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https://designers-guide.org/forum/YaBB.pl Design >> Analog Design >> Finding zero in an amplifier https://designers-guide.org/forum/YaBB.pl?num=1319559943 Message started by raja.cedt on Oct 25th, 2011, 9:25am |
Title: Finding zero in an amplifier Post by raja.cedt on Oct 25th, 2011, 9:25am hello all, while razaavi i got the follwoing doubt. For finding zero of an amplifier, generally by making Vout=0 and find the zero, i know it's simpler and explained in razaavi which is working in all cases. But at the ZERO frequency will the amplifier o/p goes to zero? up to my knoledge for zero's which are placed on jw axis only do this. Please refere page no 176 in razaavi. Thanks, raj. |
Title: Re: Finding zero in an amplifier Post by Garrett.Neaves on Oct 25th, 2011, 11:03am raja.cedt wrote on Oct 25th, 2011, 9:25am:
Yes. The Vout(s)=0 at s=gm/CGD +j0. This zero is at a complex frequency which has a non-zero real part and a zero imaginary part. If the circuit is responding to an excitation at this complex frequency, then the Vout would be zero. If the circuit is responding to a pure sinusoid with a radian frequency of w, i.e. s=0+jw, then Vout(jw) is not zero. If jw is varied from 0 to positive values, then the zero at s=gm/CGD +j0 contributes a +20dB/dec to the overall slope of the magnitude of Vout(jw)/Vin(jw) at frequencies higher than w=gm/CGD. This value of w is often referred to loosely as a zero frequency in the magnitude versus w plot. The actual zero at s=gm/CGD +j0 causes a corner frequency at w=gm/CGD in the magnitude versus w plot. |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 25th, 2011, 1:38pm hello Garrett.Neaves, To be frank i didn't understand your answer, may be i am not that good in this basics. if s=gm/CGD +j0, then there is no complex frequency, only you have attenuation, so up to my knoledge there is sinusoidal signa which can make Vout zero.... Could you please explain clearly... Thanks, raj. |
Title: Re: Finding zero in an amplifier Post by Garrett.Neaves on Oct 25th, 2011, 4:42pm raja.cedt wrote on Oct 25th, 2011, 1:38pm:
Hi Raj, s=gm/CGD +j0 is a point in the complex frequency plane. Vout(s)=0 at this frequency in the complex frequency plane. A circuit excited by pure sinusoids such as when sweeping the frequency in an ac simulation, is represented in the complex frequency plane as a sweep along the jw axis. The value of Vout(jw)/Vin(jw) as w is swept is determined by the poles and zeros of Vout(s)/Vin(s) wherever the poles and zeros lie in the complex frequency plane. That is, the poles and zeros of Vout(s)/Vin(s) effect the shape of Vout(jw)/Vin(jw) versus w even when the poles and zeros do not lie on the jw axis. I explained it as clearly as I am able in this posting type of communication. Maybe someone else can explain in a way which would be more helpful to you. Best Regards, Garrett |
Title: Re: Finding zero in an amplifier Post by buddypoor on Oct 26th, 2011, 12:53am Garrett.Neaves wrote on Oct 25th, 2011, 11:03am:
Hi Garrett. Are you able to excite a circuit with a complex frequency s that has zero imaginary part ? No, I don't think so. Thus, the output of a real circuit will be zero for one case only: If the transfer function is zero for an excitation s=jw (with real part=0). Sometimes this is called "a real zero". This may sound confusing but means: A frequency that can be created in reality (and it is purely imaginary in the s-domain). |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 26th, 2011, 1:01am @ Garrett.Neaves : thanks for your effort, i have last question, let us say if you have a TF=(s+1)/(s+2) now according to your theory at s=-1 you have zero (i also agree), but can you tell me in your magnitude plot will it go to 0, or let me ask you in another way can you give a sinusoidal signal which makes this TF magnitude 0. Thanks, raj. |
Title: Re: Finding zero in an amplifier Post by buddypoor on Oct 26th, 2011, 3:28am raja.cedt wrote on Oct 26th, 2011, 1:01am:
No, he can't. |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 26th, 2011, 3:55am hello buddypoor, then why people says that @ zero frequency gain will be 0 and @ pole frequnecy gain will be inf, up to my knoledge if you have poles or zeros on iw axis then only you get . But why this method is working in cases? Thanks, Raj. |
Title: Re: Finding zero in an amplifier Post by Garrett.Neaves on Oct 26th, 2011, 5:19am raja.cedt wrote on Oct 26th, 2011, 1:01am:
Raj, For TF=(s+1)/(s+2), there is only one zero at s=-1+j0. There is no non-zero value of w which produces a zero at s=0+jw. Note that w is the frequency of a sinusoid. Therefore there is no sinusoidal signal which can make this TF magnitude equal to 0. Garrett |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 26th, 2011, 5:37am hello Garrett.Neaves, there is Contradiction between first post and earlier post, so finally you agred that magnitude can't be zero, so why for finding Zero people are shorting o/p? for example take a simple CS amplifier there zero frequency is -gm/Cc, doesn't mean o/p at this frequency is 0 .... Thnaks, raj. |
Title: Re: Finding zero in an amplifier Post by Garrett.Neaves on Oct 26th, 2011, 7:34am raja.cedt wrote on Oct 26th, 2011, 5:37am:
