Transfer functions, startup and periodic input

Try

A vibration contains cos(t) plus half as much cos(3t). Which frequency contributes more displacement after passing through y″ + y′ + 4y = f? Applying the gain formula separately gives the periodic response

Verify the first pair by substitution before inspecting the second. The second forcing amplitude is 1/2, so omitting that multiplier doubles its output. Linearity combines correct particular solutions; it does not supply initial conditions.

Predict

Now drive the same oscillator with a triangle: f(t) = |t| on [−π,π], repeated every 2π. It has corners. Predict whether an approximation using only a few sine/cosine terms can solve the full forcing equation exactly. Also predict whether the smoother-looking output graph alone proves that its infinite series can be differentiated twice.

Experiment

Reconstruct a periodic response

Normalized teaching model y″ + c y′ + 4y = f(t). Time and displacement use fixed teaching units. Analytic values, temporary controls; no numerical time integration or saved evidence.

Damping fixed at c = 1. Forcing is |t| on [−π,π], extended with period 2π.

Largest allowed harmonic N 35917 Inspect time 00.5124816 Reset experiment

At t = 0: true forcing 0.00000, truncated forcing 0.10516, truncated displacement 0.03383. The finite sum solves the equation for its truncated forcing.

Uniform displacement error ≤ 6.1115e-3 at every time, from a proved tail bound against the full periodic response. This is a certificate, not a measured error. Bounds for velocity and acceleration: 0.04584 and 0.45837.

Fixed plot window: time 0–16, displacement −5 to 5. Solid navy: truncated periodic response. Dashed red: sinusoidal particular when defined. The periodic plot has no startup correction.

Predict before changing a control, then record an equation-based explanation in the lesson practice. The plot samples 321 times; smooth-looking samples do not prove an equation or a uniform bound.

Compare N = 3 and N = 9 at time zero and time 1. Explain why the displayed error certificate decreases by a factor 27, even though the actual pointwise error need not do so. The forcing approximation and the displacement approximation have different tail bounds. These controls show the periodic response only; adding an initial-state correction is a separate step.

Understand

Transfer means zero initial state

Let F(s) and Y(s) denote one-sided Laplace transforms of the forcing and displacement, and let y(0) = a, y′(0) = b. Assume the piecewise regularity and growth needed for these transforms in a common right half-plane. Transforming y″ + c y′ + 4y = f gives:

H is the transfer function: it maps input to output when the initial state is zero. The second term is the free response carrying initial information. Factoring the denominator links poles to characteristic roots. For c > 0 both roots have negative real part and the free response decays; repeated roots add a polynomial factor without preventing decay. For c = 0 the imaginary roots allow persistent free oscillations.

H(iω) is also the complex multiplier for a sinusoidal particular when its denominator is nonzero. This algebraic frequency evaluation does not say that a one-sided sinusoidal transform converges on the imaginary axis. For positive damping the causal impulse kernel decays and its frequency integral agrees with H(iω). The earlier convolution lesson explains how that kernel produces a zero-state response starting at time zero.

Derive the triangle’s frequencies

Evenness makes every sine coefficient vanish. Using the convention f = a₀/2 + Σ aₙ cos(nt), the mean is π/2. Integration by parts yields:

We use the Dirichlet convergence theorem: a piecewise C¹ periodic function with one-sided limits has Fourier series converging at each point to the mean of those limits. Here the periodic triangle is continuous, including its endpoints, so that mean is f(t). In addition, the coefficient magnitudes sum to a finite value; the uniform comparison test makes this particular series uniformly convergent. The theorem is stated, not reproved here; Fourier convergence separates pointwise, uniform and mean-square claims.

Apply H(in) = 1/(4 − n² + in) to each harmonic, and divide the constant input by stiffness 4. Let hₙ and jₙ be the real cosine and sine coefficients for a unit cosine input:

Every finite y_N solves the equation for the finite forcing f_N exactly. The full forcing differs from f_N. We must justify taking the limit and differentiating before claiming an infinite-series solution.

A certificate for every time, not just the plotted grid

For n ≥ 3, the transfer magnitude is at most 1/(n² − 4) ≤ 9/(5n²). Therefore each output harmonic has amplitude at most 36/(5πn⁴). Dropping the odd-only restriction increases the bound. For integer N ≥ 2, comparison with an integral gives:

This is a uniform displacement bound against the infinite periodic response. It is neither a measured error nor a bound against a physical machine. Differentiating one harmonic multiplies its amplitude by n; differentiating twice multiplies it by n². Both resulting comparison series still converge:

Uniform convergence of the derivative series, with convergence at a base point, permits termwise differentiation. Apply this theorem twice, first to y_N and then to y_N′. The limit is C². Since f_N converges uniformly to the continuous f, taking limits in the finite equations proves y″ + y′ + 4y = f. No derivative of the cusped forcing is taken. The forcing’s own uniform tail is bounded by 4/(πN), which decays more slowly than the output bound.

