# High Pass Filter Design from Transfer Function

I need to design a circuit that implements the following transfer function:

$$\G(s)= G\frac{s+z}{s+p}\$$

where, G = gain, p = pole and z = zero.

• What attempts have you made so far? – a concerned citizen Oct 22 '18 at 5:58
• tried implementing a RC parallel combination in series with a resistor of which the output voltage is measured across, however, the bode plot doesn't match up to the transfer function's - calculations as follows: TF into standard form: G(s)= 4.15 (s+530)/(s+2200) G(s)= (4.15s+2200)/(s+2200) G(s)= ((4.15/2200)s+1)/((1/2200)s+1) Circuit transfer function: V_out/V_in = ((Ts+1))/((αTs+1) ) Where, T=R_1 C_1 α=R_2/(R_1+R_2 ) Compare with standard form: T= 4.15/2200= R_1 C_1 α= 1/2200T=1/4.15= R_2/(R_1+R_2 ) Find values: R_1=3.15R_2 C_1=4.15/(R_1×2200) – user361222 Oct 22 '18 at 6:10
• What if you try R-C-R, with the output being across C-R? – a concerned citizen Oct 22 '18 at 6:28
• @aconcernedcitizen I have the same issue – user361222 Oct 22 '18 at 6:36
• C and R are in series, nothing in parallel. Think of it as a resistive divider, with a series capacitor with R2. – a concerned citizen Oct 22 '18 at 6:50

When s=0 and infinity the G(s)=1.

The transfer function at mean midpoint of sqrt(530*2220)=1070 inserted for s becomes ,

G(1030)=0.486

Thus a unity gain low pass then high pass parallel & series T filter is needed or an active equivalent.

## Hypothesis

So the transfer function resembles a passive loudness switch that cuts the audio midrange 20log0.486=-6.3dB. My Bogan stereo tube amp during the ‘60’s and ‘70’s had the -15 dB preferred solution over the bass-boom box version used since then.

One reason for this type of response to cut the loudness in the midrange only, rather increase, may be due a phone call, so we cut the midrange only then restore to flat for full loudness. It seems only Bogan got it right. Much after Baxandall invented it, stereo designers used the loudness switch boost crappy speakers bass response below 100 Hz rather than cut the midrange to match our hearing response according to the well known Fletcher-Munson curves. developed before 1933. So this transfer function is a simple approximation for these curves to listen at -6dB very slightly lower volumes.

The next clue is the passive Baxandall dual T filter fixed mid-range filter. Peter J.Baxandall invented this filter just before I was born in 1952. The variable bass treble version is still in use today in old tuners.

Can you imagine an RCR//CRC filter to do this with the R ratios being 2200/520?

If so , you may be as smart as Bax. If not learn how to find his answer. This is for your education, learning how to learn on your own without spoon-feeding.

That’s MY reason for not giving you the solution.

The answers seem to be confusing.

$$G(s) = 4.15\frac{s+530}{s+2200}$$

This transfer function at DC goes to $$\G(s) \xrightarrow{s\to 0}\ \sim1\$$, while for higher frequencies it goes to $$\G(s) \xrightarrow{s\to +\infty} 4.15\$$.

There is no way to make this filter in a passive way if the gain remains >1. I suggest looking up filters using opamps.

I suggest the following configuration:

simulate this circuit – Schematic created using CircuitLab

At low frequencies the capacitor acts like an open circuit, yielding a gain of 1. At high frequencies, the capacitor will short-circuit, producing a gain of $$\1+\frac{R_1}{R_2}\$$.

The transfer function is given as (using the EET theorem on the capacitor):

\begin{align} H(s) &= \frac{1 + \frac{Z_n}{Z}}{1 + \frac{Z_d}{Z}}\\ & = \frac{1 + (R_1+R_2)C_2s}{1 + R_2C_2s}\\ &= \left(1 + \frac{R_1}{R_2}\right)\frac{s+\frac{1}{(R_1+R_2)C_2}}{s+\frac{1}{R_2C_2}} \end{align}

We already determined that $$\1+\frac{R_1}{R_2} = 4.15\$$, and then you can choose either $$\\frac{1}{(R_1+R_2)C_2}=530\$$ or $$\\frac{1}{R_2C_2} = 2200\$$ depending on which one you want to be exact.

To make the filter perfectly, you can for example use this schematic:

simulate this circuit

The gain of this amplifier is

$$A(s) = \frac{C_1s + G_1}{C_2s + G_2}$$

which almost gives you a one on one mapping.

• Yeah I have no issue with the gain, I'm using a non-inverting amplifier with a 1k and 3.15k resistors to produce the gain necessary, it's the passive component of this which I am stuck. – user361222 Oct 22 '18 at 10:27
• @user361222 Maybe you are referring to 4.15k and 1k if you say you are using an opamp? – a concerned citizen Oct 22 '18 at 11:54
• @user361222 I have edited my post to show one possible solution. – Sven B Oct 22 '18 at 11:55
• @SvenB what do you mean by choose either ?? I need both to do so .. or do you mean choose one of them and the values for the components will sort themselves out as a result and all conditions will be satisfied? – user361222 Oct 22 '18 at 12:05
• Yes, since $G(s\to 0)\approx 1$ I took the liberty of having a gain of 1 at DC by construction. So it should sort itself out if you choose either equation. You will be off by a small margin due to the approximation. If the <0.1% difference bothers you then you can alter the circuit a little bit. – Sven B Oct 22 '18 at 12:51