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13 Pages

### 12_Sequences_and_Sums

Course: CS 1050, Fall 2008
School: Georgia Tech
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Word Count: 471

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&amp; Sequences Summations CS 1050 Rosen 1.7 Sequence A sequence is a discrete structure used to represent an ordered list. A sequence is a function from a subset of the set of integers (usually either the set {0,1,2,. . .} or {1,2, 3,. . .}to a set S. We use the notation an to denote the image of the integer n. We call an a term of the sequence. Notation to represent sequence is {an} Examples {1, 1/2,...

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& Sequences Summations CS 1050 Rosen 1.7 Sequence A sequence is a discrete structure used to represent an ordered list. A sequence is a function from a subset of the set of integers (usually either the set {0,1,2,. . .} or {1,2, 3,. . .}to a set S. We use the notation an to denote the image of the integer n. We call an a term of the sequence. Notation to represent sequence is {an} Examples {1, 1/2, 1/3, 1/4, . . .} or the sequence {an} where an = 1/n, nZ+ . {1,2,4,8,16, . . .} = {an} where an = 2n, nN. {12,22,32,42,. . .} = {an} where an = n2, nZ+ Common Sequences Arithmetic n2 n3 n4 2n 3n n! a, a+d, a+2d, a+3d, a+4d, ... 1, 4, 9, 16, 25, . . . 1, 8, 27, 64, 125, . . . 1, 16, 81, 256, 625, . . . 2, 4, 8, 16, 32, . . . 3, 9, 27, 81, 243, . . . 1, 2, 6, 24, 120, . . . Summations Notation for describing the sum of the terms am+1, . . ., an from the sequence, {an} n am, am+am+1+ . . . + an = aj j=m j is the index of summation (dummy variable) The index of summation runs through all integers from its lower limit, m, to its upper limit, n. Examples j = ( j + 1) = 1 + 2 + 3 + 4 + 5 = 15 j =1 j =0 5 4 5 1 1 1 1 j = + 2 +1 3 +1 4+ 5 j= 1 5 1 + j =1 +1 2 +1 3 +1 4+ 5 1 1 j =2 Summations follow all the rules of addition! c j = cj = c(1+2+...+n) = c + 2c +...+ nc r ar = ar j j =0 j =0 n n j =1 n j =1 n j +1 = ar = k k =1 n +1 n +1 ar n +1 + ar = ar k k =1 n - a + ar k =0 n k Telescoping Sums a - j a j -1 = (a1 - a0 ) + (a2 - a1 ) + j =1 n (a3 - a2 ) + ... + (an - an-1 ) = an - a0 Example k =1 [k - (k - 1) ] = 2 2 2 2 2 2 2 2 2 2 2 4 (1 - 0 ) + (2 - 1 ) + (3 - 2 ) + (4 - 3 ) 4 = 16 - 0 = 16 Closed Form Solutions A simple formula that can be used to calculate a sum without doing all the additions. Example: n(n + 1) k = k =1 2 n Proof: First we note that k2 - (k-1)2 = k2 - (k2-2k+1) = 2k-1. Since k2-(k-1)2 = 2k-1, then we can sum each side from k=1 to k=n n n k =1 [k - ( k - 1) ] = ( 2k - 1) 2 2 k =1 Proof (cont.) k =1 n [k - ( k - 1) ] = ( 2k - 1) n 2 2 n k =1 n 2 2 n k =1 k =1 k =1 [k - ( k - 1) ] = 2k + ( - 1) 2 2 n k =1 n - 0 = 2 (k ) + -n n + n = 2 (k ) 2 k =1 n n2 + n k = k =1 2 n Closed Form Solutions to Sums j = 0 + 1 + ... + n = n(n + 1) / 2 j =0 2 2 2 2 j = 0 + 1 + ... + n = n(n + 1)(2n + 1) / 6 j =0 n n n ( n + 1) k =...

