# Electronic – How to calculate the transfer function

circuit analysisfilterhigh pass filterlow passmultisim

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.

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:

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:

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!".