Score . (Each question Score 12points, Total Score 12points) In the analog speech digitization transmission system, using A-law 13 broken line method to encode the speech signal, and assume the minimum quantization interval is taken as a unit 4. If the input sampling value Is- -0.95 V. (1) During the A-law 13 broken line PCM coding, how many quantitative levels (intervals) in total? Are the quantitative intervals the same? (2) Find the output binary code-word? (3) What is the quantization error? (4) And what is the corresponding 11bits code-word for the uniform quantization to the 7 bit codes (excluding polarity codes)?

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Answer 1

(1) Total quantitative levels: 8192, not the same intervals.

(2) Output binary code-word: Not provided.

(3) Quantization error: Cannot be calculated.

(4) Corresponding 11-bit code-word: Not determinable without specific information.

(1) In the A-law 13 broken line PCM coding, the total number of quantization levels (intervals) is determined by the number of bits used for encoding. In this case, 13 bits are used. The number of quantization levels is given by 2^N, where N is the number of bits. Therefore, there are 2^13 = 8192 quantitative levels in total. The quantitative intervals are not the same, as they are determined by the step size of the quantization process.

(2) To find the output binary code-word, the input sampling value needs to be quantized based on the A-law 13 broken line method. However, without specific information about the breakpoints and step sizes of the A-law encoding, it is not possible to determine the exact output binary code-word.

(3) The quantization error is the difference between the actual input value and the quantized value. Since the output binary code-word is not provided, the quantization error cannot be calculated.

(4) Without the specific information about the breakpoints and step sizes for the uniform quantization to 7-bit codes, it is not possible to determine the corresponding 11-bit code-word for the uniform quantization.

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Related Questions

justify your answer about which car if either completes one trip around the track in less tame quuantitatively with appropriate equations

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To determine which car completes one trip around the track in less time, we can analyze their respective velocities and the track distance.

The car with the higher average velocity will complete the track in less time. Let's denote the velocity of Car A as VA and the velocity of Car B as VB. The track distance is given as d.

We can use the equation:

Time = Distance / Velocity

For Car A:

Time_A = d / VA

For Car B:

Time_B = d / VB

To compare the times quantitatively, we need more information about the velocities of the cars.

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An operational amplifier has to be designed for an on-chip audio band pass IGMF filter. Explain using appropriate mathematical derivations what the impact of reducing the input impedance (Zin), and reducing the open loop gain (A) of the opamp will have for the general opamps performance. What effect would any changes to (Zin) or (A) have on the design of an IGMF band pass filter?

Answers

Reducing the input impedance (Zin) and open-loop gain (A) of an operational amplifier (opamp) will have a negative impact on its general performance.

Reducing the input impedance (Zin) of an opamp will result in a higher loading effect on the preceding stages of the circuit. This can cause signal attenuation, distortion, and a decrease in the overall system gain. Additionally, a lower input impedance may lead to a higher noise contribution from the source impedance, reducing the signal-to-noise ratio.

Reducing the open-loop gain (A) of an opamp affects the gain and bandwidth of the amplifier. A lower open-loop gain reduces the overall gain of the opamp, which can limit the amplification capability of the circuit. It also decreases the bandwidth of the opamp, affecting the frequency response and potentially distorting the signal.

In the design of an on-chip audio bandpass Infinite Gain Multiple Feedback (IGMF) filter, changes to the input impedance and open-loop gain of the opamp can have significant implications.

The input impedance of the opamp determines the interaction with the preceding stages of the filter, affecting the overall filter response and its ability to interface with other components.

The open-loop gain determines the gain and bandwidth of the opamp, which are crucial parameters for achieving the desired frequency response in the IGMF filter.

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Convert the following temperatures to their values on the Fahrenheit and Kelvin scales: (b) human body temperature, 37.0°C.

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The human body temperature is 98.6 °F and 310.15 K when converted to Fahrenheit and Kelvin scales respectively

The human body temperature is 37.0°C. We can use the formulae to convert the temperature to Fahrenheit and Kelvin scales. The formulae are given below:Fahrenheit scale: F = (9/5)*C + 32

Kelvin scale: K = C + 273.15where C is the temperature in Celsius scale.On the Fahrenheit scale:F = (9/5)*37 + 32= 98.6 °FTherefore, the human body temperature is 98.6 °F.On the Kelvin scale:K = 37 + 273.15= 310.15 K.

Therefore, the human body temperature is 310.15 K. In summary, the human body temperature is 98.6 °F and 310.15 K when converted to Fahrenheit and Kelvin scales respectively.

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use the formula to calculate the relativistic length of a 100 m long spaceship travelling at 3000 m s-1.

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The relativistic length of a 100 m long spaceship traveling at 3000 m/s is approximately 99.9995 m.

The relativistic length contraction formula is given by: L=L0√(1-v^2/c^2)Where L is the contracted length.L0 is the original length. v is the velocity of the object. c is the speed of light. The formula to calculate the relativistic length of a 100 m long spaceship traveling at 3000 m/s is: L=L0√(1-v^2/c^2)Given, L0 = 100 mV = 3000 m/sc = 3 × 10^8 m/sSubstituting the values in the formula:L = 100 × √(1-(3000)^2/(3 × 10^8)^2)L = 100 × √(1 - 0.00001)L = 100 × √0.99999L = 100 × 0.999995L ≈ 99.9995 m.

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Consider a radioactive sample. Determine the ratio of the number of nuclei decaying during the first half of its halflife to the number of nuclei decaying during the second half of its half-life.

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The ratio is 2. To determine the ratio of the number of nuclei decaying during the first half of the half-life to the number of nuclei decaying during the second half of the half-life, we need to understand the concept of half-life.

The half-life of a radioactive substance is the time it takes for half of the radioactive nuclei in a sample to decay. Let's say the half-life of the radioactive substance in question is represented by "t".

During the first half-life (t/2), half of the nuclei in the sample will decay. So, if we start with "N" nuclei, after the first half-life, we will have "N/2" nuclei remaining.

During the second half-life (t/2), another half of the remaining nuclei will decay. So, starting with "N/2" nuclei, after the second half-life, we will have "N/2" divided by 2, which is "N/4" nuclei remaining.

Therefore, the ratio of the number of nuclei decaying during the first half of the half-life to the number of nuclei decaying during the second half of the half-life is:

(N/2) / (N/4)

Simplifying this expression, we get:

(N/2) * (4/N)

This simplifies to:

2

So, the ratio is 2.

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