18.03 Differential Equations, Lecture Note 7

18.03 Class 7, Feb 17, 2010 Exponential and Sinusoidal input and output [1] [2] [3] [4] [5] Sinusoidal functions Trig sum formula Integration of complex valued functions Linear equations with sinusoidal input signal Complex replacement Euler: [1] Re e^{(a+bi)t} = e^{at} cos(bt) Im e^{(a+bi)t} = e^{at} sin(bt) Sinusoids A "sinusoidal function" f(t) is one whose graph is a (co)sine wave. I drew a large general sinusoidal function, I drew the graph of f(t) . cos(theta) ; this is our model example of a sinusoid. A sinusoidal function is entirely determined by just three measurements, or parameters, which determine it in terms of cos(theta) . A = Amplitude = height of its maxima = depth of its minima P = Period = elapsed time till it repeats (or, in spatial terms, lambda = wavelength = the distance between repeats) t_0 = Time lag = time of first maximum f(t) can be written in terms of cosine. Clearly, f(t) = A cos(theta). To work out how, express theta as a function of t . I started drew a t-axis horizontally and a theta-axis vertically. When t = t_0 , theta = 0 . When t = t_0 + P , theta = 2pi . I marked these data points on all three graphs. The graph of theta as a function of t is a straight line; otherwise the cosine would get distorted, bunched up. So: theta = (2pi/P) ( t - t_0 ) and so f(t) = A cos( (2pi/P) ( t - t_0 )) The frequency is nu = the circular frequency So 1/P , measured in "cycles/unit time." More useful is: omega = 2pi/P = omega , measured in radians/unit time. theta = omega ( t - t_0 ) The "phase lag" is phi = omega t_0 = (2 pi/P) t_0 . It measures the radian measure corresponding to In terms of omega and phi , t = 0 . f(t) = A cos(omega t - phi) For example, sin(omega t) = cos(omega t - pi/2) . -- the sine lags one quarter cycle behind the cosine. Question 7.1. A graph of a sinusoidal function is displayed. The problem is to express it in the "standard form" above. 1. 2. 3. 4. 5. 6. 2 2 2 2 2 2 cos(4pi t + cos((pi/4)t cos(4pi t cos((pi/4)t cos(4t+1) cos(4t-1) pi/4) + pi/4) pi/4) - pi/4) P = 8, t_0 = -1 , A = 2 : f(t) = 2 cos( (2pi/8) ( t + 1 )) = 2 cos( (pi/4) t + pi/4 ) omega = pi/4 , phi = - pi/4 . Ans: 2. [2] Trig sum: a cos(omega t) + b sin(omega t) I showed the Trig Id applet. This sum seems always to be another sinusoidal function! How can we find its "standard form" A cos(omega t - phi) ? Recall the cosine difference identity: cos(theta - phi) = cos(phi) cos(theta) + sin(phi) sin(theta) cos(omega t - phi) = cos(phi) cos(omega t) + sin(phi) sin(omega t) Now construct a right triangle with hypotenuse the segment in the plane joining (0,0) to (a,b) . If A is the hypotenuse and phi the angle at the origin, then A = sqrt(a^2 + b^2) a = A cos(phi) b = A sin(phi) and A cos(omega t - phi) = a cos(omega t) + b sin(omega t) Question 7.2. What are the amplitude, circular frequency, and phase lag: A , omega , and phi in A cos(omega t - phi) , for the sinusoid cos(omega t) + sqrt(3) sin(omega t) 1. 2 cos(\omega t-\frac{\pi}{4}) 2. sqrt(3) cos(omega (t - pi/3)) 3. 2 cos(omega(t- pi/3)) 4. 2 cos(omega t - pi/3) 5. sqrt(3) cos(omega t - pi/3) 6. sqrt(3) cos(omega t - pi/4) Blank. Don't know. Ans: A = 2, phi = pi/3 : 4. [3] Integration Remember how to integrate Use parts twice. e^{2t} cos(t) ? Or: Differentiating a complex valued function is done separately on the real and imaginary parts. Same for integrating. e^{2t} cos(t) = Re e^{(2+i)t} so int e^{2t} cos(t) dt = Re int e^{(2+i)t} dt and we can integrate exponentials because we know how to differentiate them! int e^{(2+i)t} dt = (1/(2+i)) e^{(2+i)t} + c We need the real part. Expand everything out: 1/(2+i) = (2-i)/5 e^{(2+i)t} = e^{2t} (cos(t) + i sin(t)) so the real part of the product is (1/5) e^{2t} (2 cos(t) + sin(t)) + c More direct than the high school method! [4] Linear constant coefficient ODEs with exponential input signal Let's try x' + 2x = 4 e^{3t} We could use our integrating factor, but instead let's use the method of "optimism," or the inspired guess. The inspiration here is based on the fact that differentiation reproduces exponentials: d -- e^{rt} = r e^{rt} dt Since the right hand side is an exponential, maybe the output signal will be too: TRY x = A e^{3t} . This is not going to be the general x solution, so I'll write 2 x_p x_p for it. I don't know what A is yet, but: = 2 A e^{3t} x_p' = A 3 e^{3t} ----------------4 e^{3t} = A (3+2) e^{3t} which is OK as long as A = 4/5: x_p = (4/5) e^{3t} The general solution is this plus a transient: is one solution. x = (4/5) e^{3t} + c e^{-2t} . [6] Replacing sinusoidal signals with exponential ones Let'e go back to the original ODE x' + 2x = 2 cos(t) This equation is the real part of a complex valued ODE: z' + 2z = 2 e^{it} This is a different ODE, and I use a different variable name, We just saw how to get an exponential solution: z(t) . z_p = A e^{it} 2 zp = 2 A e^{it} zp' = i A e^{it} ----------------------2 e^{it} = A (2+i) e^{it} so z_p = 2/(i+2) e^{it} To get a solution to the original equation we should take the real part of this! Expand each factor in real and imaginary parts: z_p = (2(2-i)/5) ( cos(t) + i sin(t) ) x_p = Re(z_p) = (4/5) cos(t) + (2/5) sin(t) This is the only sinusoidal solution. To get the general solution we add a transient: x = x_p + c x_h