Electrical – How to find Magnitude of transfer function

Every circuit - even each amplifier - has frequency-dependent properties. Of course, in many cases (filters) this is a desired property. That means: Applying a voltage at any node (in most cases: input of the circuit) will cause a current distribution within the circuit that depends on the frequency of the input voltage. In particular, the signal (current or voltage) at the node which is defined as "output" will have a magnitide (and, of course, a phase shift) that depends on this frequency.

Now - the term "transfer function" is used for active or passive circuits which shall exhibit a desired frequency-dependence between well-defined input/output nodes. For example, a lowpass must provide a nearly constant output voltage if the input signal has a frequency between 0 and a certain upper limit (say: f1). For frequencies above f1 the output voltage starts to fall continuously (and the phase relation between the signals changes also). Note that this is a simplified descriptions of the real behaviour only.

The transfer function of such a circuit mathematically desribes the relation between input and output (magnitude and phase) . Therefore, the transfer function is a complex function. Example: The transfer function for a simple passive first-order RC-lowpss is

H(jw)=1/(1+jwRC)

Remark (edit): In the time domain, you can create a set of differential equations based on the state variables of the system. To solve this system (isolating the output-to-input ratio) you can make an exponential "ansatz" [(exp(st)]. It turns out

(a) that the variable "s" can be interpreted as a frequency that has a real and an imag. part (s=sigma+jw), and

(b) that the polynominal P(s) of the characteristic equation [1+P(s)=0] for solving the set of equations also appears as the denominator in the transfer function H(s)=N(s)/P(s) setting s=jw.

This is an important relation between the two domains: time and frequency domain.

First off, we transform the s domain to frequency domain with \$s=j\omega \$ which gives us:

$$T(j\omega) = \frac{1}{(j\omega+1)((j\omega)^2+j\omega+1)} = \frac{1}{(j\omega+1)(j\omega+1-\omega^2)} = \frac{1}{1-2\omega^2+j(-\omega^3+2\omega)}$$

Notice i wrote the denominator in the form \$a+bj\$. Taking the magnitude of this function gives:

$$\left |T(j\omega)\right |= \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{(1-2\omega^2)^2+(-\omega^3+2\omega)^2}} = \frac{1}{\sqrt{(4\omega^4-4\omega^2+1)+(\omega^6-4\omega^4+4\omega^2)}} =\frac{1}{\sqrt{1+\omega^6}} $$