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ECE4330 Lecture 2: Math Review (Continued) Prof ECE4330 Lecture 2: Math Review (continued) Prof. Mohamad Hassoun Trigonometric Identities (continued): 2휋 2 sin 푥 cos 푥 = sin 2푥 cos(휔푡) → 휔 = 2휋푓, 휔 = (rad/sec) 푇 1 (sin 푥)2 + (cos 푥)2 = 1 푇 = , 푓 is the Hz frequency; 푇 is the period 푓 (cos 푥)2− (sin 푥)2 = cos 2푥 1 (cos 푥)2 = (1 + cos 2푥) Note: (cos 푥)2 ≡ cos2 푥 2 1 (sin 푥)2 = (1 − cos 2푥) 2 1 (cos 푥)3 = (3 cos 푥 + cos 3푥) 4 1 (sin 푥)3 = (3 sin 푥 − sin 3푥) 4 sin(푥 ± 푦) = sin 푥 cos 푦 ± cos 푥 sin 푦 휋 휋 휋 [Example: sin(푥 ± ) = sin(푥) cos ( ) ± cos(푥) sin ( ) = ±cos (푥)] 2 2 2 cos(푥 ± 푦) = cos 푥 cos 푦 ∓ sin 푥 sin 푦 [Example: cos(푥 ± 휋) = cos(푥) cos(휋) ∓ sin(푥) sin(휋) = −cos (푥) tan 푥 ± tan 푦 tan(푥 ± 푦) = 1 ∓ tan 푥 tan 푦 1 sin 푥 sin 푦 = [cos(푥 − 푦) − cos(푥 + 푦)] 2 1 cos 푥 cos 푦 = [cos(푥 − 푦) + cos(푥 + 푦)] 2 1 sin 푥 cos 푦 = [sin(푥 − 푦) + sin(푥 + 푦)] 2 Example. Find sin(15°) using trigonometric identities (no calculator) sin(15°) = sin(45° − 30°) θ cos θ sin θ = sin(45°) cos(30°) − cos(45°) sin(30°) 휋 1 2 3 2 1 6 − 2 (30°) √3 √ √ √ √ √ 6 = − = ≈ 0.259 2 2 2 2 2 2 4 휋 (45°) √2 √2 4 2 2 Example. Express cos4(푥) as the sum of simple sinusoids. 1 (cos 푥 )4 = cos 푥 (cos 푥)3 = cos 푥 [ (3 cos 푥 + cos 3푥)] 4 3 1 = (cos 푥)2 + cos 푥 cos 3푥 4 4 3 1 1 1 = ( ) (1 + cos 2푥) + ( ) [cos(−2푥) + cos(4푥)] 4 2 4 2 3 3 1 1 = + cos 2푥 + cos 2푥 + cos 4푥 8 8 8 8 3 1 1 = + cos 2푥 + cos 4푥 8 2 8 Note that the result includes a constant term, a sinusoid at 휔0 = 2 and a sinusoid at frequency 4. These are called harmonic (multiple) frequencies. Formulas for exponentiated sinusoids as sum of pure sinusoids: n even: 푛 −1 2 푛 1 푛 2 푛 (cos 푥) = (푛) + ∑ ( ) cos[(푛 − 2푘)푥] 2푛 2푛 푘 2 푘=0 푛 −1 2 푛 푛 1 푛 2 ( −푘) 푛 (sin 푥) = (푛) + ∑(−1) 2 ( ) cos[(푛 − 2푘)푥] 2푛 2푛 푘 2 푘=0 n odd: 푛−1 2 2 푛 (cos 푥)푛 = ∑ ( ) cos[(푛 − 2푘)푥] 2푛 푘 푘=0 푛−1 2 푛−1 2 ( −푘) 푛 (sin 푥)푛 = ∑(−1) 2 ( ) sin[(푛 − 2푘)푥] 2푛 푘 푘=0 With the binomial coefficient: 0! = 1 푛 푛! ( ) = 푓표푟 0 ≤ 푘 ≤ 푛 n! = (1)(2)(3)…(n) 푘 푘! (푛 − 푘)! Example. Express cos4(푥) as a sum of pure sinusoids. 1 4! 2 4 4 (cos 푥)4 = ( ) + [( ) cos((4 − 0)푥) + ( ) cos((4 − 2)푥)] 24 2! (4 − 2)! 16 0 1 3 1 3 1 1 = + [cos 4푥 + 4 cos 2푥] = + cos 2푥 + cos 4푥 8 8 8 2 8 Your turn: Expand the following expressions as a sum of pure sinusoids, cos5 푥 = ? sin4 푥 = ? (a) Employing Trig identities and (b) Employing the formulas. Your turn: Use Mathcad to plot 푓(푥) = cos푛(푥) for 푛 = 20 and 200. Your turn (challenge): Solve for 훽 as a function of 푘 and 훼 in the equation, sin(훼) − sin(훽) + 푘푠푖푛(훼 + 훽) = 0. −2푘+(1+푘2)cos (훼) Ans. 