I need to plot this transfer function
When plotting the phase vs frequency graph, I am having problems to calculate the change in phase between 10 and 10^1.5. To my understanding, at frequency=10 the phase is -90°, then, there is a phase change and the slope should increase at a rate of 45°/dec but there is a change in frequency at 10^1.5. To calculate the degrees until the curve reaches 10^1.5 I applied simple trigonometry so that from frequency=10 to 10^1.5 the curve has moved up 10.80°. However, that answer is incorrect and the actual Bode plot shows that from frequency=10 to 10^1.5 the curve has moved up 22.5°. Could someone please explain to me how to properly calculate the phase change from frequency=10 to 10^1.5?
Best Answer
Well, we have the following transfer function:
$$\mathcal{H}\left(\text{s}\right):=\frac{\text{X}\left(\text{s}\right)}{\text{Y}\left(\text{s}\right)}=\frac{\left(1+10^{-\alpha}\cdot\text{s}\right)\left(1+10^{-\beta}\cdot\text{s}\right)}{\left(1+10^{-\gamma}\cdot\text{s}\right)\left(1+10^{-\epsilon}\cdot\text{s}\right)}\tag1$$
Where \$\alpha\in\mathbb{R}_{\ge0}\$, \$\beta\in\mathbb{R}_{\ge0}\$,\$\gamma\in\mathbb{R}_{\ge0}\$, and \$\epsilon\in\mathbb{R}_{\ge0}\$.
We can expand the RHS of the transfer function:
$$\mathcal{H}\left(\text{s}\right)=\frac{1+10^{-\beta}\cdot\text{s}+10^{-\alpha}\cdot\text{s}+10^{-\alpha}\cdot\text{s}\cdot10^{-\beta}\cdot\text{s}}{1+10^{-\epsilon}\cdot\text{s}+10^{-\gamma}\cdot\text{s}+10^{-\gamma}\cdot\text{s}\cdot10^{-\epsilon}\cdot\text{s}}=$$ $$\frac{1+\left(10^{-\alpha}+10^{-\beta}\right)\cdot\text{s}+10^{-\left(\alpha+\beta\right)}\cdot\text{s}^2}{1+\left(10^{-\gamma}+10^{-\epsilon}\right)\cdot\text{s}+10^{-\left(\gamma+\epsilon\right)}\cdot\text{s}^2}=$$ $$\frac{10^{-\left(\alpha+\beta\right)}\cdot\text{s}^2+\left(10^{-\alpha}+10^{-\beta}\right)\cdot\text{s}+1}{10^{-\left(\gamma+\epsilon\right)}\cdot\text{s}^2+\left(10^{-\gamma}+10^{-\epsilon}\right)\cdot\text{s}+1}\tag2$$
Now, when we work with sinusoidial functions we can substitute \$\text{s}=\text{j}\omega\$:
$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{10^{-\left(\alpha+\beta\right)}\cdot\left(\text{j}\omega\right)^2+\left(10^{-\alpha}+10^{-\beta}\right)\cdot\text{j}\omega+1}{10^{-\left(\gamma+\epsilon\right)}\cdot\left(\text{j}\omega\right)^2+\left(10^{-\gamma}+10^{-\epsilon}\right)\cdot\text{j}\omega+1}=$$ $$\frac{1-10^{-\left(\alpha+\beta\right)}\omega^2+\left(10^{-\alpha}+10^{-\beta}\right)\omega\text{j}}{1-10^{-\left(\gamma+\epsilon\right)}\omega^2+\left(10^{-\gamma}+10^{-\epsilon}\right)\omega\text{j}}\tag3$$
Now, we can take the argument on both sides:
$$\displaystyle\arg\left(\underline{\mathcal{H}}\left(\text{j}\omega\right)\right)=\arg\left(1-10^{-\left(\alpha+\beta\right)}\omega^2+\left(10^{-\alpha}+10^{-\beta}\right)\omega\text{j}\right)-\arg\left(1-10^{-\left(\gamma+\epsilon\right)}\omega^2+\left(10^{-\gamma}+10^{-\epsilon}\right)\omega\text{j}\right)\tag4$$
So, we get a few different cases:
Now, we can this apply this to your problem. We know that \$\alpha=2\$, \$\beta=\frac{5}{2}\$, \$\gamma=0\$, and \$\epsilon=\frac{9}{2}\$. So we can work out the different cases:
If you uses Mathematica I wrote a code that you can use:
The output, gives:
Plotting a bigger range, gives: