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Can anyone make the transfer function and magnitude of the transfer function more intuitive for me? I understand using them but I feel like there is still a gap in my understanding. I would like to get a more intuitive understanding of the transfer function. I have Googled and asked and it's still not that easy to grasp.

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  • \$\begingroup\$ This question is too broad as is. If you have a specific transfer function in mind and don't know how to derive it we can help explain it. \$\endgroup\$ – Null Oct 24 '14 at 5:02
  • \$\begingroup\$ I need help with the broad. I can derive them i don't know if I understand what it is exactly, the bottom answer did help a bit though. \$\endgroup\$ – Kyle Calica-St Oct 24 '14 at 18:01
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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.

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I guess that what you call transfer function is really the frequency response function. The transfer function is a complex function of the complex variable \$s\$. If \$H(s)\$ is your transfer function the corresponding frequency response function is \$G(\omega) = H(j\omega)\$, which is a complex function of the real variable \$\omega\$. This is a bit of nitpicking, though, but I still want to be precise, since I guess you are looking for an explanation of the "physical" meaning of G, not H.

The magnitude and the phase of G are in turn two real functions of \$\omega\$ and they are called respectively:

  • amplitude response (function): \$ ~~~ A(\omega) = |G(\omega)| \$
  • phase response (function): \$ ~~~ \Phi(\omega) = \angle G(\omega) = \arg[G(\omega)] \$

What do they mean physically? They are related to how the system responds to a sinusoidal excitation (input) \$x(t)=\sin(\omega t)\$. It can be shown that when such a signal is applied to the input, the output \$y(t)\$ will be still sinusoidal with the same frequency, but with different amplitude and phase, and these latter depends on the amplitude and phase response of the system:

\$ y(t) = A(\omega) \sin(\omega t + \Phi(w) )\$

That is, the input signal will be scaled by an amount given by the amplitude response computed at the frequency of the signal, and it will be phase-shifted by an amount given by the phase response computed at the frequency of the signal. This is true for any form of sinusoidal input, not only for a unitary-amplitude zero-phase signal, as the one I used for simplicity.

Since any periodic signal of frequency \$\omega\$ can be decomposed using Fourier series as a (possibly infinite) sum of sinusoidal signals (harmonic components) with frequency integer multiple of \$\omega\$ (\$\omega\$, 2\$\omega\$, 3\$\omega\$, ...) and since the system we are talking about is linear and thus superimposition applies, the amplitude and phase responses (usually plotted as bode diagrams) express how the harmonic components of the output will be scaled/phase-shifted with respect to the corresponding (same frequency) component of the input. Therefore the frequency response function, with its amplitude and phase, tells us how the system will alter the harmonic components of a periodic input signals.

The explanation can be generalized to aperiodic input signals, but then you must reason in terms of Fourier transforms (no more Fourier series) and the concept of harmonic component is more fuzzy (you can see an aperiodic signal as a sum of infinite harmonic components with infinitesimal amplitude and infinitely close in frequency, and so intuition must be supported by a generous amount of math calculus).

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