Detailed analysis of a transfer function

[TwinT]

Let's say I have a transfer function T(S) = Y(S) / X(S).

zeros are the roots of Y(S) and poles are the roots of X(S).
question: if you have a zero in your transfer function then at your bode plot
(20log (abs(T(S))), the function starts ro rise?
and the same with a pole: when you have one your function starts to go down instead of raisin' to infinity??

Yes, I know. The complex numbers...bla bla

Anyone can truly explain what's going on?

Thanks

 

[v_c]

Re: Bode

Well not to be picky but the you have to plot \[20 \log || T(j\omega) || \] but I think you are already aware of that.

I think you already answered how the transfer function works ... Let's say that there is a zero of form \[ 1 + s/\omega_z\] (zero at \[\omega_z\]) and a pole pole of form \[ 1 + s/\omega_p\] (pole at \[\omega_p\]). If you strictly look at these factors by themselves each of these terms increases as frequency increases. Why? Because (noting that \[s=j\omega\]) as frequency \[\omega\] gets higher and higher both \[ 1 + j\omega/\omega_z\]
and \[ 1 + j\omega/\omega_p\] increase. But since the zero is by definition in the numerator it makes the magnitude of T(s) go higher whereas the pole which is in the denominator makes the magnitude of T(s) go lower.

Does this help, or did you have some specific system in mind?
If you have something more specific, please post and I will try to answer.

Best regards,
v_c

 

[TwinT]

Re: Bode

Ok, let's go to the specific.

Common Source with derivation capacitor CS and degeneration Resistor RS. In the Sedra-Smith book it says that a zero (Id=Is=0) is found at wz=1/RSCS, which is true, and also says that this is obvious because impedance in wz goes to infinity, and then Id = Is = 0. I don't understand this, I don't see phisically a frequency that make impedance go to infinity

 

[v_c]

Re: Bode

I don't have the book in front of me here but I think you are talking about \[R_s\] in parallel with \[C_s\], right?

If this is the case then the total impedance will be
\[Z = R_s || (1/s C_s) = \frac{R_s}{1+ s C_s R_s} = \frac{R}{1+s/\omega_p}\]

It seems like that should be a pole at \[\omega_p=1/(R_s C_s)\].
Only then I can believe that the impedance goes to infinity at \[\omega_p\] since at that frequency the denominator goes to zero. It may be a typo -- I don't know. Check you version of the book and visit the web page for the book (if there is one) to see if there is any known errata for your edition.

I should add that it all depends on what transfer function you are looking at. On some transfer functions like id/vg the \[1/(R_s C_s)\] term probably appears in the numerator in which case it is a zero. So just because the this term is a zero in the impedance transfer function does not keep it from being a pole in other transfer functions. Also remember that impedance is a transfer function as well (it is v/i at the same port), whereas the id/vg transfer function is really an admittance (1/impedance), so you can see how the roles of the poles and zeros may reverse when the transfer function is inverted.

Best regards,
v_c

 

[eecs4ever]

Re: Bode

v_c,

at the pole frequency, the denominator does not go to zero?

"
Only then I can believe that the impedance goes to infinity at since at that frequency the denominator goes to zero. "

1/ (1+ jw/w_p)

when w goes to w_p, the denominator goes to 1+j.


In the book it says theres a zero at 1/RsCs because
Given Zs = Rs / (1+ jw/w_p) (Rs || Cs) , where w_p = 1/ (RsCs)

Gm (with source degeneration) = gm / ( 1+ gm*Zs)

Vout / Vin = - Gm * Rout = gm * Rout / (1+ gm Zs)
substitute in the Zs = Rs / (1+ jw/w_p)

Vout / Vin = gm * Rout / (1+ gm Rs /(1+jw/w_p) )

multiply top and bottom by (1+jw/w_p)

now this term shows up in the numerator and looks like a zero .


As for the impedance going to infinite, I don't understand that part either. = /

 

[v_c]

Re: Bode

Sorry guys ... I really screwed that up. What I meant to say is that with the Z for the resistor and parallel combination, as the frequency goes higher and higher the impedance should head toward zero. The low frequency impedance is simple Rs and at the pole frequency, the impedance should start dropping.

I think I should have reviewed that before hitting the "Submit" button.

Best regards,
v_c

 

[TwinT]

Re: Bode

OK, thanks for your help.

