Tutorial: Worked Problems

The order-\(N\) Butterworth low-pass filter with cutoff \(\Omega_c\) has the squared magnitude of Eq. (5.4),

and its \(N\) left-half-plane poles, Eq. (5.7), lie on a circle of radius \(\Omega_c\):

Here \(N\) is the order, \(\Omega_c\) the \(-3\,\text{dB}\) cutoff and \(p_k\) the poles. No frequency transformation is needed: the prototype is designed directly at the prewarped edges.

The two bounds are the linear passband and stopband gains, equivalent to \(G_p=-1\,\text{dB}\) and \(G_s=-60\,\text{dB}\):

With \(T=1\), Eq. (8.11) gives

The naive values \(\omega/T\) would be 0.9425 and 1.8850; the stopband edge moves much further because \(\tan\) grows faster than linearly.

The minimum Butterworth order, Eq. (5.5), is

With \(1/A_p^2-1\approx0.2589\), \(1/A_s^2-1=10^{6}-1\approx10^6\) and \(\Omega_s/\Omega_p=\tan(0.3\pi)/\tan(0.15\pi)\approx2.7013\):

The cutoff that meets the passband edge exactly, Eq. (5.6), is

With \(N=8\):

Substitute \(N=8\) and \(\Omega_c=1.1088\ \text{rad/s}\) into Eq. (5.4):

With \(N=8\), Eq. (5.7) places the eight left-half-plane poles at

so that

Substitute the bilinear transform with \(T=1\) into \(H_{LP}(s)\) from Step 6:

On the unit circle, \(z=e^{j\omega}\) and the bilinear transform gives \(s=j\Omega\) with \(\Omega=2\tan(\omega/2)\). The digital response therefore equals the analog response of Step 6 at the warped frequency:

Passband edge, \(\omega=0.3\pi\)

\(0.8913\ge0.89125\): the passband requirement is met, exactly at the limit, because Step 5 chose \(\Omega_c\) to make \((\Omega_p/\Omega_c)^{16}=1/A_p^2-1\).

Stopband edge, \(\omega=0.6\pi\)

\(6.93\times10^{-4}\le0.001\): the stopband requirement is met with a margin of \(3.2\ \text{dB}\).

The response decreases monotonically from 0 dB at \(\omega=0\), so it stays above \(-1\,\text{dB}\) for \(0\le\omega\le0.3\pi\) and below \(-63.2\,\text{dB}\) for \(0.6\pi\le\omega\le\pi\). Both requirements are met: the passband exactly, the stopband with a 3.2 dB margin. The \(-3\,\text{dB}\) point is at \(\omega_c=2\arctan(\Omega_c/2)\approx0.322\pi\). Figure 9.1 draws the gain against the mask.

Primes mark prototype quantities. The order-\(N\) Butterworth low-pass prototype with cutoff \(\Omega_c'\) has the squared magnitude of Eq. (5.4),

and its \(N\) left-half-plane poles, Eq. (5.7), lie on a circle of radius \(\Omega_c'\):

Here \(N\) is the order, \(\Omega_c'\) the \(-3\,\text{dB}\) cutoff and \(p_k\) the poles.

For a high-pass filter the stopband is the low band and the passband the high band. The two bounds are the linear gains, equivalent to \(G_s=-20\,\text{dB}\) and \(G_p=-1\,\text{dB}\):

With \(T=1\), Eq. (8.11) gives

The low-pass to high-pass transformation of Eq. (8.3.b), with the scaling constant set to the passband edge, is \(s\to\Omega_p/s\). On the \(j\Omega\) axis it maps \(\Omega\) to \(\Omega'=\Omega_p/\Omega\):

The gains \(A_p\) and \(A_s\) are unchanged.

The minimum Butterworth order, Eq. (5.5), is

With \(1/A_p^2-1\approx0.2589\), \(1/A_s^2-1=99\) and \(\Omega_s'/\Omega_p'=2.2361\):

The cutoff that meets the passband edge exactly, Eq. (5.6), is

With \(\Omega_p'=1\) and \(N=4\):

Mapped back with \(\Omega=\Omega_p/\Omega'\), the high-pass \(-3\,\text{dB}\) cutoff is

Substitute \(N=4\) and \(\Omega_c'=1.1840\) into Eq. (5.4):

With \(N=4\), Eq. (5.7) places the four left-half-plane poles at

so that

The high-pass filter follows from \(s\to\Omega_p/s\):

Substitute the bilinear transform with \(T=1\) into \(H_{HP}(s)\) from Step 7:

The four zeros at \(s=0\) map to four zeros at \(z=1\) (\(\omega=0\)).

On the unit circle, \(z=e^{j\omega}\) and the bilinear transform gives \(s=j\Omega\) with \(\Omega=2\tan(\omega/2)\). With \(\Omega'=\Omega_p/\Omega\) in \(|H_{LP}|^2\), the high-pass response is

Stopband edge, \(\omega=0.2\pi\)

\(0.0784\le0.1\): the stopband requirement is met with a margin of \(2.1\ \text{dB}\).

