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I know that, at steady state, the frequency response can be calculated relatively easily from the transfer function and the frequency of the input.

So if we have a system described by the transfer function \$ G(s) \$ then we have (again, at steady state): $$ u(t) = u_0 \sin(\omega t) \\ Y(s) = G(s) \cdot U(s) \\ |G(j\omega)|=\frac{y_0}{u_0} \\ \angle G(j\omega) = \phi \\ \implies y(t)=|G(j\omega)|u_0 \sin(\omega t+\phi)\\ $$

I never saw anywhere any comments in the many text books I read about a signal with a phase as an input.

Say I have this \$ u(t) \$ (the excitation) instead of the sine with a zero phase: $$ u(t)=u_0 \sin(\omega t + \theta) $$

I wonder what will happen. Does the relationships mentioned above still hold true, or not. How can I calculate \$ \phi \$ and \$ y_0/u_0 \$ while accounting for the input phase \$ \theta \$?

I would have difficulty to believe the presence of \$ \theta \$ would not have any effect on the steady state output...

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    \$\begingroup\$ The first thing came into my mind was "Circuits are time invariant systems whatever shape was the output for the sin wave as an input will be be the same but shifted when the input is a shifted sin wave" but i`m not sure if this is a correct answer \$\endgroup\$ – Elbehery Nov 1 '16 at 11:50
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Your input is a pure sine wave. Let's call it $$x(t)=A\sin(\omega t)$$

Lets further say that input \$x(t)\$ will produce output \$y(t)\$.

Now, let's change the input to $$x^{'}(t)=A\sin(\omega t +\phi)$$ Because this is a pure sine wave of the same frequency of \$x(t)\$, you can represent the new input as a time delay of the original input. Thus, $$x^{'}(t) = x(t-t_{0})$$

Time invariance now says that the output will be a time delayed version of the original output: $$y^{'}(t) = y(t-t_{0})$$

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  • \$\begingroup\$ So in other words, \$y(t) = |G(j\omega)|u_0sin(\omega t + \angle G(j\omega) + \theta)\$ ? I just want to be crystal certain about it so I can get on with my life. I'm indeed talking about LTI systems. So in other words, if the input is shifted by \$ \theta \$, the response will be equally shifted by \$ \theta \$? Correct? \$\endgroup\$ – Yannick Nov 1 '16 at 13:02
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    \$\begingroup\$ @Yannick, yes, this is how it works. \$\endgroup\$ – Scott Seidman Nov 1 '16 at 13:35
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The phase angle ϕ at the output must be considered as an additional phase shift (caused by the transfer function) if compared with the input phase θ. That´s all. For convenience, it is common practice to set set θ=0. Remember: The input phase is an arbitrary value referenced to an unknown signal phase "x". When you reference the ouput signal to the same unknown phase "x" signal, the introduced phase shift ϕ remains the same.

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