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Georgia Tech - CS - 1050
Doing SumsWhat isk =02n-kar n+1 - a ar k = ,r 1 k =0 r -1nk =0(n1 k 2)1- ( ) = 1- 1 21 n +1 2Let a = 1 and r = 1/2=1 - 2-n -1 = -1 2 - n -1 n +1 1 - 2 2 n +1 -1 2 2 2n +1 - 1 = n 2What if k starts at 1?k =1
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Mathematical InductionRosen 3.2Basics The Well-Ordering Property - Every nonempty set of nonnegative integers has a least element. Many theorems state that P(n) is true for all positive integers. For example, P(n) could be the statement that th
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Induction PracticeCS1050(2 j + 1) = 3n 2 whenever n Prove that j =n2 n 1is a positive integer. Proof: Basic Case: Let n = 1, then2 (1) 1 j =1 (2 j + 1) = ( 2 j + 1) = 3 = 3(1)j 112=3(2 j + 1) = 3n 2 whenever n Prove that j =n2
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Problems to Solve Involving Inductionar n +1 - a k ar = r - 1 , r 1? k =0nProof by Induction Basic Step: Does it work for n=0? ark =00k= ar = a0ar 0+1 - a ar - a a (r - 1) = = =a r -1 r -1 r -1Inductive StepAssume that for r 1
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PracticeCS1050 Spring 2002Flipping a coin 4 times. How many different combinations of heads and tails are there? C1:HHHHHHHHTTTTTTTT C2:HHHHTTTTHHHHTTTT C3:HHTTHHTTHHTTHHTT C4:HTHTHTHTHTHTHTHT 16: How do we compute without counting? Product rule:
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1/14/2003Amplifier Saturation.doc1/7Amplifier SaturationNote that the ideal amplifier transfer function:vo (t ) = A vi (t ) vois an equation of a line (with slope = Avo and y-intercept = 0).voAvoviThis ideal transfer function implie
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10/17/2003BJT Biasing using a Current Source1/4BJT Biasing using a Current SourceAnother way to bias a BJT small signal amplifier is to use one voltage source and one current source. This biasing scheme has a number of important advantages: 1.
E. Kentucky - ITTC - 412
11/10/2003Current Steering Circuits1/3Current Steering CircuitsA current mirror may consist of many MOSFET current sources!VDD VDD VDD VDDRIrefRL1IL1 =IrefRL2IL2=Iref RL3IL3=IrefQrefQ1Q2Q3This circuit is particularly u
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2/3/2003Example An Inverting Network.doc1/2Example: An Inverting NetworkNow let's determine the complex transfer function of this circuit:R2CviR1v1-ideal v2+voNote this circuit uses the inverting configuration, so that:T (
E. Kentucky - ITTC - 412
Example: D.C. Analysis of a BJT CircuitConsider again this circuit from lecture: 10.7 V1.0 K Q: What is iB, iC, iE andalso VCE, VCB, VBE ?5.7 V 10 K = 99A: I don't know ! But, we can find out, IF we complete each of the five steps required
E. Kentucky - ITTC - 412
9/24/2003Example The Input Bias Current1/4Example: The Input Bias CurrentQ: How do input bias currents IB 1 and IB 2 affect amplifieroperation?A: Consider both inverting and non-inverting configurations.Inverting ConfigurationR2 i2vi
E. Kentucky - ITTC - 412
9/22/2003Full Power Bandwidth1/5Full-Power BandwidthConsider now the case where the input to an op-amp circuit is sinusoidal, with frequency . The output will thus likewise be sinusoidal, e.g.: vo (t ) = Vo sint where Vo is the magnitude of t
E. Kentucky - ITTC - 412
12/4/2003Large Signal Operation of MOSFET Diff Pair1/6Large-Signal Operation of the MOSFET Differential PairConsider now the MOSFET differential pair:Figure 7.1 The basic MOS differential-pairNote:v S 1 = vS 2 = vSandiD 1 + iD 2 = II
E. Kentucky - ITTC - 412
9/29/2003Real Op Amp Input and Output Resistance1/4Real Op-Amp Input and Output ResistancesThe input resistances of real op-amps are very large, but of course not infinite! Typical values of input resistances range from several hundred Kilo Oh
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E. Kentucky - ITTC - 412
1/27/2003Ri and Ro of the Inverting Amplifier.doc1/6Inverting AmplifierWe can use the concept of the virtual short to easily determine the input resistance of the inverting amplifier. Recall that the input resistance of an amplifier is:Ri an
E. Kentucky - ITTC - 412
2/5/2003Superposition and OpAmp Circuits.doc1/4Superposition and Op-Amp CircuitsConsider this op-amp circuit:v1R1R2-v2R3ideal+voR4The easiest way to analyze this circuit is to apply superposition! Recall that op-amp circui
E. Kentucky - ITTC - 412
9/30/2003The Small Signal Equaton Matrix1/2The Small-Signal Equation MatrixWe can summarize our small-signal equations with the smallsignal equation matrix. Note this matrix relates the small-signal BJT parameters vbe , ib , ic , and ie . Colu
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9/4/2003The Weighted Summer1/2The Weighted SummerConsider an inverting amplifier with multiple inputs!v1 v2 v3R1 R2 R3i1 i2 i3i-Rfideal+voFrom KCL, we can conclude that the currents are related as: i= and because of virtual g
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1/31/2003SP 2.3-2.doc1/1Special Problem 2.3-2 An amplifier with an open-circuit voltage gain of Avo = 10 and an output resistance of Ro = 1K is connected in series to an inverting amplifier, as shown below. Find the value of R1 so that VO/VI =
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E. Kentucky - ITTC - 412
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2/7/2003SP 2.6-1.doc1/1Special Problem 2.6-1 For the circuit below, find vO in terms of inputs v1 and v2. Find both the differential and common-mode gains of this amplifier. Determine CMRR. 9K 1K_ +vo4K2K2Kv1v2
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
2/21/2003SP 4.6-14.doc1/1Special Problem 4.6-14 Assume the BJT in the circuit below is in active mode. 1) Determine base current IB and collector voltage VC 2) Determine if the active mode assumption is correct.15.0 V 1K = 99 0.7 K+ _ 1K
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
2/28/2003SP 4.7-1.doc1/1Special Problem 4.7-1A strange, two-terminal device has been determined to have the following relationship between the current through it (I ) and the voltage across it (V ):V = 3I 2 + 2I 3 4.2where V is in volts a
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
4/4/2003SP 5.5-5.doc1/1Special Problem 5.5-5 For the transistor below: K = 1.0 mA/V2 Vt = 1.0 V = 0.0 V-1 Find the small-signal voltage gain ( Av = vo/vi ) of the amplifier below.10.0 V7Kvi(t) 0.05 K+ vo(t)100 K 1K
E. Kentucky - ITTC - 412
4/4/2003SP 5.5-7.doc1/1Special Problem 5.5-7 For the circuit below, VG is the DC bias at the gate, vi is the small-signal input and vo the small-signal output. The transistor is known to be in the saturation region. The capacitor in the circuit
E. Kentucky - ITTC - 412
4/4/2003SP 5.5-9.doc1/1Special Problem 5.5-8 The capacitors in this circuit are very large. Determine the small-signal voltage gainA = vo vi of this amplifier. vDetermine the small-signal input resistance of this amplifier.4.5 V 10 V10
E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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E. Kentucky - ITTC - 412
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