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Lect_Notes_6_Leahy11

Course: EE 483, Fall 2011
School: USC
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(lecture Outline #6) The Discrete Time Fourier Transform (DTFT) (DTFT) - continued DTFT Properties LTI Systems and the Frequency Response Magnitude and Phase Response Black box = important 1 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Important Properties Shift Properties y (n M ) e jM Y (e j ) y ( n ) e j 0 n Y (e j 0 ) Proof: 2 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Important...

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(lecture Outline #6) The Discrete Time Fourier Transform (DTFT) (DTFT) - continued DTFT Properties LTI Systems and the Frequency Response Magnitude and Phase Response Black box = important 1 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Important Properties Shift Properties y (n M ) e jM Y (e j ) y ( n ) e j 0 n Y (e j 0 ) Proof: 2 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Important Properties Convolution Theorems y (k ) x(n k ) Y ( e j ) X ( e j ) k = y ( n) x ( n) 1 2 Y (e j ) X (e j )d Proof: 3 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Important Properties Differentiation nx(n) dX (e j ) j d Proof: 4 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Parsevals Theorem x ( n) y ( n) = n = * 1 2 X (e j )Y (e j )d Proof: 5 Copyright 2005, S. K. Mitra (edited: R.Leahy 2006, 2008) DTFT Properties Example - Determine the DTFT Y (e j ) of y[n] = (n + 1) n[n], < 1 x[n] = n[n], < 1 Let We can therefore write y[n] = n x[n] + x[n] From Table 3.3, the DTFT of x[n] is given th DTFT by 1 X ( e j ) = 1 e j 6 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy DTFT DTFT Properties Using the differentiation property of the DTFT given in Table DTFT given in Table 3.2, we observe that we observe that the DTFT of n x[n] is given by dX (e j ) d 1 e j j =j 1 e j = (1 e j ) 2 d d Next using the linearity property of the using the linearity property of the DTFT given in Table 3.4 we arrive at e j 1 1 + = Y (e ) = (1 e j ) 2 1 e j (1 e j ) 2 j 7 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response Consider the LTI discrete-time system with an impulse response an impulse response {h[n]} shown below shown below x[n] h[n] y[n] Its input-output relationship in the timedomain is given by the convolution sum is given by the convolution sum y[n] = h[k ] x[n k ] k = 8 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response If the input is of the form x[n] = e j n, < n < then it follows that the output is given by y[n] = h[k ] e Let j( n k ) k = = h[k ] e j k e j n k = H (e j ) = h[k ] e j k k = 9 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response Then we can write we can write y[n] = H (e j ) e j n Thus for a complex exponential input signal e j n , the output of an LTI discrete-time system is also a complex exponential signal of the same frequency multiplied by a the j complex constant H (e ) j n Thus e is an eigenfunction (or eigensequence) of the system 10 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response The quantity H (e j ) is called the frequency response of the LTI discrete-time system the discrete system H (e j ) provides a frequency-domain description of the system H (e j ) is precisely the DTFT of the impulse response {h[n]} of the system 11 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response H (e j ) can be expressed in terms of its real and imaginary parts real and imaginary parts H (e j ) = H re (e j ) + j H im (e j ) or, in terms of its magnitude and phase response: H (e j ) = H (e j ) e j() where 12 () = arg H (e j ) Copyright 2005, S. K. Mitra, 2006-2010 Frequency R.Leahy The The Response In some cases, the magnitude function is specified in decibels as G ( ) = 20 log10 H (e j ) / H (e j 0 ) 13 dB where G() is called the gain. Note that this is a ratio; here I defined it relative to the gain at =0. Be careful since gain of zero Be careful since a gain of zero corresponds to dBs. (In Matlab you need to set a minimum value if using autoscaling) Attenuation = - Gain in