EECE 301 Signals & Systems Prof. Mark Fowler1/23EECE 301 Signals & SystemsProf. Mark FowlerNote Set #9• Computing D-T Convolution• Reading Assignment: Section 2.2 of Kamen and Heck2/23Ch. 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).3/232.2 “Computing” D-T convolution-We know about the impulse response h[n]-We found out that h[n] interacts with x[n] through convolution to give the zero-state response:∑∞−∞=−=iinhixny ][][][How do we “work” this? This is needed for understanding how:(1) To analyzesystems(2) To implementsystems Don’t forget…The design process includes analysisTwo cases, depending on form of x[n]:1. x[n] is known analytically2. x[n] is known numerically or graphicallyAnalytical Convolution (used for “by-hand” analysis):Pretty straightforward conceptually:- put functions into convolution summation- exploit math properties to evaluate/simplify4/23Example:][][ nuanxn=][][ nubnhn=Recall this form from 1st -order difference equation example][nubn][nuan?][=ny∑∞−∞=−=iinhixny ][][][∑∞−∞=−−=iiniinubiua ][][)( a function of n… i gets “summed away”⎩⎨⎧<≥=0,00,1][iiiu∑∞=−−=0)(][iiniinubaNow use:⎩⎨⎧>≤=−niniinu,0,1][∑∑==−⎟⎠⎞⎜⎝⎛==niinniinibabba00)(Now use:You should beable to go here directly5/23⎪⎪⎪⎩⎪⎪⎪⎨⎧≠⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−⎟⎠⎞⎜⎝⎛−=+=+babababannyn,11,1][1“Geometric Sum”∑=⎟⎠⎞⎜⎝⎛=niinbabny0][If a = b you are adding (n + 1) 1’s and that gives n + 1So now we simplify this summation…If a ≠ b, then a standard math relationshipgives:1,1110≠−−=∑−=rrrrNNiiKnow This!!!6/23Aside: Commutativity Property of ConvolutionA simple change of variables shows that][*][][*][][][][][][nxnhinhnxiinxihinhixny∑∑∞−∞=∞−∞=−=−=So…we can use which ever of these is easier…7/23Step 1: Write both as functions of i: x[i] & h[i]Step 2: Flip h[i] to get h[-i] (The book calls this “fold”)Step 3: For each output index n value of interest, shift by n to get h[n - i](Note: positive n gives right shift!!!!)Step 4: Form product x[i]h[n –i] and sum its elements to get the number y[n]Repeat for each nGraphical Convolution StepsCan do convolution this way when signals are know numerically or by equation- Convolution involves the sum of a product of two signals: x[i]h[n – i]- At each output index n, the product changes“Commutativity” says we can flip either x[i] or h[i] and get the same answer8/23Example of Graphical Convolution. . .. . .x[n]n. . .. . .h[n]n-2 -1 1 2 3 4 5 -2 -1 1 2 3 4 5 6 7 SolutionFor this problem I choose to flip x[n]My personal preference is to flip the shorter signal although I sometimes don’t follow that “rule”… only through lots of practice can you learn how to best choose which one to flip.23Find y[n]=x[n]*h[n] for all integer values of n19/23Step 1: Write both as functions of i: x[i] & h[i]. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3. . .. . .x[-i]i-2 -1 1 2 3 4 5 6 7 2“Commutativity” says we can flip either x[i] or h[i] and get the same answer…Here I flipped x[i]. . .. . .x[i]i-2 -1 1 2 3 4 5 21Step 2: Flip x[i] to get x[-i]. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3110/23. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3x[-i] = x[0 - i]For n = 0 case there is no shift!x[0 - i] = x[-i] . . .. . .h[i]x[0 - i]i-3 -2 -1 1 2 3 4 5 6 7 32×Sum over i ⇒y[0] = 6. . .i-2 -1 1 2 3 4 5 6 7 2. . .1We want a solution for n = … -2, -1, 0, 1, 2, … so must do Steps 3&4 for all n.But… let’s first do: Steps 3&4 for n = 0 and then proceed from there.Step 3: For n = 0, shift by n to get x[n-i]Step 4: For n = 0, Form the product x[i]h[n –i] and sum its elements to give y[n]11/23Steps 3&4 for all n < 0. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3x[-i] = x[-1 - i]Negative n gives a left-shift. . .. . .h[i]x[-1 - i] = 0i-3 -2 -1 1 2 3 4 5 6 7 Sum over i ⇒. . .i-2 -1 1 2 3 4 5 6 7 2. . .1Step 3: For n < 0, shift by n to get x[n-i]Step 4: For n < 0, Form the product x[i]h[n –i] and sum its elements to give y[n]Shown here for n = -100][<∀=nny12/23So… what we know so far is that:⎩⎨⎧=<∀=0,60,0][nnny. . .y[n]n-2 -1 1 2 3 4 5 6 7 6y[n] = ???So now we have to do Steps 3&4 for n > 0…13/23. . .. . .i-2 -1 1 2 3 4 5 6 7 x[1 - i] h[i]2×3shifted to the right by oneSteps 3&4 for n = 1. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3x[-i] = x[1 - i]Positive n gives a Right-shifti-2 -1 1 2 3 4 5 6 7 2. . .. . .1Step 3: For n = 1, shift by n to get x[n-i]Step 4: For n = 1, Form the product x[i]h[n –i] and sum its elements to give y[n]y[1] = 6 + 6 = 12Sum over i ⇒14/23. . .i-2 -1 1 2 3 4 5 6 7 x[1 - i] h[i]2×3shifted to the right by twoSteps 3&4 for n = 2. . .. . .h[i]i-2 -1 1 2 3 4 5 6 7 3x[-i] = x[2 - i]Positive n gives a Right-shifti-2 -1 1 2 3 4 5 6 7 2. . .. . .1Step 3: For n = 2, shift by n to get x[n-i]Step 4: For n = 2, Form the product x[i]h[n –i] and sum its elements to give y[n]y[2] = 3 + 6 + 6 = 15Sum over i ⇒. . .1×315/23i-2 -1 1 2 3 4 5 6 7 x[1 - i] h[i]2×3shifted