controlcontrol systemphasetransfer function
I have been trying to find the phase of this transfer function:
G(jω)H(jω)=0.1695/jω(5.88jω+1)
by doing this:
Phase = -arctan(0/0.1695)-arctan(w/-5.88*w^2)
PHASE = 0-arctan(1/-5.88*w)
Is this correct ? since for some reason the result does not make sense.
First simplify the innermost feedback loop.
$$\frac{\frac{1}{s}}{1+\frac{1}{s}\ \frac{R}{L}}=\frac{L}{L s+R}$$
Now simplify the blocks in series.
$$\frac{1}{s}\frac{1}{L}\frac{L}{L s+R}=\frac{1}{s (L s+R)}$$
Simplify the remaining feedback loop.
$$ \frac{\frac{1}{s (L s+R)}}{1+\frac{1}{s (L s+R)}\frac{1}{C}}=\frac{C}{C s (L s+R)+1} $$
Finally things are in series.
$$\frac{1}{C}\frac{C}{C s (L s+R)+1} R=\frac{R}{C L s^2+C R s+1}$$
The phase change due to capacitance is not set in stone for all applied frequencies. For instance at resonance (current is maximum) the phase rapidly changes 180 degrees over a very small band of frequencies.
At higher or lower frequencies (nowhere near resonance) the phase change hardly moves at all unless the quality factor (Q) of the circuit is very small. Altering C at these extremes has very little effect on phase change.
Best Answer
A general transfer function with \$\exists\alpha\space\wedge\space\exists\beta\space\wedge\space\exists\gamma\in\mathbb{R}^+\$ is given by:
$$\mathcal{H}\left(\text{s}\right):=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\alpha}{\text{s}\left(\beta\text{s}+\gamma\right)}\tag1$$
When dealing with sinusoids we know that \$\text{s}:=\text{j}\omega\$ where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$. So we get:
$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{\underline{\text{Y}}\left(\text{j}\omega\right)}{\underline{\text{X}}\left(\text{j}\omega\right)}=\frac{\alpha}{\text{j}\omega\left(\text{j}\omega\beta+\gamma\right)}\tag2$$
Now, we know:
$$\text{j}\omega\left(\text{j}\omega\beta+\gamma\right)=\text{j}\omega\text{j}\omega\beta+\text{j}\omega\gamma=\text{j}\omega\gamma-\omega^2\beta\tag3$$
When dealing with the argument of the transfer function, we get:
$$\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=\arg\left(\frac{\underline{\text{Y}}\left(\text{j}\omega\right)}{\underline{\text{X}}\left(\text{j}\omega\right)}\right)=\arg\left(\underline{\text{Y}}\left(\text{j}\omega\right)\right)-\arg\left(\underline{\text{X}}\left(\text{j}\omega\right)\right)=$$ $$\arg\left(\frac{\alpha}{\text{j}\omega\left(\text{j}\omega\beta+\gamma\right)}\right)=\arg\left(\alpha\right)-\arg\left(\text{j}\omega\left(\text{j}\omega\beta+\gamma\right)\right)=$$ $$\arg\left(\alpha\right)-\arg\left(\text{j}\omega\gamma-\omega^2\beta\right)=0-\left(\frac{\pi}{2}+\arctan\left(\frac{\omega^2\beta}{\omega\gamma}\right)\right)=$$ $$-\frac{\pi}{2}-\arctan\left(\frac{\omega\beta}{\gamma}\right)\tag4$$
In your case we have \$\alpha=0.1695\$, \$\beta=5.88\$ and \$\gamma=1\$. So we get:
$$\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=-\frac{\pi}{2}-\arctan\left(\frac{5.88\omega}{1}\right)=-\frac{\pi}{2}-\arctan\left(5.88\omega\right)\tag5$$