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BU EECE 301 - Note Set #18

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EECE 301 Signals & Systems Prof. Mark Fowler1/12EECE 301 Signals & SystemsProf. Mark FowlerNote Set #18• C-T Systems: Frequency-Domain Analysis of Systems• Reading Assignment: Section 5.2 of Kamen and Heck2/12Ch. 1 IntroC-T Signal ModelFunctions on Real LineD-T Signal ModelFunctions on IntegersSystem PropertiesLTICausalEtcCh. 2 Diff EqsC-T System ModelDifferential EquationsD-T Signal ModelDifference EquationsZero-State ResponseZero-Input ResponseCharacteristic Eq.Ch. 2 ConvolutionC-T System ModelConvolution IntegralD-T System ModelConvolution SumCh. 3: CT Fourier SignalModelsFourier SeriesPeriodic SignalsFourier Transform (CTFT)Non-Periodic SignalsNew System ModelNew SignalModelsCh. 5: CT Fourier SystemModelsFrequency ResponseBased on Fourier TransformNew System ModelCh. 4: DT Fourier SignalModelsDTFT(for “Hand” Analysis)DFT & FFT(for Computer Analysis)New SignalModelPowerful Analysis ToolCh. 6 & 8: Laplace Models for CTSignals & SystemsTransfer FunctionNew System ModelCh. 7: Z Trans.Models for DTSignals & SystemsTransfer FunctionNew SystemModelCh. 5: DT Fourier System ModelsFreq. Response for DTBased on DTFTNew System ModelCourse Flow DiagramThe arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis).3/125.2 Response to Periodic Inputsh(t)H(ω)periodic x(t) y(t) = ?Since x(t) is periodic, write it as FS:∑∞−∞==ktjkxkectx0)(ωH(ω)tjkxkec0ωtjkxkeckH0)(0ωω(complex: magnitude & phase)Sum these to get outputSo, the input is a sum of termsLinearSystem: So… Output = Sum of Individual ResponsesBut each individual response is to a complex sinusoid input ⇒ EASY!Sum these to get input∑∞−∞==ktjkxkectx0)(ω∑∞−∞==ktjkxkeckHty0])([)(0ωωFS coefficient of y(t) Indicates “for x(t)”4/12General Insights from this Analysis1. periodic in, periodic out2. The system’s frequency response H(ω) works to modify the input FS coefficients to create the output FS coefficients:xkykckHc )(0ω=5/12Example (Ex. 5.4 with Some Injected Reality)Problem: suppose you have a circuit board that has a digital clock circuit on it. It makes the rectangular pulse train shown below:. . . . . .tx(t)1-4.5 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 4.5(Of course most digital clock circuits would run much faster)Assume:The circuit “driving” the cable has an infinitesimally small output impedance (that is good!):x(t)Thevenin of driver:y(t)x(t) Pair of wires can be modeledas an RC circuit:x(t)y(t)Suppose you need to connect this clock signal to a circuit on another circuit board using a twisted pair of wires:Q: What effect does the cable have on the clock signal at the 2ndboard???6/12Assume: The circuit being “driven” by the cable has infinite input impedance (that is good!) i.e. No loading of the RC circuitSo…x(t) y(t) (goes to driven circuit having infinite input impedance)Goal: Perform an analysis to enable you to recommend an acceptable value of cable RC time constant (Analysis Drives Design!)Step 2:Find cable’s frequency response as a function of RC:(See Ex. in section 5.1) RCjHωω+=11)(⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧=±±±=±±±=−±±±==0,21,...6,4,2,0,...11,7,3,1,...9,5,1,1kkkkkkcxkππIndicates “for x(t)”Step 1: Analytically find FS of input and compute truncated FS sum:From Ex. 3.4 we get:∑−=≈NNktjkxkoectxω)(Then plot vs. time t7/12Step 3 (optional) (But it really helps you see what is going on!)Look at frequency domain plots of Input and System (for various RC values)“stem” plot of FScoefficients’Magnitudexkc“continuous” plot ofMagnitude of system’sFrequency Resp. |H(ω)|Step 4 (optional) (This also really helps you see what is going on) Compute output FS coefficients:Look at the result → “stem” plot of ykcxkykckHc )(0ω=See plots on next 3 pages for three RC time constant values: RC = 0.01 sRC = 0.1 sRC = 1 sNote: Short RC timeconstant passes highfrequencies better than long RC time constant∑−=≈NNktjkykecty0)(ωPlot vs. time tStep 5: Compute truncated FS sum to see output signal8/12-4 -2 0 2 4-0.500.511.5time (sec)Input Signal x(t)0 5 10 1500.20.4k fo (Hz)|cxk|0 5 10 1500.20.40.60.81f (Hz)|H(f)|0 5 10 1500.20.4k fo (Hz)|cyk|-4 -2 0 2 4-0.500.511.5time (sec)Output Signal y(t)RC Circuit Analysis w/ Square Wave InputRC = 0.01 sArtifact from summing only a finite# of termsDoes Decent Job of “Passing” Most of the Significant Frequencies of the Input Input SignalInput Signal’s FS CoefficientsOutput Signal’s FS CoefficientsCable’s Freq. Resp.Output SignalDoes Decent Job of “Passing” Most of the Significant Frequencies of the Input9/12-4 -2 0 2 4-0.500.511.5time (sec)Input Signal x(t)0 5 10 1500.20.4k fo (Hz)|cxk|0 5 10 1500.20.40.60.81f (Hz)|H(f)|0 5 10 1500.20.4k fo (Hz)|cyk|-4 -2 0 2 4-0.500.511.5time (sec)Output Signal y(t)RC = 0.1 sDoes Moderate Job of “Passing” Most of the Significant Frequencies of the Input RC Circuit Analysis w/ Square Wave InputInput SignalInput Signal’s FS CoefficientsOutput Signal’s FS CoefficientsCable’s Freq. Resp.Output SignalDoes Moderate Job of “Passing” Most of the Significant Frequencies of the Input10/12-4 -2 0 2 4-0.500.511.5time (sec)Input Signal x(t)0 5 10 1500.20.4k fo (Hz)|cxk|0 5 10 1500.20.40.60.81f (Hz)|H(f)|0 5 10 1500.20.4k fo (Hz)|cyk|-4 -2 0 2 40.20.40.60.81time (sec)Output Signal y(t)RC = 1 sDoes Poor Job of “Passing” Most of the Significant Frequencies of the Input RC Circuit Analysis w/ Square Wave InputDoes Poor Job of “Passing” Most of the Significant Frequencies of the Input Input SignalInput Signal’s FS CoefficientsOutput Signal’s FS CoefficientsCable’s Freq. Resp.Output Signal11/12• We used a simple model for the cable to make it easy to analyze– But… the method would be the same even if we had a more detailed model for the cable• The input clock signal has nice sharp transitions due to its significant high frequency components• Cables that significantly suppressed the input’s high frequency components provided a low-quality clock signal to the 2ndboard• We made assumptions about the driver circuit and the driven circuit– The driver was assumed to have zero output resistance• If that were not true, its output impedance gets added to the resistor and that would further degrade the performance (in fact the driver’s output impedance may be more than the


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