SOME DSP HOWEWORK

作者: 电工李达康 | 来源:发表于2019-03-12 15:20 被阅读0次

此为DSP课程有关混叠现象(Aliasing)的几个例题，以及我的求解。

简书用得少，欢迎关注我的知乎@电工李达康~

1、Specify how many bits are needed to appropriately digitize each of the following signals. Choose from: 6 bits, 8 bits, 10 bits, 12 bits, 14 bits, or 16 bits.

a. A signal where the maximum amplitude is 1 volt and the rms noise is 1.5millivolts.
b. A signal with a signal-to-noise ratio of 900 to 1.
c. A signal with a coefficient-of-variation of 0.4%.
d. A high-fidelity audio system (hint: a jack-hammer is about 50,000 timeslouder than a pin drop).
e. A black and white digital image (hint: under the best conditions, the human eye can differentiate about 200 shades of gray between pure black and pure white).

Answer：

The random noise generated by quantization will simply add to whatever noise is already present in the analog signal. Therefore, what we should do is to digitize the signal with producing virtually no increase in the noise, which means that nothing would be lost due to quantization.

The addition noise has a mean of zero ( $\mu = 0$ ), and a standard deviation of $\frac{1}{\sqrt{12}}LSB$ ( $\sigma = \frac{1}{\sqrt{12}}LSB$ ).

Number of Bits	The $\sigma$ of Original Noise	The $\sigma$ of Quantization Noise	The $\sigma$ of Total Noise
n	$noise_{rms} \times (2^{n}-1) LSB$	$0.29LSB$	$\sqrt{{\sigma_{original}}^2+{\sigma_{quantization}}^2}$

a. A signal where the maximum amplitude is 1 volt and the rms noise is 1.5millivolts.

Number of Bits	The $\sigma$ of Original Noise	The $\sigma$ of Quantization Noise	The $\sigma$ of Total Noise
6	$0.0015 \times (2^{6}-1) LSB=0.0945 LSB$	$0.29LSB$	$0.31LSB$
8	$0.0015 \times (2^{8}-1) LSB=0.3825 LSB$	$0.29LSB$	$0.48LSB$
10	$0.0015 \times (2^{10}-1) LSB=1.5345 LSB$	$0.29LSB$	$1.56LSB$
12	$0.0015 \times (2^{12}-1) LSB=6.1425 LSB$	$0.29LSB$	$6.15LSB$
14	$0.0015 \times (2^{14}-1) LSB=24.5745 LSB$	$0.29LSB$	$24.58LSB$
16	$0.0015 \times (2^{16}-1) LSB=98.3025 LSB$	$0.29LSB$	$98.30LSB$

Therefore, we could see that the quantization noise become more and more negligible with the increase of the number of bits. If we expect nothing would be lost due to quantization, the best choice will be 12 bits.

b. A signal with a signal-to-noise ratio of 900 to 1.

We suppose that the mean value $\mu$ of signal is the half of the maximum range $M$ .

( $\mu = \frac{1}{2}M = \frac{1}{2}\times (2^n-1)LSB$ )

$SNR = \frac{\mu}{\sigma} = 900$ , so $\sigma= \frac{1}{900}\mu = \frac{1}{900}\times \frac{1}{2}M =\frac{1}{900}\times \frac{1}{2}\times (2^n-1)LSB$ .

Number of Bits	The $\sigma$ of Original Noise	The $\sigma$ of Quantization Noise	The $\sigma$ of Total Noise
6	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^6-1)LSB =0.0350LSB$	$0.29LSB$	$0.2921LSB$
8	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^8-1)LSB =0.1417LSB$	$0.29LSB$	$0.3228LSB$
10	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^{10}-1)LSB =0.5683LSB$	$0.29LSB$	$0.6380LSB$
12	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^{12}-1)LSB =2.2750LSB$	$0.29LSB$	$2.2934LSB$
14	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^{14}-1)LSB =9.1017LSB$	$0.29LSB$	$9.1063LSB$
16	$\sigma= \frac{1}{900}\times \frac{1}{2} \times (2^{16}-1)LSB =36.4083LSB$	$0.29LSB$	$36.4095LSB$

So, we'll choose 12 bits, which produce virtually no increase in the noise.

c. A signal with a coefficient-of-variation of 0.4%.

