Skip to main content

Block diagram and Vo/Vi of L-R ladder network

Short answer

The block diagram is built from I1 = (Vi - V1)/(sL1), V1 = R1(I1 - I2), I2 = (V1 - Vo)/(sL2), Vo = R2 I2. Reduction gives Vo/Vi = R1 R2 / [L1 L2 s^2 + (L1 R1 + L1 R2 + L2 R1)s + R1 R2], a second-order low-pass filter.

Draw the block diagram and reduce it to calculate Vo(s)/Vi(s) for the following network. [Figure: input Vi; series L1 to node 1; R1 from node 1 to ground; series L2 from node 1 to node 2; R2 from node 2 to ground; Vo across R2]

Let I1I_1 = current in L1L_1, V1V_1 = voltage across R1R_1 (node 1), I2I_2 = current in L2L_2 and R2R_2, VoV_o = voltage across R2R_2.

Element equations (Laplace)

I1=Vi−V1sL1V1=R1(I1−I2)I2=V1−VosL2Vo=R2I2\begin{aligned} I_1 &= \frac{V_i - V_1}{sL_1} \\ V_1 &= R_1(I_1 - I_2) \\ I_2 &= \frac{V_1 - V_o}{sL_2} \\ V_o &= R_2I_2 \end{aligned}

Block diagram

 Vi +   ┌─────┐ I1 +   ┌────┐ V1 +   ┌─────┐ I2  ┌────┐ Vo
 ──►(Σ)►│1/sL1├──►(Σ)─►│ R1 ├─┬─►(Σ)►│1/sL2├──┬─►│ R2 ├─┬─►
    ▲-  └─────┘    ▲-  └────┘ │   ▲-  └─────┘  │  └────┘ │
    │              │          │   │            │         │
    │              └──────────┼───┼────────────┘ I2      │
    └────── V1 ──────────────┘   └──────── Vo ──────────┘

Reduction step by step

  1. Last loop (I2→VoI_2 \to V_o fed back): forward R2sL2\dfrac{R_2}{sL_2}, unity feedback:
VoV1=R2/sL21+R2/sL2=R2sL2+R2,I2V1=1sL2+R2\frac{V_o}{V_1} = \frac{R_2/sL_2}{1 + R_2/sL_2} = \frac{R_2}{sL_2 + R_2}, \qquad \frac{I_2}{V_1} = \frac{1}{sL_2 + R_2}
  1. Middle loop: forward R1R_1 from (I1−I2)(I_1 - I_2) to V1V_1, feedback 1sL2+R2\dfrac{1}{sL_2 + R_2}:
V1I1=R11+R1sL2+R2=R1(sL2+R2)sL2+R1+R2\frac{V_1}{I_1} = \frac{R_1}{1 + \frac{R_1}{sL_2 + R_2}} = \frac{R_1(sL_2 + R_2)}{sL_2 + R_1 + R_2}
  1. First loop: forward 1sL1⋅V1I1\dfrac{1}{sL_1}\cdot\dfrac{V_1}{I_1}, unity feedback:
V1Vi=R1(sL2+R2)sL1(sL2+R1+R2)+R1(sL2+R2)\frac{V_1}{V_i} = \frac{R_1(sL_2 + R_2)}{sL_1(sL_2 + R_1 + R_2) + R_1(sL_2 + R_2)}
  1. Multiply by VoV1=R2sL2+R2\dfrac{V_o}{V_1} = \dfrac{R_2}{sL_2 + R_2}:
Vo(s)Vi(s)=R1R2L1L2s2+(L1R1+L1R2+L2R1)s+R1R2\frac{V_o(s)}{V_i(s)} = \frac{R_1R_2}{L_1L_2s^2 + (L_1R_1 + L_1R_2 + L_2R_1)s + R_1R_2}

At DC the gain is 1 (inductors act as shorts), so this is a second-order low-pass network.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Control Systems chapter 2 questions →