S = A sin (w t + f )

 

Where A is the amplitude, w is the frequency in radians/sec, t is time, and f is the phase shift or lag. Consider a blood pressure wave as shown in the figure (top curve). The ordinate axis is measured in units of pressure (mm Hg) or as a voltage if there has already been transduction of the biomedical signal to an electric current and voltage. The abscissa is in units of time (seconds or milliseconds). Now, if we take the Fourier transform of this signal, it means to represent the signal as a set of component sinusoidal frequencies. These are shown below the signal. They can be considered as individual signals that when summed together, form a reconstructed version of the original signal. First is shown the average signal level, or 0^th harmonic. The basic sinusoidal wave is the fundamental frequency, or 1^st harmonic. It has the same period as the signal itself, and is shown just below the 0^th harmonic. Note that its peaks are approximately aligned with the peaks of the blood pressure signal. Intuitively, this waveform alignment would most likely be useful to reconstruct the original signal, when summed with higher order frequency components. The second harmonic is shown just below the first harmonic and it has twice the frequency (half the period). Other harmonics are shown down to the eighth. Note that they tend to have diminishing amplitude with higher order harmonic and their peaks are shifted with respect to one another. These types of changes in the higher frequency harmonics are typical of cardiac signals.

Now, let us say the signal is amplified by 2x. It is the same to say: "multiply every discrete point in the signal by two". How are the Fourier harmonics changed? They are just doubled in amplitude too! They are all changed linearly, by the same amount, and so there is no change in their relative relationships. The signal has not been filtered as a result; rather it has been amplified. But what if we filter the signal? Will the relative amplitudes of the sinusoidal components change? Yes! And so will the timing of the peaks with respect to one another (called time shift or phase shift or lag - because the phase component in the equation given above changes!). Filters attenuate certain frequencies of the signal. If we use a high pass filter, high frequencies of the signal will pass through the filter unscathed, while the low frequencies will be attenuated. For high pass filters, the 0^th frequency is always completely attenuated (i.e., the average level of the signal goes to 0). If we use a low pass filter, low frequencies of the signal will pass through the filter unscathed, whilst the high frequencies will be attenuated. Let us investigate how this works.

Consider the elements of a simple high pass filter: a resistor and a capacitor. To form a high pass filter, they must be arranged so that low frequencies only are attenuated from input to output. How can we do this? Consider the arrangement shown in the diagram at left. The capacitor is labeled C and the resistor is labeled R. The capacitor has an impedance of 1/jwC, i.e., it depends on frequency. The impedance approaches 0 as the frequency approaches infinity, and vice versa. We can write Ohm's Law which is the first step in determining the transfer function:

Where V is the voltage drop across a particular impedance element (also called the potential difference), I is the current flowing through the element, and X is the impedance. We can also state Kirchoff's Current Law as follows: "The current entering a node (connection point between circuit branches) must equal the current leaving the node." Hence, i₁ must equal i₂ in the figure (since V_o is only connected to the node between C and R, it is just an extension of that node, not a distict branch). For the circuit elements shown, therefore, we can then state the relationship between the current in each branch as follows:

i₁ = (V_i - V_o) / (1/jwC) = i₂ = (V_o - 0) / R

After CAM (considerable algebraic manipulation) we would find that the transfer function (t.f.) is:

V_o(w)/V_i(w) = jwRC / (1 + jwRC)

Let us see how this acts as a high pass filter by a simple example. For low frequency components of the signal (those that approach zero), the numerator will approach 0, while the denominator will approach 1. Hence, the fraction V_o/V_i will approach 0, and therefore, for the low frequency components of V_i, the output components will be attenuated to near zero. We say "near zero" and not zero. Why? Because the fraction only actually reaches 0 when w = 0. What component of the signal has w = 0? The 0^th or DC component, which is also called the average level of the signal. It is a straight line; therefore its frequency is zero (it has no periodicity). As w → ∞, the denominator → jwRC, which becomes >>1 (where → is the symbol for "approaches" and ∞ is the symbol for infinity). Ergo, the fraction → 1, i.e., V_o (w) → V_i(w). It means that high frequencies are passed with almost no attenuation.

