Approximation of derivatives method to design IIR filters

Earlier in this digital signal processing course, we saw that we could design digital IIR filters using analog filters. One of the methods of achieving a stable digital filter is by the approximation of derivatives method.

The contention here is that a differential equation can represent an analog filter. We know that we can use transformations like the z-transform to convert a differential equation into a difference equation.

Once we have the transfer function of an analog filter, we can z-transform it to get the transfer equation of the corresponding digital equivalent filter.

The approximation by derivates method is also known as the backward difference method.

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Derivation of the transfer functions for analog and digital filters

Consider an analog filter. It’s transfer function will be of the differential equation form as shown below.

$y(t)\quad =\quad \frac { dx(t) }{ dt }$

Taking Laplace transform on both ends:

$Y(s)\quad =\quad sX(s)$

The transfer function of the analog filter can be given by:

$\frac { Y(s) }{ X(s) } \quad =\quad H(s)\quad =\quad s$

Note that this is just a generic transfer function. To get the transfer function of a digital filter we will dive into some specifics.

approximation of derivatives method to design a digital filter — Consider an analog input to the analog filter

Next, we are going to input a sample analog signal to this analog filter and find the transfer function in terms of this analog signal. We will z-transform this transfer function to get the digital filter’s transfer function.

$y(nT)\quad =\quad \frac { x(nT)-x(nT-T) }{ T }$

Thus,

$y(n)\quad =\quad \frac { x(n)-x(n-1) }{ T }$

Taking z-transform on both sides:

$Y(Z)\quad =\quad \frac { X(Z)-{ Z }^{ -1 }X(Z) }{ T }$

Solving for the transfer function of the digital filter:

$\frac { Y(Z) }{ X(Z) } \quad =\quad H(Z)\quad =\quad \frac { 1-z^{ -1 } }{ T }$

Now that we have the transfer function for the digital filter we can do the following:

Equate this with the transfer function of the generic analog filter’s transfer function.
Replace this value in the place of ‘s’ in any analog filter’s transfer function to get a digital version of that particular analog filter.

The mapping between S and Z planes

To map from s to z-plane we need to find the values of σ and Ω in s = σ + jΩ. We have the relationship between s and z from above as:

$s\quad =\quad \frac { 1-{ z }^{ -1 } }{ T }$

Rewriting the above equation for z:

$z\quad =\quad \frac { 1 }{ 1-sT }$

Substituting s = σ+jΩ in the above equation and solving gives us:

$z\quad =\quad \frac { 1-\sigma T+j\Omega T }{ { (1-\sigma T) }^{ 2 }+{ (\Omega T) }^{ 2 } }$

Separating the real and imaginary parts we get:

$z\quad =\quad \frac { 1-\sigma T }{ { (1-\sigma T) }^{ 2 }+{ (\Omega T) }^{ 2 } } \quad +\quad j\frac { \Omega T }{ { (1-\sigma T) }^{ 2 }+{ (\Omega T) }^{ 2 } }$

For σ=0

$z\quad =\quad \frac { 1 }{ 1+{ (\Omega T) }^{ 2 } } \quad +\quad j\frac { \Omega T }{ 1+{ (\Omega T) }^{ 2 } }$

When we vary Ω from -∞ to +∞, the corresponding locus of points in the z-plane is a circle with radius 1/2 and with its center at z=1/2.

Similarly, when we map the equation $s\quad =\quad \frac { 1-{ z }^{ -1 } }{ T }$ the left half-plane of the s-domain maps inside the circle with 0.5 radians. Moreover, the right half-plane of the s-domain is mapped outside the unit circle.

Mapping of s-plane into the z-plane by the approximation of derivatives method (Source)

Thus we can say that this transformation of an analog filter results in a stable digital filter.

What are the limitations of the approximation of derivatives method of designing a digital IIR filter?

As we can see from the mapping above, the location of poles in the z-domain is confined to smaller frequencies. Thus the approximation of derivatives method is limited to designing low pass and bandpass IIR filters with small resonant frequencies only. It can’t be used to develop high pass and band-reject filters.

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