Simplifying Units in Ezunits Package

Hello everyone

I am very impressed with the capabilities of Maxima with the ezunits
package when it comes to practical engineering tasks. However, I still
encounter certain problems that I am unable to solve satisfactorily on my
own.
Please take a look:

(%i4)   C_2: 10 `nF;
        C_1: 100 `pF;
        R_1: 10 `kOhm;

(C_2)   10 ` nF
(C_1)   100 ` pF
(R_1)   10 ` kOhm

Function fundamental_units returns proper unit value, but Maxima can't
simplify the fraction using ``.

(%i7)   1/(R_1*(C_2 + C_1));
        fundamental_units(%);
        1/(R_1*(C_2 + C_1)) ``%;

(%o5)   (1/10 ` 1/kOhm)/(100 ` pF+10 ` nF)
(%o6)   1/s
(%o7)   (1/10 ` 1/kOhm)/(100 ` pF+10 ` nF) `` 1/s

I can solve this using some manipulations: force denominator to be
converted first or use subst to convert capacitance units to be the same:

(%i10)  1/(R_1*(C_2 + C_1)) `` kHz;
        1/((R_1*(C_2 + C_1)) ``s) `` kHz;
        subst(nF = 1000 * pF, 1/(R_1*(C_2 + C_1))) ``kHz;

(%o8)   (1/10 ` 1/kOhm)/(100 ` pF+10 ` nF) `` kHz
(%o9)   1000/101 ` kHz
(%o10)  1000/101 ` kHz

The problem is that these aren't very versatile and clean solutions, and
this example is very simple.

Here is another one I have a problem with:

(%i15)  H(s, C, R_1, R_2) := (1/(s * C)) / (R_1 + R_2 + 1/(s * C));
    H(s, 18`nF, 1`kOhm, 1`kOhm);
    f: dimensionally(subst(%i*2*%pi*50`Hz, s, %));
    fundamental_units(f);
    f ``%;

(%o11)  H(s,C,R_1,R_2):=(1/(s*C))/(R_1+R_2+1/(s*C))
(%o12)  (s^(-1)/18 ` 1/nF)/(s^(-1)/18 ` 1/nF+2 ` kOhm)
(f)     (-(%i/(1800*%pi)) ` 1/(Hz*nF))/(-(%i/(1800*%pi)) ` 1/(Hz*nF)+2 `
kOhm)
(%o14)  1
(%o15)  (-(%i/(1800*%pi)) ` 1/(Hz*nF))/(-(%i/(1800*%pi)) ` 1/(Hz*nF)+2 `
kOhm) `` 1

Is there perhaps a simple and universal way to simplify units like in
MathCad or other engineering tools? I would appreciate any help.

Best regards,
Mateusz Mróz

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