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## Lecture 25

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**Lecture 25**Introduction to steady state sinusoidal analysis Overall idea Qualitative example and demonstration System response to complex inputs Complex arithmetic review Phasor representation of sinusoids Related educational modules: Section 2.7.0, 2.7.1, 2.7.2**Steady state sinusoidal response – overview**• We have examined the natural response and step response of electrical circuits • We now consider the forced response of a circuit to sinusoidal inputs • We will only consider the steady state response to the sinusoidal input • Apply a sinusoidal input and let t • The steady state sinusoidal response • Corresponds to the particular solution**Why is this important?**• Sinusoidal signals are very common • Power signals commonly sinusoidal (AC signals) • Carrier signals in communications often sinusoidal • The mathematics is considerably simpler • Differential equations become algebraic • System behavior often specified in terms of the system’s steady-state sinusoidal response • Example: Audio system specifications • It’s “natural” – our senses often work this way**System response to sinusoidal input**• Apply a sinusoidal input, beginning at t = 0 • u(t) = Acos(t + ), t>0**Sinusoidal response of linear systems**• The steady state response of a linear system, to a sinusoidal input, will be a sinusoid of the same frequency (particular solution of same form as input) • The amplitude and phase can change • These changes are, in general, a function of the frequency**-**• Demo system response • Sinusoidal input to tower • Indicate response: transient, steady-state, frequency dependence**RL circuit steady state sinusoidal response**• Apply a sinusoidal input to RL circuit: u(t) = Acos(t + ) • Governing equation (t):**-**• Note on previous slide that di/dt is NOT zero for steady-state sinusoidal response!**Determining steady state sinusoidal responses**• Obtaining solutions in terms of sines and cosines is tedious! • Try a “trick” involving complex exponentials: Acos(t + ) = Re{Aej(t+)} = Re{Aejejt} • Look at the response of the system to a complex exponential input, Aejejt • Results in a complex exponential response, Bejejt • The actual input is the real part of the complex input • The actual output is the real part of the complex output**-**• Note: Complex exponentials previously discussed in lecture 21. • We’ll do a little review, but it may be worthwhile for you to review lecture 21, if you are insecure about complex exponentials – we’ll be using them a LOT now • Point out that complex input not physically realizable!**RL circuit response – revisited**• Apply a complex exponential input: u(t) = Aej ejt • Governing equation (t) • Assume form of solution:**-**• Annotate last bullet of previous slide, to show di/dt and where terms go in governing differential equation**RL circuit response to complex input**• Substitute assumed solution into governing equation: • We can cancel ejt : • Since [Lej+R] is simply a complex number:**-**• Note in previous slide: • In equation 1, we no longer have a differential equation – it’s algebraic! • In equation 2, the governing equation is no longer even a function of time! (The coefficients are, however, functions of frequency) • The drawback: complex numbers are now involved. (Point out in equation 2) • We will do a little complex arithmetic review in the next few slides.**Complex numbers – review**• Rectangular coordinates: • Polar coordinates: • Relationships:**Review of complex arithmetic**• Given two complex numbers: • Addition: • Subtraction:**Review of complex arithmetic – continued**• Same two complex numbers: • Multiplication: • Division:**Review of complex arithmetic**• Same two complex numbers, but in polar form: • Multiplication: • Division:**-**• Annotate first “division” equation to note that 1/exp(phi) = exp(-phi)**Complex arithmetic – summary**• Addition, subtraction generally easiest in rectangular coordinates • Add or subtract real and imaginary parts individually • Multiplication, division generally easiest in polar coordinates • Multiplication: multiply magnitudes, add phases • Division: divide magnitudes, subtract phases**Phasors**• Recall: • Complex exponentials can be used to represent sinusoids • Complex exponentials can be written as a complex number multiplying a time-varying complex exponential • The complex number Aejprovides the magnitude and phase of the original sinusoidal signal**Phasors – definition**• The complex number (in polar form) providing the magnitude and phase of a sinusoidal signal is called a phasor.**Example**• Use phasors to determine the current i(t) in the circuit below if Vs(t) = Vmcos(100t).**-**• In previous slide, show: • Change of input to complex exponential • Derivation of governing differential equation.**Example – continued**• The governing differential equation is: • Since the input has the form , where is a phasor representing the input magnitude and phase, the output must be of the form , where is a phasor representing the output magnitude and phase.**-**• In previous slide, emphasize concepts. Do substitution and write algebraic equation.**-**• In previous slide, do complex arithmetic to solve for current phasor; convert back to time domain.**Example – Time domain signals**• Input: • Response: • The response lags the input by 45