dB Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response Example - Consider the M-point moving average filter with an impulse response average filter with an impulse response given by h[n] = 1 / M , 0 n M 1 0, otherwise Its frequency response is then given by H ( e j ) = = 14 1 M M 1 j n e n =0 1 sin( M / 2) j ( M 1) / 2 e M sin( / 2) Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response Thus, the magnitude response of the M-point moving average filter is given by H ( e j ) = 1 M sin( M / 2) sin( / 2) and the phase response is given by ( M 1) M/2 2 k + ( M ) () = 2 k =0 15 note second term gives -discontinuties as gain passes through zero. More on phase soon . Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy The The Frequency Response Magnitude 12 10 Phase 3 M=11 2 8 1 6 0 4 -1 2 -2 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -3 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Magnitude and phase response of the moving average filter. Note linear phase with -discontinuities. 16 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy DTFT, DTFT, LTI and Frequency Response Consider the linear constant coefficient difference equation: difference equation: N M k =0 k =0 d k y[n k ] = pk x[n k ] Multiply each side by e jn and sum over n: N d n = k = 0 k y[n k ]e j n = M p x[n k ]e n = k = 0 17 j n k Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy DTFT, DTFT, LTI and Frequency Response Gives the relationship: N Y (e j ) d k e jk = X (e k =0 By the convolution theorem y ( n) = h( k ) x ( n k ) j M ) p k e jk k =0 Y (e j ) = H (e j )( X (e j ) k = M j Y (e ) = X ( e j ) 18 p e j k d e j k k =0 N k =0 k = H ( e j ) k Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy DTFT DTFT Computation Using MATLAB The function freqz can be used to compute the Frequency response of an LTI compute the Frequency response of an LTI system from the coefficients of the LDE: p0 + p1e j + .... + pM e jM H (e ) = d 0 + d1e j + .... + d N e jN j at a prescribed set of discrete frequency points = 19 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy DTFT DTFT Computation Using MATLAB For example, the statement H = freqz(num,den,w) returns the frequency response values as a vector H of a DTFT defined in terms of the vectors num and den containing the coefficients { pi } and {di } , respectively at a prescribed set of frequencies between 0 and given by the vector w 20 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Example A digital Chebyshev filter: [b,a]=cheby1(5,.5,.5); freqz(b,a); 21 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy Example A digital IIR Butterworth filter [b,a]=butter(30,.5); freqz(b,a); 22 Copyright 2005, S. K. Mitra, 2006-2010 R.Leahy
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USC - EE - 483
OutlineOutline (lecture #7) The Discrete Time Fourier Transform(DTFT)(DTFT) - continued The importance of phase in signals andsystem Phase response and discontinuities Linear phase filters Group delayBlack box All pass filters=important1Cop
USC - EE - 483
OutlineOutline (lecture #8) The Discrete Fourier Transform (DFT)Forward and inverse DFT, definition and proofDFTExamplesDFT in matrix formBasis function of the DFTThe FFT AlgorithmBlack box=important1Copyright 2005, S. K. Mitra; 2006-2010 R.L
USC - EE - 483
OutlineOutline (lecture #9) The DFT and Other Unitary TransformsThThe unitary transformDiscrete Cosine TransformThe Haar Transform2D Transforms Relationship between DFT and DTFTbetween DFT and DTFT (coming soon .)Black box=important1Copyrig
USC - EE - 483
OutlineOutline (lecture #10) Relationship between the Fouriertransforms the Samping Theorem andtransforms, the Samping Theorem andSpectral Analysis I 1. Relationship between DTFT and DFT Relationship between FT and DTFT (samplingtheorem) Analog s
USC - EE - 483
OutlineOutline (lecture #11) Relationship between the Fouriertransforms the Samping Theorem andtransforms, the Samping Theorem andSpectral Analysis IIBlack box=important1Copyright 2005, S. K. Mitra; 2006-2010, R.LeahyPractical Spectral Analysis
USC - EE - 483
