We aren't endorsed by this school 
EEL 3123  UCF Study Resources
 UCF
 Unknown
 Find Textbooks

Microelectronics 4th Neaman Chpt4
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 4 By D. A. Neamen Problem Solutions _ Chapter 4 4.1 (a) (i) g m = 2 kn W I DQ 2 L 0.1 W W 0.5 = 2 (0.5) = 2.5 2 L L k W (ii) I DQ = n (VGSQ VTN )2 2 L 0.1 2 0. 5 = (2.5)(VG

Microelectronics 4th Neaman Chpt7
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 7 By D. A. Neamen Problem Solutions _ Chapter 7 7.1 a. T ( s) = T (s) = V0 ( s ) Vi ( s ) = 1/ ( sC1 ) 1/ ( sC1 ) + R1 1 1 + sR1C1 b. fH = 1 1 = f H = 159 Hz 3 2 R1C1 2 (10 )(106 ) c. V0

Fourier
School: UCF
Course: Networks And Systems
! " # ! ! $%& ' () , = * $ / ) () % = = % = = +% % +% < < % % ( ) ( % ) ( ) = ( ) = $ , % % +% = % ) ( ) % ( ) = ( ) % % % ( ( % % + % () () ) + + % ( ( % % ) ) ( ) ( % ) % ( % ) % % ( ) % % = < < + < < + & = = + = + % = + ) = ( + = %< < < < <

MT1Fall05
School: UCF
Course: Networks And Systems
! " #$ %& '()*+% ,  $., = $*, + = + + + + ! # "# $ $% %& % # ' $ , () . # #$ (#)% # () ! %* +" /% % 0% # + * 1 2 &

MT2Fall05
School: UCF
Course: Networks And Systems
! " # % $ &'( )(* + , ! # &+ " $ ! %& '( ) *%& '(+ , + ) $ $ ! + % . ,#) ) = + . /0%( 1&/ $%2+ 34( 1&/ $+ & / + & + & & . ) , ) )+ , ) + %. 0 0*  *

SolMT2EEL3123.01
School: UCF
Course: Networks And Systems
!" # $ &,+ &+ &.+ % & ' () * + ! ! "#$ %& '(% ) * , ( = ( ! #' +  . ' +/. ' + = + + = +( / +(/ . / ' += 0 & + + +(/ / ( / + +/ / +(/ ( + 1 + ( 2 & 3 4 , $ & 3  54 , $ 2 2 # ' 7 2 2 6 54 4 2 2 2 2 + 2 2 , , # 2" 5 = = =% = 4 = (/ = = = = (= = ( %

SolMT2EEL3123.02
School: UCF
Course: Networks And Systems
!" # $ &+* &,* &* % & '( ) * ! ! "#$ %& '(% ) * , ! #' +  . ' +/. ' + . ' += + 0 & ( + + +( ( + , $ & 3  54 , 2 $ $ 6 2 2 2 2 # , , + 2 54 2 + 1 & 3 4 $ +(/ / 2" + 4 7 2 2 2 2 22 ( = = =% = 4 = (/ = = = = (= = ( % : = = / = ( '( % : ( = ( % :( 9

SolMT1Fall05
School: UCF
Course: Networks And Systems
! = cfw_ = = = cfw_ > = + cfw_ = = = + "! ! = + + ! = + + + + + + = + = + #= +# cfw_ =# + % ( )* + + = = ! + + + $#+ "! + = # = = = + ! + = + " # +# = + + + + # +# =# = =" &' ( ) *+ * ( ,  )( ) * = = + + + + = = + + + + = . . = # . . / )  01 5 4 ( *

SolMT2Fall05
School: UCF
Course: Networks And Systems
! " # % $ &'( )(* + , ! # &+ " $ ! %& '( ) *%& '(+ , + ) $ $ ! + % . ( .(* /*(01( 2 = %& = * = = *%& = /3+1 = * * = 34 8595 .&* : , 34 )5 %6 %.7 = %0&+2 = = * = = +&/ /3+1 * &+ & 1 = = 1+/0 %0&+2 &+ & 1 + = /3+1 = %0&+2 =