We suppose that the mean value $\mu$ of signal is the half of the maximum range $M$ .
( $\mu = \frac{1}{2}M = \frac{1}{2}\times (2^n-1)LSB$ )

$SNR = \frac{\mu}{\sigma} = \frac{1}{0.4\%} = 250$ , so $\sigma= \frac{1}{900}\mu = \frac{1}{250}\times \frac{1}{2}M =\frac{1}{250}\times \frac{1}{2}\times (2^n-1)LSB$ .

Number of Bits	The $\sigma$ of Original Noise	The $\sigma$ of Quantization Noise	The $\sigma$ of Total Noise
6	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^6-1)LSB =0.126LSB$	$0.29LSB$	$0.3162LSB$
8	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^8-1)LSB =0.510LSB$	$0.29LSB$	$0.5867LSB$
10	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^{10}-1)LSB =2.046LSB$	$0.29LSB$	$2.0664LSB$
12	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^{12}-1)LSB =8.190LSB$	$0.29LSB$	$8.1951LSB$
14	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^{14}-1)LSB =32.766LSB$	$0.29LSB$	$32.7673LSB$
16	$\sigma= \frac{1}{250}\times \frac{1}{2} \times (2^{16}-1)LSB =131.070LSB$	$0.29LSB$	$131.0703LSB$

So, we'll choose 10 bits, which produce virtually no increase in the noise.

d. A high-fidelity audio system (hint: a jack-hammer is about 50,000 times louder than a pin drop).

The maximum of signal may 50,000 bigger than the minimum. Therefore, the maximum M should be great enough.

$\because M = (2^n-1)LSB > 50,000LSB$ , $15<log_2(50000-1) <16$

$\therefore n =16$

In conclusion, 16 bits is what we need in this high-fidelity audio system.

e. A black and white digital image (hint: under the best conditions, the human eye can differentiate about 200 shades of gray between pure black and pure white).

The black and white digital image just need about 200 levels to form a grayscale image.

$\because M = (2^n-1)LSB > 200LSB$ , $7<log_2(200-1) <8$

$\therefore n =8$

Therefore, every pixel would be a 8-bit data, which stores the image information we need.

2、An analog electronic signal is composed of three sine waves: 1 kHz @ 1 volt amplitude, 3 kHz @ 2 volts amplitude, and 4 kHz @ 5 volts amplitude (all voltage readings are peak-to-peak). The signal is digitized with 12bits, spread over the range of -5 volts to +5 volts. For each sampling rate below, describe the frequency components that exist in the digital signal.Be sure to specify three things for each component: its digital frequency (a number between 0 and 0.5), its amplitude (in digital numbers, peak-to-peak),and its phase relative to the original analog signal (either 0 degrees or 180 degrees).

a. Sampling rate = 100 kHz.

b. Sampling rate = 10 kHz.

c. Sampling rate = 7.5 kHz.

d. Sampling rate = 5.5 kHz.

e. Sampling rate = 5 kHz.

f. Sampling rate = 1.7 kHz.

Answer：

The maxinum frequency of analog electronic signal is $4kHz$ . Therefore, the Nyquist frequency should be greater than $8kHz$ ( $= 4kHz \times 2$ ) . If not, there is going to be a alising.

a. Sampling rate = 100 kHz.

$\because f_s =100kHz > 8kHz$ $\therefore$ No aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency : $f_1 = 1k \div 100k = 0.01$
- Amplitude : $A_1 = 1$
- Phase : $0^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency : $f_2 = 3k \div 100k = 0.03$
- Amplitude : $A_2 = 2$
- Phase : $0^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency : $f_3 = 4k \div 100k = 0.04$
- Amplitude : $A_3 = 5$
- Phase : $0^{\circ}$

b. Sampling rate = 10 kHz.