Lecture #12 OutlineZ-transform definitiondefinitionRegions of convergenceRational Z-transforms: poles and zerosInverse Z-transformBlack box=important1Copyright 2005, S. K. Mitra; 2006-2010, R.LeahyThe z-Transform A generalization of the DTFT d
USC - EE - 483
Lecture #13 Outline LTI systems: the impulse response and thesystems: the impulse response and thesystem function. Rational z-transforms and LTI systems Causality and stability in LTI systems System functions and the frequencyfunctions and the freq
USC - EE - 483
Lecture #14 OutlineA couple of random examplesGeneralized linear phase (again)Minimum and all pass filtersDifferentiators and Hilbert transformsDesign of general purpose filters .Black box=important1Copyright 2005, S. K. Mitra/R. Leahy 2006-2010
USC - EE - 483
Lecture #15 Outline FIR Filter design develop a numericalFilter designnumericaltechnique for determing FIR filter coefficientsto optimally approximate the desired frequencyresponse. Least squares design Design using windows Equiripple filter desi
USC - EE - 483
Lecture #16 Outline An extended, complete FIR Filter designextended complete FIR Filter designexampleBlack box=important#Copyright 2005, S. K. Mitra/ RM Leahy 2008-2010FIR Design Example Well conclude our study of FIR filter design withconclude
USC - EE - 483
Lecture #17 Outline IIR filter Designfilter Design Analog prototype filters and their designBlack box=important#Copyright 2005, S. K. Mitra/ RM Leahy 2008Analog Prototypes The set of analog filters that are mostset of analog filters that are mo
USC - EE - 483
Lecture #18 Outline IIR filter Designfilter Design The Bilinear transform design method Spectral transformations IIR design by numerical optimizationBlack box=importantCopyright 2005, S. K. Mitra/ RM Leahy 2009Bilinear Transformation Method Ide
USC - EE - 483
Lecture #19An Introduction to AdaptiveFilteringMotivationThe ideal transfer functionFiltering by power minimizationSteepest descentdescentLMS algorithmApplications#Copyright 2010 R. LeahyMotivationMotivation+-G(z)#Copyright 2010 R. Leahy
USC - EE - 483
Lecture #20 Outline: Quanztizationand Digital Filter Structures Quantization: Oversampling DACs Representation of binary signals and quantization noise Sigma-delta converters. Issues in implementation of digital filters: Efficiency (minimize number
USC - EE - 483
EE483: Digital Signal ProcessingInstructor: Professor LeahySAMPLE MIDTERM, FALL 2006Time Allowed: 80 MinutesPlease answer all questions. For partial credit you must show how you reach your solutions.Make sure that any sketches you draw are labelled a
Texas A&M - MATH - 410
Math 410.500Exam 1Solutions1. (15 pts.)(a) What is the Uniform Cauchy Criterion for uniform convergence of asequence of functions fn (x) on a set E ? Answer: For every &gt; 0there exists N so that m, n N implies that |fn (x) fm (x)| &lt; holds for all x
Texas A&M - MATH - 410
Math 410.500Exam 2, version ASolutions1. (20 pts.) Determine the interval of convergence of the power seriesn=1(1)n n (x 3)3n .n8Solution: Its probably easiest to use the ratio test:(1)n+1 (x 3)3(n+1) / n + 18n+1|x 3|3lim= limnn8(1)n (x 3)
Texas A&M - MATH - 410
Math 410.500Exam 2, version BSolutions1. (20 pts.) Determine the interval of convergence of the power seriesn=1(1)n n (x 1)3n .n6Solution: Its probably easiest to use the ratio test:(1)n+1 (x 1)3(n+1) / n + 16n+1|x 1|3lim= limnnn6(1)n (x
Texas A&M - MATH - 410
Math 410.500: answers to exam 11. (a) f is analytic on (a, b) i for each x0 (a, b) there exists a power serieskak (x x0 ) and a number R &gt; 0 so that f (x) =k=0ak (x x0 )kk=0on (x0 R, x0 + R). (It's too much to ask for that f equals a single1powe
Texas A&M - MATH - 410
Math 410.500: answers to exam 21. (a) U Rn is open if and only if for every a U there exists r &gt; 0 sothat Br (a) U .(b) The boundary of U is the set of all points x Rn so that r &gt; 0,both Br (x) U = and Br (x) U c = .(c) E Rn is connected if there doe
Texas A&M - MATH - 410
Math 410.500: answers to exam 31. (a) True. (This is Theorem 11.13.)(b) False. (Problem 2, section 11.2 has rst order partials, but is notdierentiable.)(c) False. (See example 11.11.)(d) True. (This is Theorem 11.15.)(e) False. (See example 11.18.)
