2 Pages

Decisions

Course: PSYC 430, Fall 2008
School: Idaho
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Decisions Classifying and Test Utility Whenever we use a test to make decisions, there will be some errors Increasing the correlation between test g and criterion will reduce errors Can manipulate the kinds of correct or incorrect decisions by altering our "cutoff" scores Correct Decisions: True Positives (TP) score well on both the True Negatives (TN) score poorly on both test and the...

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Decisions Classifying and Test Utility Whenever we use a test to make decisions, there will be some errors Increasing the correlation between test g and criterion will reduce errors Can manipulate the kinds of correct or incorrect decisions by altering our "cutoff" scores Correct Decisions: True Positives (TP) score well on both the True Negatives (TN) score poorly on both test and the criterion the test and the criterion Incorrect Decisions: Positives False (FP) score well on the test, False Negatives (FN) score poorly on the but perform poorly on the criterion test, but perform well on the criterion 1 Actual Performance Success Failure Reject (predict failure) False Negatives True Negatives A...

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Idaho - PSYC - 430
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Psychology 430Brief History of Psychological Testing1History of TestingBrass Instruments Era SirFrancis Galtonand sensory discrimination RT JamesMcKeen Cattell Bodilyenergy and mental energy are inseparable2Modern Mental Testi
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Psychology 430Norms and StandardsSetting Standards Norm-referenced tests e.g., personality tests Criterion-referenced tests e.g., certification/licensure exams Establishment of critical (or cut) score may be difficultNorms Selecting a no
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Psychology430ReliabilityClassicalTestTheory(CTT) Truescore=truelevelofability Differencebetweentruescoreandobservedscoreresultsfrom measurementerror X=T+eWhereX=observedscore,T=true score,ande=errorCTT(cont.)Assumesthaterrorsof
Idaho - PSYC - 430
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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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Idaho - PSYC - 430
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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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Idaho - PSYC - 430
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Idaho - PSYC - 430
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Idaho - PSYC - 430
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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
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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
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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
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
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E. Kentucky - ITTC - 412
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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
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E. Kentucky - ITTC - 412
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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
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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