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Lecture 22

Lecture 21 | Lecture 23

Response to Everlasting Exponentials

Suppose we have a linear time-invariant system which has forcing function tex:$$e^{\omega t}$$ where tex:$$\omega$$ is a frequency

and a transfer function is tex:$$H(s)$$.

Normally, we would have to calculate the Laplace transform of the forcing function, in this case, tex:$$e^{\omega t} \leftrightarrow \frac{1}{s - \omega}$$ and then calculate the inverse Laplace transform of tex:$$\frac{H(s)}{s - \omega}$$; however, note the following:

Assume we have the impulse response tex:$$h(t)$$. In this case, the response to an exponential function is tex:$$h(t) * e^{\omega t} = \int_0^\infty h(\tau) e^{\omega(t - \tau)} d\tau = e^{\omega t} \int_0^\infty h(\tau) e^{-\omega \tau} d\tau$$. But this integral is nothing more than the Laplace transform of the function tex:$$h(t)$$ and by definition, the Laplace transform of the impulse function is the transfer function. Therefore, we have the very important result

tex:$$h(t) * e^{\omega t} =  H(\omega) e^{\omega t}$$.

The response of a sum of exponential functions tex:$$\sum_{k = 0}^\infty e^{\omega_k t}$$ is the sum of the responses to the exponential functions and the response of each exponential function is the exponential multiplied by the transfer function evaluated at the frequency: tex:$$\sum_{k = 0}^\infty H(\omega_k) e^{\omega_k t}$$.


What is the response of the circuit with transfer function tex:$$H(s) = \frac{1}{s + 1}$$ when the input/forcing function is

  • tex:$$x(t) = 2.321e^{-4t}$$
  • tex:$$x(t) = e^{-t}$$
  • tex:$$x(t) = 7.9 + 4.5e^{-2t} - 6.1e^{-3t}$$
  • tex:$$x(t) = cos(2t)$$

In the first case, the response is tex:$$y(t) = H(-4) 2.321e^{-4t} = \frac{1}{-4 + 1} 2.325e^{-4t} = -0.775e^{-4t}$$.

In the second case, the transfer function is undefined: tex:$$H(-1) = \frac{1}{0}$$. This is the resonant frequency of the system and such a system would blow up.

In the third case, we may consider tex:$$7.9 = 7.1e^{0t}$$ and therefore the response is tex:$$H(0) 7.9 + 4.5 H(-2)e^{-2t} - 6.1H(-3)e^{-3}  = 7.9 - 4.5e^{-2t} + 3.05e^{-3t}$$ .

In the fourth case, we note that this function is the real part of the complex exponential tex:$$e^{j2t}$$. The response of the complex exponential is tex:$$H(j2)e^{j2t} = (0.2 - j0.4)e^{j2t}$$. The real part of the response is tex:$$y(t) = 0.2cos(2t) + 0.4\sin(2t)$$.