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Not sure where to start on this one:

$$H(s)={(s-j\omega_0)(s+j\omega_0)\over(s+\omega_0\cos\theta+j\omega_0\sin\theta)\left(s+\omega_0\cos\theta-j\omega_0\sin\theta\right)}$$

Sketch the frequency response for $\theta=60^o, 75^o, 87^o$

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  • $\begingroup$ I'm not sure of what I am going to say, but maybe it will help: looks like you have a division of Laplace transforms. Shifting the frequency response seems like shifting the Laplace transforms and using relevant properties. $\endgroup$ Commented Jan 21, 2014 at 18:35

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The frequency response is the behavior of $|H(i\omega)|$ (amplitude response) and $\angle H(i\omega)$ (phase response) with respect to $\omega$.

You may observe that the results are dependent on $\omega/\omega_0$ which is especially handy since you dont know $w_0$. So assume $w_0=1$ evaluate the above quantities at several $\omega$ values and plot them in graphs with $\omega/\omega_0$ as x-axis.

Also note that for these kind of plots a log-log (for $|H(i\omega)|$) or a log-linear (for $\angle H(i\omega)$) are usually preferred (you will see why, when you plot it).

The interesting variation will be probably from $\omega/\omega_0 = 0.01 ... 100$.

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