Continue to Site

# simple question about deriving phase from transfer function

Status
Not open for further replies.

#### hoki222

##### New Member
Hi can somebody explain, how to derive the phase function from the following transfer function:

H(s)=s/(s^2+s+1)

I found the Magnitude already to be

|H(jw|=w/root[w^2+(1-w^2)^2]

I know the relationship Angle=arctan(Im/Re) but still cant find the right phase function. Plz Help!

so you have the magnitude correct. how did you derive that? I'm assuming that you set s=jw and then plugged and chugged - but did you set up the equation for H(jw) so that there were no imaginary terms in the denominator? If you do that, you should end up with:

H(jw) = [w*w + jw*(1-w*w)]/[w*w + (1-w*w)^2]

to get the phase, its just arctan[w(1-w*w)/(w*w)] = arctan[(1-w*w)/w]

Thank you for the reply, but this is not 100% correct. I did the same multiplying the complex conjugate to the transferfct. and got arctan[(1-w^2)/w] too.. but is only the half truth.. I found that this equation is only true for phases 0° to -90°.. for -90° to 0° u must add 180°

It is easier and quicker to derive the phase with this relationship i found on wikipedia arg(H)=arg(y)-arg(x)

with y = jw and x = (1-w^2)+jw

phase=90°-arctan(w/(1-w^2))

i found that this only true for phases 90° to 0° and u must add 180° to get the correct phase from 0 to -90°.. then u can plot the whole phase correctly

Why do you think the solution is wrong for 0 t0 -90? The angular frequency, w, can only take on positive values, so the solution should be unique.

are you sure you understand what you are solving for here? I had no problems plotting either the gain or the phase. The plot makes complete sense - the zero is forcing the gain to increase at a rate of +20dB/decade until it hits the double pole at around 0.15 or 0.16 degrees (1 radian!) where it then decreases at -20dB/decade. The phase plot shows the double pole by shifting -180 degrees.

#### Attachments

• plot.jpg
87.7 KB · Views: 520
Phase/equation with arctan function

maybe i make an error in reasoning, i would be glad if u can correct me.

thx for ur comment.. i dont know for sure if my problem is clear for u my Problem was not to analyse the phase and amplitude curves, i just wanted to derive the phase equation without the computer..

i think outtolunch u used the bode function in matlab but if i plot just the phase equation directly [phase=90°-arctan(w/(1-w^2))] i get the plot a lil different (than the real phase like in ur attachment) which has a pole at 1 and shifts 180°.

If i use w greater 1 in the phaseequation the phase is not equal anymore with the true phase since the arctan function has a pole at 1.

thats why i said u must add 180° (shifting y) in order to get the right phase above w=1 and is only true for 90° to 0°.

Of course i could missed something here, ill be glad to be corrected or adviced.

#### Attachments

• phase.jpg
35.6 KB · Views: 382
Hi hoki,

You apparently know how to proceed yourself (check accuracy) so it's only

The saving grace comes in the form of the two argument inverse tangent function.
The two argument itan function, as the name implies, takes two arguments rather
than just one like itan(45). The difference is that the two arg itan function can
distinguish between angles such as when imag is negative and real is negative.
The single arg itan can not do that because of course a negative divided by a
negative is a positive, so it thinks that the angle is in the first quadrant rather
than the third.
You can probably find the two argument itan(a,b) on the web, but if not,
just think of the real and imaginary parts as being positive (take abs() of each)
and figure out the angle yourself by noting the position in the complex
plane and figuring how the result of the single argument itan() should be
modified.
As you probably know, we can form a little table:

By using this table, you can correct the angle returned with arctan(abs(y/x)).
The x and y signs however should come from the complete x and y, which in this
case is the real and imaginary:

real=w^2/(w^4-w^2+1)
imag=(w-w^3)/(w^4-w^2+1)

If we inspect the real part of this solution, we find that it stays positive,
but the imag part can swing plus or minus. This means that the solutions will
all lie between -90 and 90 degrees (first and fourth quadrants) so the
single argument arctan() function will work for this particular function.
There are functions that this will not work with however, such as:
H(s)=s^2/(s^3+s^2+s+1)
in which case we are forced to either use the two argument itan function or
figure out the angle from inspection of the imag and real parts.

Status
Not open for further replies.