Digital Circuits is a compact, high-scoring sub-topic of “Basics of Mechatronics” (Section A.2) in the GATE Robotics and Automation (RA) 2027 syllabus. Its questions test three areas: Boolean algebra and simplification, combinational blocks (multiplexers, encoders, decoders), and sequential circuits (flip-flops and counters). This guide reviews each with its key identities and solves five GATE-style problems in full.
TABLE OF CONTENTS
Boolean algebra & simplification
Boolean algebra manipulates logic expressions using AND, OR and NOT. The identities you must know cold are the absorption law A + AB = A, the redundancy/consensus form A + A′B = A + B, and De Morgan’s theorems (A·B)′ = A′ + B′ and (A + B)′ = A′·B′. Karnaugh maps give the minimal sum-of-products form.
Combinational circuits
Combinational blocks compute outputs purely from present inputs. A multiplexer (MUX) selects one of 2n inputs using n select lines; a decoder activates one of 2n outputs from an n-bit code; an encoder does the reverse. The select-line count n = log2(inputs) is the recurring calculation.
Sequential circuits & counters
Sequential circuits have memory: their output depends on present inputs and past state, stored in flip-flops (SR, D, JK, T). A counter built from k flip-flops has 2k states and divides the clock frequency by 2k. A mod-N counter needs the smallest k with 2k ≥ N.
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Worked examples (GATE-style)
Example 1 — Boolean simplification
Simplify F = A + A′B.
Solution. Using the consensus identity, A + A′B = (A + A′)(A + B) = 1·(A + B).
F = A + B.
Example 2 — multiplexer select lines
How many select lines does a 16-to-1 multiplexer need?
Solution. n = log2(16) = 4 select lines (since 24 = 16 inputs).
Example 3 — decade counter flip-flops
How many flip-flops are needed to build a mod-10 (decade) counter?
Solution. We need the smallest k with 2k ≥ 10. 23 = 8 < 10, but 24 = 16 ≥ 10.
So 4 flip-flops are required.
Example 4 — frequency division
A 4-bit ripple counter is clocked at 16 kHz. What is the frequency at the output of the last flip-flop?
Solution. Each flip-flop halves the frequency, so 4 flip-flops divide by 24 = 16.
fout = 16 kHz / 16 = 1 kHz.
Example 5 — number of counter states
How many distinct states does a 3-bit binary counter cycle through?
Solution. 3 flip-flops give 23 = 8 states (000 through 111).
Key results
KEY RESULTS
Consensus: A + A′B = A + B
De Morgan: (AB)′ = A′ + B′
MUX: 2n inputs → n select lines
Counter: k flip-flops → 2k states, ÷2k clock
Mod-N counter: smallest k with 2k ≥ N
Full adder: S = A ⊕ B ⊕ Cin
Common mistakes to avoid
- Mis-applying De Morgan’s theorem — complement each variable and swap AND with OR.
- Confusing 2n inputs with n inputs — a 16-to-1 MUX needs 4, not 16, select lines.
- Off-by-one on mod-N counters — take the smallest k with 2k ≥ N.
- Assuming a counter divides by k — it divides by 2k.
- Mixing combinational and sequential logic — only sequential circuits store past state.
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Basics of Mechatronics — full solved-problem series
This guide is one part of the Section A.2 (Basics of Mechatronics) solved-problem series. Work through every sub-topic:
See also the umbrella guide, Basics of Mechatronics — Important Questions, and the complete GATE RA 2027 Syllabus.
Frequently asked questions
How many select lines does a multiplexer need?
A multiplexer with 2n data inputs needs n select lines. So a 4-to-1 MUX needs 2, an 8-to-1 needs 3, and a 16-to-1 needs 4 select lines.
What is the difference between combinational and sequential circuits?
Combinational circuits produce outputs that depend only on the present inputs (adders, multiplexers, decoders). Sequential circuits also depend on stored past state held in flip-flops (counters, registers, state machines).
How many flip-flops does a mod-N counter need?
A mod-N counter needs the smallest number of flip-flops k such that 2k ≥ N. For a decade (mod-10) counter, 24 = 16 ≥ 10, so four flip-flops are required.
This solved-problems guide is part of the complete GATE RA 2027 Syllabus overview and the Basics of Mechatronics syllabus guide.
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