Texas A&M - MATH - 410
Math 410.500Exam 12/16/051. (20 pts.) Short answer:(a) Dene: f is analytic on an open interval (a, b).(b) State Dirichlet's test for uniform convergence offk (x) gk (x)k=1on a set E .(c) State the Cauchy criterion for uniform convergence of a seq
Texas A&M - MATH - 410
Math 410.500Exam 23/30/05There are problems on both sides of this sheet!1. (18 pts.) Dene the term in italics :(a) U Rn is an open set.(b) What is the boundary of a set U Rn ?(c) E Rn is a connected set. (Either the denition from the bookor the de
Texas A&M - MATH - 410
Math 410.500Exam 34/29/05There are problems on both sides of this sheet!1. (10 pts.) True or false? (no explanation is needed)(a) If f is dierentiable at a Rn then f is continuous at a.(b) If all rst order partials of f exist at a then f is dierenti
Texas A&M - MATH - 410
Solutions to homework #1 a) False. You can use a divergent p series as a counterexample, or6.1.01k=1 k+ k+1 , which I showed in class was a diverging telescoping series.b) False. As a counterexample, take any convergent series whose termsare nonzer
Texas A&M - MATH - 410
Solutions to homework # 2 6.3.2d. A routine ratio test:limk(1 3 (2k 1) (2k + 1) / (2k + 2)!)(1 3 (2k 1) / (2k )!==limk2k + 1(2k + 1) (2k + 2)0,hence the series converges absolutely and therefore converges. Since everything is obviously positi
Texas A&M - MATH - 410
Solutions # 3x 7.1.1 a) On any interval [a, b], we have |x| max (|a| , |b|), so that n 0 cfw_xmax(|a|,|b|)max(|a|,|b|)x. For any &gt; 0, if&lt; n, then n 0 &lt; , thus nnconverges uniformly to zero on [a, b]. b) There are two things to show.1First, fn
Texas A&M - MATH - 410
Solutions to assignment # 4 7.3.1b: Writing out the series as 1+ x2 +32 x4 + x6 +34 x8 + x10 + , we seethat the sequence of roots of the coecients is 1, 0, 1, 0, 31/2 , 0, 1, 0 , 31/2 ,and so on, so the limit supremum is 3, and the radius of convergenc
Texas A&M - MATH - 410
Solutions to assignment 5 8.1.1: a)x y = x z + z y x z + z y &lt; 2 + 3 = 5 b)|x y x z| = |x (y z)| x y z x y + (z) x (y + z)&lt; 2 (3 + 4) = 14,where I used the Cauchy-Schwartz inequality in the rst step, andthen the triangle inequality. c)|x (
Texas A&M - MATH - 410
Assignment # 6Solutions: 8.3.5 a) The sketch is below: the region in question is the intersection ofthe two disks.b) U is relatively open in E1 , since its the intersection of an open set withE1 . c) U is relatively closed in E2 , since its the inter
Texas A&M - MATH - 410
Solutions to suggested problems for exam 2 9.1.1a: Obviously, we suspect that the limit is (0, 1). To prove thisfrom the denition, we need to produce N () so that k N () implies(1)1k , 1 k2 (0, 1) &lt; . This is equivalent to()1111=,+ 4 &lt; .k
Texas A&M - MATH - 367
Denition 38 Triangular region is the union of a triangle and its interior. Denition 39 A polygonal region is the union of a nite number of triangular regions such that if two triangular regions intersect, their intersection is an edge or vertex of both. D
Texas A&M - MATH - 367
Homework 7 Identify the congruent triangles and show they are congruent. Note a picture can have more than one pair of triangles.1.2.3.4.5.6.7.8.12
Texas A&M - MATH - 367
An axiomatic system example. Assume that a club of two or more students is organized into committees in such a way that each of the following conditions are satised. a) Every committee is a set of one or more students. b) For each pair of students, there
Texas A&M - MATH - 367
