Mathematics Advanced • Year 11 • Module 1 • Lesson 14
Further Transformations & Synthesis
Apply |f|, f(|x|), and 1/f to physics, electronics, optics and modelling contexts.
Problem 1 — Full-wave rectifier (electronics)
An AC voltage v(t) varies as v(t) = 240 sin(t), where t is in seconds. A full-wave rectifier outputs |v(t)| (negative parts flipped above zero). The lesson notes that |f(x)| is mathematically equivalent to a rectifier.
Set up: What are we solving for?
(i) Sketch v(t) and y = |v(t)| on the same axes for 0 ≤ t ≤ 2π. Label the new "cusp" points and the peak voltage. 3 marks
(ii) At t = 4 s, v(4) ≈ 240 sin(4) ≈ −181.6 V. State the rectified value |v(4)| and explain in one sentence the electrical meaning. 2 marks
(iii) Confirm the rectifier never outputs a negative value, by reference to the algebraic definition of |x|. State the range of |v(t)|. 2 marks
Stuck? Lesson § Audio clipping and symmetry effects.Problem 2 — Symmetric signal generator (synthesiser)
A music synthesiser takes a sawtooth wave f(t) = t (for −1 ≤ t ≤ 1) and applies the transformation y = f(|t|) to produce an even waveform (useful for generating "even harmonics").
Set up: What are we solving for?
(i) Sketch f(t) and y = f(|t|) on separate axes for −1 ≤ t ≤ 1. Comment on the shape of f(|t|) and why it is "even". 3 marks
(ii) State the value of y = f(|t|) at t = −0.5 and t = 0.5. Confirm they are equal and explain why this is guaranteed for any f(|t|). 2 marks
(iii) The original sawtooth has the point (−0.5, −0.5). What happens to this point in y = f(|t|)? Explain in one line. 2 marks
Problem 3 — Reciprocal intensity (physics)
The brightness B of a star (in arbitrary units) as a function of its distance d (in light-years) from Earth follows an inverse-square pattern. A particular model uses
B(d) = 1 / f(d), where f(d) = d² − 100, d > 0
(The "− 100" represents instrument calibration relative to a reference star.)
Set up: What are we solving for?
(i) State the vertical asymptote of B(d) for d > 0 and explain what it means physically. 2 marks
(ii) Find the value of d (d > 0) for which B = 1, and explain why this corresponds to f = 1 on the reciprocal graph (invariant point). 2 marks
(iii) For d very large, what is the long-term brightness behaviour? Use the reciprocal HA result to justify. 2 marks
Stuck on (ii)? Solve 1/f(d) = 1, i.e. f(d) = 1.Problem 4 — Depth sounder (absolute value of a parabola)
A boat's depth sounder records ocean-floor elevation y (in metres relative to mean sea level) along a transect of distance x (in km) from the shore. A modelled section gives
y = x² − 6x + 5, for 0 ≤ x ≤ 6
Negative y corresponds to below-sea-level (i.e. underwater). For display, the captain wants the absolute depth |y|.
Set up: What are we solving for?
(i) Find the x-intercepts of y, i.e. the points along the transect where the ocean floor crosses sea level. 2 marks
(ii) Find the minimum of y (the deepest point) and state the depth |y_min| that the rectified display will show. 3 marks
(iii) Sketch the rectified display |y| for 0 ≤ x ≤ 6, marking the two cusps (where the floor crosses sea level) and the peak depth. 3 marks
Problem 5 — Reciprocal of a piecewise function (engineering)
A heating element has resistance R (ohms) modelled by f(T) = T − 5 for T > 5 and f(T) = 5 − T for T < 5 (so f is V-shaped about T = 5, with f(5) = 0). The conductance G is the reciprocal: G = 1/R for R ≠ 0.
Set up: What are we solving for?
(i) Identify the parent function of f and describe its key features. 2 marks
(ii) State the vertical asymptote of G(T) and explain the engineering meaning: what is happening at T = 5? 2 marks
(iii) Find G(7) and G(3). Compare and explain why the conductance is "even" about T = 5 even though the model splits into two pieces. 3 marks
Stuck on (i)? The two pieces glue into y = |T − 5|.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Rectifier
Set up. We are checking that the absolute-value transformation produces the physical effect of a rectifier and reading the range.
