Discussion #7 Example ProblemThis problem illustrates how Fourier series are helpful tools for analyzing electronic circuits. Often in electronic circuits we need sinusoids of various frequencies…But we may already have circuitry in the system that generates a clock signal…Can we avoid having to build a sinusoidal oscillator by exploiting the clock signal?)(txtThe box labeled “oscillator” in the figure below produces this signal.Assume that the output impedance of the oscillator is very low so that the rest of the circuit doesn’t load it (recall our previous discussions of loading). Assume that the resistor is 800Ω, the capacitor is 1μF, and the inductor is 250μHLet’s imagine that our job is to design circuitry to convert a square wave into a sinusoid of 10000 Hz. Suppose we have a square wave oscillator that generates the following signal with a period of 100μs:(a) Find the Fourier Series for the square wave signal x(t) that gets applied to the input of the RLC circuit. (b) Find the frequency response H(ω) of the RLC circuit and use a computer to plot |H(f)| vs. f ≥0 (with f in Hz) (c) Find the FS spectrum at the output of the circuit and verify that it represents approximatelya 10000 Hz sinusoid(d) Find the total harmonic distortion of our resultant sinusoid. Total Harmonic Distortion (THD) is used in practice to measure the quality of a sinewave generator (a perfect one has 0% THD).THD is also used measure the quality of a linear system (e.g., an audio amplifier): you put a sinusoid in and should get a sinusoid of the same frequency out… but if the system is nonlinear you’ll get power at other frequencies… THD measures how much power is at these other frequencies relative to how much is at the input frequency. What we need to do for this design:For our case T0= 100μs 002Tπω=⇒kHzTf 1012/000===⇒πωSo… as we have found before… the FS of this signal is:()()()...000,502sin54000,302sin34000,102sin4)(tAtAtAtx×−×−×−=ππππππ(b) Now we find the frequency response H(ω) of the circuit by asking: How ejωtgoes through the system (where we imagine that ω is arbitrary)?tjetxω=)(tjeHtyωω)()( =(a)This is just phasor-based analysis of the circuit!!!!Recall: Phasor-Analysis Steps:1. Convert the Circuit into an Equivalent “Complex DC Circuit”a) Write Sinusoidal Source as a Phasor (phasor = “complex DC source”)b) Write Ls/Cs as impedances (impedance = complex resistance)i. Capacitor: ZC(ω) = 1/jωC Inductor: ZL(ω) = jωL2. Solvethe resulting “Complex-Valued DC Circuit” using appropriate combinations of DC analysis methods:a) Fundamental Rules: KVL & KCL i. Systematic Schemes based on KVL & KCL: Loop, Nodal, Meshii. Simple Resulting “Tricks”: Voltage Divider & Current Dividerb) Equivalent Circuit Rules: a) Thevenin/Norton, b) Series/Parallel Combinations3. Convertoutput phasor back into real-valued sinusoida) Or… for frequency response analysis… just identify H(ω)The impedances are:CjZLjZCLωωωω1)(&)( ==Step 1: In our case we are already part way to a phasor because we are asking about ejωt, so technically our phasor is just 1.XrBut let’s just keep the input notated as an arbitrary phasor: Now convert the capacitor and inductor into impedances at the arbitrary frequency ω:XHYrr)(ω=XrCjω1LjωThe “Complex-DC” circuit to analyze is this:Parallel combo:()LCLjCjLjCjLjZZZZZCLCL2||111ωωωωωω−=+⎟⎟⎠⎞⎜⎜⎝⎛=+=Step 2: Now we solve this Complex DC Circuit:XHYrr)(ω=XrCjω1LjωFor this circuit we don’t need the powerful “Systematic Schemes”…We can use “Parallel Combo” together with “Voltage Divider”XHYrr)(ω=XrCjω1LjωLCLj21ωω−LjLCRLjHωωωω+−=⇒)1()(2The plot of H(ω) on the next page (for the given RLC values) shows that this circuit is a narrow “bandpass” filter – it passes a narrow band of frequencies (in this case centered at 10kHz, the sinusoidal oscillator frequency we want!)Now Voltage Divider gives:()XLjLCRLjXLCLjRLCLjXZRZYrrrr⎥⎦⎤⎢⎣⎡+−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−+−=⎥⎥⎦⎤⎢⎢⎣⎡+=ωωωωωωω)1(11222||||)(ωHXHYrr)(ω=XrCjω1LjωLCLj21ωω−This Circuit’s Frequency ResponseZoom-In Around 10 kHz(c) Each term in the FS of the input looks like this:)000,102sin(,...5,3,1)sin(0tnborntnbnn××=πωSo we get the following frequencies: 10kHz30kHz50kHz etc. Each of these sine waves goes through the circuit according to our general result to give:)(ωH)sin(0tnbnω))(sin()(000ωωωnHtnbnHn∠+So the output is:))(sin()(000ωωωnHtnbnHn∠+Now from the plot of |H(ω)| we see that the sinusoidal term at 10kHz in x(t) will get multiplied by:|H(2π×10,000)| ≈ 1But all the other frequencies (20 kHz, 30 kHz, 40 kHz, etc.) get multiplied by very small numbers. The input magnitude spectrum, the frequency response, and the output magnitude spectrum are shown in the plots on the next page.⇒ The output spectrum shows one significant sinusoid with all the other terms negligible⇒ Output ≈ single sinusoid @ 10kHzThe bottom plot is found by multiplying the top plot by the middle plot(d) Total Harmonic Distortion is the power at the desired frequency relative to the power at all other frequencies (as %) ()()()⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡×=+++×=∑∞=2/)(2/%100%10020122010@40@30@20@ωωHbnHbPPPPTHDnnkHzkHzkHzkHzLWe’ll use the computer…so we’ll only compute this sum forsome large enough # termsFrom computer calculation:Power @ 10kHz = 0.1406 W Power Elsewhere = 0.33 x 10-4W!%02.0THD1406.01033.0%100THD4LowVery=⎟⎟⎠⎞⎜⎜⎝⎛××=−⇒ Output is a good quality sinusoidSome practical things to worry about:Note that our peak doesn’t fall exactlyat 10 kHz !!!1. There are only certain typical R, L, C values that are availablea. Standard 1% resistors are available having Rs have first two digits of 10 11 12 13 15 16 18 20 22 24 27 30 33 36 39 43 47 51 56 62 68 75 82 91b. Capacitors and Inductors are also available in certain standard valuesc. So… we may not be able to choose standard values to get a peak exactly at 10 kHz!!2. All components have tolerances: accurate to 1%, 5%, 10%, 20%, etc. of the nominal valuea. So… we should choose the tolerance needed to ensure that the peak will not move too much over the possible component value