Esercizi sui Quadripoli

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Esercizi sui Quadripoli

Un quadripolo è un modello matematico utilizzato per rappresentare circuiti elettrici lineari a quattro terminali. I quadripoli possono essere utilizzati per analizzare circuiti complessi e sono fondamentali nell'ingegneria elettrica e nelle telecomunicazioni. I quadripoli possono essere classificati in base al loro comportamento, come quadripoli passivi e attivi.

Parametri dei Quadripoli

I quadripoli possono essere descritti attraverso vari parametri, tra cui:

  1. Parametri Z (impedenza):

    • V_1 = Z_{11} I_1 + Z_{12} I_2V1=Z11I1+Z12I2V_1 = Z_{11} I_1 + Z_{12} I_2
    • V_2 = Z_{21} I_1 + Z_{22} I_2V2=Z21I1+Z22I2V_2 = Z_{21} I_1 + Z_{22} I_2

  2. Parametri Y (admittanza):

    • I_1 = Y_{11} V_1 + Y_{12} V_2I1=Y11V1+Y12V2I_1 = Y_{11} V_1 + Y_{12} V_2
    • I_2 = Y_{21} V_1 + Y_{22} V_2I2=Y21V1+Y22V2I_2 = Y_{21} V_1 + Y_{22} V_2

  3. Parametri H (ibridi):

    • V_1 = H_{11} I_1 + H_{12} V_2V1=H11I1+H12V2V_1 = H_{11} I_1 + H_{12} V_2
    • I_2 = H_{21} I_1 + H_{22} V_2I2=H21I1+H22V2I_2 = H_{21} I_1 + H_{22} V_2

  4. Parametri T (trasmissione):

    • V_1 = T_{11} V_2 + T_{12} I_2V1=T11V2+T12I2V_1 = T_{11} V_2 + T_{12} I_2
    • I_1 = T_{21} V_2 + T_{22} I_2I1=T21V2+T22I2I_1 = T_{21} V_2 + T_{22} I_2

Esercizi

Esercizio 1: Calcolo dei parametri Z

Un quadripolo ha i seguenti parametri di impedenza:

  • Z_{11} = 2 \, \OmegaZ11=2 ΩZ_{11} = 2 \, \Omega
  • Z_{12} = 1 \, \OmegaZ12=1 ΩZ_{12} = 1 \, \Omega
  • Z_{21} = 0.5 \, \OmegaZ21=0.5 ΩZ_{21} = 0.5 \, \Omega
  • Z_{22} = 3 \, \OmegaZ22=3 ΩZ_{22} = 3 \, \Omega

Se la corrente I_1 = 2 \, \text{A}I1=2 AI_1 = 2 \, \text{A} e I_2 = 1 \, \text{A}I2=1 AI_2 = 1 \, \text{A}, calcola le tensioni V_1V1V_1 e V_2V2V_2.

Soluzione:
Utilizziamo le equazioni dei parametri Z:

V_1 = Z_{11} I_1 + Z_{12} I_2 V1=Z11I1+Z12I2 V_1 = Z_{11} I_1 + Z_{12} I_2

V_2 = Z_{21} I_1 + Z_{22} I_2 V2=Z21I1+Z22I2 V_2 = Z_{21} I_1 + Z_{22} I_2

Calcoliamo V_1V1V_1:

V_1 = 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \, \text{V} V1=2⋅2+1⋅1=4+1=5 V V_1 = 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \, \text{V}

Calcoliamo V_2V2V_2:

V_2 = 0.5 \cdot 2 + 3 \cdot 1 = 1 + 3 = 4 \, \text{V} V2=0.5⋅2+3⋅1=1+3=4 V V_2 = 0.5 \cdot 2 + 3 \cdot 1 = 1 + 3 = 4 \, \text{V}

Esercizio 2: Calcolo dei parametri Y

Un quadripolo ha i seguenti parametri di ammittanza:

  • Y_{11} = 0.1 \, \text{S}Y11=0.1 SY_{11} = 0.1 \, \text{S}
  • Y_{12} = 0.05 \, \text{S}Y12=0.05 SY_{12} = 0.05 \, \text{S}
  • Y_{21} = 0.02 \, \text{S}Y21=0.02 SY_{21} = 0.02 \, \text{S}
  • Y_{22} = 0.15 \, \text{S}Y22=0.15 SY_{22} = 0.15 \, \text{S}

Se le tensioni V_1 = 10 \, \text{V}V1=10 VV_1 = 10 \, \text{V} e V_2 = 5 \, \text{V}V2=5 VV_2 = 5 \, \text{V}, calcola le correnti I_1I1I_1 e I_2I2I_2.