Anyway I don't totally figured out what's going on there. Is clear the question ?
Let's isolate the problem: suppose we have a circuit as in the figure.
Doing Y(w) = Io(w)/Vi(w) , we obtain Y(w)= 1/R (1 + SCR). Obviusly this function has a zero in S = -1/CR= -wz ,then wz = 1/CR. Okay, now we are getting somewhere. If the admitance has a "zero" in wz the the impedance has a pole in wz, right? Then in wz the impedance goes to infinity and Io is null.
My question is: I don't see this phisically. For w = 0 the circuit is pure resistive. For w infinitum the current goes to hell. Therefore, in the middle, the current is increasing from Io= Vi/R to infinitum. When is zero?? That's what I don't Understand. Thanks

 

[demodb]

Re: Bode

i think what really is going on when you have the zero is that the impedance of the resistor equals the impedance of the capacitor. Increasing frequencies will lead to a smaller capacitor impedance, so more current. What is going on is at the zero the impedance is going from being resistive to becoming capacitive.

At least that's how i see it.

 

[VVV]

Re: Bode

Math will provide the answer, whether we like or not.

Let's take the example of the RC filter circuit, having one pole: ωp=1/RC
We will resort to math, but let's just start from the physics. You apply a signal with frequency f at the input. What is the output amplitude?

The circuit behaves just like a divider, one of the parts being a capacitor.
Therefore, the output voltage can be written:

Vout=Vin*(1/jωC)/(R+1/jωC)=Vin*1/(1+jωCR)

With the notation ωp=1/RC, we can write:

Vout=Vin*1/(1+jω/ωp) (you should remark that ωp is just a notation; yes that is the pole, but we do not care about that yet)

Now we can write:
Vout/Vin=1/(1+jω/ωp)

So now the question comes: how is this ratio going to change as the frequency is varied? This is called the attenuation of the circuit, expressed in dB. It is plotted on "log paper", basically the abscissa is the log of the grequency, while the y-axis is the attenuation in dB. This is the Bode plot.
Imagine you are plotting this on a paper where the y-axis has 10dB/cm.
Let's plot.

A=20*log|Vout/Vin|=20*log|1/(1+jω/ωp)|=20*log(1/√(1+(ω/ωp)²)

Now consider what happens when ω=0.01*ωp (the input frequency is 100 times lower than ωp, or two decades below)

A=-0.0004 dB. If you plot that on your paper with 10dB/cm, it looks as 0dB.

Now try ω=0.1*ωp (one decade below)
A=-0.04 dB If you plot that, it still looks like 0dB

Now try ω=0.5*ωp
A=-0.97 dB Now you can see it, about 1mm on your paper, still not much

Now try ω=ωp
A=-3 dB That is about 3mm below the 0dB line; remember this point

Now try ω=10*ωp (one decade above ωp)
A=-20 dB Oops! What happened? When you were at one decade below, the attenuation was only 0.04dB!

Now try ω=100*ωp (two decades above ωp)
A=-40 dB

Now try ω=1000*ωp (three decades above ωp)
A=-60 dB

Ahaaa... Above ωp, whenever you go up in frequency by one decade, the amplitude drops by 20dB. But below ωp, you could go as far down as you wanted, there was hardly any change.

Well, ωp is the pole location and the corresponding frequency is fp=1/(2ΠRC)
To find it, you measure the amplitude and see when it drops by 3dB (remember from above, at ω=ωp, the attenuation is -3dB)

So there you have it, some math is involved, but the approach is more "practical". A similar development can be done for a zero and you will find that the zero will make the amplitude INCREASE by 20dB for every decade increase in frequency.

 

[eecs4ever]

Re: Bode

"Doing Y(w) = Io(w)/Vi(w) , we obtain Y(w)= 1/R (1 + SCR). Obviusly this function has a zero in S = -1/CR= -wz ,then wz = 1/CR. Okay, now we are getting somewhere. If the admitance has a "zero" in wz the the impedance has a pole in wz, right? Then in wz the impedance goes to infinity and Io is null."

When you operate your circuit, it never goes to the zero that you described. Admitance never reaches 0 , even at the "zero" and impedance does not reach infinite.

Y(w) = 1/R (1 + SCR)

When you talk about the frequency response of your circuit, is the reponse
of your circuit along the jw axis.

"zero" is not when the numerator == 0
at the "zero" S is not equal to -1/RC.
at the "zero" S= j / CR . This is the point when Y(w_z) looks like 1/R (1+ j)
this is when the admittance starts to look more capactive than resistive.
after this point, the admittance will look even more capacitive.
Y(10*w_z) = 1/R (1+ 10 * j)
higher admittance => lower impedance.

impedance of the resistor looks like real numbers
impedance of capacitors look like j*something
impedance of inductors look like 1/j * something

 

[jayc]

Re: Bode

impedance of capacitors look like j*something
impedance of inductors look like 1/j * something

You have it backwards.

\[Z_C = \frac{1}{j\omega C}\rightarrow Y_C = j\omega C\]
\[Z_L = j\omega L\rightarrow Y_L = \frac{1}{j\omega L}\]

 

[eecs4ever]

Bode

haha, jayc is right =).

what i meant to say was that the admittance of capacitors look like j*something.