Denition 28 Parallel Two lines are parallel in a plane if they dont intersect. Postulate 16 Parallel Postulate Through a point not on a given line there is exactly one parallel to the line. Postulate 17 Lobachevsky and Bolya Postulate Through a point not
Texas A&M - MATH - 367
Theorem 29 If A and B are equidistant from P and Q then every point between A and B has the same property. Theorem 30 If a line L contains the midpoint of P Q and contains another point which is equidistant from P and Q, then L P Q.Theorem 40 Through a g
Texas A&M - MATH - 367
Denition 33 polygon A polygon is the union of n segments in a plane, intersecting at and only at their endpoints, such that exactly two segments contain each endpoint and no two consecutive segments are on the same line. We are going to assume that any po
Texas A&M - MATH - 367
Denition 12 Segment If A and B are two points then the segment between them is the set of points A X B , for all X in S , along with A and B . It is written AB Denition 13 Ray Is the set of points C on AB with A not between B and C . It is written AB . De
Texas A&M - MATH - 367
Theorem 14 Segment construction [3.6.C-2] Given a segment AB and a ray CD, there is exactly one point E on CD such that AB CE . = Theorem 15 Segment Addition [3.6.C-3] If A B C and D E F and AB DE and = BC EF then AC DF . = = Denition 22 Convex A set G is
Texas A&M - MATH - 367
Postulate 12 Space separation postulate [4.5.ss-1] Given a plane in space. The set of all points that do not lie in the plane is the union of two sets S1 , S2 (the book uses H1 and H2 we will reserve those for half-planes) such that each of the sets is co
Texas A&M - MATH - 367
Denition 7 Logical System consists of undened terms, denitions, assumptions and theorems. The undened terms for geometry are set, point, line, plane. Denition 8 One-to-one correspondence is if two sets have the same number of elements. If two sets have a
Texas A&M - MATH - 367
Denition 1 Negation If p is a statement, the statement p is the negation of p. Example 1 Form the negation of the following statements. a The moon is rising. b ABC is a remote interior angle. c Point C is between points A and B . d m 3 = 25 Solution a The
Texas A&M - MATH - 251
Answers to exam 2, version A1. In general, the directional derivative of f at (a, b) in the direction of a unitvector u is Du f (a, b) = f (a, b) u. So,=exy + xyexy , x2 exy=f2e, eat (1, 1). The vector going from (1, 1) to (2, 3) is 1, 2 , but thi
Texas A&M - MATH - 251
Answers to exam 3, version A 1. If D is the region bounded by x and x2 , put in = ky to getx = =which integrates out to 5 . 8Dx dA dA D1 x xky dy dx 0 x2 , 1 x 2 ky dy dx 0 x2. The solid that we're integrating over has its top as part of a spherical
Texas A&M - MATH - 251
Exam 1, version A 9/19/08 1. For the function f (x, y ) = 16 4x2 + 9 y 2 , (a) (6 pts.) sketch the domain. (b) (6 pts.) nd the range. 2. (8 pts.) Find two unit vectors perpendicular to both 2, 1, 1 and 1, 2, 1 . 3. (10 pts.) Find the equation of the plane
Texas A&M - MATH - 251
Math 251.512Exam 2, version A10/15/081. (11 pts.) If f (x, y ) = xexy , nd the directional derivative of f at (1, 1) in thedirection from (1, 1) to (2, 3).2. (11 pts.) FindRxdA, where R is the rectangle 1 x 2, 0 y 4.1 + 2yx, use dierentials to
Texas A&M - MATH - 251
Exam 3, version A 11/12/08 Conversion from spherical to cartesian coordinates:Math 251.512x = sin cos y = sin sin z = cos dV =2 sin d d d1. (10 pts.) Determine x of the center of mass of a plate bounded by y = x and y = x2 , if density is proportional
