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I have The following circuit network (link to multisim) and i am asked to find the transfer function, the max and min value, the cutoff frequency, angle and the type of the filter.

[the actual problem][2]][2]

I have done the calculations and found that the circuit is a low pass filter, and its cutoff frequency is 26842 Hz which is awfully high and therefore i am very worried that i have gotten something wrong. I also found that when w=0 the value for the transfer function = 0.6803 and when w approaches infinity the value for the transfer function =0 and hence i concluded that it was a low pass filter. here is my calculate for the transfer function:

$$H_w=\frac{1}{(((R1+R3)*(j*w*C))+(((R3*R1)/(R2))*(j*w*C))+(R1/R2)+1)}$$

here is how it looks after i plug in the values for the resistors: $$H_w= \frac{1}{(1.47 + j*w*0.000054562299999999998836925505310975)}$$

j represents the imaginary number. Basically you can see that if w=0 i get 1/1.47= 0.6803

Anyone can varify this is correct for me? and if so any chance that that "anyone" knows how verify that with multisim as i cant read the graph it produce. Please help.

[2]: https://i.stack.imgur.com/kcbNM.pngenter image description here

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  • \$\begingroup\$ If you can do a small signal ac simulation, you can get the behaviour with frequency. I was expecting H=0.5 for \$\omega=0\$ and H=0 for \$\omega\to\infty\$. \$\endgroup\$ – nidhin Nov 22 '18 at 19:47
  • \$\begingroup\$ so is my transfer function correct and how do i do that ac simulation? \$\endgroup\$ – Reddevil Nov 22 '18 at 19:48
  • \$\begingroup\$ I don't know this tool. But most spice tools would have that option. \$\endgroup\$ – nidhin Nov 22 '18 at 19:52
  • \$\begingroup\$ What tool are you comfortable with? would you please simulate the circuit for me? \$\endgroup\$ – Reddevil Nov 22 '18 at 19:55
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    \$\begingroup\$ Input, Output and reference nodes need to be specified. Don't skimp on information. \$\endgroup\$ – Andy aka Nov 22 '18 at 19:55
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Applying the FACTs is the fastest way to go for this circuit. It is a first-order filter (one energy-storing element) and its transfer function obeys the following expression:

\$H(s)=H_0\frac{1+s\tau_2}{1+s\tau_1}\$

The terms \$\tau_1\$ and \$\tau_2\$ respectively designate the time constants involving the considered energy-storing element (here it is \$C_1\$) when the circuit is observed with a zeroed stimulus (\$V_{in}=0\;V\$, short the source) and when the response \$V_{out}\$ is nulled (0 V despite the stimulus presence). Here, there is no zero and \$\tau_2=0\$.

In this expression, \$H_0\$ represents the quasi-static gain obtained for \$s=0\$. To determine the dc transfer function for \$s=0\$, open the capacitor and redraw the circuit:

enter image description here

The dc gain is immediate and equal to \$H_0=\frac{R_2}{R_2+R_1}\$

Now, for the time constant, simply reduce the stimulus to 0 V and replace \$V_{in}\$ by a short circuit. Then, "look" into the capacitor's connections to determine the resistance. This is the arrow followed by R? in the drawing. You see a resistance equal to: \$R=(R_1||R_2)+R_3\$ leading to a time constant \$\tau_1\$ equal to \$\tau_1=[(R_1||R_2)+R_3]C_1\$. And this is it!

The transfer function is obtained by inspecting the circuit and immediately appears in a low-entropy form:

\$H(s)=H_0\frac{1}{1+\frac{s}{\omega_p}}\$ with \$H_0=\frac{R_2}{R_2+R_1}\$ and \$\omega_p=\frac{1}{[(R_1||R_2)+R_3]C_1}\$

This is the correct way of writing this transfer function: a leading term and a pole clearly factored. The paralleled terms must not be developed: this is what provides insight in this expression and lets you immediately see how the time constant evolves if one of the resistance goes down or approaches infinity. The plot is given below:

enter image description here

You can see how swift it is to get to the result which is expressed in a meaningful form in one shot. Nothing wrong with the matrix form shown below but I feel it is a bit "oversized" for this simple circuit. Lemmy would have said "overkill!".

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Reddevil - yes, the given transfer function (general form) is correct. Perhaps, it is helpful to combine both expressions with (jwC) as a common factor. In this case, the transfer function assumes the classical form H=1/(c+jwT) with c=const. and T=time constant.

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Transfer functions are expressions that represent an output quantity divided by an input quantity. You do not specify an input or output quantity here.

You also need a reference voltage. Voltages are specified as potential differences between points. Without a reference, you would not be able to say node 4 is 5 or 12 or 10000 volts; you could only say that node 4 is 5 or 10 or 1000 volts greater than node 3.

Once you find out what your input, output, reference point, and reference voltage are, you can proceed with finding the transfer function. There are a few different techniques you can use, but I like to use the mesh current method. I sketched an example of the mesh current method for your circuit below. An internet search of mesh current method will tell you how to do it. The mesh current method will give you a system of N linear equations where N is the number of current loops in your network.

You can solve this system of linear equations by hand, or you can write them in matrix form Z*I = V where Z contains all of your R's, C's, and L's; I is your current loops; and V is your voltage. You can then solve for the currents I with a matrix solver like numpy in Python or MATLAB using I = inv(Z)*V. This is a good way to check your answer or solve larger networks. The link below describes this method.

http://www.analyzemath.com/applied_mathematics/electric_circuit_1.html

Once you have solved for your mesh currents, you can find any voltage difference. In this example, the voltage across R2 is R2*(I1-I2). Now, if we know that the reference voltage is 0 volts and the reference point is node 2, then we can say that the voltage at node 3 is 0 + R2*(I1-I2).

Once you know the quantity (a voltage in this case) at your output node, you can find the transfer function by dividing by your input quantity (a voltage in this case). This will be a rational function, and the roots of the denominator are called the poles and the roots of the numerator are called the zeros. Your cutoff frequency is the pole of your transfer function. If you have multiple poles, you will have multiple cutoff frequencies if the poles are unique. This makes sense for a bandpass or notch filter. If the poles are the same, i.e. the denominator of the rational function has repeated roots, then you will only have 1 cutoff frequency but you will have more attenuation after the cutoff frequency compared to if there was only 1 pole at the cutoff frequency.

I get the same transfer function you did

\$H(s)=\frac{R2}{R1 + R2 + sC_1(R_1R_2+R_1R_3+R_2R_3)}\$

enter image description here

syms R1 R2 R3 C1 s v1

% write mesh current equations
Z = [-(R2+R1) R2;
    R2 -(R2+R3+1/(s*C1))];
V = [-v1; 0];
I = inv(Z)*V;

% identify transfer function
tf = (I(2) * 1/(s*C1)) / v1;

% solve poles and zeros
[num,den] = numden(tf);
zeros = solve(num,s); % there are no zeros
poles = solve(den,s);

% numerical evaluation
vars = [R1 R2 R3 C1];
numVars = [100 1e3 1.24e3 1e-9];
cutoff = vpa(subs(poles(1), vars, numVars));
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  • \$\begingroup\$ I see you have now specified inputs and outputs for your problem. I am not going to edit my answer because I think the information is useful for someone just learning about transfer functions. \$\endgroup\$ – DavidG25 Nov 23 '18 at 20:44
  • \$\begingroup\$ Have you solved it fully? Did you get similar values or at least equation to the ones I did ? What's your cutoff frequency then? Thanks a lot for your time \$\endgroup\$ – Reddevil Nov 23 '18 at 22:07

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