Lecture 171 Higher Order Circuits

Lecture 172 Higher Order Circuits The text has a chapter on 1st order circuits and a chapter on 2nd order circuits. The text has no chapter on 3rd order circuits. Why?

Lecture 173 Higher Order Circuits are Boring! The behavior of a higher order (3rd or higher order) circuit is not qualitatively different than that of a 1st or 2nd order circuit. Particular solutions are similar, especially for constant and sinusoidal sources.

Lecture 174 More on Higher Order Circuits The natural response is a sum of decaying exponentials and/or exponentially decaying sinusoids. The responses of higher order circuits have the same sort of characteristics as 1st and 2nd order circuits There are more terms in the solution.

Lecture 175 Mathematical Justification Any voltage or current in an nth order circuit is the solution to a differential equation of the form

Lecture 176 Particular Solution The particular solution v p (t) is typically a weighted sum of f(t) and its first n derivatives. If f(t) is constant, then v p (t) is constant. If f(t) is sinusoidal, then v p (t) is sinusoidal.

Lecture 177 Complementary Solution The complementary solution is the solution to Complementary solution has the form

Lecture 178 Characteristic Equation s 1 through s n are the roots of the characteristic equation

Lecture 179 Time Waveforms If s i is a real root, it corresponds to an exponential term If s i is a complex root, there is another complex root that is its complex conjugate, and together they correspond to an exponentially decaying sinusoidal term

Lecture 1710 Example A 3rd order circuit has the following characteristic equation: s 3 + 6s s + 6 = 0 What terms would we expect in the complementary solution?

Lecture 1711 Finding Roots of Polynomials with MATLAB We have a polynomial with coefficients 1, 6, 11, and 6. The following MATLAB command finds its roots: roots([ ])

Lecture 1712 Answer The roots of the characteristic equation are -1, -2, and -3 The complementary solution is Initial conditions will determine the values of the constants.

Lecture 1713 Example A 4th order circuit has the following characteristic equation: s 4 + s 3 - 2s 2 + 2s + 4 = 0 What terms would we expect in the complementary solution?

Lecture 1714 Answer The roots of the characteristic equation are -1, -2, 1+j, and 1-j The complementary solution is Initial conditions will determine the values of the constants.

Lecture 1715 Summary In an nth order linear circuit, any voltage or current is the solution to an nth order linear constant coefficient differential equation. The particular solution is usually –Constant for constant sources (DC SS) –Sinusoidal for sinusoidal sources (AC SS)

Lecture 1716 Summary (cont.) The complementary solution is usually a sum of decaying exponentials and exponentially decaying sinusoids. –Time constant –Damping ratio and natural frequency

Lecture 1717 Summary (cont.) Transients usually are associated with the complementary solution. The actual form of transients usually depends on initial capacitor voltages and inductor currents. Steady state responses usually are associated with the particular solution.

Lecture 1718 Summary (cont.) You should be able to work problems in which capacitors in first order circuits charge and discharge. You should be able to find the damping ratio and natural frequency in second order circuits and determine if they are under damped, over damped, or critically damped.

Lecture 1719 Summary (cont.) You should be able to identify the transient and steady state portions of a waveform. You should be able to describe the role of initial conditions in the transient portion of the waveform. You should be able to explain why 3rd and higher order circuits don’t act qualitatively different than 1st and 2nd order circuits.