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.

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

### Given:
- **μ = 255**
- **Maximum input voltage \(V_{\text{max}} = 4V\)**
- **Input voltage \(V_{\text{in}} = 0.75V_{\text{max}}\)**

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

\[
V_{\text{out}} = \frac{\ln(1 + \mu |V_{\text{in}} / V_{\text{max}}|)}{\ln(1 + \mu)}
\]

where:
- \(V_{\text{out}}\) is the output voltage after compression,
- \(V_{\text{in}}\) is the input voltage,
- \(\mu\) is the compression parameter (255 in this case),
- \(V_{\text{max}}\) 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:

\[
G_{\text{compressed}} = \frac{V_{\text{out}}}{V_{\text{in}}}
\]

### Part (b) – Compressed Output Voltage for \(V_{\text{in}} = 4V\)
This is essentially asking for the output voltage when the input is at its maximum (\(V_{\text{in}} = V_{\text{max}} = 4V\)).

Using the μ-law equation:

\[
V_{\text{out}} = \frac{\ln(1 + \mu |V_{\text{in}} / V_{\text{max}}|)}{\ln(1 + \mu)}
\]

Since \(V_{\text{in}} = V_{\text{max}}\), the equation simplifies to:

\[
V_{\text{out}} = \frac{\ln(1 + \mu)}{\ln(1 + \mu)}
\]

\[
V_{\text{out}} = 1
\]

Thus, the output voltage is **\(V_{\text{out}} = V_{\text{max}}\)** 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:

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

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 \(V_{\text{in}} = 0\):

\[
V_{\text{out}} = \frac{\ln(1)}{\ln(1 + \mu)} = 0
\]

The output dynamic range is then:

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

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

---

### Summary of Results:
1. **Compressed Voltage Gain** \(G_{\text{compressed}}\): To be calculated using the given input of \(0.75V_{\text{max}}\).
2. **Compressed Output Voltage** for maximum input of 4V: \(V_{\text{out}} = 4V\).
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 \(0.75V_{\text{max}}\)? 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 \(V_{\text{in}} = 2V\)?
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.

Learn how to calculate the compressed voltage gain, output voltage for a 4V input, and the input/output dynamic ranges for a compressor with μ = 255. This step-by-step solution covers signal compression, μ-law compression formulas, and dynamic range analysis. Suitable for students in Electrical Engineering and Signal Processing.

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