AnsHW2
School: UCF
Course: Networks And Systems
+ = = + = cfw_ = + + = = + + cfw_ + + = = cfw_ + = = cfw_ + = + + cfw_+ = + + + + + + + !" $ % ! = & cfw_ ' !" ( $ % ! * = = cfw_ + !" & + = # ' + ) ' & = *& + !" + +

SolHW1
School: UCF
Course: Networks And Systems
Solutions to HW 1 1. Find the Laplace Transform of the function f (t ) = te t (t 1) cfw_ Lcfw_ f (t ) = L te t (t 1) = te t (t 1)e st dt 0 t g (t ) = te e therefore st Lcfw_ f (t ) = g (t ) (t 1)dt 0 t2 Shifting property f (t ) (t t o )dt = f (t o ), t1

SolHW2
School: UCF
Course: Networks And Systems
! + ! + = = = = # = + % + + # + = + + = $+ +$ = %+ + $+ + $+ $+ & = + $+ $+ = # " + % +' + $' + $% $ + $+ +$ + = + $ + + $ = + " + = = +$ + + = +$ = # +$ % +# = + = + + +$ + $ = = +$ + + = $+ = = + $+ + $+ # = +$ $ + # + $+ +& + $ = + $+

SolHW3
School: UCF
Course: Networks And Systems
=! ! = ! + !+ ! = ! + = += + + + ! + + ]= =[ +! cfw_ = # + + + = $% + + = + ! != + + + + + + ! + +! + = +! + + + = ] =! = ! = + + +! ! =! = +! [ = += = != + = ! ! + + ! cfw_! + ! = + ! + = = = ! + + ! + +! + !" + + + +! + =! + =! + = +! + " & ' (

SolHW4
School: UCF
Course: Networks And Systems
! "# ' $% & " $ () & ' , $ &$ $) '  ) . " / 01 ' " $* $* $* $ 20 $ & $ $+ + $ &$ $) ) 0$ 3 3 $ * () & ' $* , $ &$ $) ' $*  ) . " # 1 $ " $* / 1 / $ & $ $+ + $ &$ $) ) 0 / 4 * 5 ' 6 1 * ) 1 ) $ $ , + 4 * $ ) /1 * 8 ) .7 . 7 64 *) / = = + + =/ = = / =

SolHW5
School: UCF
Course: Networks And Systems
= + + + ! & " ' $ # % ( ) !$ $ * + # !$ $ * , . +/ . 0 / 0 !" # $% !& 0 1! &1 ,2/ / . 4&5 ( ( ! +/ ! ) ! .! .& . ., .3 , 0 ,& & 5 4 , & & , 4 45 4 . ! . & . . . , 3 , + + = ! + + = + ! = + * ! ! + * = & + * ! + * / 6 . +/ & ( ( 3 ) ,! .! .& . ., .3 , 0 0

Microelectronics 4th Neaman Chpt6
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 6 By D. A. Neamen Problem Solutions _ Chapter 6 6.1 I CQ (a) (i) g m = VT VT r = ro = I CQ = = 0.5 = 19.23 mA/V 0.026 (180)(0.026) = 9.36 k 0.5 V A 150 = = 300 k I CQ 0.5 2 = 76.92 mA/V

Microelectronics 4th Neaman Chpt3
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 By D. A. Neamen Problem Solutions _ Chapter 3 3.1 Kn = k n W 120 10 2 = 0.75 mA/V 2L 2 0. 8 (a) (i) I D = 0 [ ] = (0.75)[2(2 0.4)(0.1) (0.1) ] = 0.2325 mA = (0.75)[2(3 0.4)(0.1) (0.1)

