6.003: Signals and SystemsConvolutionMarch 2, 2010Mid-term Examination #1Tomorrow, Wednesday, March 3, 7:30-9:30pm, 34-101.No recitations tomorrow.Coverage: Representations of CT and DT SystemsLectures 1–7Recitations 1–8Homeworks 1–4Homework 4 will not collected or graded. Solutions are posted.Closed book: 1 page of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Multiple Representations of CT and DT SystemsVerbal descriptions: preserve the rationale.Difference/differential equations: mathematically compact.y[n] = x[n] + z0y[n − 1] ˙y(t) = x(t) + s0y(t)Block diagrams: illustrate signal flow paths.+Rz0X Y+As0X YOperator representations: analyze systems as polynomials.YX=11 − z0RYX=A1 − s0ATransforms: representing diff. equations with algebraic equations.H(z) =zz − z0H(s) =1s − s0ConvolutionRepresenting a system by a single signal.Responses to arbitrary signalsAlthough we have focused on responses to simple signals (δ[n], δ(t))we are generally interested in responses to more complicated signals.How do we compute responses to a more complicated input signals?No problem for difference equations / block diagrams.→ use step-by-step analysis.Check YourselfExample: Find y[3]+ +R RX Ywhen the input isx[n]n1. 1 2. 2 3. 3 4. 4 5. 50. none of the aboveResponses to arbitrary signalsExample.+ +R R00 00x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R10 01x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R11 02x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R11 13x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R01 12x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R00 11x[n]ny[n]nResponses to arbitrary signalsExample.+ +R R00 00x[n]ny[n]nCheck YourselfWhat is y[3]? 2+ +R R00 00x[n]ny[n]nAlternative: SuperpositionBreak input into additive parts and sum the responses to the parts.nx[n]y[n]n=++++=n−1 0 1 2 345nnnn−1 0 1 2 345nSuperpositionBreak input into additive parts and sum the responses to the parts.nx[n]y[n]n=++++=n−1 0 1 2 345nnnn−1 0 1 2 345nSuperposition works if the system is linear.LinearityA system is linear if its response to a weighted sum of inputs is equalto the weighted sum of its responses to each of the inputs.Givensystemx1[n] y1[n]andsystemx2[n] y2[n]the system is linear ifsystemαx1[n] + βx2[n] αy1[n] + βy2[n]is true for all α and β.SuperpositionBreak input into additive parts and sum the responses to the parts.nx[n]y[n]n=++++=n−1 0 1 2 345nnnn−1 0 1 2 345nSuperposition works if the system is linear.SuperpositionBreak input into additive parts and sum the responses to the parts.nx[n]y[n]n=++++=n−1 0 1 2 345nnnn−1 0 1 2 345nReponses to parts are easy to compute if system is time-invariant.Time-InvarianceA system is time-invariant if delaying the input to the system simplydelays the output by the same amount of time.Givensystemx[n] y[n]the system is time invariant ifsystemx[n − n0] y[n − n0]is true for all n0.SuperpositionBreak input into additive parts and sum the responses to the parts.nx[n]y[n]n=++++=n−1 0 1 2 345nnnn−1 0 1 2 345nSuperposition is easy if the system is linear and time-invariant.Structure of SuperpositionIf a system is linear and time-invariant (LTI) then its output is thesum of weighted and shifted unit-sample responses.systemδ[n] h[n]systemδ[n − k] h[n − k]systemx[k]δ[n − k] x[k]h[n − k]systemx[n] =∞Xk=−∞x[k]δ[n − k] y[n] =∞Xk=−∞x[k]h[n − k]ConvolutionResponse of an LTI system to an arbitrary input.LTIx[n] y[n]y[n] =∞Xk=−∞x[k]h[n − k] ≡ (x ∗ h)[n]This operation is called convolution.NotationConvolution is represented with an asterisk.∞Xk=−∞x[k]h[n − k] ≡ (x ∗ h)[n]It is customary (but confusing) to abbreviate this notation:(x ∗ h)[n] = x[n] ∗ h[n]NotationDo not be fooled by the confusing notation.Confusing (but conventional) notation:∞Xk=−∞x[k]h[n − k] = x[n] ∗ h[n]x[n] ∗ h[n] looks like an operation of samples; but it is not!x[1] ∗ h[1] 6= (x ∗ h)[1]Convolution operates on signals not samples.Unambiguous notation:∞Xk=−∞x[k]h[n − k] ≡ (x ∗ h)[n]The symbols x and h represent DT signals.Convolving x with h generates a new DT signal x ∗ h.Structure of Convolutiony[n] =∞Xk=−∞x[k]h[n − k]−2−1 0 1 2 345nx[n] h[n]∗−2−1 0 1 2 345nStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]−2−1 0 1 2 345nx[n] h[n]∗−2−1 0 1 2 345nStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]−2−1 0 1 2 345kx[k] h[k]∗−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]−2−1 0 1 2 345kx[k] h[k]h[−k]∗flip−2−1 0 1 2 345k−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]−2−1 0 1 2 345kx[k] h[k]h[0 − k]∗shift−2−1 0 1 2 345k−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]−2−1 0 1 2 345kx[k] h[k]h[0 − k]h[0 − k]∗multiply−2−1 0 1 2 345k−2−1 0 1 2 345k−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]kx[k] h[k]h[0 − k]h[0 − k]x[k]h[0 − k]∗multiply−2−1 0 1 2 345kk−2−1 0 1 2 345k−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]kx[k] h[k]h[0 − k]h[0 − k]x[k]h[0 − k]∗sum∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345kStructure of Convolutiony[0] =∞Xk=−∞x[k]h[0 − k]kx[k] h[k]h[0 − k]h[0 − k]x[k]h[0 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 1Structure of Convolutiony[1] =∞Xk=−∞x[k]h[1 − k]kx[k] h[k]h[1 − k]h[1 − k]x[k]h[1 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 2Structure of Convolutiony[2] =∞Xk=−∞x[k]h[2 − k]kx[k] h[k]h[2 − k]h[2 − k]x[k]h[2 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 3Structure of Convolutiony[3] =∞Xk=−∞x[k]h[3 − k]kx[k] h[k]h[3 − k]h[3 − k]x[k]h[3 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 2Structure of Convolutiony[4] =∞Xk=−∞x[k]h[4 − k]kx[k] h[k]h[4 − k]h[4 − k]x[k]h[4 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 1Structure of Convolutiony[5] =∞Xk=−∞x[k]h[5 − k]kx[k] h[k]h[5 − k]h[5 − k]x[k]h[5 − k]∗∞Xk=−∞kk−2−1 0 1 2 345k−2−1 0 1 2 345k= 0Check Yourself∗Which plot shows the result of the convolution above?1. 2.3. 4.5. none of the aboveCheck Yourself∗Express mathematically:23nu[n]∗23nu[n]=∞Xk=−∞ 23ku[k]!× 23n−ku[n
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