BMEN90033 · WEEK 8 · CASCADING FILTERS

Cascading first-order low-pass stages

A first-order low-pass filter has one pole and rolls off at $20\,\text{dB}$ per decade. Chaining $N$ stages multiplies their transfer functions in the ideal case and yields an $N$th-order response with steeper stopband attenuation. The product rule holds only when the stages do not interact. In a passive chain each stage loads the previous one, splitting the repeated pole and degrading the passband. A unity-gain buffer between stages restores the ideal product.

low-pass filter cascade loading buffered stages

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01first-order low-pass

The first-order RC low-pass transfer function

A series RC network with the output taken across the capacitor forms a passive first-order low-pass filter. The impedance divider at complex frequency $s$ gives

$$H(s) \;=\; \frac{V_{\text{out}}(s)}{V_{\text{in}}(s)} \;=\; \frac{1/(sC)}{R + 1/(sC)} \;=\; \frac{1}{1 + sRC}.$$

The denominator has a single root at $s = -1/RC$. Setting $s = j\omega$ gives the magnitude and phase on the imaginary axis:

$$\small \begin{aligned} |H(j\omega)| &= \frac{1}{\sqrt{1 + (\omega/\omega_c)^2}}, \\ \angle H(j\omega) &= -\arctan\!\left(\frac{\omega}{\omega_c}\right), \\ \omega_c &= \frac{1}{RC}. \end{aligned}$$

Magnitude and phase response

Below $\omega_c$ the magnitude is near unity and the phase near zero. At $\omega = \omega_c$ the magnitude is $1/\sqrt{2}$ ($-3\,\text{dB}$) and the phase is $-45^\circ$. Above $\omega_c$ the magnitude falls at $20\,\text{dB}$ per decade and the phase asymptotes to $-90^\circ$. The slider moves the pole along the negative real axis: smaller $RC$ raises $\omega_c$, larger $RC$ lowers it.

A first-order stopband slope is fixed at $20\,\text{dB}$ per decade. Steeper attenuation requires a higher-order filter or a different topology. Chaining first-order stages is the simplest route to higher order.

bode response · H(s) = 1 / (1 + sRC)

02ideal cascade

Cascade response under stage independence

If the stages act independently, with the output of stage $k$ depending only on its own input, the overall transfer function is the product of the stage transfer functions:

$$H_{\text{total}}(s) = \prod_{k=1}^{N} H_k(s).$$

Cascading $N$ identical first-order stages with the same time constant therefore gives

$$H_N(s) = \frac{1}{(1 + sRC)^N}, \qquad |H_N(j\omega)| = \frac{1}{\bigl(1 + (\omega/\omega_c)^2\bigr)^{N/2}}.$$

The asymptotic high-frequency slope is $-20N\,\text{dB}$ per decade and the phase tends to $-90N^\circ$. The pole at $s = -1/RC$ acquires multiplicity $N$.

Composite $-3\,\text{dB}$ frequency

The composite $-3\,\text{dB}$ frequency is not $\omega_c$. Setting $|H_N(j\omega_{-3})| = 1/\sqrt{2}$ gives

$$\omega_{-3} = \omega_c\sqrt{\,2^{1/N} - 1\,}.$$

For $N = 2$ this is $0.64\,\omega_c$; for $N = 4$, $0.43\,\omega_c$. Each added stage contributes a further $-3\,\text{dB}$ at the original cutoff, so the passband edge contracts as the stopband slope steepens. The dashed marker tracks the composite cutoff.

The product rule assumes the stages do not interact. In hardware this fails unless the stages are isolated, because the input impedance of stage $k+1$ draws current from the output of stage $k$.

bode response · H_N(s) = 1 / (1 + sRC)^N

03loaded passive cascade

Response of an N-stage passive cascade

In a passive cascade the stages are wired directly together and each stage loads the previous one. The product rule no longer applies, and the composite poles differ from the isolated-stage pole. The exact response follows from the product of ABCD matrices of the individual sections:

$$\small \begin{aligned} \begin{bmatrix} A_{\text{tot}} & B_{\text{tot}} \\ C_{\text{tot}} & D_{\text{tot}} \end{bmatrix} &= \prod_{k=1}^{N}\begin{bmatrix} 1 + sR_kC_k & R_k \\ sC_k & 1 \end{bmatrix}, \\[4pt] H_{\text{loaded}}(s) &= \frac{1}{A_{\text{tot}}(s)}. \end{aligned}$$

