6.003: Signals and SystemsFeedback, Poles, and Fundamental ModesFebruary 9, 2010Last Time: Multiple Representations of DT SystemsVerbal descriptions: preserve the rationale.“To reduce the number of bits needed to store a sequence oflarge numbers that are nearly equal, record the first number,and then record successive differences.”Difference equations: mathematically compact.y[n] = x[n] − x[n − 1]Block diagrams: illustrate signal flow paths.−1Delay+x[n] y[n]Operator representations: analyze systems as polynomials.Y = (1 − R) XLast Time: Feedback, Cyclic Signal Paths, and ModesSystems with signals that depend on previous values of the samesignal are said to havefeedback.Example: The accumulator system has feedback.Delay+X YBy contrast, the difference machine does not have feedback.−1Delay+X YLast Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.Last Time: Feedback, Cyclic Signal Paths, and ModesThe effect of feedback can be visualized by tracing each cyclethrough the cyclic signal paths.Delay+p0X Y−1 0 1 2 34nx[n] = δ[n]−1 0 1 2 34ny[n]Each cycle creates another sample in the output.The response will persist even though the input is transient.Geometric Growth: PolesThese unit-sample responses can be characterized by a single number— thepole — which is the base of the geometric sequence.Delay+p0X Yy[n] =pn0, if n >= 0;0, otherwise.−1 0 1 2 34ny[n]−1 0 1 2 34ny[n]−1 0 1 2 34ny[n]p0= 0.5 p0= 1 p0= 1.2Check YourselfHow many of the following unit-sample responses can berepresented by a single pole?n nn nnCheck YourselfHow many of the following unit-sample responses can berepresented by a single pole? 3n nn nnGeometric GrowthThe value of p0determines the rate of growth.y[n]y[n]y[n]y[n]−1 0 1zp0< −1: magnitude diverges, alternating sign−1 < p0< 0: magnitude converges, alternating sign0 < p0< 1: magnitude converges monotonicallyp0> 1: magnitude diverges monotonicallySecond-Order SystemsThe unit-sample responses of more complicated cyclic systems aremore complicated.RR1.6−0.63+X Y−1 0 1 2 345 678ny[n]Not geometric. This response grows then decays.Factoring Second-Order SystemsFactor the operator expression to break the system into two simplersystems (divide and conquer).RR1.6−0.63+X YY = X + 1.6RY − 0.63R2Y(1 − 1.6R + 0.63R2) Y = X(1 − 0.7R)(1 − 0.9R) Y = XFactoring Second-Order SystemsThe factored form corresponds to a cascade of simpler systems.(1 − 0.7R)(1 − 0.9R) Y = X+0.7R+0.9RX YY2(1 − 0.7R) Y2= X (1 − 0.9R) Y = Y2+0.9R+0.7RX YY1(1 − 0.9R) Y1= X (1 − 0.7R) Y = Y1The order doesn’t matter (if systems are initially at rest).Factoring Second-Order SystemsThe unit-sample response of the cascaded system can be found bymultiplying the polynomial representations of the subsystems.YX=1(1 − 0.7R)(1 − 0.9R)=1(1 − 0.7R)| {z }×1(1 − 0.9R)| {z }=z}| {(1 + 0.7R+ 0.72R2+ 0.73R3+ ···) ×z }| {(1 + 0.9R + 0.92R2+ 0.93R3+ ···)Multiply, then collect terms of equal order:YX= 1 + (0.7 + 0.9)R + (0.72+ 0.7 ×0.9 + 0.92)R2+ (0.73+ 0.72× 0.9 + 0.7 × 0.92+ 0.93)R3+ ···Multiplying PolynomialGraphical representation of polynomial multiplication.YX= (1 + aR + a2R2+ a3R3+ ···) × (1 + bR + b2R2+ b3R3+ ···)1aRa2R2a3R3+1bRb2R2b3R3+... ... ... ...X YCollect terms of equal order:YX= 1 + (a + b)R + (a2+ ab + b2)R2+ (a3+ a2b + ab2+ b3)R3+ ···Multiplying PolynomialsTabular representation of polynomial multiplication.(1 + aR + a2R2+ a3R3+ ···) × (1 + bR + b2R2+ b3R3+ ···)1 bR b2R2b3R3···1 1 bR b2R2b3R3···aR aR abR2ab2R3ab3R4···a2R2a2R2a2bR3a2b2R4a2b3R5···a3R3a3R3a3bR4a3b2R5a3b3R6······ ··· ··· ··· ··· ···Group same powers of R by following reverse diagonals:YX= 1 + (a + b)R + (a2+ ab + b2)R2+ (a3+ a2b + ab2+ b3)R3+ ···−1 0 1 2 345 678ny[n]Partial FractionsUse partial fractions to rewrite as a sum of simpler parts.RR1.6−0.63+X YYX=11 − 1.6R + 0.63R2=1(1 − 0.9R)(1 − 0.7R)=4.51 − 0.9R−3.51 − 0.7RSecond-Order Systems: Equivalent FormsThe sum of simpler parts suggests a parallel implementation.YX=4.51 − 0.9R−3.51 − 0.7R+0.9R4.5+−3.5R0.7+X YY1Y2If x[n] = δ[n] then y1[n] = 0.9nand y2[n] = 0.7nfor n ≥ 0.Thus, y[n] = 4.5(0.9)n− 3.5(0.7)nfor n ≥ 0.Partial FractionsGraphical representation of the sum of geometric sequences.−1 0 1 2 345 678ny1[n] = 0.9nfor n ≥ 0−1 0 1 2 345 678ny2[n] = 0.7nfor n ≥ 0−1 0 1 2 345 678ny[n] = 4.5(0.9)n− 3.5(0.7)nfor n ≥ 0Partial FractionsPartial fractions provides a remarkable equivalence.RR1.6−0.63+X Y+0.9R4.5+−3.5R0.7+X YY1Y2→ follows from thinking about system as polynomial (factoring).PolesThe key to simplifying a higher-order system is identifying its poles.Poles are the roots of the denominator of the system functionalwhen R →1z.Start with system functional:YX=11 − 1.6R+0.63R2=1(1−p0R)(1−p1R)=1(1−0.7R)| {z }p0=0.7(1−0.9R)| {z }p1=0.9Substitute R →1zand find roots of denominator:YX=11 −1.6z+0.63z2=z2z2−1.6z+0.63=z2(z−0.7)| {z }z0=0.7(z−0.9)| {z }z1=0.9The poles are at 0.7 and 0.9.Check YourselfConsider the system described byy[n] = −14y[n − 1] +18y[n − 2] + x[n − 1] −12x[n − 2]How many of the following are true?1. The unit sample response converges to zero.2. There are poles at z =12and z =14.3. There is a pole at
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