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Mathematical Methods

Laplace in practice: differential equations, circuits and the transfer function

Solving differential equations, impedances in the s domain, the transfer function and feedback, the second-order system, overshoot and settling time.

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With the Laplace transform a differential equation with its initial conditions becomes an algebraic equation, solved with basic arithmetic. In this lesson we use it to solve differential equations, to analyse circuits directly in the s domain and to study the transfer function and the step response of second-order systems, the most common in electronics and control.

Solving a differential equation

The method has three steps: transform the equation, solve for Y(s), invert.

Example.

    \[ y'' + 3y' + 2y = 4,\qquad y(0) = 0,\quad y'(0) = 1 \]

Transforming, with \mathcal L\{y''\} = s^2Y - sy(0) - y'(0) and \mathcal L\{y'\} = sY - y(0):

    \[ s^2Y - 1 + 3sY + 2Y = \frac4s \quad\Rightarrow\quad Y(s) = \frac{s + 4}{s(s + 1)(s + 2)} \]

Partial fractions: \frac{A}{s} + \frac{B}{s + 1} + \frac{C}{s + 2}, with A = \frac{4}{2} = 2, B = \frac{3}{(-1)(1)} = -3, C = \frac{2}{(-2)(-1)} = 1. So

    \[ y(t) = 2 - 3e^{-t} + e^{-2t} \]

Check: y(0) = 0, y'(0) = 3 - 2 = 1, and in steady state y\to 2, as the final value theorem says: \lim_{s\to 0}sY = \frac{4}{2} = 2. No constants to find at the end: the initial conditions entered the calculation from the start.

Circuits in the s domain

Transforming the component laws, with zero initial conditions:

ComponentTimes domainImpedance
resistorv = RiV = RIR
inductorv = L\,di/dtV = sL\,IsL
capacitori = C\,dv/dtI = sC\,V\dfrac{1}{sC}

They are the impedances of the AC lesson with s in place of i\omega. The circuit is solved with the generalised Ohm’s law, dividers and Kirchhoff’s laws, as if it were made of resistors. Non-zero initial conditions are represented by extra sources: a capacitor charged to v_0 becomes the impedance 1/(sC) in series with a source v_0/s.

Example: the RC low-pass. The divider between R and 1/(sC) gives

    \[ H(s) = \frac{V_{out}}{V_{in}} = \frac{1/(sC)}{R + 1/(sC)} = \frac{1}{1 + sRC} \]

With a step input, V_{in} = 1/s, one gets V_{out} = \frac{1}{s(1 + s\tau)} and therefore v_{out} = 1 - e^{-t/\tau}: the capacitor charging, in three lines.

The transfer function

For a linear system with zero initial conditions, the ratio of the output and input transforms does not depend on the input:

    \[ H(s) = \frac{Y(s)}{X(s)} \]

It is the transfer function, and it is the Laplace transform of the impulse response. It holds the whole system:

  • its poles give the natural modes and stability;
  • evaluated at s = i\omega it gives the frequency response: |H(i\omega)| is the gain and \arg H(i\omega) the phase shift at each frequency, from which Bode plots are drawn;
  • multiplied by the input transform it gives the output.

Cascading systems means multiplying transfer functions. With negative feedback, a block G(s) with feedback K(s) becomes

    \[ H(s) = \frac{G(s)}{1 + G(s)K(s)} \]

the formula behind feedback op-amp amplifiers and every control system.

The second-order system

A great many systems (the RLC circuit, a suspension, a hard disk motor) behave like

    \[ H(s) = \frac{\omega_0^2}{s^2 + 2\zeta\omega_0 s + \omega_0^2} \]

where \omega_0 is the natural angular frequency and \zeta the damping ratio. The poles are

    \[ s = -\zeta\omega_0 \pm \omega_0\sqrt{\zeta^2 - 1} \]

real and distinct if \zeta \gt 1, coincident if \zeta = 1, complex conjugate if \zeta \lt 1.

Step response of a second-order system for five values of the damping ratio
Step response for \zeta_1 = 0.15, \zeta_2 = 0.4, \zeta_3 = 0.7, \zeta_4 = 1 and \zeta_5 = 2. With low damping the response overshoots the final value and rings for a long time; with \zeta = 1 it arrives without overshooting; with high damping it is slow.

For \zeta \lt 1 the step response is

    \[ y(t) = 1 - \frac{e^{-\zeta\omega_0 t}}{\sqrt{1 - \zeta^2}}\sin\!\left(\omega_d t + \varphi\right),\qquad \omega_d = \omega_0\sqrt{1 - \zeta^2},\quad \cos\varphi = \zeta \]

The parameters found in specifications:

  • the maximum overshoot, e^{-\pi\zeta/\sqrt{1 - \zeta^2}}: 37% with \zeta = 0.3, 4.6% with \zeta = 0.7, zero with \zeta \ge 1;
  • the 2% settling time, about 4/(\zeta\omega_0);
  • the rise time, inversely proportional to \omega_0.

The value \zeta \approx 0.7 is a classic compromise: a fast response with an overshoot of a few per cent. It is the damping of a second-order Butterworth filter (\zeta = 1/\sqrt2), the one with the flattest possible frequency response.

Example. A series RLC with the output across the capacitor has H(s) = \frac{1}{LCs^2 + RCs + 1}, so \omega_0 = \frac{1}{\sqrt{LC}} and \zeta = \frac R2\sqrt{\frac CL}. With L = 1 mH and C = 1 µF, \omega_0 = 31\,600 rad/s (5 kHz); to get \zeta = 0.7 you need R = 1.4\sqrt{L/C} \approx 44 Ω.

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