18.03SC Practice Problems 7 Sinusoids Solution Suggestions 1. Write each of the following functions (of t) in the form A cos(ωt − φ). In each case, begin by drawing a right triangle with sides a and b. (a) cos(2t) + sin(2t). √ (b) cos(πt) − 3 sin(πt). eit (c) Re . 2 + 2i Recall the geometric derivation of the general sinusoid formula −− −−−−−−−−−−−−→ � a cos(ωt) + b sin(ωt) = ( a, b) · (cos(ωt), sin(ωt)) = a2 + b2 cos(ωt − φ), where φ is the angle of the ﬁrst vector. This is where the triangle comes in - we will draw the vector of coefﬁcients to determine its magnitude and direction. (Equiva lently, we are determining the polar coordinates of the complex number a + bi.) (a) cos(2t) + sin(2t): ( a, b) = (1, 1). √ √ √ Here, the right triangle has hypotenuse 1 + 1 = 2, so A = 2. Both summands have angular frequency √ 2, so ω = 2. φ is the angle of the triangle, which is π/4, so cos(2t) + sin(2t) = 2 cos(2t − π/4). √ √ (b) cos(πt) − 3 sin(πt) : ( a, b) = (1, − 3). � √ This right triangle has hypotenuse 12 + (− 3)2 = 2 and angle −π/3. So cos(πt) − √ 3 sin(πt) = 2 cos(πt − (−π/3)) = 2 cos(πt + π/3). (c) eit = cos(t) + i sin(t) and 1 2+2i = 1 2+2i take the real part of the product to obtain 2−2i 2 − 2i = eit Re 2+ 2i = · 2 − 2i i = 1− 4 . Multiply 22 +22 1 1 4 cos( t ) + 4 sin( t ). out and For 14 cos(t) + 14 sin(t), ( a, b) = ( 14 , 14 ), which gives the same triangle as in (a), except scaled by 1/4. it e So Re 2+ 2i = 1 4 cos(t) + 14 sin(t) = √ 2 4 cos(t − π/4).

18.03SC Practice Problems 7: Sinusoids

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