6.003: Signals and Systems Lecture 11 October 20, 200916.003: Signals and SystemsFrequency ResponseOctober 20, 2009Mid-term Examination #2Wednesday, October 28, 7:30-9:30pm, Walker Memorial.No recitations on the day of the exam.Coverage: cumulative with more emphasis on recent materiallectures 1–12homeworks 1–7Homework 7 will include practice problems for mid-term 2.However, it will not collected or graded. Solutions will be posted.Closed book: 2 page of notes (812× 11 inches; front and back).Designed as 1-hour exam; two hours to complete.Review sessions during open office hours.Conflict? Contact [email protected] by Friday, October 23, 5pm.ReviewLast time, we saw how a linear, time-invariant (LTI) system can becharacterized by its unit-sample/impulse response.DT: y[n] = (x ∗ h)[n] =∞Xk=−∞x[k]h[n −k]CT: y(t) = (x ∗ h)(t) =Z∞−∞x(τ)h(t − τ)dτCharacterizing a system by its unit-sample/impulse response is es-pecially insightful for some systems.MicroscopeBlurring can be represented by convolving the image with the optical“point-spread-function” (3D impulse response).targetimage∗=Blurring is inversely related to the diameter of the lens.Hubble Space TelescopeHubble images before and after upgrading the optics.before afterhttp://hubblesite.orgFrequency ResponseToday we will investigate a different way to characterize a system:thefrequency response.Many systems are naturally described by their responses to sinusoids.Example: audio systems6.003: Signals and Systems Lecture 11 October 20, 20092Check YourselfHow were frequencies modified in following music clips?HF: high frequencies ↑: increasedLF: low frequencies ↓: decreasedclip 1 clip 21. HF↑ HF↓2. LF↑ LF↓3. HF↑ LF↓4. LF↑ HF↓5. none of the aboveFrequency Response PreviewIf the input to a linear, time-invariant system is an eternal sinusoid,then the output is also an eternal sinusoid:• same frequency• possibly different amplitude, and• possibly different phase angle.x(t) = cos(ωt)ty(t) = M cos(ωt + φ)tLTIsystemThe frequency response is a plot of the magnitude M and angle φas a function of frequency ω.DemonstrationMeasure the frequency response of a mass, spring, dashpot system.x(t)y(t)Frequency ResponseCalculate the frequency response.Methods• solve differential equation→ find particular solution for x(t) = cos ω0t• find impulse response of system→ convolve with x(t) = cos ω0tNew method• use eigenfunctions and eigenvaluesEigenfunctionsIf the output signal is a scalar multiple of the input signal, we referto the signal as an eigenfunction and the scale multiplier as theeigenvalue.systemx(t) λx(t)eigenvalueeigenfunctionCheck Yourself: EigenfunctionsConsider the system described by˙y(t) + 2y(t) = x(t).Determine if each of the following functions is an eigen-function of this system. If it is, find its eigenvalue.1. e−tfor all time2. etfor all time3. ejtfor all time4. cos(t) for all time5. u(t) for all time6.003: Signals and Systems Lecture 11 October 20, 20093Complex ExponentialsComplex exponentials are eigenfunctions of LTI systems.If x(t) = estand h(t) is the impulse response theny(t) = (h ∗ x)(t) =Z∞−∞h(τ)es(t−τ )dτ = estZ∞−∞h(τ)e−sτdτ = H(s)estestH(s) estLTIh(t)Furthermore, the eigenvalue is H(s) !Rational System FunctionsIf a system is represented by a linear differential equation with con-stant coefficients, then its system function is a ratio of polynomialsin s.Example:¨y(t) + 3 ˙y(t) + 4y(t) = 2¨x(t) + 7 ˙x(t) + 8x(t)ThenH(s) =2s2+ 7s + 8s2+ 3s + 4≡N(s)D(s)Vector DiagramsThe value of H(s) at a point s = s0can be determined graphicallyusing vectorial analysis.Factor the numerator and denominator of the system function tomake poles and zeros explicit.H(s0) = K(s0− z0)(s0− z1)(s0− z2) ···(s0− p0)(s0− p1)(s0− p2) ···z0z0s0− z0s0s-planes0Each factor in the numerator/denominator corresponds to a vectorfrom a zero/pole (here z0) to s0, the point of interest in the s-plane.Vector DiagramsExample: Find the response of the system described byH(s) =1s + 2to the input x(t) = e2jt(for all time).−2s0− p0s-planes0= 2jThe denominator of H(s)|s=2jis 2j + 2, a vector with length 2√2 andangle π/4. Therefore, the response of the system isy(t) = H(2j)e2jt=12√2e−jπ4e2jt.Vector DiagramsThe value of H(s) at a point s = s0can be determined by combiningthe contributions of the vectors associated with each of the polesand zeros.H(s0) = K(s0− z0)(s0− z1)(s0− z2) ···(s0− p0)(s0− p1)(s0− p2) ···The magnitude is determined by the product of the magnitudes.|H(s0)| = |K||(s0− z0)||(s0− z1)||(s0− z2)|···|(s0− p0)||(s0− p1)||(s0− p2)|···The angle is determined by the sum of the angles.∠H(s0) = ∠K + ∠(s0−z0) + ∠(s0−z1) + ···−∠(s0−p0) −∠(s0−p1) −···Frequency ResponseResponse to eternal sinusoids.Let x(t) = cos ω0t (for all time). Thenx(t) =12ejω0t+ e−jω0tand the response to a sum is the sum of the responses.y(t) =12H(jω0) ejω0t+ H(−jω0) e−jω0t6.003: Signals and Systems Lecture 11 October 20, 20094Conjugate SymmetryThe complex conjugate of H(jω) is H(−jω).The system function is the Laplace transform of the impulse re-sponse:H(s) =Z∞−∞h(t)e−stdtwhere h(t) is a real-valued function of t for physical systems.H(jω) =Z∞−∞h(t)e−jωtdtH(−jω) =Z∞−∞h(t)ejωtdt ≡H(jω)∗Frequency ResponseResponse to eternal sinusoids.Let x(t) = cos ω0t (for all time), which can be written asx(t) =12ejω0t+ e−jω0tThe response to a sum is the sum of the responses,y(t) =12H(jω0)ejω0t+ H(−jω0)e−jω0t= RenH(jω0)ejω0to= Ren|H(jω0)|ej∠H(jω0)ejω0to= |H(jω0)|Renejω0t+j∠H(jω0)oy(t) = |H(jω0)|cos (ω0t + ∠ (H(jω0))) .Frequency ResponseThe magnitude and phase of the response of a system to an eternalcosine signal is the magnitude and phase of the system functionevaluated at s = jω.H(s)cos(ωt) |H(jω)|cosωt + ∠H(jω)Vector Diagramss-planeσω5−55−5H(s) = s − z1−5 0 55|H(jω)|−5 5π/2−π/2∠H(jω)Vector Diagramss-planeσω5−55−5H(s) =9s − p1−5 0 55|H(jω)|−5 5π/2−π/2∠H(jω)Vector Diagramss-planeσω5−55−5H(s) = 3s − z1s − p1−5 0 55|H(jω)|−5 5π/2−π/2∠H(jω)6.003: Signals and Systems Lecture 11 October 20, 20095Example: Mass, Spring, and Dashpotx(t)y(t)F = M a = M ¨y(t) = K(x(t) − y(t)) − B ˙y(t)M¨y(t) + B ˙y(t) + Ky(t) = Kx(t)(s2M + sB +
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