Question 2.3.3: The Sine and Cosine Functions Derive the Laplace transforms ...
The Sine and Cosine Functions
Derive the Laplace transforms of the exponentially decaying sine and cosine functions, e^{−at} sin 𝜔t and e^{−at} cos 𝜔t, for t ≥ 0, where a and 𝜔 are constants.
Note that from the Euler identity, e^{j\theta}=\cos \theta + j \sin \theta, with 𝜃 = 𝜔t, we have
e^{-at}(\cos \omega t +j \sin \omega t)= e^{-at}e^{j\omega t}=e^{-(a-j \omega)t} \quad (1)
Thus, the real part of e^{-(a-j \omega )t} is e^{−at} cos 𝜔t and the imaginary part is e^{−at} sin 𝜔t. However, from the result of Example 2.3.2, with a replaced by a − j𝜔, we have
\mathcal{L}[e^{-a-j \omega}t]= \frac{1}{s+a-j \omega} \quad (2)
In this form, we cannot identify the real and imaginary parts. To do so, we multiply the numerator and denominator by the complex conjugate of the denominator and use the fact that (x − jy)(x + jy) = x² + y²
(see Table 2.1.3); that is,
| Table 2.1.3 Roots and complex numbers. |
| The quadratic formula |
| The roots of as² + bs + c = 0 are given by |
| s=\frac{-b \pm \sqrt{b^2-4ac}}{2a} |
| For complex roots, s = −𝜎 ± j𝜔, the quadratic can be expressed as |
| as² + bs + c = a [ (s +𝜎)² + 𝜔²] = 0 |
| Complex numbers |
| Rectangular representation: |
| z = x + jy, j = \sqrt{-1} |
| Complex conjugate: |
| \overset{-}{z} = x − jy |
| Magnitude and angle: |
| |z| = \sqrt{x^2+y^2} \quad \theta = \angle z = \tan ^{-1} \frac{y}{x} (See Figure 2.1.11) |
| Polar and exponential representation: |
| z = |z| \angle \theta = |z| e^{j \theta} |
| Equality: If z_1 = x_1 + jy_1 and z_2 = x_2 + jy_2, then |
| z_1 = z_2 if x_1 = x_2 and y_1 = y_2 |
| Addition: |
| z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2) |
| Multiplication: |
| z_1z_2 = |z_1||z_2|∠(\theta_1 + \theta_2) |
| z_1z_2 = (x_1x_2− y_1y_2) + j(x_1y_2 + x_2y_1) |
| Complex-conjugate multiplication: |
| (x + jy)(x − jy) = x² + y² |
| Division: |
| \frac{1}{z}= \frac{1}{x+yj} = \frac{x-jy}{x²+y²} |
| \frac{z_1}{z_2}=\frac{|z_1|}{|z_2|} \angle (\theta_1-\theta_2) |
| \frac{z_1}{z_2}=\frac{x_1+jy_1}{x_2+jy_2}=\frac{x_1+jy_1}{x_2+jy_2} \frac{x_2-jy_2}{x_2-jy_2}=\frac{(x_1+jy_1)(x_2-jy_2)}{x_2^2+y_2^2} |
\frac{1}{x-jy}=\frac{x+jy}{(x-jy)(x+jy)}=\frac{x+jy}{x^2+y^2}
Thus, with x = s + a and y = 𝜔, equation (2) becomes
\mathcal{L}[e^{-(a-j \omega)t}]= \frac{1}{s+a-j \omega}= \frac{s+a+j \omega}{(s+a-j \omega)(s+a+j \omega)} \\ =\frac{s+a+j \omega}{(s+a)^2+\omega^2}= \frac{s+a}{(s+a)^2+\omega^2}+j\frac{\omega}{(s+a)^2+\omega^2}
From equation (1), we see that the real part of this expression is the transform of e^{−at} cos 𝜔t and the imaginary part is the transform of e^{−at} sin 𝜔t. Therefore,
\mathcal{L}[e^{-at} \cos \omega t]=\frac{s+a}{(s+a)^2+ \omega^2}
and
\mathcal{L}[e^{-at} \sin \omega t]=\frac{\omega}{(s+a)^2+ \omega^2}
Note that the transforms of the sine and cosine can be obtained by letting a = 0. Thus,
\mathcal{L}{(\cos \omega t)}=\frac{s}{s^2+\omega^2} \quad \text{and} \quad \mathcal{L}{(\sin \omega t)}=\frac{\omega}{s^2+\omega^2}
