difference between the two adjacent value is all 0.000326, which means that the interface voltage step q = 0.000326 q=0.000326 q=0.000326
16. We can find that 20 0.000326 = 61349.69 \frac{20}{0.000326}=61349.69 0.00032620=61349.69, which is approximate to 65536 = 2 16 65536=2^{16} 65536=216, so we can deduce that the number of bits of the interface is 16 bits.