Understanding Circuit Analysis from a Bode Plot

[AbbeAbyss]

Homework Statement

http://www.wifstrand.se/Albert/stuff/p13.3-6.png

Homework Equations

From circuit analysis I've found the network function to be

H(ω) = -(10^(-4) +jωC₁)/(1/R₂ + jωC₂)

(could be wrong)

The Attempt at a Solution

I don't know how to determine the network function from the Bode plot but my attempt is

H(ω) = (1 + jω/4000)/(1 + jω/400)

If I equate the first H(ω) with the second H(ω) from which I can get C₁ = 25 nF, but I can't find R₂ and C₂. In a solutions manual I read that k = 32 dB = 40 = R₂/R₁ from which I can find C₂. I can tell the information is gathered from the Bode plot but I don't know how to arrive at this answer.

 

[aralbrec]

The information you should spot from the Bode Diagram is that there is a single pole (-20dB rolloff at w=400 rad/s) and a single zero (+20dB at w=4k rad/s). There is also a dc gain of 32dB at w=0. The pole and the zero locations will be decided by the two RC products.

The thing with the pole and zero locations is that before the response is affected by the pole or zero (ie before the 20dB slope begins), the associated capacitors have essentially no effect on the circuit. For this Bode plot, this means when w<<400 rad/s you could analyze the circuit without any capacitors in it and know the transfer function. This is what they are talking about when they decide to use a low/mid/high frequency model for an equivalent circuit -- they are deciding ahead of time what capacitors matter in the response within a certain frequency range. So, in this circuit, at w=0 the transfer function will be H(j0)=-R₂/R₁=32dB

But you don't have to depend on your intuition. The circuit is simple enough that finding the complete transfer function doesn't take much time. I think the function you posted is probably correct (I haven't plugged in the values) so I'll just give you the transfer function with symbols in:

$H(s)=-\frac{Z_2}{Z_1}=-\frac{R_2 \frac{1}{sC_2}}{R_2 +\frac{1}{sC_2}}\frac{R_1 +\frac{1}{sC_1}}{R_1 \frac{1}{sC_1}}=-\frac{R_2}{R_1}\frac{sR_1 C_1 +1}{sR_2 C_2+1}$

It is much easier to understand the circuit if symbols are left in rather than immediately substituting values.

The form of this equation also makes it easy to spot the zero and pole locations (eg, $s_z = -\frac{1}{R_1 C_1}$ ). You can now find the gain at dc H(j0) to see if the intuitive guess was right.

 

[rude man]

When you see a capacitor in parallel with a resistor it's easier to use G = 1/R for the resistors. Then Y = G + sC for the parallel combination where Y = 1/Z.

It's also easier to use the formula V0/Vi = Yi/Yf instead of Zf/Zi where here Yi = G1 + sC1 and Yf = G2 + sC2.

So here Vo/Vi = (G1+sC1)/(G2 + sC2). It's then easy to put this in standard engineering form, Vo/Vi = (G1/G2)(1+sC1/G1)/(1+sG2C2) = (R2/R1)(1+sR1C1)/(1+sCR2C2).