# How is this transfer function generated using this LTIC circuit diagram?

I am studying for my signals and systems final in 3 hours and cannot seem to wrap my head around one of the homework/practice problems. We are given the following figure: For each figures (1, 2, and 3) we are asked to prove that the 3 dB bandwidth of the system is:

1. Wc
2. 10Wc
3. Wc/10

I understand how to go about doing the first system. I first calculate |H(w)| and then get |H(0)| and |H(Wc)| and then run the results through 20log(|H(Wc)|/|H(0)|) (which comes out to -3 dB).

However I cannot figure out how to calculate the transfer function for the next two systems. The solutions provide me with this formula: Which for the second system ends up being However I would love to understand how they got to that point. I guess I do not understand how to translate the circuit diagram into a new equation.

Can anyone provide some hints in this regard? Thanks!

• You get an equation of the form y(t) = f(t) - 9y(t) for the second one, you have to do some fiddling to expand and get all the y(t)'s on one side and all the f(t)'s on the other, and yes feedback networks can be a pain to analyse sometimes. – Sam Dec 18 '16 at 18:51
• The equation for G'(s) is the standard expression for a basic negative feedback closed loop system: G(s) is the forward path; H(s) is the feedback path. In fig 2, G(s)=wc/(s+wc); H(s)=9 – Chu Dec 18 '16 at 19:18
• EE.SE supports MathJax, could be helpful to improve readability of your equations :) – uint128_t Dec 18 '16 at 20:34

For example: The third system with positive feedback - let's call $$\H(s)\$$ the resulting system:

$$\frac{Y(s)}{F(s)} = H(s) = \frac{\omega_c}{s+0.1\omega_c}$$

Putting the pole in the standard form $$\\frac{s}{\omega}+1\$$ (in order to highlight the pole frequency):

$$H(s) = \frac{10}{\frac{s}{\frac{\omega_c}{10}}+1}$$

So, the -3 dB frequency for this first order system is $$\\omega_0=\frac{\omega_c}{10}\$$ rad/s.

Alternatively, there is a long form to derive this result. Begin doing $$\s = j\omega\$$:

$$H(j\omega)= \frac{10}{\frac{j\omega}{\frac{\omega_c}{10}}+1}$$

The module is:

$$\left |H(j\omega) \right |= \frac{10}{\sqrt{\left ( \frac{10\omega}{\omega_c} \right )^2 +1}}$$

If $$\\omega_0\$$ is the -3dB cutoff frequency (in other words, the frequency when the amplitude reaches 0.707 of maximum one), we can write $$\\left | H(j\omega_0) \right |= \frac{1}{\sqrt{2}}\left | H(0) \right |\$$.

As $$\\left | H(0) \right | = 10\$$:

$$\frac{\left |H(j\omega_0)\right |}{\left | H(0) \right |} = \frac{1}{\sqrt{\left ( \frac{10\omega_0}{\omega_c} \right )^2 +1}} = \frac{1}{\sqrt{2}}$$

Resolving for $$\\omega_0\$$:

$$\omega_0 = \frac{\omega_c}{10}$$

For the 2nd system, the output from the summing junction is: -

$-9y(t) + f(t)$ therefore the final output y(t) equals: -

$(-9y(t) + f(t))\times(\dfrac{\omega}{s+\omega})$

Solving we get: -

$y(t)(1 + \dfrac{9\omega}{s+\omega})= f(t)\dfrac{\omega}{s+\omega}$

And avoiding a few dreary steps....

$\dfrac{y(t)}{f(t)} = \dfrac{\omega}{s+10\omega}$