Finalexamsample1
School: UCF
Course: Networks And Systems
EEL 3123C TEST 3  PART A  MANDATORY DURATION: 60 minutes Dec 7, 2007 1. The voltage across a 50 resistor is v (t) = 4t exp ( jtj) : What is the total energy dissipated in the resistor? What percentage of the total energy dissipated in the resistor can b

Midterm2samplequestion
School: UCF
Course: Networks And Systems
EEL 3123C TEST 3  PART A  MANDATORY DURATION: 60 minutes 1. The current through a 50 resistor is i (t) = 4t exp ( t) u (t) : What percentage of the total energy dissipated in the resistor can be associated p with the the frequency band 0 ! 3 rad/s? (50)

HW1_fall_2010
School: UCF
Course: Networks And Systems
Networks and Systems EEL 3123, Section 1 HOMEWORK 1 Assigned Aug 31, 2010, Due on Sep 7, 2010 Covers Chapter 12. If there are doubts, you are welcome to see me and discuss your problems. Your notes and the textbook should be ample material to solve these

HW2_Fall_2010
School: UCF
Course: Networks And Systems
Networks and Systems EEL 3123, Section 1 HOMEWORK 2 Assigned Sep 9, 2010 Due on Sep 16, 2010 Covers Chapter 13. Topics include circuit analysis using Laplace transform. If there are doubts, you are welcome to see me and discuss your problems. Your notes a

HW3_Fall_2010
School: UCF
Course: Networks And Systems
Networks and Systems EEL 3123, Section 1 HOMEWORK 3 Assigned Sep 23, 2010 Due on Sep 30, 2010 Covers Chapter 14. Topics include low pass, high pass, band pass and band reject filter design. If there are doubts, you are welcome to see me and discuss your p

HW4_Fall_2010
School: UCF
Course: Networks And Systems
Networks and Systems EEL 3123, Section 1 HOMEWORK 4 Assigned Oct 21, 2010 Due in class on Oct 28, 2010 Covers Chapter 16. If there are doubts, you are welcome to see me and discuss your problems. Your notes and the textbook should be ample material to sol

HW5_Fall_2010
School: UCF
Course: Networks And Systems
Networks and Systems EEL 3123, Section 1 HOMEWORK 5 Assigned November 10, 2010, Due in class on Nov 23, 2010 Covers Chapter 17 & 18. If there are doubts, you are welcome to see me and discuss your problems. Your notes and the textbook should be ample mate

Electric Networks Lab Manual
School: UCF
LABORATORY MANUAL EEL 3123 NETWORKS AND SYSTEMS DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING UNIVERSITY OF CENTRAL FLORIDA Prepared by Dr. PARVEEN WAHID Ms. YA SHEN FALL 2011 PREFACE This lab manual for EEL 3123  Networks and Systems is an updated v

Microelectronics 4th Neaman Chpt5
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 5 By D. A. Neamen Problem Solutions _ Chapter 5 5.1 (a) i E = (1 + )i B 1 + = 325 = 116 = 115 2.8 115 = = 0.9914 1 + 116 iC = i E i B = 325 2.8 = 322 A = (b) 1 + = 1.80 = 90 = 89 0.020 89

Microelectronics 4th Neaman Chpt1
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions _ Chapter 1 1.1 ni = BT 3 / 2 e (a) Silicon Eg / 2 kT 1.1 exp 2 ( 86 106 ) ( 250 ) 19 = 2.067 10 exp [ 25.58] ni = 1.61 108 cm 3 (i) ni = ( 5.23 101

Microelectronics 4th Neaman Chpt2
School: UCF
Course: Electronics
Microelectronics: Circuit Analysis and Design, 4th edition Chapter 2 By D. A. Neamen Problem Solutions _ Chapter 2 2.1 1000 (a) For I > 0.6 V, O = ( I 0.6 ) 1020 For I < 0.6 V, O = 0 1000 (b) (ii) O = 0 = [10 sin ( t )1 0.6] 1020 0. 6 Then sin (