How is the block diagram of the RC circuit built?

Write one equation per element, each gives one block, then join them

What does the R1C2 term in the denominator show?

The second stage loads the first, so the result is not a simple product of two first-order lags

2078 Kartik · 4 marks

Block diagram of a two-stage RC circuit

Short answer

Block diagram uses three summing junctions with feedback loops. Overall transfer function is V_o(s)/V_i(s) = 1/[R₁R₂C₁C₂ s² + (R₁C₁ + R₁C₂ + R₂C₂)s + 1], which includes loading term R₁C₂.

Draw the block diagram of circuit shown below. [Figure: input Vi; series R1 to a node; capacitor C1 from that node to ground; series R2 to the output node; capacitor C2 from output node to ground; Vo across C2; mesh currents i1(t), i2(t)]

Write one equation per element so each gives one block, then join the blocks.

Equations (Laplace domain)

Let $V_1$ = voltage across $C_1$ .

\begin{aligned} I_1(s) &= \frac{V_i(s) - V_1(s)}{R_1} \\ V_1(s) &= \frac{I_1(s) - I_2(s)}{sC_1} \\ I_2(s) &= \frac{V_1(s) - V_o(s)}{R_2} \\ V_o(s) &= \frac{I_2(s)}{sC_2} \end{aligned}

Block diagram

 Vi─►(Σ)►[1/R1]─►(Σ)►[1/sC1]─┬─►(Σ)►[1/R2]─┬─►[1/sC2]─┬─►Vo
      ▲-          ▲-         │   ▲-        │          │
      │           └── I2 ────┼───┼─────────┘          │
      └── V1 ────────────────┘   └── Vo ──────────────┘

$V_1$ is fed back (negative) to the first summer, $I_2$ to the second and $V_o$ to the third.

Resulting transfer function

Reducing the three interacting loops (or by Mason's rule) gives

\frac{V_o(s)}{V_i(s)} = \frac{1}{R_1R_2C_1C_2s^2 + (R_1C_1 + R_1C_2 + R_2C_2)s + 1}

The term $R_1C_2$ shows the loading of the second stage on the first; the result is not simply the product $\dfrac{1}{(1+sR_1C_1)(1+sR_2C_2)}$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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Block diagram of a two-stage RC circuit

Equations (Laplace domain)

Block diagram

Resulting transfer function

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