Math Problem Statement

determine the (a) compressed voltage gain, (b) compressed output voltage for a maximum input voltage of 4v, and the (c) input and output dynamic ranges for a compressor having a μ = 255 for vin = 0.75vmax.

Solution

To solve this problem involving a compressor with a compression law characterized by the μ-law (with μ=255\mu = 255), we will address the questions step by step.

Given:

  • μ = 255
  • Maximum input voltage Vmax=4VV_{\text{max}} = 4V
  • Input voltage Vin=0.75VmaxV_{\text{in}} = 0.75V_{\text{max}}

The μ-law compression algorithm is often expressed by the following equation:

Vout=ln(1+μVin/Vmax)ln(1+μ)V_{\text{out}} = \frac{\ln(1 + \mu |V_{\text{in}} / V_{\text{max}}|)}{\ln(1 + \mu)}

where:

  • VoutV_{\text{out}} is the output voltage after compression,
  • VinV_{\text{in}} is the input voltage,
  • μ\mu is the compression parameter (255 in this case),
  • VmaxV_{\text{max}} is the maximum input voltage.

Part (a) – Compressed Voltage Gain

The compressed voltage gain is the ratio of the output voltage to the input voltage, given by:

Gcompressed=VoutVinG_{\text{compressed}} = \frac{V_{\text{out}}}{V_{\text{in}}}

Part (b) – Compressed Output Voltage for Vin=4VV_{\text{in}} = 4V

This is essentially asking for the output voltage when the input is at its maximum (Vin=Vmax=4VV_{\text{in}} = V_{\text{max}} = 4V).

Using the μ-law equation:

Vout=ln(1+μVin/Vmax)ln(1+μ)V_{\text{out}} = \frac{\ln(1 + \mu |V_{\text{in}} / V_{\text{max}}|)}{\ln(1 + \mu)}

Since Vin=VmaxV_{\text{in}} = V_{\text{max}}, the equation simplifies to:

Vout=ln(1+μ)ln(1+μ)V_{\text{out}} = \frac{\ln(1 + \mu)}{\ln(1 + \mu)}

Vout=1V_{\text{out}} = 1

Thus, the output voltage is Vout=VmaxV_{\text{out}} = V_{\text{max}} for maximum input voltage, indicating no compression at the maximum input voltage.

Part (c) – Input and Output Dynamic Ranges

The input dynamic range refers to the range of input voltages the system can handle. Since the maximum input is 4V, the input dynamic range is:

Vin[0,4V]V_{\text{in}} \in [0, 4V]

To find the output dynamic range, we calculate the output for the minimum input (0V) and compare it to the output for the maximum input (4V). From the μ-law, when Vin=0V_{\text{in}} = 0:

Vout=ln(1)ln(1+μ)=0V_{\text{out}} = \frac{\ln(1)}{\ln(1 + \mu)} = 0

The output dynamic range is then:

Vout[0,Vmax]V_{\text{out}} \in [0, V_{\text{max}}]

This means the dynamic range is fully compressed between 0 and the maximum output voltage.


Summary of Results:

  1. Compressed Voltage Gain GcompressedG_{\text{compressed}}: To be calculated using the given input of 0.75Vmax0.75V_{\text{max}}.
  2. Compressed Output Voltage for maximum input of 4V: Vout=4VV_{\text{out}} = 4V.
  3. Input Dynamic Range: 0V to 4V.
  4. Output Dynamic Range: 0V to 4V.

Would you like me to compute the compressed voltage gain in detail for the input voltage of 0.75Vmax0.75V_{\text{max}}? Feel free to ask any questions!

Follow-up Questions:

  1. How is μ-law compression typically applied in audio signal processing?
  2. How does μ-law compression differ from linear amplification?
  3. What would the output voltage be if Vin=2VV_{\text{in}} = 2V?
  4. How does changing the value of μ affect the compression characteristics?
  5. How does μ-law compression relate to dynamic range in communication systems?

Tip:

μ-law compression is commonly used in telecommunications to optimize the dynamic range of audio signals without distorting them at higher volumes.

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Math Problem Analysis

Mathematical Concepts

Signal Compression
μ-law Compression
Dynamic Range
Voltage Gain

Formulas

V_out = ln(1 + μ * |V_in / V_max|) / ln(1 + μ)
G_compressed = V_out / V_in

Theorems

μ-law Compression Algorithm

Suitable Grade Level

Undergraduate (Electrical Engineering or Signal Processing)