EECE 301 Signals & Systems Prof. Mark Fowler1/20EECE 301 Signals & SystemsProf. Mark FowlerNote Set #26• D-T Systems: DTFT Analysis of DT Systems• Reading Assignment: Sections 5.5 & 5.6 of Kamen and Heck2/20Ch. 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 Signal ModelPowerful 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).3/205.5: System analysis via DTFT][nh][nx∑∞−∞=−=∗=iinxihnxnhny][][][][][Recall that in Ch. 5 we saw how to use frequency domain methods to analyze the input-output relationship for the C-T case. We now do a similar thing for D-T Define the “Frequency Response” of the D-T systemWe now return to Ch. 5 for its DT coverage!Back in Ch. 2, we saw that a D-T system in “zero state” has an output-input relation of:∑∞−∞=Ω−=ΩnnjenhH ][)(DTFT of h[n]Perfectly parallel to the same idea for CT systems!!!4/20From Table of DTFT properties:)()(][][ΩΩ↔∗HXnhnxSo we have:)(][ΩHnh)(][ΩXnx][][][ nxnhny∗=)()()( ΩΩ=ΩHXY)()()()()()(Ω∠+Ω∠=Ω∠ΩΩ=ΩHXYHXYSo…So…in general we see that the system frequency response re-shapes the input DTFT’s magnitude and phase. ⇒ System can:-emphasize some frequencies-de-emphasize other frequenciesPerfectly parallel to the same ideas for CT systems!!!The above shows how to use DTFT to do general DT system analyses …virtually all of your insight from the CT case carries over!5/20Now lets look at the special case: Response to Sinusoidal Input,...3,2,1,0,1,2,3)cos(][0−−−=+Ω= nnAnxθFrom DTFT Table:[]⎪⎩⎪⎨⎧<Ω<−Ω−Ω+Ω+Ω=Ω−elsewhere periodic)()()(00ππδδπθθjjeeAXΩ)(ΩX0Ω0Ω−ππ2π2−π−We only need to focus our attention here∫−ΩΩΩΩ=ΩΩ=ΩπππdeXHnyXHYnj)()(21][)()()(6/20So what does Y(Ω) look like?[]⎪⎩⎪⎨⎧<Ω<−Ω−ΩΩ+Ω+ΩΩ−=Ω−elsewhere periodic)()()()()(0000ππδδπθθjjeHeHAY)(00)(0000)()()()( NowΩ∠−Ω∠Ω=Ω−Ω=ΩHjHjeHHeHHUsed symmetry properties[]⎪⎩⎪⎨⎧<Ω<−Ω−Ω+Ω+ΩΩ=ΩΩ∠++Ω∠+−elsewhere periodic)()()()(0))((0))((000ππδδπθθHjHjeeHAYFrom DTFT Table we see this is the DTFT of a cosine signal with:)(Phase)(Amplitude00Ω∠+=Ω=HHAθ7/20So…))(cos()(][000Ω∠++ΩΩ= HnAHnyθ)(ΩH)cos(0θ+Ω nA))(cos()(000Ω∠++ΩΩ HnAHθSystem changes amplitude and phase of sinusoidal inputPerfectly parallel to the same ideas for CT systems!!!8/20ImReExample 5.8 (error in book)Suppose you have a system described byΩ−+=ΩjeH 1)(⎟⎠⎞⎜⎝⎛+=⎟⎠⎞⎜⎝⎛+= nnnnx2sin2)0cos(22sin22][ππAnd you put the following signal into itCosine with Ω = 0Find the output.So we need to know the system’s frequency response at only 2 frequencies.212ππjeH−+=⎟⎠⎞⎜⎝⎛ImRej−121)0(0=+=− jeH421πjej−=−=9/20⎟⎠⎞⎜⎝⎛−=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛∠+⎟⎠⎞⎜⎝⎛==⋅=⋅=42sin2222sin22][422)0cos(2)0(][21πππππnHnHnynHnySince the system is linear we can consider each of the input terms separately….And then add them to get the complete response…⎟⎠⎞⎜⎝⎛−+=42sin224][ππnny10/20Note: In the above example we used )(ΩH)sin(0θ+Ω nA))(sin()(000Ω∠++ΩΩ HnAHθQ: Why does that follow?A: It is a special case of the cosine result that is easy to see:- convert sin(Ω0n + θ) into a cosine form- apply the cosine result- convert cosine output back into sine form…which is the “sine” version of our result above for a cosine input11/20Analysis of Ideal D-T lowpass Filter (LPF)π2π3π4π4−π3−π2−filter lowpass Ideal)( =ΩHπ1π−-BBJust as in the CT case… we can specify filters. We looked at the ideal lowpass filter for the CT case… here we look at it for the DT case.Cut-off frequency = B rad/sampleAs always with DT… we only need to look here12/20D-T Ideal LPFDACADC @Fs = 1/TF~2Fs @ sample~2~let == FBπThis slide shows how a DT filter might be employed… but ideal filters can’t be built in practice. We’ll see later a few practical DT filters.x(t) x[n] y[n] y(t)π< off-cut B)(ΩXππ2π2−π−Ω)(ωXωB~B~−)(ΩYππ2π−π2−BB−Ω)(ωYωTB /TB /−Whole System (ADC – D-T filter – DAC) acts like an equivalent C-T system13/20We know the frequency response of the ideal LPF… so find its impulse response: From DTFT Table :⎟⎠⎞⎜⎝⎛= nBBnhππsinc][Why can’t an ideal LPF exist in practice??][nhn……Key Point: h[n] is non-zero here⇒ starts beforethe impulse that “makes it” is even “applied”!⇒ Can’t build an Ideal LPF(Same thing is true in C-T)14/20Causal Lowpass Filter (But not Ideal)In practice, the best we can do is try to approximate the ideal LPF If you go on to study DSP you’ll learn how to design filters that do a good job at this approximationHere we’ll look at two “seat of the pants” approaches to get a good LPFApproach #1Truncate & Shift Ideal h[n] ⎥⎦⎤⎢⎣⎡−=2][Nnhnhtruncapprox][nhapproxn……ShiftTruncate][nhtruncn……⎪⎩⎪⎨⎧≤≤−=)(,022],[][evenNotherwiseNnNnhnhidealtruncN+1 non-zero samples15/20-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 100.20.40.60.811.2Ω/π|H(Ω)|-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 100.20.40.60.811.2Ω/π|H(Ω)|-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 100.20.40.60.811.2Ω/π|H(Ω)|N = 20 N = 60 N = 120 Some general insight: Longer lengths for the truncated impulse responseGives better approximation to the ideal filter response!!Let’s see how well these work…Frequency Response of a filter truncated to 21