Passband edge, \(\omega=0.4\pi\)

\(0.8913\ge0.89125\): the passband requirement is met, exactly at the limit, because Step 6 chose \(\Omega_c'\) to make \((\Omega_c/\Omega_p)^{8}=1/A_p^2-1\).

\(|H(e^{j\omega})|\) increases monotonically from 0 at \(\omega=0\) to 1 at \(\omega=\pi\), because \(\tan(\omega/2)\) increases monotonically on \(0\le\omega<\pi\). The response is therefore below 0.0784 throughout \(0\le\omega\le0.2\pi\) and above 0.89125 throughout \(0.4\pi\le\omega\le\pi\). The \(-3\,\text{dB}\) point is at \(\omega_c=2\arctan(\Omega_c/2)\approx0.350\pi\). Figure 9.2 draws the response against the mask.

Upsampling by \(U\) raises the rate by 2 and downsampling by \(D\) lowers it by 3, so

Upsampling divides every digital frequency by \(U\); downsampling multiplies it by \(D\). The input edge \(\pi/3\) therefore moves to

This is below \(\pi\), so the converted signal fits in the output band without aliasing.

Inserting \(U-1=1\) zero between samples compresses the spectrum by 2:

Within \(-\pi\le\omega\le\pi\) there are two kinds of copy, each of peak 1: the baseband copy on \(|\omega|\le\pi/6\), and the image centred at \(\pm\pi\), occupying \(5\pi/6\le|\omega|\le\pi\).

The filter must keep the baseband copy and remove the nearest image. The passband edge is the baseband edge, and the stopband edge is where the image begins:

The ideal cutoff sits at the centre of the transition band, Eq. (7.20):

After filtering, the signal occupies \(|\omega|\le\pi/6<\pi/D=\pi/3\), so the downsampler does not alias it. The transition band \(\pi/6<|\omega|<5\pi/6\) contains no signal, so its shape does not matter.

The filter has gain \(G=U=2\) in the passband and removes the image:

with peak 2. The image remains only at the stopband level, at most \(2\delta_s\); Step 7 makes this meet the specification.

Keeping every third sample stretches the spectrum by 3 and scales it by \(1/D\):

with peak \(2/3\). The gain of 2 restores the time-domain amplitude: the area under one period of the spectrum is \(\tfrac{1}{2}\cdot\tfrac{2\pi}{3}\cdot1=\tfrac{\pi}{3}\) for \(X\) and \(\tfrac{1}{2}\cdot\pi\cdot\tfrac{2}{3}=\tfrac{\pi}{3}\) for \(Z\), so \(z[0]=x[0]=1/6\). Figure 9.3.b draws all four stages.

Follow the recipe of Section 7.5.1. The ripple and attenuation are given as positive dB numbers, so \(G_p=-0.1\,\text{dB}\) and \(G_s=-60\,\text{dB}\). Eq. (7.18) gives

A window design produces a unit-gain filter within \(1\pm\delta_p\) in the passband and below \(\delta_s\) in the stopband. Multiplying the coefficients by \(G=2\) scales both deviations by 2. The passband ripple is relative to the nominal gain, so \(\delta_p\) is unchanged. The stopband limit is an absolute level, so the unit-gain design must satisfy \(2\delta_s\le0.001\):

The window must meet the tighter tolerance, Eq. (7.22):

This is the stopband gain the window must reach, \(G_s\approx-66.02\,\text{dB}\).

From the table, the stopband gains are \(-21\) (rectangular), \(-26\) (Bartlett), \(-44\) (Hann) and \(-53\,\text{dB}\) (Hamming): none reaches \(-66.0\,\text{dB}\). Blackman gives \(G_s=-74\,\text{dB}\le-66.0\,\text{dB}\) and \(G_p=-0.0017\,\text{dB}\ge-0.1\,\text{dB}\), so it is the first window that meets both. Blackman has \(K=12\).

Eq. (7.15) sets the main-lobe width \(K\pi/L\) equal to \(\Delta\omega\):

Choose a Type I filter: round \(M\) up to the next even integer, so that the impulse response has a centre sample and an integer group delay.

at the filter's sampling rate \(2F_x\).

The linear tolerances are those of Step 7, \(\delta_p\approx0.0114\) and \(\delta_s=0.0005\), so \(-G_s=66.02\,\text{dB}\). Since \(-G_s>50\), the top case of Eq. (7.17) applies:

Round up to the next even integer for a Type I filter:

The coefficients are \(h[n]=2\,\dfrac{\sin(\omega_c(n-7))}{\pi(n-7)}\,w_K[n]\), \(0\le n\le14\), with \(w_K[n]\) from Eq. (7.16) at \(\beta=6.32\). After scaling by 2, the stopband stays below \(2\times0.0005=0.001\), as required.

The Kaiser design uses \(L=15\) coefficients against Blackman's 19, which is 4 fewer (21% fewer). Both are symmetric Type I filters, so the number of distinct multipliers is \((L+1)/2\): 8 against 10. The delay falls from 9 to 7 samples at the rate \(2F_x\), that is, from 4.5 to 3.5 input sample periods. Blackman is longer because its fixed \(-74\,\text{dB}\) stopband exceeds the \(-66.0\,\text{dB}\) requirement, while \(\beta\) tunes the Kaiser window to the requirement.