Polar plot for $\frac{j\omega}{1 + j\omega \tau}$

Question

Okay I'm from electrical engineering background and we represent imaginary number as $j$. I'm supposed to draw polar plot for transfer function given as $$F(j\omega) = \frac{j\omega}{1 + j\omega \tau}$$ as $\omega$ varies from zero to infinity, and $\tau$ is constant. I tried calculating magnitude and phase for $\omega = 0$, which both comes out to be zero. Was I right? And I'm having lots of trouble calculating it for $\omega = \infty$, can anyone guide me pls.

Answer 1

Since you are looking at large $\omega$, just divide both numerator and denominator by $j\omega$. What you get is $$F(j\omega)=\frac{1}{\tau+\frac{1}{j\omega}}$$ For large $\omega$, the $1/(j\omega)$ part goes to zero (imaginary 0, but still 0). So $$\lim_{\omega\rightarrow\infty}F(j\omega)=\frac{1}{\tau}$$

In general, you can write $$F(j\omega)=\frac{j\omega}{1+j\omega\tau}=\frac{j\omega(1-j\omega\tau)}{(1+j\omega\tau)(1-j\omega\tau)}\\=\frac{\omega^2\tau+j\omega}{1+\omega^2\tau^2}\\=\frac{\omega^2\tau}{1+\omega^2\tau^2}+j\frac{\omega}{1+\omega^2\tau^2}$$ The magnitude is then $$|F(j\omega)|=\frac{\omega\sqrt{1+\omega^2\tau^2}}{1+\omega^2\tau^2}=\frac{\omega}{\sqrt{1+\omega^2\tau^2}}$$ The phase is given by $$\tan\psi=\frac{\omega}{\omega^2\tau}=\frac{1}{\omega\tau}$$