How could i determine a transfer function (A(s)) which responds to a symmetrical 1Vp-p 1kHz square-wave input signal, and gives a 5Vp-p square-wave output signal
with tilt==> δ=10% and rise/fall time ==>tr=tf=10μs
Electronic – Square Wave transfer function
circuit-designtransfer function
Related Solutions
There isn't an explicit formula for n cascaded RC low pass filters (as you have shown) because as you move from left to right, R2 and C2 load the output of R1 and C1 thus changing R1&C1's response and because R3 and C3 are doing the same to R2 and C2 this in turn reflects on the loading of R1 and C1.
Using 2 port networks and matrix maths does help you solve more easily for several sections but the formulas do tend to get long after a few sections and solving for the cut-off frequency and Q is very tiresome.
The best solution is to use a simulator like LTSpice.
Given that you have an op-amp in your circuit acting as a buffer it wouldn't be unreasonable to use this to help. This sallen-key calculator is very good and gives you the response and formulas for a 2nd order low pass filter: -
Cascading 2 of these will probably get you a better response than cascading five passive RC filters and the beauty of this method is that you can cascade them without interaction between components because the op-amp acts as a buffer.
Assuming when you say transfer function you are not referring to an S-domain "transfer function" for the behaviour (as this is actually really hard and relies on alot of heaviside functions) but more of a relationship of input to output.
ASUMMUMING the supply is an ideal supply with each voltage sources being 120degree's separated and their amplitudes & freq are all the same and that there is no supply feeder impedance
With the firing angle = 0 and thus the SCR's act like diodes, we know that:
Vd = \$\frac{3\sqrt{2}}{\pi}V_L\$
Where Vd = DClink voltage and \$V_L\$ = the Line-Line (rms) voltage from the supply
We want to know what Vd is with regards to some arbitary firing angle \$\alpha\$
\$V_\alpha = Vd - \frac{A_\alpha}{\pi/3} \$
\$A_\alpha\$ is the volts-second area that occurs every 60degrees which reduces the average DClink.
We know that: \$V_a = \sqrt{2} V_L Sin(\omega t)\$
Thus
\$A_\alpha = \int\limits_0^\alpha \sqrt{2}V_LSin(\omega t) d(\omega t) \$
\$ = \sqrt{2}V_L(1-cos\alpha)\$
Thus
\$ V_\alpha = \frac{3\sqrt{2}}{\pi} V_Lcos\alpha \$
Best Answer
A 1st order approximation using :
For a 1kHz square wave with T/2 sag or tilt by 10%
\$\tau_1=5ms =0.5ms/ 0.1(sag)= 0.5ms/0.1 = 1/\omega_1\$ = HPF
Sine tr=0.35/f_{-3dB} BW for rise time 10 to 90%
\$\tau_2=4.55us=1/\omega_2 =t_r/(2\pi *0.35)=10us/(2\pi *0.35)\$ = LPF assuming 10~90% for rise time
Then the transfer function can be made with gain constant =5
\$H(s)=5\dfrac{0.005s+1}{(4.55*10^{-6}) s+1}\$