ELECTENG291 · Fundamentals of Electrical Engineering
Linearity & Superposition
Module 1.1 and 1.4 of University of Auckland ELECTENG 291 pin down what makes a circuit linear — it must be both homogeneous and additive — and then exploit that property with the superposition principle, analysing a circuit one independent source at a time. Linearity proofs are a signature examinable theme (a level-shifter or averaging circuit turns up almost every year in the tests), and superposition is one of the standard multi-source solving methods rewarded on the tests and the final exam.
What this chapter covers
- 01Linear iff BOTH homogeneous AND additive; otherwise non-linear
- 02Homogeneity (scaling): f{a·x(t)} = a·f{x(t)} for every constant a
- 03Additivity: f{x1(t) + x2(t)} = f{x1(t)} + f{x2(t)}
- 04Proof-by-counter-example: one violation of either property disproves linearity
- 05Worked cases: level-shifter y = x + K (non-linear for K ≠ 0), averaging/low-pass (linear), y = max(x) and the exponential diode (non-linear)
- 06The superposition principle: total response = sum of responses to each independent source acting alone
- 07Zeroing sources correctly: voltage source → short circuit, current source → open circuit; dependent sources stay active
Is a level-shifter linear? (a signature linearity proof)
- +1State the test: the block is linear only if it is BOTH homogeneous and additive. Check each; a single failure means non-linear.
- +1Homogeneity. Scale the input by a: f{a·x} = a·x + K. Compare with a·f{x} = a(x + K) = a·x + aK. These are equal for all a only if K = aK for every a, which forces K = 0.
- +1Additivity. Feed x1 + x2: f{x1 + x2} = x1 + x2 + K. Compare with f{x1} + f{x2} = (x1 + K) + (x2 + K) = x1 + x2 + 2K. Equal only if 2K = K, i.e. K = 0.
- +1Conclude: both properties hold only when K = 0. For any K ≠ 0 at least one property fails.
- +1Answer: the block is non-linear for every K ≠ 0, and linear only in the trivial case K = 0. The constant offset is exactly what breaks both homogeneity and additivity.
Key terms
- Linearity
- A circuit or block is linear if and only if it is both homogeneous and additive. Only linear circuits obey superposition; the presence of any constant offset, product, max/min or exponential relation typically breaks it.
- Homogeneity (scaling)
- The property f{a·x(t)} = a·f{x(t)} for every constant a — scaling the input scales the output by the same factor. Failing it (e.g. an added constant) makes the block non-linear.
- Additivity
- The property f{x1 + x2} = f{x1} + f{x2} — the response to a sum of inputs equals the sum of the individual responses. This is precisely the property superposition relies on.
- Superposition principle
- For a linear circuit, the total response equals the sum of the responses produced by each independent source acting alone, with all other independent sources set to zero.
- Zeroing a source
- When applying superposition, an independent voltage source set to zero becomes a short circuit and an independent current source set to zero becomes an open circuit. Dependent sources are never zeroed — they remain active throughout.
- Counter-example test
- The efficient way to prove non-linearity: exhibit a single input (or scaling) for which homogeneity or additivity fails. One counter-example is a complete disproof.
Linearity & Superposition FAQ
How do I prove a circuit is linear or non-linear?
Test the two properties. For homogeneity, feed a·x and check the output equals a times the original output. For additivity, feed x1 + x2 and check the output equals f{x1} + f{x2}. If both hold in general, it is linear; if either fails for even one input, it is non-linear. Marks are awarded for setting up each property clearly and stating the concluding condition — not just the yes/no.
Why does superposition only work for linear circuits?
Superposition is a direct consequence of additivity and homogeneity: it says the response to several sources equals the sum of the responses to each alone. That is only valid when the circuit is additive and homogeneous — i.e. linear. Apply it to a non-linear element and the individual responses will not add up to the true response.
How do I zero sources when applying superposition?
Set independent voltage sources to zero by replacing them with a short circuit, and independent current sources to zero by replacing them with an open circuit. Crucially, leave all dependent (controlled) sources active — they are part of the circuit's linear behaviour, not independent excitations.
Can Sia help me with linearity and superposition proofs?
Yes, as a study aid. Sia can walk a homogeneity/additivity proof line by line, show why a level-shifter or an exponential diode fails, and check that you zeroed sources correctly in a superposition solve. It explains and drills; it does not do graded assessment for you, and University of Auckland academic-integrity rules apply — confirm what is permitted on Canvas.
Exam move
Learn the linearity test as a fixed two-step ritual: write the homogeneity check f{a·x} vs a·f{x}, then the additivity check f{x1+x2} vs f{x1}+f{x2}, and let the required condition fall out. Memorise the standard verdicts the course reuses — a constant offset (level-shifter) and the exponential diode are non-linear, an averaging/low-pass block is linear — so you recognise the pattern fast under exam time. For superposition, practise the mechanics until zeroing is automatic: voltage source → short, current source → open, dependent sources stay in. Always solve for one source at a time, label each partial response (e.g. V_o due to the 15 V source, V_o due to the 6 mA source), then add. Cross-check a superposition answer against a second method (nodal or Thévenin) where you can. Confirm assessment details on Canvas.
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