High pass RL circuit angular frequency

Question

I am answering this question as shown in the image.

I had no trouble deriving \$\frac{V_{out}}{V_{in}}\$ in the first question(answer is \$\frac{j\omega L}{R + j\omega L}\$

I also had no trouble with doing part b). I just plugged in \$\omega\$=0 into the equation derived in a), and got zero.

I am having trouble with c). When I plug in \$\omega\$=0 into the equation derived, I get \$\frac{V_{out}}{V_{in}}=0\$, and thus the angle of \$\frac{V_{out}}{V_{in}}\$ is zero... However, this is wrong. I then tried to multiply the equation derived by the complex conjugate, to which I then separated the real and imaginary terms, however, I still got the \$\frac{V_{out}}{V_{in}}\$ is 0... I have found that the angle of \$\frac{V_{out}}{V_{in}}\$ when \$\omega\$=0 is 90^o, but I do not understand how this is, because when I substitute this value in the equation it does not make any sense to me how the angle is 90^o...

Thank you.

PS. When multiplying the equation by the complex conjugate, I got:

$$\frac{\omega^2 L^2}{R^2 + \omega^2 L^2}+j\frac{\omega RL}{R^2 + \omega^2 L^2}$$

Answer 1

Despite the TF being zero at 0 Hz there is still a phase angle. At 0 Hz, the current taken from the input is wholly determined by R. This is because XL, at 0 Hz, is 0 ohms. Hence the current taken from the input is Vin/R. In other words, the current through the resistor and the inductor is in phase with the input voltage.

In an inductor, the voltage leads the current by 90 degrees hence, despite the voltage being zero at 0 Hz, there is a leading phase angle of 90 degrees. If you plotted a graph for phase angle versus frequency, as frequency got lower and lower you would see the output voltage tending to have a leading 90 degrees phase angle relative to the input voltage.

If you go to this site you can model an RL high-pass filter to confirm your findings. The picture below is for 1 ohm and 1 henry: -

Answer 2

I have found that the angle of \$\frac{Vout}{Vin}\$ when ω=0 is \$90^o\$, but I do not understand how this is.

Take the limit of the phase angle of the TF for ω going to zero from a positive value. You'll find \$\pi / 2\$.

Answer 3

You're overcomplicating. You have already found out the transfer function:

$$H(s)=\frac{sL}{sL+R}=\frac{s}{s+\frac{R}{L}}$$

and all you have to do now is to apply the phase formula:

$$\Phi(\omega)=\arctan\frac{\Im(H(s))}{\Re(H(s))}$$

whose real and imaginary parts you also correctly derived:

$$\Phi(\omega)=\arctan\left(\frac{\omega RL}{\omega^2L^2+R^2}\frac{\omega^2L^2+R^2}{\omega^2L^2}\right)=\arctan\left(\frac{R}{\omega L}\right)$$

and, since you're talking a four quadrant response, and you have to evaluate it at \$\omega\$=0, it reduces to:

$$\arctan2(\tau, 0)=\frac{\pi}{2}$$

where \$\tau\$ is whatever time constant you have there. That's what I hinted at in the comment.