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.
Transfer function and its magnitude
filterimpedance-matchingtransfer function
Related Solutions
The poles in your \$H(s)\$ are \$s = -3\$ and \$s = -2\$ because they make the denominator zero. I'm not sure why you think the Bode plots suggest the poles are positive, but perhaps your confusion has to do with the fact that a Bode plot uses \$j\omega\$ as the \$x\$-axis where \$\omega\$ is the angular frequency. The poles are on the real (\$x\$) axis in the \$s\$-plane so they are symmetric about the imaginary (\$y\$) axis, meaning the Bode plot is the same whether the frequency \$\omega\$ is positive or negative.
The source of the confusion may also be due to the fact that there is an error in the second link you posted. The author uses the form
$$H(s)=A\frac{(s/z_0+1)(s/z_1+1)\cdots(s/z_n+1)}{(s/p_0+1)(s/p_1+1)⋯(s/p_n+1)}$$
for the transfer function but claims that the poles are at \$s = p_0\$, etc. This is incorrect in general because at \$s = p_0\$ the relevant term of the denominator is \$p_0/p_0 + 1 = 1 + 1 = 2 \neq 0\$. The pole is actually at \$s = -p_0\$ so that the relevant term of the denominator is \$-p_0/p_0 + 1 = -1 + 1 = 0\$. The author either meant to say that the poles were at \$s = -p_0\$, etc., or use the form \$s/p_0 - 1\$ for each term.
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Best Answer
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.