Convert from complex exponentials to sinusoids

Question

I'm working through some notes on signals and systems, and got stuck trying to fill in the missing steps in converting the left hand side to the right hand side of the following equality:

$$ \alpha_i v_i e^{\lambda_i t} + \alpha_i^* v_i^* e^{\lambda_i^* t} = K_i e^{\sigma_i t} \left[ u_i \cos(\omega_i t + \theta_i) - w_i \sin(\omega_i t + \theta_i) \right] $$ where $$ \lambda_i = \sigma_i + j \omega_i \mbox{ and } \lambda_i^* = \sigma_i - j \omega_i \\ v_i = u_i + j w_i \mbox{ and } v_i^* = u_i - j w_i \\ $$ I was trying with Euler's formula, but got hung up with combining terms. Any help is appreciated!

Answer 1

Here's the nice trick that's used on Feynman's lectures to derive an analogous formula. The keyword is "factor out the average frequency". That is, write

$$ \alpha_i v_i\exp(\lambda_i t) + \alpha_i^\star v_i^\star\exp (\lambda_i^\star t) = \exp(\frac{\lambda_i+\lambda_i^\star}{2}t) \left(\alpha_iv_i \exp(\frac{\lambda_i -\lambda_i^\star}{2}t) + \alpha_i^\star v_i^\star \exp(-\frac{\lambda_i -\lambda_i^\star}{2}t) \right) $$ Now use the fact that $w+w^\star=2\Re w$.

Answer 2

Try using $$ \cos z = \frac{e^{iz}+e^{-iz}}{2} \quad \sin z = \frac{e^{iz}-e^{-iz}}{2i} $$

Answer 3

Thanks Giuseppe Negro and mvw for your suggestions. Using your tips, I've been able to work through the steps. Here is my work:

\begin{equation} \lambda_i = \sigma_i + j\omega_i, \lambda_i^* = \sigma_i - j\omega_i \end{equation}

\begin{equation} \mathbf{v}_i = \mathbf{u}_i + j\mathbf{w}_i, \mathbf{v}_i^* = \mathbf{u}_i - j\mathbf{w}_i \end{equation}

\begin{equation} \alpha_i\mathbf{v}_ie^{\lambda_it} + \alpha_i^*\mathbf{v}_i^*e^{\lambda_i^*t} = \mathbf{K}_ie^{\sigma_it}[\mathbf{u}_icos(\omega_it+\theta_i) - \mathbf{w}_isin(\omega_it+\theta_i)] \end{equation}

How do we translate from the weighted sum of complex conjugate eigenvectors to the sinusoidal representation on the right? First, as Giuseppe pointed out, utilize the trick from the Feynman lectures of factoring out the "average frequency".

\begin{eqnarray} (\alpha_i\mathbf{v}_ie^{\lambda_it} + \alpha_i^*\mathbf{v}_i^*e^{\lambda_i^*t}) \cdot \frac{e^{\frac{1}{2}(\lambda_i + \lambda_i^*)t}}{e^{\frac{1}{2}(\lambda_i + \lambda_i^*)t}} = e^{\frac{1}{2}(\lambda_i + \lambda_i^*)t}[\alpha_i\mathbf{v}_ie^{\frac{1}{2}(\lambda_i - \lambda_i^*)t} + \alpha_i^*\mathbf{v}_i^*e^{-\frac{1}{2}(\lambda_i - \lambda_i^*)t}] \nonumber \end{eqnarray}

Next, utilize the definition of $\lambda_i$ and $\lambda_i^*$ and simplify

\begin{equation} = e^{\sigma_it}[\alpha_i\mathbf{v}_ie^{j\omega_it} + \alpha_i^*\mathbf{v}_i^*e^{-j\omega_it}] \end{equation}

Next, plug in definition of $\mathbf{v}_i$ and $\mathbf{v}_i^*$ and group terms by $\mathbf{u}_i$ and $\mathbf{w}_i$

\begin{equation} = e^{\sigma_it}[\mathbf{u}_i(\alpha_ie^{j\omega_it} + \alpha_i^*e^{-j\omega_it}) + j\mathbf{w}_i(\alpha_ie^{j\omega_it} - \alpha_i^*e^{-j\omega_it})] \end{equation}

Re-write the complex numbers $\alpha_i$ and $\alpha_i^*$ in terms of magnitude and angle: \begin{equation} \alpha_i = |\alpha_i|(cos\theta_i + jsin\theta_i), \alpha_i^* = |\alpha_i|(cos\theta_i - jsin\theta_i) \end{equation}

Now, plug this in, factor out$|\alpha_i|$, and group by sine and cosine terms

\begin{eqnarray} = |\alpha_i|e^{\sigma_it}[\mathbf{u}_i(cos\theta_i(e^{j\omega_it} + e^{-j\omega_it}) + jsin\theta_i(e^{j\omega_it} - e^{-j\omega_it})) + \nonumber\\ j\mathbf{w}_i(cos\theta_i(e^{j\omega_it} - e^{-j\omega_it}) + jsin\theta_i(e^{j\omega_it} + e^{-j\omega_it}))] \nonumber \end{eqnarray}

Next, use Euler's formula to convert the exponential pairs: \begin{equation} cos\phi = \frac{1}{2}(e^{j\phi} + e^{-j\phi}), sin\phi = \frac{1}{2j}(e^{j\phi} - e^{-j\phi}) \end{equation}

\begin{eqnarray} = |\alpha_i|e^{\sigma_it}[\mathbf{u}_i(2cos(\omega_it) \cdot cos \theta_i - 2sin(\omega_it) \cdot sin\theta_i) - \nonumber\\ \mathbf{w}_i(2sin(\omega_it) \cdot cos\theta_i + 2cos(\omega_it) \cdot sin\theta_i)] \nonumber \end{eqnarray}

Finally, use trig identities to convert products of sin, cos terms into sums of sin, cos terms plus phase offset: \begin{eqnarray} 2 cos\theta cos\phi = cos(\theta - \phi) + cos(\theta + \phi) \\ 2sin\theta sin\phi = cos(\theta-\phi) - cos(\theta+\phi) \\ 2sin\theta cos\phi = sin(\theta + \phi) + sin(\theta - \phi) \\ 2cos\theta sin\phi = sin(\theta + \phi) - sin(\theta - \phi) \end{eqnarray}

\begin{equation} = 2|\alpha_i|e^{\sigma_it}[\mathbf{u}_icos(\omega_it + \theta_i) - \mathbf{w}_isin(\omega_it + \theta_i)] \end{equation} Let $K_i = 2|\alpha_i|$, and put it all together:

\begin{equation} \alpha_i\mathbf{v}_ie^{\lambda_it} + \alpha_i^*\mathbf{v}_i^*e^{\lambda_i^*t} = \mathbf{K}_ie^{\sigma_it}[\mathbf{u}_icos(\omega_it+\theta_i) - \mathbf{w}_isin(\omega_it+\theta_i)] \end{equation}