Soluzione:
Utilizziamo le equazioni dei parametri Y:

I_1 = Y_{11} V_1 + Y_{12} V_2 I1=Y11V1+Y12V2 I_1 = Y_{11} V_1 + Y_{12} V_2

I_2 = Y_{21} V_1 + Y_{22} V_2 I2=Y21V1+Y22V2 I_2 = Y_{21} V_1 + Y_{22} V_2

Calcoliamo I_1I1I_1:

I_1 = 0.1 \cdot 10 + 0.05 \cdot 5 = 1 + 0.25 = 1.25 \, \text{A} I1=0.1⋅10+0.05⋅5=1+0.25=1.25 A I_1 = 0.1 \cdot 10 + 0.05 \cdot 5 = 1 + 0.25 = 1.25 \, \text{A}

Calcoliamo I_2I2I_2:

I_2 = 0.02 \cdot 10 + 0.15 \cdot 5 = 0.2 + 0.75 = 0.95 \, \text{A} I2=0.02⋅10+0.15⋅5=0.2+0.75=0.95 A I_2 = 0.02 \cdot 10 + 0.15 \cdot 5 = 0.2 + 0.75 = 0.95 \, \text{A}

Esercizio 3: Conversione tra parametri Z e parametri H

Un quadripolo ha i seguenti parametri Z:

  • Z_{11} = 4 \, \OmegaZ11=4 ΩZ_{11} = 4 \, \Omega
  • Z_{12} = 2 \, \OmegaZ12=2 ΩZ_{12} = 2 \, \Omega
  • Z_{21} = 1 \, \OmegaZ21=1 ΩZ_{21} = 1 \, \Omega
  • Z_{22} = 3 \, \OmegaZ22=3 ΩZ_{22} = 3 \, \Omega

Calcola i parametri H corrispondenti.

Soluzione:
I parametri H possono essere calcolati dai parametri Z utilizzando le seguenti relazioni:

H_{11} = Z_{11} \quad \text{e} \quad H_{12} = \frac{Z_{12}}{Z_{22}} H11=Z11eH12=Z12Z22 H_{11} = Z_{11} \quad \text{e} \quad H_{12} = \frac{Z_{12}}{Z_{22}}

H_{21} = \frac{Z_{21}}{Z_{22}} \quad \text{e} \quad H_{22} = \frac{1}{Z_{22}} H21=Z21Z22eH22=1Z22 H_{21} = \frac{Z_{21}}{Z_{22}} \quad \text{e} \quad H_{22} = \frac{1}{Z_{22}}

Calcoliamo i parametri H:

  1. H_{11} = Z_{11} = 4 \, \OmegaH11=Z11=4 ΩH_{11} = Z_{11} = 4 \, \Omega
  2. H_{12} = \frac{Z_{12}}{Z_{22}} = \frac{2}{3}H12=Z12Z22=23H_{12} = \frac{Z_{12}}{Z_{22}} = \frac{2}{3}
  3. H_{21} = \frac{Z_{21}}{Z_{22}} = \frac{1}{3}H21=Z21Z22=13H_{21} = \frac{Z_{21}}{Z_{22}} = \frac{1}{3}
  4. H_{22} = \frac{1}{Z_{22}} = \frac{1}{3}H22=1Z22=13H_{22} = \frac{1}{Z_{22}} = \frac{1}{3}

Quindi, i parametri H sono:

  • H_{11} = 4H11=4H_{11} = 4
  • H_{12} = \frac{2}{3}H12=23H_{12} = \frac{2}{3}
  • H_{21} = \frac{1}{3}H21=13H_{21} = \frac{1}{3}
  • H_{22} = \frac{1}{3}H22=13H_{22} = \frac{1}{3}

Esercizio 4: Calcolo della risposta di un quadripolo

Un quadripolo ha i seguenti parametri T:

  • T_{11} = 0.5T11=0.5T_{11} = 0.5
  • T_{12} = 0.2T12=0.2T_{12} = 0.2
  • T_{21} = 0.3T21=0.3T_{21} = 0.3
  • T_{22} = 0.4T22=0.4T_{22} = 0.4

Se la tensione in ingresso V_2 = 12 \, \text{V}V2=12 VV_2 = 12 \, \text{V} e la corrente in ingresso I_2 = 1 \, \text{A}I2=1 AI_2 = 1 \, \text{A}, calcola la tensione in uscita V_1V1V_1 e la corrente in uscita I_1I1I_1.

Soluzione:
Utilizziamo le equazioni dei parametri T:

V_1 = T_{11} V_2 + T_{12} I_2 V1=T11V2+T12I2 V_1 = T_{11} V_2 + T_{12} I_2

I_1 = T_{21} V_2 + T_{22} I_2 I1=T21V2+T22I2 I_1 = T_{21} V_2 + T_{22} I_2

Calcoliamo V_1V1V_1:

V_1 = 0.5 \cdot 12 + 0.2 \cdot 1 = 6 + 0.2 = 6.2 \, \text{V} V1=0.5⋅12+0.2⋅1=6+0.2=6.2 V V_1 = 0.5 \cdot 12 + 0.2 \cdot 1 = 6 + 0.2 = 6.2 \, \text{V}

Calcoliamo I_1I1I_1:

I_1 = 0.3 \cdot 12 + 0.4 \cdot 1 = 3.6 + 0.4 = 4.0 \, \text{A} I1=0.3⋅12+0.4⋅1=3.6+0.4=4.0 A I_1 = 0.3 \cdot 12 + 0.4 \cdot 1 = 3.6 + 0.4 = 4.0 \, \text{A}

English version

Quadripole Exercises

A quadrupole is a mathematical model used to represent linear four-terminal electrical circuits. Quadrupoles can be used to analyze complex circuits and are essential in electrical engineering and telecommunications. Quadrupoles can be classified based on their behavior, as passive and active quadrupoles.

