Using Matlab to elegantly calculate frequency response

Steven O. skrev:

Usually, nothing is done "at once". One gets an expression H(w) that varies with angular frequency w, and computes the parameters one at the time.

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So you get a computed result that matches the measuerements?

Hmm... depends on what you mean by "elegant." There are program packages that do what you want. (P)SPICE used to be one, for all I know it still is.

However, doing things by pen and paper is a way better method of teaching than just plugging some numbers into a simulator. For some reason I get the impression you take some intro course in electronics.

Well, I am sure somebody do things that way. I am not so sure if that's a very good way of learning the trade.

Rune

Ever heard of APL? You could get a demo copy from IBM's site. APL2 is probably the most powerful array processing language and it supports complex numbers so no problem there. There's a bit of a learning curve but once you have these expressions expressed for f{is}2500, you simply let f be a bunch of frequencies (a vector) e.g. for f from about 12Hz to

Ted

Forget about Matlab, learn first to get the Transfer function. You started right, the currents into each node are zero.

Wrong. Do you remember that the impedance of a capacitor is 1/jwC? So in the second term the current would be: (V3-V2)/(1/jwC2). The third current can not flow into the opamp, so it flows first through R4 and then C1 to gnd. Now a series of R and C are? V2/(R4 + 1/jwC1) Do not forget the j being the imaginary operator sqrt(-1).

We have now to make a loop where te sum of the voltages is zero: V1 + (V2-V1) + (V3-V2) + V3 = 0

You can now rearrange those two equations, substitute V2 and finally calculate V3/V1. This is called the transfer function and gives you values for all frequencies. This is the goal.

Everything below is unnecessary, and in fact you have wasted your time with it. When you think you got the transfer function right, post it here and we can do the Laplace transform to get an easier way of writing it.

Here I made a mistake, the voltages have to be in the same direction as the loop, so if V1 is positive, then the last term(V3) must be negative

This is your problem- line above tells MATLAB to make w a row vector.

I would say the same thing if asked to do that crap with w a row vector.

Set up a loop with index ranging from 1 to Len(f), make w a scalar and build Vm index by index, where Vm[i]=blahblah( some algebra with f[i]).

Ban,

Thanks for taking the time to reply. However, you read a little too hastily. In my full Matlab calculations, included in the original posting, c2 is clearly *defined* as 1/(j*w*). I'm quite sure I got the transfer function correct, both because I checked my algebra carefully, and because the resulting plot is identical to the plot generated by a Multisim simulation of the circuit.

The problem -- again, I refer you to the original posting -- is that the transfer function is complicated enough that extracting w as a function of all the circuit parameters is not even remotely trivial.

Steve O.

"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.

Steven O. skrev:

Eh... might this be part of the problem? There is no need to "extract" w from the transfer function. It is the variable of the tranfer function H(w). All the components make up the parameters.

Rune

Obviously you're missing the point that realization of elegance can only be obtained through a distant non-trivialization of this extraction.

The goal of the exercise is to say, "In order for H(w) to equal 0.707, w must be ." So what I want is w = f (Vin, Vout, component values).

Steve O.

"Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.

Clarification:

1. The DC biases on the amp (V2 = 12V on pin 7 and V3 = -12V on pin 4) have the same labels (V2,V3) as other nodes in the circuit. Since this is an AC problem, lets call the DC biases V20 and V30 to eliminate that confusion.
2. The OP Amp is a voltage amplifier with a large negative gain. Looking at the voltage gain between the OP amp output and inputs yields

Vout = -A*(V3-Vout) = A*V3 / (A-1) ~ V3 (for A >> 1).

1. If you assume Vout = V3, then R4 and C2 are in parallel and the same current flows through R2, R4&C2, and C1. Therefore

(V2-Vin)/R2 = (Vout-V2)/(R4 + 1/jwC2) = jwC1 * Vout

Consequently

( 1/(R4+1/jwC2) ) * Vout - ( (1/R2)+ 1/(R4 + 1/jwC2)) * V2 = -Vin/R2 ( 1/(R4+1/jwC2) -jwC1) * Vout - ( 1/(R4 + 1/jwC2)) * V2 = 0

Which is not the same as mine.

No!

Never again try to calculate a matrix inverse or ratio of determinants to solve linear equations. Enter the command

help slash

then use

V = V\\Vr.

See if this solves your problem.

Hope this helps.

Greg

Best Match for a high school student's needs.

Steven O. skrev:

I don't understand what you mean. If you design the circuit from stratch, there is a spec somewher that says "H(w0) == 0.707" for some specified frequency w0. If you are given a cirquit, with components in place, what you want to do is to plot H(w) over a range of w.

If you want to take a cirquit diagram and solve for the w where H(w)==0.0707, I think you are in for a big job.

Rune

Steve, do not quarrel with me, look what is happening. You have made up one loop with your equations, and you are able to express V2 as a function of V3, but when you enter this into your first equation , you get what I wrote for the loop 0 = 0 which is true, but doesn't help you to solve the transfer function. :-(

In order to solve the thing you have to set up another equation to solve the dependency of V2 from V1. We need another loop to determine the current through C2 and then we can find the current through R2 as the sum of those other two currents.

And please do not write c2 = 1/jwC2 which is misleading and sloppy, write Xc2 or X2 instead, so your professor can understand you.

Tell us what your professor answered to you, maybe it was not so far away from what I said?

You are still wasting your time. Matlab is only able to solve those equations numerically, it doesn't have intelligence. This requires *your* understanding.

Since you still think it is the problem with Matlab, I looked a little closer. What you have is only of first order and that is clearly wrong. So go back to your circuit and draw it on a paper. Look at the buffer. This is a dependent voltage controlled voltage source. Make two loops where you can superimpose the two voltage sources, treat the dependent source as you learned it. Then derive the transfer function V3/V1(w) and finally replace s=jw/wg (the sinosoidal steady state).

It can be done by solving a quartic equation. See below.

Define absH2 = H*conj(H) and h2 = 1/2. Then you are looking for the frequency at which absH2 = h2.

H(jw) = Vout/Vin will be complex and the ratio of two polynomials in jw. Since Vout is the voltage across a capacitor, H(0) = 1 and H(j*inf) = 0. Since there are two independent storage devices in the circuit, the order of the polynomial in the denominator will be

1. Therefore, absH2 will be real and the ratio of two polynomials in w. The order of the polynomial in the denominator will be 4.

Method1: Convert the equation absH2 = h2 into a quartic equation that can be solved for w. This can be solved analytically or numerically.

Method2: Use one of MATLABs minimization routines to minimize (absH-h0)^2

Method3: Define a vector of frequencies. Next find the indices for which max(find(absH2 >= h2)) and min(find(absH2 G4/G4 as w --> 0 --> jwC2/(jwC1)(jwC2)R2 as w --> inf

Hope this helps.

Greg