$\because f_s =10 kHz > 8kHz$ $\therefore$ No aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency : $f_1 = 1k \div 10k = 0.1$
- Amplitude : $A_1 = 1$
- Phase : $0^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency : $f_2 = 3k \div 10k = 0.3$
- Amplitude : $A_2 = 2$
- Phase : $0^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency : $f_3 = 4k \div 10k = 0.4$
- Amplitude : $A_3 = 5$
- Phase : $0^{\circ}$

c. Sampling rate = 7.5 kHz.

$\because f_s =7.5 kHz < 8kHz$ $\therefore$ It caused aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency : $f_1 = 1k \div 7.5k = 0.13$
- Amplitude : $A_1 = 1$
- Phase : $0^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency : $f_2 = 3k \div 7.5k = 0.4$
- Amplitude : $A_2 = 2$
- Phase : $0^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency :
  - $\because 4k \div 7.5k = 0.53 > 0.5$
  - $\therefore f_3 = 1 - 0.53 =0.47$
- Amplitude : $A_3 = 5$
- Phase : $180^{\circ}$

d. Sampling rate = 5.5 kHz.

$\because f_s =5.5 kHz < 8kHz$ $\therefore$ It caused aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency : $f_1 = 1k \div 5.5k = 0.18$
- Amplitude : $A_1 = 1$
- Phase : $0^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency :
  - $\because 3k \div 5.5k = 0.55 > 0.5$
  - $\therefore f_2 = 1 - 0.55 =0.45$
- Amplitude : $A_2 = 2$
- Phase : $180^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency :
  - $\because 4k \div 5.5k = 0.73 > 0.5$
  - $\therefore f_3 = 1 - 0.73 =0.27$
- Amplitude : $A_3 = 5$
- Phase : $180^{\circ}$

e. Sampling rate = 5 kHz.

$\because f_s =5kHz < 8kHz$ $\therefore$ It caused aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency : $f_1 = 1k \div 5k = 0.20 = f_3$
- Amplitude : $A_1 = A_3 =1-5=-4$
- Phase : $0^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency :
  - $\because 3k \div 5k = 0.60> 0.5$
  - $\therefore f_2 = 1 - 0.60 =0.40$
- Amplitude : $A_2 = 2$
- Phase : $180^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency :
  - $\because 4k \div 5k = 0.80 > 0.5$
  - $\therefore f_3 = 1 - 0.80 =0.20 =f_1$
- Amplitude : $A_3 = A_1 = 1-5=-4$
- Phase : $180^{\circ}$

f. Sampling rate = 1.7 kHz.

$\because f_s =1.7Hz < 8kHz$ $\therefore$ It caused aliasing.

The component one (1 kHz @ 1 volt amplitude) :
- Digital frequency :
  - Digital frequency :
  - $\because 1k \div 1.7k = 0.59> 0.5$
  - $\therefore f_2 = 1 - 0.59 =0.41$
- Amplitude : $A_1 = 1$
- Phase : $180^{\circ}$
The component two (3 kHz @ 2 volt amplitude) :
- Digital frequency :
  - $\because 3k \div 1.7k = 1.76> 0.5$
  - $\therefore f_2 = 2-1.76=0.24$
- Amplitude : $A_2 = 2$
- Phase : $180^{\circ}$
The component three (4 kHz @ 5 volt amplitude) :
- Digital frequency :
  - $\because 4k \div 1.7k = 2.35 > 0.5$
  - $\therefore f_3 = 2.35 - 2 =0.35$
- Amplitude : $A_3 =5$
- Phase : $0^{\circ}$

3、On television, rotating objects such as wagon wheels and airplane propellers often appear to be moving very slowly or even backwards. This is a result of aliasing, caused by the sampling rate of the video (30 frames per second) being less than twice the frequency of the rotational motion. To understand this, imagine we paint one of the blades of an airplane propeller so that we can identify it from the other blades. We will then turn the propeller at 33 rotations per second, in a clockwise direction. In frame number 1 of our video sequence, the marked blade happens to be exactly at the top of the propeller.

