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Item Response Theory
Course: PSYC 430, Fall 2008School: Idaho
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Idaho - PSYC - 430
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Idaho - PSYC - 430
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Idaho - PSYC - 430
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Georgia Tech - CS - 1050
Rosen 1.3Propositional Functions Propositional functions (or predicates) are propositions that contain variables. Ex: Let P(x) denote x > 3 P(x) has no truth value until the variable x is bound by either assigning it a value or by quantifying
Georgia Tech - CS - 1050
Rosen 1.4, 1.5Basic Definitions Set - Collection of objects, usually denoted by capital letter Member, element - Object in a set, usually denoted by lower case letter Set Membership - a A denotes that a is an element of set A Cardinality of a
Georgia Tech - CS - 1050
More Functions and SetsRosen 1.6Inverse Image Let f be a function from set A to set B. Let S be a subset of B. We define the inverse image of S to be the subset of A containing all pre-images of all elements of S. f-1(S) = {aA | f(a) S}A Ba
Georgia Tech - CS - 1050
Sequences & SummationsCS 1050 Rosen 1.7Sequence 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
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
Georgia Tech - CS - 1050
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
Georgia Tech - CS - 1050
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
Georgia Tech - CS - 1050
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
Georgia Tech - CS - 1050
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:
E. Kentucky - ITTC - 412
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
E. Kentucky - ITTC - 412
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
E. Kentucky - ITTC - 412
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
E. Kentucky - ITTC - 412
9/25/2003Reducing the Effect of Input Bias Current1/5Reducing the Effect of Input Bias CurrentWe found that the input bias current will cause an offset in the output voltage. There is a solution to this problem-place a resistor (R3) on the non
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
E. Kentucky - ITTC - 412
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
E. Kentucky - ITTC - 412
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 =
E. Kentucky - ITTC - 412
1/31/2003SP 2.4-1.doc1/1Special Problem 2.4-1 The input VA() is a signal expressed as its Fourier Transform. You of course recall that the impedance of an inductor is Z = j L . Note both resistors have the same value. Determine VO() in terms o
E. Kentucky - ITTC - 412
1/31/2003SP 2.4-2.doc1/1Special Problem 2.4-2 Using two ideal op-amps, design a circuit which takes two inputs (VA(t) and VB (t) ) and produces at an opencircuit output the signal:d VA (t ) VO (t ) = - 2VB (t ) dtThe only capacitor that you
E. Kentucky - ITTC - 412
1/31/2003SP 2.5-1.doc1/1Special Problem 2.5-1 For the following circuit, determine vO in terms of vI. 4K 1V+ _ +1KvovI1K 4K
E. Kentucky - ITTC - 412
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
E. Kentucky - ITTC - 412
2/7/2003SP 2.6-3.doc1/1Special Problem 2.6-3 Determine the open-circuit voltage gain of this amplifier.R4 =24K i2 vi R1 =1K R3 =4K-idealR2 =4K+vo
E. Kentucky - ITTC - 412
2/14/2003SP 2.7-2.doc1/1Special Problem 2.7-2 The amplifier below has a 3 dB bandwidth of 100 kHz.9K1K_vovi+1) 2) 3)What is the gain-bandwidth product of the op-amp used in this amplifier? At what frequency is the magnitude of th
E. Kentucky - ITTC - 412
2/14/2003SP 2.7-3.doc1/1Special Problem 2.7-3 Homer S. constructed the following circuit in his garage. He determined that the 3dB frequency (f3dB) of this circuit is 1 MHz - exactly the bandwidth he wanted. In order to generate more gain, Home
E. Kentucky - ITTC - 412
2/14/2003SP 2.8-2.doc1/1Special Problem 2.8-2 Consider the following amplifier:9K1K_15V vovi+-15VThe Slew Rate of the Op-Amp is 10V/s. The D.C. power supply voltages are +/- 15 Volts. The input signal vi is a triangle wave with p
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-2.doc1/1Special Problem 4.6-2 Assume that both transistors are in active mode. Find then the base currents and collector voltages for each transistor. Determine whether the active assumption for each transistor was correct.15.7
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-4.doc1/1Special Problem 4.6-4 Assume the transistor is in the active region and determine collector voltage VC. Was the active assumption correct? Clearly state why or why not.10.0 V1K VC 3.7 V = 49700500
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-5.doc1/1Special Problem 4.12 -5 It is known that the emitter voltage of the transistor in this circuit is at 2 volts with respect to ground (i.e., VE = 2.0 V). What then, is the value of resistor R?10.2 VR10 K 3.7 V = 100
