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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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 7 '17 at 13:48
  • \$\begingroup\$ I see....but then how is my formula wrong. And when you mean 1 decade above, what does that mean for w? \$\endgroup\$ – user4826575 May 7 '17 at 13:52
  • \$\begingroup\$ Your formula is correct and gives -arctan(0)=-180deg,0 deg, +180deg (for w>>infinite). Remember that the arctan- function is not unambiguous. \$\endgroup\$ – LvW May 7 '17 at 14:58
  • \$\begingroup\$ 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? \$\endgroup\$ – user4826575 May 7 '17 at 15:03
  • \$\begingroup\$ tan=sin/cos. Try this ratio for 0 deg and try it for 180 deg. Are the results the same? \$\endgroup\$ – LvW May 7 '17 at 19:34
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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.

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  • \$\begingroup\$ So, is it always the case that a single s shifts by 90 degrees? \$\endgroup\$ – user4826575 May 7 '17 at 18:41
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    \$\begingroup\$ 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. \$\endgroup\$ – Andy aka May 7 '17 at 19:10
  • \$\begingroup\$ Hmmm. I see.....what about the fact s is in the denominator. Does that change if the funtion becomes K*s^2 \$\endgroup\$ – user4826575 May 7 '17 at 21:22
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    \$\begingroup\$ 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 \$\endgroup\$ – Andy aka May 8 '17 at 7:16
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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°.

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  • \$\begingroup\$ 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. \$\endgroup\$ – user4826575 May 7 '17 at 14:41
  • \$\begingroup\$ 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 \$\endgroup\$ – Simus994 May 7 '17 at 14:48
  • \$\begingroup\$ It is not always -180... It is - 180 for high frequencies, almost 10 times far from the second pole \$\endgroup\$ – Simus994 May 7 '17 at 14:50
  • \$\begingroup\$ I apologize.....I still dont understand how we came up with the -180. Is it from a EE standpoint or trig standpoint? \$\endgroup\$ – user4826575 May 7 '17 at 15:08
  • \$\begingroup\$ 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? \$\endgroup\$ – Simus994 May 8 '17 at 4:59
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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.

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

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

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