Frequency at the MSB of a 4-bit binary up counter A 4-bit binary up counter is clocked at 20 kHz. What output frequency appears at the most significant bit (MSB)?

Introduction / Context:Binary counters divide frequency by powers of two at successive bit positions. Knowing the division at each bit is essential for generating timing references and subclocks from a master clock.

Given Data / Assumptions:

• 4-bit binary up counter with input clock f_in = 20 kHz.
• Bit 0 (LSB) toggles every clock edge, dividing by 2.
• Each more significant bit divides the input frequency by an additional factor of 2.

Concept / Approach:For an n-bit binary counter, the frequency at bit k (starting at k = 0 for LSB) is f_in / 2^(k+1). The MSB of a 4-bit counter is bit 3, so its frequency is f_in / 2^4.

Step-by-Step Solution:Compute divisor for MSB (bit 3): 2^4 = 16.Calculate f_out = 20 kHz / 16.20,000 Hz / 16 = 1,250 Hz = 1.25 kHz.

Verification / Alternative check:A short truth-table or timing sketch shows the LSB toggling each edge (10 kHz), next bit at 5 kHz, then 2.5 kHz, and MSB at 1.25 kHz, confirming the calculation.

Why Other Options Are Wrong:2.50 kHz corresponds to bit 2, not the MSB. 160 kHz and 320 kHz exceed the input frequency and are impossible for a divider.

Common Pitfalls:Off-by-one errors in the power of two; confusing toggle rate with full period; mixing Hz and kHz units.

Final Answer:1.25 kHz