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-7.doc1/1Special Problem 4.6-7 Assume the transistor below is in active region ( = 49). 49)1) Determine the emitter current and voltage collector to emitter.2) Check the assumption and determine if it is valid.15.7V1K1K
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-9.doc1/1Special Problem 4.6-9 For the circuit below, assume that the BJT is in the active region. Determine the base current and the collector voltage. Determine whether the active assumption was correct.10.7 V 1K 0.7 K = 78
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-10.doc1/1Special Problem 4.6-10 Assume the BJT is in active mode.1) Based on this assumption, find collector current iC, collector voltage VC and emitter voltage VE . 2) Determine if the original assumption was correct.15.7 V
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-11.doc1/1Special Problem 4.6-11 Assume both transistors are in active mode. 1) Based on these assumptions, determine the collector currents iC, and collector voltages VC for each transistor. 2) Determine if the original assumpti
E. Kentucky - ITTC - 412
2/21/2003SP 4.6-13.doc1/1Special Problem 4.6-13 Assume the BJT is in active mode.1) Based on this assumption, find collector current iC, collector voltage VC and emitter voltage VE . 2) Determine if the original assumption was correct.1K=
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
2/21/2003SP 4.6-15.doc1/1Special Problem 4.6-15 Assume that both BJTs are in active mode. Find the collector voltage for each BJT. Check to see if the active assumption is correct.10.5 V 10.7 VQ22K =101 100 K 5.0 VQ1=100 100 K 1K- 0.7
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
11/6/2003SP 5.4-21/1 /1Special Problem 5.4-2 The gate voltage of this circuit is at 1.0 volt (VG = -1.0 V). Note the MOSFET is a depletion type. Find the value of resistor R.10R4K9K-10K = 0.25 mA/V2 Vt = - 3.0 VVG = -1.0 V
E. Kentucky - ITTC - 412
11/6/2003SP 5.4-41/1 /1Special Problem 5.4-4 In the circuit below, find the values of the drain-to-source voltage VDS and the gate-to-source voltage VGS. Note that the MOSFET in the circuit is a depletion type.11.0 V11.0 V2K 1KVt = - 2.
E. Kentucky - ITTC - 412
11/6/2003SP 5.4-51/1Special Problem 5.4-5 The PMOS transistor in the circuit below is known to have a gate-tosource voltage of VGS = -4 V. 1) Find the value of resistor R. 2) Find the value of VDS .15 V R 1K K = 0.75 mA/V2 Vt = - 2 V 1K 1K
E. Kentucky - ITTC - 412
11/6/2003SP 5.4-61/1 /1Special Problem 5.4-6 The MOSFET in the circuit below is a depletion type. Determine the drain current and voltage drain to source.10 V K = 0.25 mA/V2 Vt = - 4 V 1K1K 1K
E. Kentucky - ITTC - 412
4/4/2003SP 5.5-3.doc1/1Special Problem 5.5 - 3 The circuit below has been designed as an amplifier, so that we know it is saturation. . The threshold voltage of the transistor is Vt = 1.0 Volts, K = 1.0 mA/V2, and = 0.0 . The capacitors are ve
E. Kentucky - ITTC - 412
4/4/2003SP 5.5-4.doc1/1Special Problem 5.5 - 4 The threshold voltage of the transistor is Vt = 2.0 Volts, K = 0.125 mA/V2. The capacitors are very large. Find the small-signal gain Av = vo/vi of this amplifier.15.0 V15.0 V15 K16 K+ 10
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
4/21/2003SP 5.6-1.doc1/1Special Problem 5.6-1 In the circuit below, Q1 has K1=2 mA/V2 and Vt = 2.0 V. But, Q2 has K2 = 1 mA/V2 and Vt = 2.0 V. In other words Q1 and Q2 are not identical! 1) Determine R so that the drain current of Q1 is 8 mA. 2
E. Kentucky - ITTC - 412
11/10/2003SP 5.6-41/1 /1Special Problem 5.6-4 In the circuit below, Q1 has K=1.0 mA/V2 and Vt1 = 1.0 V. The transistor Q2 likewise has K = 1 mA/V2, but has a threshold voltage of Vt2 = 2.0 V. In other words Q1 and Q2 are not identical! The resi
E. Kentucky - ITTC - 412
11/25/2003SP 5.7-11/1Special Problem 5.7-1 In the circuit below: a) the DC drain current isID = 4.0 mA .b) the DC gate voltage of Q1 is VG 1 = 10.0 V . c) the DC source voltage of Q1 is VS 1 = 7.0 V . d) MOSFETs Q1 and Q2 are in saturation
E. Kentucky - ITTC - 412
10/16/2003SP 5.9-11/1 /1Special Problem 5.9-1 The circuit below uses a special transistor that is unrelated to either a MOSFET or a BJT. The transfer function of this circuit is likewise shown below. 15.0V 15.0 RvOCircuit Transfer Function
E. Kentucky - ITTC - 412
11/10/2003SP 6.6-11/1Special Problem 6.6-1 For the differential amplifier shown below, the DC components of vI1 and vI2 are equal. Both transistors Q1 and Q2 are in saturation. The transistors are identical, with K= 1mA/V2 and Vt = 1.0 V Find t
SUNY Buffalo - CSE - 620
Optical Switching NetworksPresentation byJoaquin CarbonaraReferences Work by Ngo,Qiao,Pan, Anand, Yang Chu/Liu/Zhang Pippinger/Feldman/Friedman Winkler/Haxell/Rasala/WilfongIntroductionStatement of the ProblemAbout Optical Networks Wa