 

Now then, let us consider the precise relationship of V_o(w)/V_i(w) to w. We can graph it as shown in the figure. At w = 0, V_o/V_i = 0. The lower frequencies are substantially attenuated until the corner (cutoff) frequency, w_c. How is the cutoff expressed quantitatively? It is simply 1/RC, i.e.:

V_o(w)/V_i(w) = (jw/w_c) / (1 + jw/w_c)

Where RC is also known as the time constant, symbolized as t . The units of resistance (ohms) times capacitance (farads) gives time in seconds. Such information is very helpful in designing the filter, because using it we can adjust the cutoff frequency of the filter by simply adjusting the value of R and/or C. If for example we would like a cutoff frequency at 1 HZ, then we can use the relationship between frequency and RC to choose the appropriate circuit elements:

RC = 1/w_c = 1/ 2pf_c

Since resistors come in many resistance values using the same package size, often a readily available capacitor is first selected, and then the value of R is chosen to attain the appropriate cutoff frequency. The design of the filter must not only include the cutoff frequency, which thereby, determines the passband, but also the rolloff (degree to which the signal is attenuated as frequency varies). Additional quantitative definitions of the cutoff frequency and the rolloff will be discussed in the section on Bode plots, below.

 

Bode Plots

This is a graphical representation of the filter function (plot of the transfer function versus frequency). A first-order filter has a single pole, with a slope of -1 when plotted on log/log coordinates (i.e., -20db/decade of frequency on a Bode plot). A value of -20db/decade means that there is a drop of -20 decibels (= -20log₁₀(V_o/V_i), where V_o/V_i = 10) per decade change in frequency (per order of magnitude change). For example, this is shown at left for a low pass filter. Note both axes in this plot are log base 10. The 1^st order transfer function, similar to the equation for 1^st order high pass filter given above, is:

V_o(w)/V_i(w) = 1 / (1 + jw/w_c)

|V_o(w)/V_i(w)| = 1 / sqrt[(1 + (w/w_c)²)]

20 log₁₀ |V_o(w)/V_i(w)| = 20 log₁₀ {1 / sqrt[(1 + (w/w_c)²)]}

When w = 0, then 20 log₁₀ |V_o(w)/V_i(w)| = 0dB. When w = w_c (i.e., when the frequency of the signal is at the corner frequency of the filter), then 20 log₁₀ |V_o(w)/V_i(w)| = -3dB. When w = 10 w_c, V_o(w)/V_i(w) = 1/10, and 20log |V_o(w)/V_i(w)| = -20dB. Hence, the slope of this filter is -20dB/decade frequency change along the rolloff.

 

 

RLC (2^nd order) low-pass filter

The filter at left is a second order low pass type. It is a very common filter and can be made of discrete elements. Let us compute the transfer function of this filter:

(V_i - V_o) / (R + jwL) = (V_o - 0) / (1/jwC)

(V_i - V_o) / (V_o) = (R + jwL ) jwC

V_i / V_o = (jwRC - w²LC ) + 1

V_o / V_i = 1 / [1 + jwRC + (jw)² LC]

Where w_c² = 1/LC (the corner or cutoff frequency) and Q = w₀L/R (called the quality factor). The jw term can also be written as 's'. Taking the magnitude of the last equation:

|V_o / V_i| = 1 / sqrt {[1 + jwRC + (jw/w_c)² ]²}

20 log₁₀ |V_o / V_i| = 20 log₁₀ {1 / sqrt {[1 + wRC + (w/w_c)² ]²}}

Therefore, when w changes by 1 decade (w/w_c = 10), then (w/w_c)² = 1/100, 1 / sqrt {[1 + wRC + (w/w_c)² ]² µ 1/100, and the transfer function changes by -40dB per decade. Hence the sharpness of the rolloff has doubled in units of decibels. This is a 2-pole filter, because of the quadratic term, which characteristically has a rolloff of -40dB per decade. Similarly, 3-pole filters have rolloffs of -60dB per decade, and so on for higher order filters. High pass filters have the same rolloff properties, except that the low frequencies are attenuated rather than the high frequencies.

Concerning the phasic properties of the filters, each pole of the filter results in a phase shift of -45° at the cutoff frequency and above. Hence the 1-pole filter shifts the phase by -45°, and the 2-pole filter shifts it by -90° along the rolloff. For more information on Bode Plots, see this link. We could do an additional lecture on it, if the class would like.

We have spoken about frequency components of signals. Is it possible to construct a signal from its component frequencies using discrete analog hardware circuitry? Yes! Devices called analog multipliers can take two signals at the inputs and multiply them to form the output signal. We know from trigonometry that

sin²(w ) = 1/2 - 1/2 cos(2w )

cos(2w ) = sin(2w - 90°)

We can use these equations to create integer multiple frequencies of the original frequency w . Firstly, we can input sin(w) to both inputs of the multiplier and scale and shift appropriately to form sin(2w). It would be possible, therefore, to adjust a series of integer multiple frequency components in hardware to construct models of various biological signals, and then to show how the signals are affected by filtering at later points in the circuitry. One might use this, for example, to show how the blood pressure signal is filtered as it is transmitted from heart to peripheral vessels.

More info on higher-order filter properties