Maximum and minimum value

Question

The transfer function of a system is given by $$ \frac{V_o(s)}{V_i(s)} = \frac{1-s}{1+s} $$ Let the output of the system be $v_o(t)= V_m \sin(\omega t + \phi)$ for the input $v_i(t) = V_m \sin(\omega t)$. Then the minimum and maximum values of $ \phi$(in radians) are respectively

1. $ \frac{-\pi}{2}$ and $\frac{\pi}{2}$
2. $ \frac{-\pi}{2}$ and $0$
3. $ 0$ and $\frac{\pi}{2}$
4. $ {-\pi}$ and $0$

I'm getting angle of transfer function as $$ \angle H(j \omega) = -2 \tan^{-1} \omega$$ So minimum and maximum value must be $ -\pi $ to $ \pi$. But none of the option matches. So, should I consider option (4) as the right answer. But could the maximum value of phase difference be 0 ?

Answer 1

Well, when we use that:

$$\text{s}=\omega\text{j}\tag1$$

Where $\text{j}^2=-1$

We get for the argument of the transfer function:

$$\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=\arg\left(\frac{1-\omega\text{j}}{1+\omega\text{j}}\right)=\arg\left(\frac{1-\omega^2}{1+\omega^2}-\frac{2\cdot\omega}{1+\omega^2}\cdot\text{j}\right)\tag2$$

Now, when $\omega\in\mathbb{R}^+$ we get that:

1. The real part of the transfer function: $\Re\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=\frac{1-\omega^2}{1+\omega^2}$ can be positive, $0$ or negative.
2. The imaginary part will always be negative:$$\Im\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=-\frac{2\cdot\omega}{1+\omega^2}<0\tag3$$

So, for the argument we get that:

1. Situation 1, when the real part is positive (that happens when $\omega<1$): $$\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=\arctan\left(\frac{-\frac{2\cdot\omega}{1+\omega^2}}{\frac{1-\omega^2}{1+\omega^2}}\right)=\arctan\left(\frac{2\cdot\omega}{\omega^2-1}\right)\tag4$$
2. Situation 2, when the real part is $0$ (that happens when $\omega=1$): $$\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=-\frac{\pi}{2}\space\text{radians}=\frac{3\pi}{2}\space\text{radians}\tag5$$
3. Situation 3, when the real part is negative (that happens when $\omega>1$): $$\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=\arctan\left(\frac{-\frac{2\cdot\omega}{1+\omega^2}}{\frac{1-\omega^2}{1+\omega^2}}\right)-\pi=\arctan\left(\frac{2\cdot\omega}{\omega^2-1}\right)-\pi\tag6$$

Now, in order to find the maximum and minimum of the first and third situation:

$$\frac{\text{d}}{\text{d}\omega}\arg\left(\underline{\mathcal{H}}\left(\omega\text{j}\right)\right)=0\space\Longleftrightarrow\space-\frac{2}{1+\omega^2}=0\tag7$$

And there is no solution to that equation.