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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 ?

1 Answers 1

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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.

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    Shouldn't $2/(1 + \omega^2) = 0$ ? Because $dH/d\omega = 1/(1 + 4\omega^2/(\omega^2 - 1)^2) = 0$, so $\Rightarrow 2 \omega/(\omega^2 - 1) = \infty \Rightarrow \omega^2 -1 = 0 \Rightarrow \omega = \pm 1$ – 2017-02-14
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    @AnshKumar I do not see what you're doing, but in order to find the maximum and minimum set the derivative of the argument equal to zero. – 2017-02-14
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    I just differentiated $arctan$ which is $1/(1 + \omega^2)$ and equated it to zero and I'm getting $\omega = \pm 1$. Which is giving me solution to equation (7) – 2017-02-14
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    @AnshKumar No, when you do: $$\frac{\text{d}}{\text{d}\omega}\left(\arctan\left(\frac{2\cdot\omega}{\omega^2-1}\right)\right)=\frac{\text{d}}{\text{d}\omega}\left(\arctan\left(\frac{2\cdot\omega}{\omega^2-1}\right)-\pi\right)=-\frac{2}{1+\omega^2}$$ – 2017-02-14
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    Isn't $$\frac{d }{dx}tan^{-1} x = \frac{1}{1 + x^2}$$? – 2017-02-14
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    @AnshKumar Sure it is, but notice that there is a fraction inside the arctan and the top and bottom contain the variable where you have to differentiate with respect to. Look here http://www.wolframalpha.com/input/?i=D%5BArcTan%5B(2*w)%2F(w%5E2-1)%5D,w%5D – 2017-02-14
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    Oh yes, there is my mistake. So, should I claim to the university that the given options are wrong ? – 2017-02-14
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    So the maximum will be when $\omega=0$ then the argument is just $0$ and when $\omega\to-\infty$ then the argument $\to-\pi$ so the answer is answer 4. – 2017-02-14