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The input, \$x(t)\$, and output, \$y(t)\$ of a linear time invariant, dissipative system are related via the differential equation

$$d^4(y(t)) + 6d^3(y(t)) + 14d^2(y(t))+14d(y(t)) + 5y(t) = - d^2(x(t)) - 9x(t)$$

If \$x(t) = \cos(t)-\sin(t)\$, determine the steady state response.

Now, I know

$$H(iw) = \frac{-(iw)^2 - 9}{(iw)^4 + 6(iw)^3 + 14(iw)^2 + 14(iw) + 5}$$

But I'm having a hard time understanding how this transfer function will help find the steady state response. Especially since there is no circuit given.

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2 Answers 2

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If a sinusoidal signal is applied to an LTI system, the output will be having same frequency as input but will have different phase and amplitude. If the input is \$\cos(wt)\$, then output will be: $$|H(jw)|\cos\left(wt+\angle H(jw)\right)$$ Where \$H(jw)\$ is the frequency response. \$\angle H(jw)\$ is the argument of \$H(jw)\$.

In your case, the input is sum of two sinusoids with \$w=1\$. Find the output for each one after evaluating magnitude and phase of \$H(jw)\$ at \$w=1\$ and add them to get the result.

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Use \$\small j\$ instead of \$\small i\$. \$\small j\$ is used in engineering to avoid confusion with current.

Express the TF as a complex number: \$\small A+jB\$, then the modulus and argument give the steady-state frequency domain gain and phase angle as functions of frequency, \$\omega\: \small rad \:s^{-1}\$

To find the response to \$\small cos(t)-sin(t)\$, you could use the double angle representation: \$\small cos(t)-sin(t)=\sqrt2\:sin(t+3\pi/4)\$. Hence, multiply the gain, calculated from the above analysis, by \$\small \sqrt2\$ and add \$\small 135^o\$ to the phase angle.

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