Texas A&M - MATH - 251
Math 251.504Exam 1, version ASolutions1. (11 pts.) Find the equation of the plane containing the points (0, 1, 1), (2, 1, 2), and (3, 0, 1).Answer: To nd the equation of a plane, we need a point in the plane and a vector normal tothe plane. Weve alre
Texas A&M - MATH - 251
Math 251.504Exam 1, version BSolutions1. (11 pts.) Find the equation of the plane containing the points (1, 1, 1), (1, 1, 0), and (3, 0, 1).Answer: To nd the equation of a plane, we need a point in the plane and a vector normal tothe plane. Weve alre
Texas A&M - MATH - 251
Math 251.504Exam 2, version ASolutions1. (10 pts.) Evaluate D x dA, where D is the triangular region withvertices (0, 0), (1, 1), and (1, 4). Solution: In the order dy dx, thedouble integral becomes1 1 4xxy |y=4x dxx dy dx =y =x0x0 1=3x2 dx
Texas A&M - MATH - 251
Math 251.504Exam 2, version BSolutions1. (10 pts.) Evaluate D x dA, where D is the triangular region withvertices (0, 0), (1, 1), and (1, 3). Solution: In the order dy dx, thedouble integral becomes1 1 3xxy |y=3x dxx dy dx =y =x0x0 1=2x2 dx
Texas A&M - MATH - 151
11.1: VectorsDenitions:vector:additionscalar multiplicationsubtractionmagnitude:unit vector:i and j:1Examples:Given the vectors a =&lt; 3, 5 &gt; and b starts at the point (1, 1) and ends at the point (1, 3),write i + j in terms of a and b.Given a
Texas A&M - MATH - 151
11.2: Dot ProductDenitions:The dot product of the vectors a and b is given byDot Product computation formulaFrom the denition, it follows that the angle between two vectors is given bya and b are orthogonal if and only ifOrthogonal complementsScal
Texas A&M - MATH - 151
11.3: Vector Functions and Parametrized CurvesDenitions:(Recall) function:Vector Valued function:Parametrized Curve:Eliminating the ParameterVector and Parametric Equations of a Line1Examples:Given the curve parametrized by r(t) = (t2 + 1)i + (t
Texas A&M - MATH - 151
12.1/2.2: Intro to Calculus and LimitsGoal #1: To nd the slope of a line tangent to a curve at a given point.Concept of a Limit (Maplet):Innite Limits and Vertical Asymptotes:1Examples:x2 + 1x 1 x 1limOn Beyond Average: Find the vertical asympto
Texas A&M - MATH - 151
12.3: Analytic Computation of LimitsProperties of Limits: (pp 91-93. Basis for the techniques used in the following examples.)Examples:lim x3 3x2 + 1x 1limx42x + 8x2 + x 12lim r(t) where r(t) =t25t3 + 4t3i+t2 4t2j1Squeeze Theorem: If g
Texas A&M - MATH - 151
12.5: ContinuityDenitions:f is continuous at x = aRemovable DiscontinuitiesSource for understanding: Maplet Left and Right Hand Limits and Continuities, located athttp:/calclab.math.tamu.edu/maple/maplets/ (NetID login)1Theorems:Limits inside Con
Texas A&M - MATH - 151
12.6: Limits at InnityIn 2.2, we learned that if y as x a, then the graph of f has a vertical asymptote atx = a. Similarly, if y L as x , then the graph of the function has a horizontal asymptoteat y = L.Key Limit:limx1=xComputing Limits at Inn
Texas A&M - MATH - 151
12.7: Tangents, Velocities, and Rates of ChangeWe are now ready to nd a formal way of computing the slope of the line tangent to the curve. Re-viewthe animation from 2.1 posted on my webpage. What happens as the second xcoordinate moves closerto the g
Texas A&M - MATH - 151
13.1: The DerivativeNow that we can nd the slope of the line tangent to a curve at any point (provided the limit of theslope exists), we can talk about a new function based on this calculation.Denition: The derivative function of a function f (or the