Digital Audio Amplifiers

Question

Can anyone tell me why a filter that blocks DC signal in an Amplifier would have a cutoff frequency of 4 Hz . Why would someone choose a frequency as low as 4Hz to filter out DC content before processing the audio signal further ?

Answer 1

In order to block the DC component in a signal, the high-pass filter can be used.

In the figure above, a basic high-pass filter which is known as the passive RC high-pass filter can be seen. The working principle of it depends heavily on the capacitor’s behaviour at varying frequencies as the formula depicts:

$$X_{C}(j\omega) = \frac{1}{j\omega{C}} \Omega$$

The formula is named as the capacitive reactance which is actually a capacitor’s resistance (not the internal resistance) by which an alternating current is attenuated while it passes through it.

So, let’s analyse the given circuit. If the Kirchhoff’s voltage law is applied on that circuit;

$$-x(t) + {\frac{1}{C}\int_{t_{0}}^{t} i(t) \, dt} + V_{C}(t_{0}) + i(t)R = 0$$ $$-x(t) + {\frac{1}{C}\int_{0}^{t} i(t) \, dt} + V_{C}(0) + i(t)R = 0$$

Let’s assume that the capacitor’s inital voltage $V_{C}(0) = 0V$.

$$-x(t) + {\frac{1}{C}\int_{0}^{t} i(t) \, dt} + i(t)R = 0$$ $${\frac{1}{C}\int_{0}^{t} i(t) \, dt} + i(t)R = x(t)$$

If the last equation is differentiated with respect to time and the relation $y(t) = i(t)R$ is used according to the circuit;

$${\frac{1}{C}i(t)} + R\frac{di(t)}{dt} = \frac{dx(t)}{dt}$$ $${\frac{1}{RC}y(t)} + \frac{dy(t)}{dt} = \frac{dx(t)}{dt}$$

By rearranging the last equation, we obtain;

$${\frac{dy(t)}{dt} + \frac{1}{RC}y(t)} = \frac{dx(t)}{dt}$$

Interpreting what is happening in a high-pass filter, low-pass filter, etc. by looking at their differential equations can be demanding. Since these circuits are used to make changes on signals with regards to changes in frequencies, analysing these circuits in the frequency domain can be quite useful. In order to do this, the Fourier transforms of the both sides of the last differential equation can be taken.

$${j\omega}Y(j\omega) + {\frac{1}{RC}}Y(j\omega) = {j\omega}X(j\omega)$$ $${Y(j\omega)}(j\omega + \frac{1}{RC}) = {j\omega}X(j\omega)$$

Consequently, the transfer function $H(j\omega)$ becomes;

$$H(j\omega) = \frac{Y(j\omega)}{X(j\omega)} = \frac{j\omega}{j\omega + \frac{1}{RC}}$$

As it can be seen from the formula, basically, the transfer function results in a complex number. In order to obtain its magnitude, the absolute value of it can be calculated.

$$|H(j\omega)| = |\frac{Y(j\omega)}{X(j\omega)}| = |\frac{j\omega}{j\omega + \frac{1}{RC}}|$$ $$|H(j\omega)| = \frac{|j\omega|}{|j\omega + \frac{1}{RC}|}$$ $$|H(j\omega)| = \frac{\omega} {\sqrt{ {\omega^{2}} + \frac{1}{R^{2} C^{2}} }}$$

This equation relates the input voltage of the passive RC high-pass filter to the output voltage by the input voltage’s frequency. Actually, this equation explains the main working principle of the passive RC high-pass filter by itself.

In order to make the transfer function’s behaviour i.e. the high-pass filter’s behaviour over frequency perceptible, let’s create an example circuit with defined values of $R$ and $C$ and illustrate the magnitude graph of the transfer function. I’ve chosen the values arbitrarily and assumed that $R = 20 k\Omega$ and $C = 1 \mu{F}$.

The MATLAB code for drawing the magnitude graph of the transfer function is given below:

f = 0: 0.01: 20;  % Frequency range in Hz

R = 20000;  % Resistance value in Ohms
C = 0.000001;  % Capacitance value in Farads

omega = 2 * pi * f;  % Angular frequency in rad/s

mtf = omega ./ sqrt((1 / power(R * C, 2)) + power(omega, 2));

plot(f, mtf, "r", "LineWidth", 2);
grid on;
xlabel("Frequency (Hz)");
ylabel("|H(f)|");
title("Frequency Response of the Passive RC High-Pass Filter");

I’ve plotted the graph with respect to the frequency as most of the time, we are dealing with classic frequency rather than angular frequency.

The resultant plot is given below:

As it can be seen from the plot above, the corner/cut-off frequency of the example high-pass filter is around 8 Hz which can be calculated by the formula;

$$f_{C} = \frac{1}{2\pi{RC}} Hz$$

Corner/cut-off frequency is the frequency at which the output signal’s power which is transferred to the load (in our case, it is the resistor $R$) is half of that of the input signal. Alternatively, it is the point at which the output voltage’s magnitude is approximately 0.707 times the input voltages’s magnitude.

So, for our passive RC high-pass filter; if the input signal’s frequency is chosen lower than the corner/cut-off frequency of the filter, the input signal will be attenuated dramatically.

Since a DC signal has the frequency of $f_{DC} = 0 Hz$ ideally, it can be removed from the input signal.

However, I don’t have any idea about why the manufacturer has chosen the cut-off frequency as 4 Hz but choosing it lower than the minimum frequency value of the human hearing range which is 20 Hz is obvious as we don’t want to miss a frequency component while we are listening to music, conversations, etc. via a loudspeaker.

This specific cut-off frequency rating could be about the designer’s state of mind or the circuitry’s needs.