Raj, Would you precisely document the contradiction? Garrett |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 26th, 2011, 7:51am hello, i donno how to document it. In your first post you said Vout will go to Zero and in one post you told it's won't go. Did you understand my Question? Any how i am decribing here for you 1.take a simple source follower, it gives left half zero @gm/Cgs, does it means o/p is zero at this frequency (just tell me yes/no) 2. In razaavi book he calculated this frequency by shorting the o/p. Thanks, Raj. |
Title: Re: Finding zero in an amplifier Post by Garrett.Neaves on Oct 26th, 2011, 8:18am raja.cedt wrote on Oct 26th, 2011, 7:51am:
Raj, I believe I clearly understand your question. I have been consistently saying that there is a complex frequency at which Vout(s)=0 (i.e. Vout(s)=0 when s=gm/CGD + j0 and that there is no pure sinusoid s=0+jw for which Vout(jw)=0. This is a necessary point for us to agree on if we are to take the next step to satisfy your question. Best Regards, Garrett |
Title: Re: Finding zero in an amplifier Post by Ricky Chen on Oct 29th, 2011, 10:23pm hi raja.cedt, Fundamental things u have to bear in mind:- 1) Real frequency i.e jw and "s" (complex frequency) are two different things 2) T.F. must be zero or infinite at the ZEROs ( in s form) and at POLES ( in s form) 3) If poles or zeros happens to be REAL frequency (i.e. s=jwz or jwp) then there exists sinusoids which makes T.F. infinity or zero For example a LC parallel circuit:- in that case the denominator is LCs^2+1 . So poles are +/- j (1/(LC)^2). So real frequency poles. So there exists a sinusoid at this frequency which makes the LC ckt to go to infinity. i.e. one get o/p at this frequency without any input (i.e it oscillate at the Resonance frequency). Similarly you can find REAL zero for a series LC ckt. 4) Complex poles and ZEROS give 20 dB/decade roll off in bode plot. I hope this statement you can understand and prove urself why it is so . Regards, -Ricky |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 30th, 2011, 12:35am hello chen, we both are on the same line man, i too agree that if you have poles or zero's on j*w axis then only we get 0 or infinite gain (this is simple from the math as well). My question is simple. Let us say if you have (s+1)/(den), then zero will be at -1+(j*0) means there is sinusoidal signal frequency which can give you 0 gain, and source follower also belonged to this type but still razaavi made o/p short, found the zero frequency. Thanks, raj. |
Title: Re: Finding zero in an amplifier Post by buddypoor on Oct 30th, 2011, 2:22am Hi raj, At first, I completely agree with Ricky's explanation. Secondly, there may be some confusion using the term "zero". Therefore, I'll try to give my explanation: Taking your simple example H(s)=(1+sRC)/D(s). This is a transfer function with the complex variable s. I have added a time constant RC due to consistency in units. At s=-1/RC there is a zero, which means that the transfer function output is zero at this "point" (it is not a frequency!). It is simply the negative real part of s. This property of H(s) cannot be measured (it is just a mathematical fiction) because one cannot produce a "frequency" which is a negative constant. Referring to measurements, we can, however,transfer H(s) into a frequency response A(jw) by simply replacing s by jw. And we can ask ourself: What can be measured at a "frequency" that equals the above given "zero": w=1/RC ? As a result, we can compute the magnitude (and phase) for A(jw) and we see that at this frequency the BODE diagram exhibits a change in the slope. Thus, we can conclude that H(s) exhibits a zero at a certain value for the variable s (in our case negative-real). If we, however, interpret this variable as a frequency we see that the frequency response (Bode diagram) at this point changes the slope. The point of confusion is that in both cases we speak of a "zero frequency". Unfortunately, Razavi is not consistent because on page 176 this zero is called "wz" and on page 177 he uses "sz". Thus, on page 176 he implicitely referres to A(jw) and on page 177 he referres to H(s). A similar case exists, of course, for poles. We speak of pole frequencies, thereby knowing that the magnitude of A(jw) goes not up to infinity. This can be observed only in the 3D-plot for H(s). ________________ Regards |
Title: Re: Finding zero in an amplifier Post by Ricky Chen on Oct 30th, 2011, 5:33am Let us say if you have (s+1)/(den), then zero will be at -1+(j*0) means there is sinusoidal signal frequency which can give you 0 gain The above statement is WRONG. You can't have a sine wave which have ZERO o/p for this T.F. |
Title: Re: Finding zero in an amplifier Post by raja.cedt on Oct 30th, 2011, 5:50am hello Ricky Chen i mean to say that there is no sin signal which gives zero gain..it's a typo. Thanks, raj. |
Title: Re: Finding zero in an amplifier Post by buddypoor on Oct 30th, 2011, 6:09am Hi Ricky and Raj, in addition to my comments above I like to point to the following: The frequency response A(jw) of a circuit shows a slope change at frequencies that correspond with singularities (zeros/poles) of H(s) in the s-domain. And this fact demonstrates the usefulness of the s-domain. This can be part of the answer to the question (that very often can be read): What is the purpose and the advantage of introducing the complex frequency s ? |
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