훽 = 푐표푠−1 { } 1−2푘푐표푠(훼)+푘2 Has the same frequency Example. Show that 푎 cos 푥 + 푏 sin 푥 = 퐶 cos(푥 + 휃 ) where −푏 퐶 = √푎2 + 푏2 푎푛푑 휃 = tan−1 푎 Employing the trig identity cos(푥 + 푦) = cos 푥 cos 푦 − sin 푥 sin 푦, we may write: C cos(푥 + 휃) = (퐶 cos 휃) cos 푥 + (− 퐶 sin 휃) sin 푥 Let 푎 = 퐶푐표푠(휃) and 푏 = −퐶푠푖푛(휃), then 푎2 + 푏2 = 퐶2(cos 휃)2 + (−퐶)2(sin 휃)2 = 퐶2[(cos 휃)2 + (sin 휃)2] = 퐶2 Therefore, 퐶 = √푎2 + 푏2 −푏 퐶 푠푖푛휃 −푏 and = = 푡푎푛 휃 → 휃 = 푡푎푛−1 푎 퐶 푐표푠휃 푎 Example. Express 푓(푡) = cos(푡) + sin(푡) as a single sinusoid. Solution: a = 1 , b = 1 퐶 = √12 + 12 = √2 −푏 −1 −휋 휃 = tan−1 = tan−1 ( ) = tan−1(−1) = 푎 1 4 휋 푓(푡) = √2 cos (푡 − ) 4 −푏 Caution when evaluating tan−1 ( ) 퐰퐢퐭퐡 풂 < 0 using a calculator: 푎 If you use your calculator to find the inverse tangent and if 푎 < 0, then you must add (or subtract) 휋 to (from) the answer. Example. Express −2 cos(푡) − 3 sin(푡) as a single sinusoid. Here,푎 = −2; 푏 = −3. Then, −푏 3 tan−1 ( ) = tan−1 ( ) = −0.983 radians (calculator answer) 푎 −2 Since 푎 < 0, we need to add (or subtract) 휋, 휃 = −0.983 ± 3.1416 = 2.159 (or − 4.1246) 퐶 = √(−2)2 + (−3)2 = √4 + 9 = √13 푓(푡) = √13 cos(푡 + 2.159); we choose the smaller angle (in absolute value) Matlab and Mathcad have the atan2 function that automatically adjusts the angle. Here is the syntax. Mathcad syntax: atan2(denominator, numerator) Example: atan2(-2,3) = 2.159 Matlab syntax: atan2(numerator, denominator) Example: atan2(3,-2) = 2.159 Your turn: Express the following functions as a single cosine function. Verify your answer by plotting the functions. 휋 푓(푥) = cos(푥) − sin(푥) , 푓(푥) = − cos(푥) + sin(푥) , 푓(푥) = cos(푥) + sin (푥 + ) 3 Complex Numbers Representations: Cartesian, polar, phasor, and exponential z ∈ C, 푗 ≜ √−1 (or 푗2 = −1) Re (z) = a Im (z) = b Cartesian representation: z = a + jb We may represent the complex number in polar form as Polar representation: z = (r, θ) where, r = |z| is the magnitude of z θ = ∠푧 is the angle of z Phasor notation: z = |z| ∠푧 Employing the Pythagorean Theorem we may write, 2 2 푟 = √푎 + 푏 a 푏 푏 tan 휃 = → 휃 = tan−1 ( ) 푎 푎 푎 = 푟 cos 휃 푏 = 푟 sin 휃 Therefore, we have the following relation between the Cartesian and polar representations: z = a + jb = r (cos 휃 + j sin 휃) Euler’s Identity: 푒푗휃 = 푐표푠 휃 + 푗 푠푖푛 휃 Leonhard Euler(1707 – 1783) was a pioneering Swiss mathematician and physicist. (Euler’s Identity will be proved in the next lecture) 푗휋 1 − −휋 −휋 Example: = 푒 2 = cos + 푗 sin = 0 + 푗(−1) = −푗 푗휋 2 2 푒 2 Exponential form of complex numbers: z = a + jb = r (cos 휃 + j sin 휃) = r푒푗θ = |푧|푒푗∠푧 Example. Express z = 2 + j3 in exponential form. 