Question:
In the block diagram shown below, an analog signal, \( V_{IN} = \sin(2\pi \cdot 10^6 t) \) is quantized by a 10-bit Nyquist ADC. Later, 4 LSBs are dropped and 6 MSBs are converted to an analog signal \( V_{OUT} \) while using a 6-bit DAC. Assume uniform distribution for the quantization noise. The peak SQNR at the output of DAC is \(\underline{\hspace{2cm}}\) dB.
In the block diagram shown below, an analog signal, \( V_{IN} = \sin(2\pi \cdot 10^6 t) \) is quantized by a 10-bit Nyquist ADC. Later, 4 LSBs are dropped and 6 MSBs are converted to an analog signal \( V_{OUT} \) while using a 6-bit DAC. Assume uniform distribution for the quantization noise. The peak SQNR at the output of DAC is \(\underline{\hspace{2cm}}\) dB.
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For calculating SQNR, use the formula \( SQNR = 6.02N + 1.76 \) dB where \(N\) is the number of bits.
Updated On: Dec 24, 2025
- 61.96
- 25.84
- 49.92
- 37.88
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The Correct Option is D
Solution and Explanation
In this problem, we need to calculate the peak Signal-to-Quantization Noise Ratio (SQNR) for a 6-bit DAC output from a 10-bit ADC with 4 LSBs dropped. The formula for SQNR is given by:
\[
SQNR = 6.02N + 1.76 \text{ dB},
\]
where \(N\) is the number of bits.
Step 1: Calculate the SQNR for a 6-bit DAC.
After 4 LSBs are dropped, the system operates as a 6-bit DAC. Therefore,
\[
N = 6.
\]
Using the formula for SQNR:
\[
SQNR = 6.02 \times 6 + 1.76 = 36.12 + 1.76 = 37.88 \text{ dB}.
\]
Final Answer: 37.88 dB
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