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©R. C. Jaeger and T. N. Blalock 7/05/10 16-4 16.7 V GS 1  V TN  2 I DS 1 K n ' W L       1  V TN  2 I DS 1 K n ' W L       1 1   W / L   2 W / L    V TN  2 I DS 1 K n ' W L       1   W / L   4 W / L           V GS 2  V TN  2 I DS 2 K n ' W L       2  V TN  2 I DS 2 K n ' W L       1 1   W / L   2 W / L    V TN  2 I DS 2 K n ' W L       1   W / L   4 W / L           I DS 2  I DS 1 : V GS 2  V GS 1  2 I DS 2 K n ' W L        W / L   2 W / L    

Snell_s_law_.pdf-AM Snell's law - Wikipedia, the

ECE-231 Secon 101 Fall 2010 Quiz #1 September 21, 2010 Soluon Set Problem #1 (soluon a) 35 W i 2 node e 12 volts i 3 node b - + 20 W 25 W node a i 4 100 W node c Use Kirchhoﬀ’s laws and Ohm’s law to produce a set of equaons that could be solved , using algebra, to correctly determine the actual value of V - + i 1 V a) KVL: 25 W x i 1 + 20 W x i 2 - 12V = 0 b) KVL: 35 W x i 3 + 100 W x i 4 - 20 W x i 2 = 0 c) KVL: V - 100 W x i 4 =0; e) Ohm’s Law: V be = 20 W x i 2 f) KCL: i 1 - i 2 - i 3 = 0 [ 6 equaons, 6 unknowns] Controlled voltage source: d) V= 2 x V be Problem #1 (soluon b) 35 W i 2 12 volts i 3 node b - + 20 W 25 W node a i 4 100 W node c Controlled voltage source: d) V= 2 x V be Use Kirchhoﬀ’s laws and Ohm’s law to produce a set of equaons that could be solved , using algebra, to correctly determine the actual value of V - + i 1 V a) KVL: 25 W x i 1 + 20 W x i 2 - 12V = 0 b) KCL: i 1 - i 2 - i 3 =0 c) KVL: 35 W x i 3 + V - 20 W x i 2 =0 e) Ohm’s law: V be = 20 W x i 2 [ 5 equaons, 5 unknowns] Note: in this soluon, we don’t have to consider i 4 or the 100 W resistor node e Problem #2 For each of box A, box B, box C, and box D: determine the magnitude of the power ﬂowing into or out of that box; and determine whether power is ﬂowing into the box or out of the box. box B i2 node e 25 volts (voltage source) i3 node b + - 100 W node a i1 node c i4 100 W 100 W i6 box D box C box A i5 i1= 2A; i2= -0.5A ; i3= -0.5A ; i4= 0.5A; i5= -1.5A ; v ab = 25V; v be = -50V ; v cd = 25V; v de = 25V node d 50W 37.5W 37.5W 25W Problem #2 (Including all elements) For each of box A, box B, box C, and box D: determine the magnitude of the power ﬂowing into or out of that box; and determine whether power is ﬂowing into the box or out of the box. box B i2 node e 25 volts (voltage source) i3 node b + - 100 W node a i1 node c i4 100 W 100 W i6 box D box C box A i5 i1= 2A; i2= -0.5A ; i3= -0.5A ; i4= 0.5A; i5= -1.5A ; v ab = 25V; v be = -50V ; v cd = 25V; v de = 25V node d 50W 37.5W 37.5W 25W 25W 100W 50W 25W Problem #2 (What’s in the boxes?) For each of box A, box B, box C, and box D: determine the magnitude of the power ﬂowing into or out of that box; and determine whether power is ﬂowing into the box or out of the box. 1 0 0 W box B i2 node e 25 volts (voltage source) i3 node b + - 100 W node a i1 node c i4 100 W 100 W i6 + 2 5 V - box D box C box A i5 node d + 25V - + 25V- i1= 2A; i2= -0.5A ; i3= -0.5A ; i4= 0.5A; i5= -1.5A ; v ab = 25V; v be = -50V ; v cd = 25V; v de = 25V

