0
\$\begingroup\$

So I have this transfer function: $$\begin{align} \dfrac{g_m-sC_{gd}}{s(C_{gs}+C_{gd})}=\dfrac{-C_{gd}\left(s-\dfrac{g_m}{C_{gd}}\right)}{(C_{gd}+C_{gs})\left(s-\dfrac{0}{C_{gd}+C_{gs}}\right)}=-\dfrac{C_{gd}}{C_{gd}+C_{gs}}\dfrac{s-\dfrac{g_m}{C_{gd}}}{s} \end{align}$$

Then im supposed to find the zero and pole frequency expression based from my transfer function and this is what I got: $$ ω_{z} = \frac{-gm}{C_{gd}} $$ $$ ω_{p} = \frac{1}{C_{gd} + C_{gs}} $$

But then that two is wrong. But I dont understand why. Then we are asked for the gain to when ω = ωz and this is my formula:

$$ A_{v},dB = 20\log\left|\frac{C_{gd}*\sqrt{1+(\frac{\omega_{z}}{\omega_{z}})^2}}{ \sqrt{(-\frac{ω_{z}}{w_{p}})^2}}\right| $$

The above formula for my gain is also wrong but i guess thats because my formula for the zero and pole frequency is also wrong. So where are my mistakes here? Help would be much appreaciated!

\$\endgroup\$
13
  • \$\begingroup\$ Why are you introducing the zero factor in your first line of equations? \$\endgroup\$
    – Andy aka
    Dec 12, 2020 at 13:21
  • 2
    \$\begingroup\$ A pole or a zero is the inverse of a time constant whose unit is time [s]. You can see that in the pole expression, you don't have a time constant in the denominator. \$g_m\$ has the dimension of [A]/[V] and must also appear in the pole expression. It is \$g_m\$ that you must factor in the numerator (and bring back to the denominator then) to have \$g_m(1-\frac{s}{\omega_z})\$. \$\endgroup\$ Dec 12, 2020 at 13:24
  • 1
    \$\begingroup\$ @Rein Well you've just edited that line to remove one of them so I think you don't really need this explaining. The pole frequency is zero from what I can see. \$\endgroup\$
    – Andy aka
    Dec 12, 2020 at 13:25
  • 1
    \$\begingroup\$ The confusion often comes from the factor affected to \$s\$ in the denominator. In this example, the pole is at \$s=0\$, it is located at the origin. However, there is a factor associated to it and I usually call it the 0-dB crossover pole meaning that when \$s=\omega_{po}\$ then the gain is 1 or 0 dB. \$\endgroup\$ Dec 12, 2020 at 13:29
  • 1
    \$\begingroup\$ @Rein the pole frequency is zero period. That is what you asked for. I don't know what you are modelling and therefore I cannot say anything else. \$\endgroup\$
    – Andy aka
    Dec 12, 2020 at 13:35

1 Answer 1

4
\$\begingroup\$

When you write a transfer function, it is important to respect a so-called low-entropy format where the poles and the zeroes clearly appear. In your case, for instance, the correct expression should be in the form of: \$H(s)=\frac{1-\frac{s}{\omega_z}}{\frac{s}{\omega_{po}}}\$.

The numerator hosts a right-half-plane zero (a RHPZ) while the denominator shows a pole located at the origin for \$s=0\$. By factoring \$g_m\$ in the numerator, you can determine the zero and what I call the 0-dB crossover pole. Rearranging, you have \$\omega_z=\frac{g_m}{C_{gd}}\$ and \$\omega_{po}=\frac{g_m}{C_{gd}+C_{gs}}\$. Please note that you always need to verify the homogeneity of your formulas: a pole or a zero is the inverse of a time constant expressed in seconds [s]. As \$g_{m}\$ is expressed in [A]/[V], we are ok.

Now, this expression is not the best we can do because it brings no insight in what is going on with this formula. The most compact formula uses what is called an inverted zero: from your first expression, format \$sC_{gd}\$, simplify by \$s\$ and you now have \$H(s)=H_{inf}(1-\frac{\omega_z}{s})\$ naturally highlighting the gain of this expression when \$s\$ approaches infinity: \$H_{inf}=\frac{C_{gd}}{C_{gd}+C{gs}}\$.

I have gathered these different expressions in the below Mathcad sheet and they are identical:

enter image description here

Many times, I have seen people confused with transfer functions that are not optimally written. Properly rearranging a transfer function with a leading term (when it exists) followed by a dimensionless quotient is the way to go.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy