Question 1.2: a) Using the complex exponential representation, evaluate th...
a) Using the complex exponential representation, evaluate the sum of the simple harmonic functions u_1 = A \cos (ωt) \text {and} u_2 = B \sin(ωt) . b) Derive the same result using the phasor diagram.
a) The functions u_1 and u_2 are respectively the real parts of \underline{u}_1=A e^{iωt} \text { and } \underline{u}_2 =B e^{i(ωt-\pi /2)} =-iBe^{iωt} . Their sum is \underline{u} =\underline{u}_1 + \underline{u}_2 = \underline{C} e^{iωt} ,where the complex amplitude is \underline{C} = A – iB which we may cast in the form C e^{i\phi } with
C=\sqrt{A^2+B^2} \ , \ \ \ \ \ \ \ \cos\phi =A/C \ \ \ \ \ \text {and} \ \ \ \ \ \ \sin \phi = -B/C .
The complex function u can then be written as \underline{u} =Ce^{i(ωt+\phi )} and, by taking its real part, we get u = C cos(ωt + Φ).
b) The functions u_1 \text {and} u_2 are respectively represented by the phasors \overrightarrow{u_1} and \overrightarrow{u_2} of lengths A and B, and making the angles ωt and (ωt – π/2) with the x-axis (Figure 1.4a). Thus \overrightarrow{u_1} is orthogonal to \overrightarrow{u_2} . Their vector sum u is the hypotenuse of a right triangle. Pythagoras’ theorem gives its magnitude C = \sqrt{A^2+B^2} The phasor \overrightarrow{u} along with \overrightarrow{u_1} makes an angle Φ such that tan Φ = -B/A. This angle is equal to the phase lead of u over u_{1}. For instance, if A and B are positive, Φ lies between -π/2 and 0; then u effectively has a phase lag over u_{1}. As u_{1} and u_{2} have the same frequency, we may draw the phasor diagram at t = 0 as can be seen in Figure 1.4b.