Answer：

a. How many rotations does the marked blade make between two successive frames?

$\because$ The Angular Velocity $\omega = 2\pi \times n= 66\pi$ $rad/s$ and the time between two successive frames $t_{sample} = \frac{1}{30} s = 0.033s$

$\therefore$ Between two successive frames, the angle it turn is $\theta = \omega t_{sample} = \frac{11\pi}{5} = 6.912rad = 2\pi +0.629rad$ .

$\therefore$ The rotations are $\theta/2\pi = 1.1$ in actural, but in video it would be $0.629rad$ ( $0.1$ circle )

b. Draw a sketch of how the propeller would appear in frames 1, 2, 3 and 4.

In frames 1, 2, 3, 4, it just like that the propeller just turn 0.628rad, 1.257rad, 1.885rad, 2.513rad.

c. How many frames does it take for the marked blade to again appear at the top?

$\because$ The rotations are $\theta/2\pi = 1.1$ between two successive frames.

$\therefore$ Just between 10 frames, the rotations will be 11. In other words, you will find that the marked blade again appear at the top in the 11th frame.

d. What rotational frequency is (c) in rotations per second?

In c, we know that the cycle of it is $T = 10 \times t_{sample} = 10 \times \frac{1}{30} = \frac{1}{3}s$ . Therefore, the rotational frequency is $f= \frac{1}{T} = 3Hz$

e. Is this apparent rotation clockwise or counterclockwise?

$\because$ The propeller turns at 33 rotations per second in a clockwise direction. Between two successive frames, the angle it turn is $\theta = \omega t_{sample} = \frac{11\pi}{5} = 6.912 rad = 2\pi +0.629rad$ .

$\because$ Between two successive frames,the angle we see is 0.629rad. Meanwhile, $0<0.629<\pi$ .

$\therefore$ The rotation direction in video will be clockwise.

f. Explain using Fig. 3-4 how the marked blade's actual frequency, the frame rate, and the marked blade's observed frequency are related.

$\because$ The marked blade's actual frequency $f_{actual}= 33Hz$

$\therefore$ The Nyquist frequency $f_n = 2 \times f_{actual} = 66Hz$

$\therefore$ The frame rate $f_{sample} = 30Hz < f_n$ . The aliasing is caused.

$\because$ $33 \div 30 = 1.1$

$\therefore$ The digital frequency of the marked blade $f_{digital} = 1.1-1 =0.1$ and the phase is $0^{\circ}$ .

It means that its digital frequency will be like sampled from $0.1 \times 30 = 3Hz$ . Meanwhile, the phase of it is $0^{\circ}$ , which means that the rotation direction in video will be the same as actural, clockwise.

g. Repeat (a) to (f) when the propeller is turning at 57 rotations per second.

In fact, we could use the analysis of (f), which will help us calculate and understand it in depth.

$\because$ The marked blade's actual frequency $f_{actual}= 57Hz$ instead of $33Hz$ .

$\therefore$ The Nyquist frequency $f_n = 2 \times f_{actual} = 114Hz$ .

$\therefore$ The frame rate $f_{sample} = 30Hz < f_n$ . The aliasing is caused again.

$\because$ $57 \div 30 = 1.9$

$\therefore$ The digital frequency of the marked blade $f_{digital} = 2-1.9 =0.1$ and it is worth noting that the phase is $180^{\circ}$ .

$\therefore$ The rotational frequency in video is $0.1 \times 30 = 3Hz$ . But it is because the phase is $180^{\circ}$ that the rotation direction in video will be counterclockwise.

In conclusion:

The angle between two successive frames $\theta = \omega t_{sample} = \frac{2\pi \times 57}{30} = 11.938 rad = 4\pi - 0.628 rad$ . (a)
The marked blade again appear at the top in the $11th$ frame. (c)
The rotational frequency is $f= \frac{1}{T} = 3Hz$ . (d)
The rotation direction in video will be counterclockwise. (e)