(i) v(t) = 240 sin(t) oscillates between −240 and 240, crossing zero at t = 0, π, 2π. The rectified |v(t)| coincides with v on [0, π] (where sin ≥ 0) and is the reflection of v on [π, 2π]. Cusps appear at the zero crossings (0, 0), (π, 0), (2π, 0). Peak voltage 240 V at t = π/2 and t = 3π/2.
(ii) |v(4)| = |−181.6| = 181.6 V. Physical meaning: the rectifier outputs the magnitude of the AC voltage, regardless of polarity, so a 181.6 V "negative" peak becomes a 181.6 V positive display.
(iii) By the definition |x| ≥ 0 for all real x, the rectified signal can never be negative. Range of |v(t)| is 0 ≤ |v| ≤ 240.
Problem 2 — Symmetric signal
Set up. We are taking a non-even sawtooth and forcing evenness by composing with |t|.
(i) f(t) = t is a straight line from (−1, −1) to (1, 1). f(|t|) = |t|, which is a V-shape with vertex at (0, 0), reaching (1, 1) at both t = ±1. It is even because |−t| = |t|.
(ii) f(|−0.5|) = f(0.5) = 0.5; f(|0.5|) = f(0.5) = 0.5. Equal, as guaranteed: for any f, f(|−t|) = f(|t|), making f(|t|) automatically even.
(iii) The original point (−0.5, −0.5) is discarded. In f(|t|), at t = −0.5 we get f(0.5) = 0.5, so the new point is (−0.5, +0.5). The original negative-y point is replaced by the mirror of the positive-y point.
Problem 3 — Reciprocal intensity
Set up. We are reading off VA, invariant points, and HA for a reciprocal transformation in a physics context.
(i) f(d) = d² − 100 = 0 at d = 10 (taking d > 0). So VA at d = 10 light-years. Physically, the model "blows up" at the calibration distance — the model is not valid right at d = 10.
(ii) B = 1 ⇒ f(d) = 1 ⇒ d² − 100 = 1 ⇒ d² = 101 ⇒ d = √101 ≈ 10.05 light-years. This is the invariant point because B(d) = 1/f(d), so when f = 1, B = 1 as well (the reciprocal of 1 is 1).
(iii) As d → ∞, f(d) → ∞, so B(d) = 1/f(d) → 0. HA y = 0: the star appears dimmer and dimmer with distance, approaching zero brightness.
Problem 4 — Depth sounder
Set up. We are using |f(x)| to convert a signed seafloor profile into a positive-only display.
(i) x² − 6x + 5 = 0 ⇒ (x − 1)(x − 5) = 0 ⇒ x = 1 km or x = 5 km.
(ii) Vertex of y = x² − 6x + 5: complete the square: y = (x − 3)² − 4. So min at (3, −4), i.e. the deepest point is at x = 3 km, with y = −4 m. Display depth |y_min| = 4 m.
(iii) Outside [1, 5] the floor is above sea level (y ≥ 0), so |y| = y. Inside [1, 5] the floor dips below; reflect those values above. Cusps at (1, 0) and (5, 0), peak display depth (3, 4) is now a local maximum on the rectified plot. Sketch shows a parabola going to 0 at x = 1, rising to 4 at x = 3, returning to 0 at x = 5, and beyond x = 5 (or before x = 1) rising upward.
Problem 5 — Resistance / conductance
Set up. We are recognising a piecewise definition as an absolute value, then reading the reciprocal.
(i) The two pieces define f(T) = |T − 5|, parent y = |T|. Key features: V-shape, vertex at (5, 0), positive everywhere except T = 5.
(ii) G(T) = 1/f(T) has VA at T = 5. Engineering meaning: at T = 5 the resistance is 0 (short circuit), so the conductance "diverges to infinity" — the model is unphysical right at T = 5.
(iii) G(7) = 1/|7 − 5| = 1/2 S. G(3) = 1/|3 − 5| = 1/2 S. Equal: G is even about T = 5 because |T − 5| = |(5 − T)|, so the reciprocal is also symmetric about T = 5. The piecewise split happens to glue into a single absolute-value function whose reciprocal inherits that symmetry.