푟 = √22 + 32 = √13 3 휃 = tan−1 ( ) = 0.983 푟푎푑푖푎푛푠 2 z = (√13 , 0.983) = √13푒푗 (0.983) Polar form exponential form or: ∠0.983 in phasor notation √13 Conjugate of a complex number: z = a + jb = r푒푗휃 푧∗ = 푎 − 푗푏 = 푟푒−j휃 Properties of the imaginary unit (푗): 2 푗2 = (√−1) = −1 , 푗3 = 푗푗2 = −푗, 푗4 = (−1)(−1) = +1 , 푒푡푐. 1 1 푗 푗 푗 Also, 푗−1 = = = = = −푗 or, 푗−1 = 푗−1(1) = 푗−1푗4 = 푗3 = −푗 푗 푗 푗 푗2 −1 Algebra of complex numbers z = a + jb z + 푧∗= (a + jb) + (a – jb) = 2a So, we may write: z + 푧∗ = 2 Re{푧} 푧푧∗ = (푎 + 푗푏) ∙ (푎 − 푗푏) = 푎2 − 푗푎푏 + 푗푎푏 − 푗2푏2 = 푎2 + 푏2 = |푧|2 = 푟2 Therefore, we may write: 푧푧∗ = |푧|2 Dividing by the conjugate: 푧 푟푒푗휃 = = 푒푗(휃−(−휃)) = 푒2푗휃 = 1∠2휃 푧∗ 푟푒−푗휃 1 −푗휃 Note that we used the fact that = 푒 푒푗휃 Matlab: Cartesian to polar conversion (built-in functions) Let 푧 = −2 + 푗 Alternative Matlab method: Mathcad: Cartesian to polar conversion: Summary of Matlab/Mathcad/TI-89 Complex number instructions: Assume z = a + jb (j is the imaginary unit i) Mathcad: (use i from the menu or define j as √−1 ) Re(z), Im(z), 푧̅, |z|, arg(z) [also: angle(a,b), atan2(a,b)] Note: Conjugate operation is obtained by holding down the Shift key and then pressing the ” key Matlab: (i is built-in) real(z), imag(z), conj(z), abs(z), angle(z), atan2(b,a), [angle, r] = cart2pol(a,b), [a, b] = pol2cart(angle, r) (angle in radians) TI-89 (syntax is similar to Matlab): real(z), imag(z), conj(z), abs(z), angle(z) (Here, the imaginary unit is i located above the “CATALOG” key) Exponential Form Matlab: Computer Example CB.1, Page 10 of Lathi’s Text. Part(a): Part(b): Reciprocal: Given (z = a + jb) 1 1 1 (푎 − 푗푏) 푎 − 푗푏 푧∗ = = = = 푧 푎 + 푗푏 (푎 + 푗푏)(푎 − 푗푏) 푎2 + 푏2 |푧|2 1 푧∗ ∴ = 푧 |푧|2 Reciprocal with the exponential representation: 1 1 1 푧 = 푟푒푗휃 → = = 푒−푗휃 푧 푟푒푗휃 푟 Division of two complex numbers: 푗휃1 푗휃2 푧1 = 푟1푒 ; 푧2 = 푟2푒 푗휃1 푧1 푟1푒 푟1 푗(휃1− 휃2) = 푗휃 = 푒 푧2 푟2푒 2 푟2 (recall, phasors: 푟1∠휃1 , 푟2∠휃2) 푟1∠휃1 푟1 = ∠(휃1 − 휃2) 푟2∠휃2 푟2 Squaring: 2 푧2 = (푟푒푗휃) = 푟2푒푗2휃 Eulers′ Formula → 푒푗휋 = cos(휋) + 푗푠푖푛(휋) = −1 + 푗0 = −1 푒푗휋 + 1 = 0 The most beautiful formula (theorem) in mathematics! In the fall 1988 issue of Mathematical Intelligencer, a scholarly journal of mathematics (published by Springer-Verlag), there was the call for a vote on the most beautiful theorem in mathematics. The readers, consisting of mostly mathematicians, voted Euler’s formula as the most beautiful. In fact, three out of the five most beautiful theorems of mathematics in that survey were contributed by Euler (they ranked first second and fifth).
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