For $N$ identical stages with the same $R$ and $C$, the middle coefficients of $A_{\text{tot}}$ grow faster than those of the binomial expansion of $(1+sRC)^N$. Writing $x = sRC$:

$$\begin{aligned} A_2(x) &= 1 + 3x + x^2, \\ A_3(x) &= 1 + 6x + 5x^2 + x^3, \\ A_4(x) &= 1 + 10x + 15x^2 + 7x^3 + x^4. \end{aligned}$$

Splitting of the repeated pole

The repeated pole at $s = -1/RC$ splits into $N$ distinct real poles along the negative real axis. The slowest pole drifts toward the origin and pulls the composite $-3\,\text{dB}$ frequency below the ideal value; the fastest pole moves outward but does not compensate, since both contribute to the same magnitude. The asymptotic slope remains $-20N\,\text{dB}$ per decade, but the transition broadens, the passband sags, and the phase rolls off earlier. The plot compares the loaded cascade (red) against the ideal product $1/(1+sRC)^N$ (blue).

The error grows with $N$. One stage is exact; two stages already deviate measurably; at four or five stages the loaded response differs substantially from the ideal product, with a slower passband drop-off and a lower effective cutoff. The next section isolates the two-stage case and examines impedance scaling as a partial mitigation.

bode response · loaded N-stage passive cascade vs ideal

s-plane · poles of H_loaded(s)

04the loading problem

Loading in a two-stage passive cascade

Connecting two RC sections directly violates the independence assumption. The input impedance of the second section is finite at all frequencies and draws current from $C_1$. Solving the two-node network with $R_1 = R_2 = R$ and $C_1 = C_2 = C$ gives

$$H_{\text{real}}(s) = \frac{1}{1 + 3sRC + s^2 R^2 C^2},$$

compared with the ideal product

$$H_{\text{ideal}}(s) = \frac{1}{(1 + sRC)^2} = \frac{1}{1 + 2sRC + s^2 R^2 C^2}.$$

The middle coefficient shifts from $2$ to $3$. The repeated pole at $s = -1/RC$ splits into two real poles at $s = (-3 \pm \sqrt{5})/(2RC)$, approximately $-0.382/RC$ and $-2.618/RC$. The slower pole pulls the passband edge inward; the faster pole moves outward but does not compensate, since both contribute to the same magnitude.

Mitigation by impedance scaling

Loading depends on $R_2$ relative to the source impedance at the output of stage one. Raising $R_2/R_1$ reduces the current drawn and recovers the ideal response, provided $C_2$ is reduced in the same proportion to preserve the stage cutoff. The slider sweeps $k = R_2/R_1 = C_1/C_2$; the real and ideal curves converge as $k$ grows.

Even at $k = 100$ the passband sags by a fraction of a decibel. Adding stages compounds the error: a four-stage passive chain at $k = 1$ differs substantially from $1/(1 + sRC)^4$. Loading is intrinsic to passive networks; the general remedy is to isolate the stages.

bode response · two-stage passive cascade vs ideal product

05buffered cascade

Stage isolation with a unity-gain buffer

An ideal voltage follower has infinite input impedance and zero output impedance. Inserting one between successive RC stages forces the input current of stage $k+1$ to zero, so stage $k$ drives only the buffer and its output voltage is unaffected by subsequent stages. The product rule applies exactly:

$$H_N(s) = \frac{1}{(1 + sRC)^N}.$$

The op-amp passes voltage at unity gain. In the non-inverting follower configuration the inverting input is tied to the output, so $V_{\text{out}} = V_+$ and the input draws no current.

Comparison of buffered and passive cascades

The plot shows the buffered $N$-stage response (blue) against the unbuffered passive cascade (red) for the same $N$. The buffered curve follows the ideal $-20N\,\text{dB}$ per decade slope. The passive curve sits below it in the passband and has a broader transition, since its poles are no longer coincident.

The buffer fixes the loading problem at the cost of an active component and a power supply. The same isolation principle underlies Sallen-Key and other active topologies, where the op-amp additionally shapes pole locations to produce Butterworth, Chebyshev, or Bessel responses rather than the repeated-pole product.

bode response · buffered vs unbuffered cascade