With zero starting values the transformed equation is Y = H(s)·ℒ[q], where H(s) = 1/(a·s² + b·s + c) depends on the equation and not on the forcing. The values of s where H has a zero denominator are its poles, and they are the characteristic roots. A pole with negative real part contributes a term that dies out; a pole with positive real part contributes one that grows without bound. That is the stability criterion.
Core ideas
Three ideas: the equation and the forcing separate in the transform, the poles are roots you already know how to find, and the sign of a real part decides the long run.
With y(0) = 0 and y′(0) = 0 the initial terms vanish and (a·s² + b·s + c)·Y = ℒ[q]. So Y is H(s) times ℒ[q]. The factor H is built from the left side alone, so it describes the system, and the second factor describes what is done to it.
A pole is a value of s at which a fraction has a zero denominator. The denominator of H is the characteristic polynomial with s in place of r, so its zeros are exactly the characteristic roots found earlier in the Subject. Nothing new is computed; the roots are read from the denominator.
A real pole p contributes a term built on e^(pt), and a complex pair α ± βi contributes terms built on e^(αt) times a cosine or a sine of βt. In both cases the term dies out when the real part is negative and grows when it is positive. The imaginary part only sets the frequency.
The material
The transfer function, three equations from rest, and the criterion that sorts them.
Take a·y″ + b·y′ + c·y = q(t) with y(0) = 0 and y′(0) = 0. The transformed equation of the previous lesson loses its starting terms, and what is left is one product.
The fraction H(s) = 1/(a·s² + b·s + c) is the transfer function of the equation. Its poles are the values of s at which a·s² + b·s + c = 0.
These are the characteristic roots of “The Characteristic Equation” in Second-Order Linear Equations, with s as the unknown.
The first equation is y″ + 3y′ + 2y = 1 from rest. The transform of the forcing 1 is 1/s, so Y = 1/(s·(s² + 3s + 2)), and the denominator factors as s·(s + 1)·(s + 2). The transfer function has the poles −2 and −1. Substituting roots gives the coefficients, and Y = (1/2)/s − 1/(s + 1) + (1/2)/(s + 2).
Inverting term by term gives y = 1/2 − e^(−t) + (1/2)·e^(−2t). Each pole became one term. The poles −1 and −2 of H became the two exponentials that die out, and the pole at 0 came from the forcing, not from the equation: it is the constant 1/2 that the solution settles at. The poles of H decide whether the transient dies; the poles of ℒ[q] decide what the response settles into.
Let H be the transfer function of a constant-coefficient equation. Every term built on a pole of H tends to zero as t grows, whatever the forcing and the starting values, if and only if every pole of H has negative real part. One pole with positive real part gives a term that grows without bound.
A real pole p contributes a multiple of e^(pt), and a complex pair α ± βi a multiple of e^(αt) times a cosine or a sine. A repeated pole adds a factor t, as in the previous lesson.
The exponential e^(pt) shrinks to zero when p < 0 and grows without bound when p > 0; the same holds for e^(αt), and the sine and cosine only stay between −1 and 1. So the sign of the real part decides.
□
For y″ + 2y′ + 5y = 5 from rest, H has the denominator s² + 2s + 5 = (s + 1)² + 4. The poles are −1 ± 2i. The real part −1 is negative, so the transient dies, and the imaginary part 2 is the frequency. With the forcing 5/s the solution is y = 1 − e^(−t)·cos(2t) − (1/2)·e^(−t)·sin(2t), which oscillates about 1 with an envelope shrinking like e^(−t).
For y″ − y′ − 2y = 2 from rest, the denominator is s² − s − 2 = (s − 2)(s + 1), and the poles are 2 and −1. The pole 2 has positive real part. Its term is (1/3)·e^(2t), and the full solution is y = −1 + (1/3)·e^(2t) + (2/3)·e^(−t); the lab prints the same three terms with the growing one first. It starts at 0, as it must, and then the exponential takes over. At t = 5 the factor e^(10) is ≈ 22026.5, rounded, and every further unit of time multiplies the term by e², which is ≈ 7.38906, rounded.
One edge of the criterion is worth stating. A pole exactly on the imaginary axis, such as the pair ±2i of y″ + 4y = 0, has real part 0. It gives a term that neither shrinks nor grows, so the criterion as stated, with its strict inequality, does not call that equation stable.
Interactive
Transforms, partial fractions and solutions, exactly
Method
No decomposition is needed to classify the long run. The denominator of the transfer function is enough.
Replace y″, y′ and y in the left side by s², s and 1 to get a·s² + b·s + c. The transfer function is its reciprocal.
Solve a·s² + b·s + c = 0. Factor it, or use the quadratic formula for a complex pair.
Write down the real part of each pole. Any positive one means growth. If all are negative, the transient dies.
Transform the forcing. Its poles appear in Y but not in H, and they give the steady state the response settles into.
Decompose, invert, and check that each term has the behaviour its pole predicted.
The first preset: the poles, then the solution, then the long run.
Both poles of H are negative, so both exponentials die out and y settles at 1/2. That is also the value the equation gives when y″ and y′ are zero: 2y = 1.
Change the middle coefficient to make s² − s − 2 and the pole 2 appears, which is the third preset. The only difference in the arithmetic is one sign; the difference in the long run is bounded against unbounded.
You should be able to say which poles belong to the equation and which to the forcing, read a complex pair as a decay rate and a frequency, and confirm the prediction against the lab's solution.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.