Positive damping makes this periodic solution unique: the difference of two periodic solutions would be a free solution that both decays and repeats, so must be zero. Every initialized solution approaches it after a decaying free correction. Uniqueness and attraction require that damping argument, not simply successful summation.

Repair a claim after damping is removed

For y″ + 4y = f with the same triangle, the input contains no harmonic at the resonant frequency 2. There is still a bounded periodic particular obtained by dividing the odd coefficients by 4 − n². But adding A cos(2t) + B sin(2t) gives other 2π-periodic solutions, and initial differences never decay. Thus “bounded,” “unique periodic” and “attracting” are distinct properties. If the input is changed to f + ε cos(2t), ε ≠ 0, it acquires the secular term εt sin(2t)/4. The previously missing harmonic changes the conclusion completely.

Discontinuous forcing: match the state, not the derivative

A square wave is a useful changed input because the triangle argument’s twice-uniformly-differentiable conclusion cannot simply be reused. Consider the stable first-order system y′+a y=q(t), with a>0, where q=1 on (0,h), q=−1 on (h,2h), repeated with period 2h. Values assigned at the switching instants do not affect its integral solution. Seek a continuous, piecewise C¹ periodic state satisfying the equation away from the switches and almost everywhere.

Put r=exp(−ah). If y(0)=y₀, integrating factors give y(t)=1/a+(y₀−1/a)exp(−at) on [0,h]. Write y₁=y(h). On [h,2h], y(t)=−1/a+(y₁+1/a)exp(−a(t−h)). Substitution gives the period map y(2h)=r²y₀−(1−r)²/a. Solving y(2h)=y₀ yields

$$y_0=-\frac{1-r}{a(1+r)},\qquad y_1=\frac{1-r}{a(1+r)}.$$

The denominator is nonzero because 0<r<1. These formulas construct the periodic solution directly, including its endpoint matching. Any two initialized solutions differ by C exp(−at), so the periodic solution is unique and every other state approaches it. At t=h, the forcing jumps by −2 while y remains continuous, hence the one-sided derivatives differ by −2. The state is not globally C¹. A classical equation at every switching point is not the right solution concept; the integral identity y(t)=y(0)+∫₀ᵗ(q(s)−a y(s))ds holds at every t.

For a=1 and h=ln 2, r=1/2: y₀=−1/3 and y₁=1/3. Immediately before the downward switch y′=2/3; immediately after it y′=−4/3. The jump is −2, not a jump in y itself. Copying a smooth-forcing derivative assumption would erase this distinction.

Match a discontinuous forcing

Predict the state and one-sided derivative before moving across the switch at phase 1. The formula evaluates a periodic state, not a numerical time-step simulation.

Damping a: 1.0 Half-period h: 2.0 Time divided by h: 0.50

State 0.351946; forcing 1; selected one-sided derivative 0.648054. At a switch the control chooses the right-hand branch. State amplitude: 0.761594.

Temporary practice only. No saved response, automatic proof grade or mastery award.

What Fourier truncation does and does not establish

The square wave’s Fourier coefficients are 4/(πn) on odd sine harmonics n, when h=π. Its series converges to the midpoint value zero at switches. Transfer by 1/(a+in) makes the response coefficients O(1/n²), so the response series converges uniformly. The derivative coefficient bound is only O(1/n), not summable; the earlier absolute-uniform derivative argument fails. The explicit piecewise solution above supplies existence and the correct regularity without pretending a finite harmonic approximation proves them. Gibbs overshoot in the input approximation is not a discontinuity of the exact state.

Changed case: remove stability and audit the conclusion

Before opening this feedback, solve y′=q(t) for the same square wave and decide which conclusions survive. Its integral rises by h and then falls by h, so every initial value gives a 2h-periodic continuous triangular state. There is no unique periodic state and no attraction. Now change the negative half to −1/2: the integral over one period is h/2, so every solution drifts upward by h/2 and no periodic state exists. A complete response integrates both halves, checks state matching, identifies the derivative jumps and distinguishes periodic existence from asymptotic stability. Retain the first argument and its correction; this public feedback is not a saved assessment.

This original switch-by-switch example extends the historical MIT18.03 SC periodic-input scope. It proves the stated first-order result, not a general theorem for every discontinuous nonlinear or distribution-valued forcing.

Apply

For displacement error at most 0.001, solve 12/(5πN³) ≤ 0.001. The sufficient integer cutoff is N = 10; using the displayed choices requires N = 17. This conservative certificate does not imply N = 9 actually fails. Next explain why an error estimate for displacement alone would not justify differentiating a different series twice. Complete the practice derivations before reading their solutions.

Continue

Use the changed forcing and output mixed task to change the stiffness, measured output and resonant harmonic. Return to Laplace basics for transform initial terms or gain and practical resonance for the denominator derivation.

Source scope: MIT 18.03SC syllabus, Fall 2011 and its final assessment materials, consulted 28 October 2026. This original sequence targets frequency/periodic-response outcomes; it does not claim complete university-course coverage or independently verified learning.