ECE | 231 | Circuits and Systems 1 | homework | ECE-231 Secon 101 Fall 2010 Q

Differential Equations © 2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx Differential Equations Periodic Functions, Even/Odd Functions and Orthogonal Functions This is going to be a short section. We just need to have a brief discussion about a couple of ideas that we’ll be dealing with on occasion as we move into the next topic of this chapter. Periodic Function The first topic we need to discuss is that of a periodic function. A function is said to be periodic with period T if the following is true, ( ) ( ) for all f x T f x x + = The following is a nice little fact about periodic functions. Fact 1 If f and g are both periodic functions with period T then so is f g + and fg . This is easy enough to prove so let’s do that. ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) f g x T f x T gx T f x gx f g x fg x T f x T gx T f xgx fg x + + = + + + = + = + + = + + = = The two periodic functions that most of us are familiar are sine and cosine and in fact we’ll be using these two functions regularly in the remaining sections of this chapter. So, having said that let’s close off this discussion of periodic functions with the following fact, Fact 2 ( sin ) x ω and ( cos ) x ω are periodic functions with period 2 T π ω = . Even and Odd Functions The next quick idea that we need to discuss is that of even and odd functions. Recall that a function is said to be even if, ( ) ( ) f x f x − = and a function is said to be odd if, ( ) ( ) f x f − =− x The standard examples of even functions are ( ) 2 f x x = and ( ) ( ) cos gx x = while the standard examples of odd functions are ( ) 3 f x x = and ( ) ( ) x sin gx = . The following fact about certain integrals of even/odd functions will be useful in some of our work. © 2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx Differential Equations Fact 3 1. If ( ) f x is an even function then, ( ) ( ) 0 2 L L L f x dx f x dx − = ∫ ∫ 2. If ( ) f x is an odd function then, ( ) 0 L L f x dx − = ∫ Note that this fact is only valid on a “symmetric” interval, i.e. an interval in the form [ ] , L L − . If we aren’t integrating on a “symmetric” interval then the fact may or may not be true. Orthogonal Functions The final topic that we need to discuss here is that of orthogonal functions. This idea will be integral to what we’ll be doing in the remainder of this chapter and in the next chapter as we discuss one of the basic solution methods for partial differential equations. Let’s first get the definition of orthogonal functions out of the way. Definition 1. Two non-zero functions, ( ) f x and ( ) gx , are said to be orthogonal on a x b ≤ ≤ if, ( ) ( ) 0 b a f xgxdx = ∫ 2. A set of non-zero functions, ( ) { } i f x , is said to be mutually orthogonal on (or just an a x b ≤ ≤ orthogonal set if we’re being lazy) if ( ) i f x and ( ) j f x are orthogonal for every i . In other j ≠ words, ( ) ( ) 0 0 i j b a i j f x f x dx c i j ≠ ⎧ = ⎨ > = ⎩ ∫ Note that in the case of i j = for the second definition we know that we’ll get a positive value from the integral because, ( ) ( ) ( ) 2 0 i i i b b a a f x f x dx f x dx = > ⎡ ⎤ ⎣ ⎦ ∫ ∫ Recall that when we integrate a positive function we know the result will be positive as well. Also note that the non-zero requirement is important because otherwise the integral would be trivially zero regardless of the other function we were using. Before we work some examples there are a nice set of trig formulas that we’ll need to help us with some of the integrals. © 2007 Paul Dawkins 4 http://tutorial.math.lamar.edu/terms.aspx

Physical Dimensions inches (millimeters) unless otherwise noted Small Outline Package Order Number DAC0808LCM NS Package Number M16A Dual-In-Line Package Order Number DAC0808 NS Package Number N16A DAC0808 www.national.com 11 Notes LIFE SUPPORT POLICY NATIONAL’S PRODUCTS ARE NOT AUTHORIZED FOR USE AS CRITICAL COMPONENTS IN LIFE SUPPORT DEVICES OR SYSTEMS WITHOUT THE EXPRESS WRITTEN APPROVAL OF THE PRESIDENT AND GENERAL COUNSEL OF NATIONAL SEMICONDUCTOR CORPORATION. As used herein: 1. Life support devices or systems are devices or systems which, (a) are intended for surgical implant into the body, or (b) support or sustain life, and whose failure to perform when properly used in accordance with instructions for use provided in the labeling, can be reasonably expected to result in a significant injury to the user. 2. A critical component is any component of a life support device or system whose failure to perform can be reasonably expected to cause the failure of the life support device or system, or to affect its safety or effectiveness. National Semiconductor Corporation Americas Email: support@nsc.com National Semiconductor Europe Fax: +49 (0) 180-530 85 86 Email: europe.support@nsc.com Deutsch Tel: +49 (0) 69 9508 6208 English Tel: +44 (0) 870 24 0 2171 Français Tel: +33 (0) 1 41 91 8790 National Semiconductor Asia Pacific Customer Response Group Tel: 65-2544466 Fax: 65-2504466 Email: ap.support@nsc.com National Semiconductor Japan Ltd. Tel: 81-3-5639-7560 Fax: 81-3-5639-7507 www.national.com DAC0808 8-Bit D/A Converter National does not assume any responsibility for use of any circuitry described, no circuit patent licenses are implied and National reserves the right at any time without notice to change said circuitry and specifications.

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