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.

Question

The Sine and Cosine Functions

Derive the Laplace transforms of the exponentially decaying sine and cosine functions, [latex]e^{−at}[/latex] sin 𝜔t and [latex]e^{−at}[/latex] cos 𝜔t, for t ≥ 0, where a and 𝜔 are constants.

Accepted Answer

Note that from the Euler identity, [latex]e^{j heta}=\cos heta + j \sin heta[/latex], with 𝜃 = 𝜔t, we have

[latex]e^{-at}(\cos \omega t +j \sin \omega t)= e^{-at}e^{j\omega t}=e^{-(a-j \omega)t} \quad (1)[/latex]

Thus, the real part of [latex]e^{-(a-j \omega )t}[/latex] is [latex]e^{−at}[/latex] cos 𝜔t and the imaginary part is [latex]e^{−at}[/latex] sin 𝜔t. However, from the result of Example 2.3.2, with a replaced by a − j𝜔, we have

[latex]\mathcal{L}[e^{-a-j \omega}t]= \frac{1}{s+a-j \omega} \quad (2)[/latex]

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

[latex]s=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/latex]

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 = [latex]\sqrt{-1}[/latex]

Complex conjugate:

[latex]\overset{-}{z} = x − jy[/latex]

Magnitude and angle:

[latex]|z| = \sqrt{x^2+y^2} \quad heta = \angle z = an ^{-1} \frac{y}{x}[/latex] (See Figure 2.1.11)

Polar and exponential representation:

[latex]z = |z| \angle heta = |z| e^{j heta}[/latex]

Equality: If [latex]z_1 = x_1 + jy_1[/latex] and [latex]z_2 = x_2 + jy_2[/latex], then

[latex]z_1 = z_2[/latex] if [latex]x_1 = x_2[/latex] and [latex]y_1 = y_2[/latex]

Addition:

[latex]z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)[/latex]

Multiplication:

[latex]z_1z_2 = |z_1||z_2|∠( heta_1 + heta_2)[/latex]

[latex]z_1z_2 = (x_1x_2− y_1y_2) + j(x_1y_2 + x_2y_1)[/latex]

Complex-conjugate multiplication:

(x + jy)(x − jy) = x² + y²

Division:

[latex]\frac{1}{z}= \frac{1}{x+yj} = \frac{x-jy}{x²+y²}[/latex]

[latex]\frac{z_1}{z_2}=\frac{|z_1|}{|z_2|} \angle ( heta_1- heta_2)[/latex]

[latex]\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}[/latex]

[latex]\frac{1}{x-jy}=\frac{x+jy}{(x-jy)(x+jy)}=\frac{x+jy}{x^2+y^2}[/latex]

Thus, with x = s + a and y = 𝜔, equation (2) becomes

[latex]\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}[/latex]

From equation (1), we see that the real part of this expression is the transform of [latex]e^{−at}[/latex] cos 𝜔t and the imaginary part is the transform of [latex]e^{−at}[/latex] sin 𝜔t. Therefore,

[latex]\mathcal{L}[e^{-at} \cos \omega t]=\frac{s+a}{(s+a)^2+ \omega^2}[/latex]

and

[latex]\mathcal{L}[e^{-at} \sin \omega t]=\frac{\omega}{(s+a)^2+ \omega^2}[/latex]

Note that the transforms of the sine and cosine can be obtained by letting a = 0. Thus,

[latex]\mathcal{L}{(\cos \omega t)}=\frac{s}{s^2+\omega^2} \quad ext{and} \quad \mathcal{L}{(\sin \omega t)}=\frac{\omega}{s^2+\omega^2}[/latex]

Question 2.3.3: The Sine and Cosine Functions Derive the Laplace transforms ...

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