Discussion #7 Example Problem This problem illustrates how Fourier series are helpful tools for analyzing electronic circuits. Often in electronic

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1 Discussion #7 Example Poblem This poblem illustates how Fouie seies ae helpful tools fo analyzing electonic cicuits. Often in electonic cicuits we need sinusoids of vaious fequencies But we may aleady have cicuity in the system that geneates a clock signal Can we avoid having to build a sinusoidal oscillato by exploiting the clock signal?

2 et s imagine that ou job is to design cicuity to convet a squae wave into a sinusoid of Hz. Suppose we have a squae wave oscillato that geneates the following signal with a peiod of 100μs: x(t) The box labeled oscillato in the figue below poduces this signal. Assume that the output impedance of the oscillato is vey low so that the est of the cicuit doesn t load it (ecall ou pevious discussions of loading). Assume that the esisto is 800Ω, the capacito is 1μF, and the inducto is 250μH t

3 What we need to do fo this design: (a) Find the Fouie Seies fo the squae wave signal x(t) that gets applied to the input of the RC cicuit. (b) Find the fequency esponse H() of the RC cicuit and use a compute to plot H(f) vs. f 0 (with f in Hz) (c) Find the FS spectum at the output of the cicuit and veify that it epesents appoximately a Hz sinusoid (d) Find the total hamonic distotion of ou esultant sinusoid. Total Hamonic Distotion (THD) is used in pactice to measue the quality of a sinewave geneato (a pefect one has 0% THD). THD is also used measue the quality of a linea system (e.g., an audio amplifie): you put a sinusoid in and should get a sinusoid of the same fequency out but if the system is nonlinea you ll get powe at othe fequencies THD measues how much powe is at these othe fequencies elative to how much is at the input fequency.

4 (a) Fo ou case T 0 100μs 0 2π T 0 1 f0 0 / 2π 10kHz T 0 So as we have found befoe the FS of this signal is: x( t) 4 A π sin 4 A 3π 4 A sin 5π ( 2π 10,000 t ) sin ( 2π 30,000 t ) ( 2π 50,000 t )... (b) Now we find the fequency esponse H() of the cicuit by asking: How e jt goes though the system (whee we imagine that is abitay)? x( t) e jt y( t) H () e jt This is just phaso-based analysis of the cicuit!!!!

5 Recall: Phaso-Analysis Steps: 1. Convet the Cicuit into an Equivalent Complex DC Cicuit a) Wite Sinusoidal Souce as a Phaso (phaso complex DC souce ) b) Wite s/cs as impedances (impedance complex esistance) i. Capacito: Z C () 1/jC Inducto: Z () j 2. Solve the esulting Complex-Valued DC Cicuit using appopiate combinations of DC analysis methods: a) Fundamental Rules: KV & KC i. Systematic Schemes based on KV & KC: oop, Nodal, Mesh ii. Simple Resulting Ticks : Voltage Divide & Cuent Divide b) Equivalent Cicuit Rules: a) Thevenin/Noton, b) Seies/Paallel Combinations 3. Convet output phaso back into eal-valued sinusoid a) O fo fequency esponse analysis just identify H()

6 Step 1: In ou case we ae aleady pat way to a phaso because we ae asking about e jt, so technically ou phaso is just 1. But let s just keep the input notated as an abitay phaso: X Now convet the capacito and inducto into impedances at the abitay fequency : The impedances ae: Z ( ) j & Z ( ) C 1 jc The Complex-DC cicuit to analyze is this: X j 1 jc Y H () X

7 Step 2: Now we solve this Complex DC Cicuit: X j 1 jc Y H () X Fo this cicuit we don t need the poweful Systematic Schemes We can use Paallel Combo togethe with Voltage Divide Paallel combo: Z Z Z ZC + Z C ( j) j + 1 jc 1 jc j 2 1 C X j j 1 j2 C 1 C Y H () X

8 j C R j H + ) (1 ) ( 2 The plot of H() on the next page (fo the given RC values) shows that this cicuit is a naow bandpass filte it passes a naow band of fequencies (in this case centeed at 10kHz, the sinusoidal oscillato fequency we want!) Now Voltage Divide gives: ( ) X j C R j X C j R C j X Z R Z Y ) ( H () X H Y () X C j 1 j C j 2 1

9 This Cicuit s Fequency Response Zoom-In Aound 10 khz

10 (c) Each tem in the FS of the input looks like this: b b n n sin( n0t) n 1,3,5,... o sin( n 2π 10,000t) So we get the following fequencies: 10kHz 30kHz 50kHz etc. Each of these sine waves goes though the cicuit accoding to ou geneal esult to give: b n sin( n t) 0 H () H ( n0) bn sin( n0t + H ( n0)) So the output is: H n ) bn sin( n t + H ( n )) ( 0 0 0

11 Now fom the plot of H() we see that the sinusoidal tem at 10kHz in x(t) will get multiplied by: H(2π 10,000) 1 But all the othe fequencies (20 khz, 30 khz, 40 khz, etc.) get multiplied by vey small numbes. The input magnitude spectum, the fequency esponse, and the output magnitude spectum ae shown in the plots on the next page. The output spectum shows one significant sinusoid with all the othe tems negligible Output single 10kHz

12 The bottom plot is found by multiplying the top plot by the middle plot

13 (d) Total Hamonic Distotion is the powe at the desied fequency elative to the powe at all othe fequencies (as %) P THD 20kHz n n 2 100% 1 Fom compute calculation: + ( b H ( n )) ( b H ( ) ) 0 P 0 2 P 2 / 2 / 2 10kHz kHz Powe Elsewhee 0.33 x 10-4 W + Output is a good quality sinusoid We ll use the compute so we ll only compute this sum fo some lage enough # tems THD 100% THD 0.02% Vey ow!

14 Some pactical things to woy about: Note that ou peak doesn t fall exactly at 10 khz!!! 1. Thee ae only cetain typical R,, C values that ae available a. Standad 1% esistos ae available having Rs have fist two digits of b. Capacitos and Inductos ae also available in cetain standad values c. So we may not be able to choose standad values to get a peak exactly at 10 khz!! 2. All components have toleances: accuate to 1%, 5%, 10%, 20%, etc. of the nominal value a. So we should choose the toleance needed to ensue that the peak will not move too much ove the possible component value anges!!!