# Phase shift at very high frequencies

I was given a transfer function:

$$H(s)= \frac{5 \cdot 10^8}{s^2+6\cdot10^4\cdot s+5\cdot10^8}$$

The question I was asked was , what was the phase relationship between input and output voltage at high frequencies.

My approach was to find the phase of the transfer function, which I got to be:

$$-\tan^{-1}\Bigg(\frac{6\cdot10^4w}{5\cdot10^8-w^2}\Bigg)$$

I assumed that "very high frequencies" meant w = ∞, plugging it in, I get the answer to be 45°.

However, the answer from my teacher was that it was -180, because the transfer function had 2 poles, each at -90. While I can understand this, I wanted to know why my method was wrong.

• High frequency here means at least 1 decade above the break point then rounding the phase. You don't have to go all the way to ∞. If you got 45° then your formula is wrong. – Tony Stewart Sunnyskyguy EE75 May 7 '17 at 13:48
• I see....but then how is my formula wrong. And when you mean 1 decade above, what does that mean for w? – user4826575 May 7 '17 at 13:52
• Your formula is correct and gives -arctan(0)=-180deg,0 deg, +180deg (for w>>infinite). Remember that the arctan- function is not unambiguous. – LvW May 7 '17 at 14:58
• I am sorry. I am not the best at math. I still don't understand where -180 comes from. Is it from an EE standpoint or a trig standpoint? – user4826575 May 7 '17 at 15:03
• tan=sin/cos. Try this ratio for 0 deg and try it for 180 deg. Are the results the same? – LvW May 7 '17 at 19:34

At "high frequencies" i.e. those tending to infinity, the original H(s) formula reduces to: -

$\dfrac{K}{s^2}$

At that means a 180 degrees phase shift because a single s shifts by 90 degrees and s squared by a further 90 degrees.

• So, is it always the case that a single s shifts by 90 degrees? – user4826575 May 7 '17 at 18:41
• Yes, because s is a complex number involving "j" (or "i" if you are math dude). The operator j is a shift of 90 degrees and j^2 is a shift of 180 degrees hence this is why we say j is the square root of minus 1. – Andy aka May 7 '17 at 19:10
• Hmmm. I see.....what about the fact s is in the denominator. Does that change if the funtion becomes K*s^2 – user4826575 May 7 '17 at 21:22
• 1/j = -j so with an "s" in the numerator, the phase shift is +90 and in the denominator it's -90: girlsangle.files.wordpress.com/2012/07/blog_072412_08.jpg – Andy aka May 8 '17 at 7:16

Try to calculate the limit for w - > infinity of you phase function. The result is 0. Note that your Re() part of the denominator for high frequencies is negative, so you have to add 180° (or - 180°, the same) to the phase. So 0° - 180 = 180°.

• Thank you for the awnser. This was what I was considering. However, I dont understand how you can just - 180°. Why is it 180, or is it always 180? And do we minus because the limit is 1/-w? Sorry for all the questions. – user4826575 May 7 '17 at 14:41
• Note that mathematically all trigonometric functions show the same results to multiples of 2pi angles so for example sin(x - pi) = sin(x - pi +2pi) = sin(x + pi). In this case we say -180 to put a stress on the monotony of the function that decrease from 0 to - 90 to - 180 – Simus994 May 7 '17 at 14:48
• It is not always -180... It is - 180 for high frequencies, almost 10 times far from the second pole – Simus994 May 7 '17 at 14:50
• I apologize.....I still dont understand how we came up with the -180. Is it from a EE standpoint or trig standpoint? – user4826575 May 7 '17 at 15:08
• It is from mathematics standpoint. Do you know that when you calculate the phase you have to add (+ or -) 180 degrees if the real part is negative? – Simus994 May 8 '17 at 4:59

I think it all starts with rewriting the transfer function in a so-called low-entropy format:

$H(s)=\frac{5.10^8}{s^2+6.10^4s+5.10^8}=\frac{1}{1+\frac{6.10^4}{5.10^8}s+\frac{s^2}{5.10^8}}$

This is the typical form of a second-order network in which there is no zero (zeros are the roots of the numerator $N(s)$ while poles are the roots of the denominator $D(s)$) and a gain of 1 when $s=0$:

$H(s)=H_0\frac{1}{1+b_1s+b_2s^2}=H_0\frac{1}{1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2}$

from which you can identify the resonant frequency $\omega_0$ and the quality factor $Q$.

Because the order is 2, you have two roots in the denominator $D(s)$. If you solve $D(s)=0$, then you will find the poles expressions and realize how they move in relationship with $Q$. A pole in the left-half plane contributes a 90° phase lag. Two poles as in here would contribute twice this value or -180° as $s$ increases beyond the resonance.

To calculate the phase response at any point, replace $s$ by $j\omega$, expand, collect real and imaginary parts. As correctly pointed, for $s$ approaching infinity, the equation reduces to $H_{inf}=\frac{\omega_0^2}{s^2}$. When there is no real part in the complex number $z=x+jy$, meaning $x=0$, then the argument $arg(z)=tan^{-1}\frac{y}{x}$ returns $90°$. Because you have two poles ($s^2$) in the denominator, then $argH(s)=argN(s)-argD(s)=0-180=-180°$.

The key is truly to properly write the transfer function so that gains, poles and zeros appear in a well-ordered form. From that, you can infer the phase and magnitude response quickly as long as there is no delay and poles/zeros are in the left-half plane. Have a look at http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf to learn how Fast Analytical Circuits Techniques (FACTs) make use of low-entropy expressions.

Probably you don't know the mathematical theory behind the phase concept. Let's study this: Analog and Digital Signal and Systems - Complex Numbers

• This is link-only, and should be a comment. – Jashaszun Jun 13 '17 at 22:25
• Oh my God sorry for my mortal sin – Simus994 Jun 13 '17 at 22:29