Quadripole Parameters

Quadripoles can be described by various parameters, including:

  1. Z (impedance) parameters:
  • V_1 = Z_{11} I_1 + Z_{12} I_2V1=Z11I1+Z12I2V_1 = Z_{11} I_1 + Z_{12} I_2
  • V_2 = Z_{21} I_1 + Z_{22} I_2V2=Z21I1+Z22I2V_2 = Z_{21} I_1 + Z_{22} I_2
  1. Y (admittance) parameters:
  • I_1 = Y_{11} V_1 + Y_{12} V_2I1=Y11V1+Y12V2I_1 = Y_{11} V_1 + Y_{12} V_2
  • I_2 = Y_{21} V_1 + Y_{22} V_2I2=Y21V1+Y22V2I_2 = Y_{21} V_1 + Y_{22} V_2
  1. H (hybrid) parameters:
  • V_1 = H_{11} I_1 + H_{12} V_2V1=H11I1+H12V2V_1 = H_{11} I_1 + H_{12} V_2
  • I_2 = H_{21} I_1 + H_{22} V_2I2=H21I1+H22V2I_2 = H_{21} I_1 + H_{22} V_2
  1. T (hybrid) parameters (transmission):
  • V_1 = T_{11} V_2 + T_{12} I_2V1=T11V2+T12I2V_1 = T_{11} V_2 + T_{12} I_2
  • I_1 = T_{21} V_2 + T_{22} I_2I1=T21V2+T22I2I_1 = T_{21} V_2 + T_{22} I_2

Exercises

Exercise 1: Calculating Z parameters

A quadrupole has the following impedance parameters:

  • Z_{11} = 2 \, \OmegaZ11=2 ΩZ_{11} = 2 \, \Omega
  • Z_{12} = 1 \, \OmegaZ12=1 ΩZ_{12} = 1 \, \Omega
  • Z_{21} = 0.5 \, \OmegaZ21=0.5 ΩZ_{21} = 0.5 \, \Omega
  • Z_{22} = 3 \, \OmegaZ22=3 ΩZ_{22} = 3 \, \Omega

If the current I_1 = 2 \, \text{A}I1=2 AI_1 = 2 \, \text{A} and I_2 = 1 \, \text{A}I2=1 AI_2 = 1 \, \text{A}, calculate the voltages V_1V1V_1 and V_2V2V_2.

Solution:
We use the Z-parameter equations:

V_1 = Z_{11} I_1 + Z_{12} I_2 V1=Z11I1+Z12I2 V_1 = Z_{11} I_1 + Z_{12} I_2

V_2 = Z_{21} I_1 + Z_{22} I_2 V2=Z21I1+Z22I2 V_2 = Z_{21} I_1 + Z_{22} I_2

Let's calculate V_1V1V_1:

V_1 = 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \, \text{V} V1=2⋅2+1⋅1=4+1=5 V V_1 = 2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5 \, \text{V}

Let's calculate V_2V2V_2:

V_2 = 0.5 \cdot 2 + 3 \cdot 1 = 1 + 3 = 4 \, \text{V} V2=0.5⋅2+3⋅1=1+3=4 V V_2 = 0.5 \cdot 2 + 3 \cdot 1 = 1 + 3 = 4 \, \text{V}

Exercise 2: Calculating Y-parameters

A quadrupole has the following admittance parameters:

  • Y_{11} = 0.1 \, \text{S}Y11=0.1 SY_{11} = 0.1 \, \text{S}
  • Y_{12} = 0.05 \, \text{S}Y12=0.05 SY_{12} = 0.05 \, \text{S}
  • Y_{21} = 0.02 \, \text{S}Y21=0.02 SY_{21} = 0.02 \, \text{S}
  • Y_{22} = 0.15 \, \text{S}Y22=0.15 SY_{22} = 0.15 \, \text{S}

If the voltages V_1 = 10 \, \text{V}V1=10 VV_1 = 10 \, \text{V} and V_2 = 5 \, \text{V}V2=5 VV_2 = 5 \, \text{V}, calculate the currents I_1I1I_1 and I_2I2I_2.

Solution:
We use the Y-parameter equations:

I_1 = Y_{11} V_1 + Y_{12} V_2 I1=Y11V1+Y12V2 I_1 = Y_{11} V_1 + Y_{12} V_2

I_2 = Y_{21} V_1 + Y_{22} V_2 I2=Y21V1+Y22V2 I_2 = Y_{21} V_1 + Y_{22} V_2

Let's calculate I_1I1I_1:

I_1 = 0.1 \cdot 10 + 0.05 \cdot 5 = 1 + 0.25 = 1.25 \, \text{A} I1=0.1⋅10+0.05⋅5=1+0.25=1.25 A I_1 = 0.1 \cdot 10 + 0.05 \cdot 5 = 1 + 0.25 = 1.25 \, \text{A}

Let's calculate I_2I2I_2:

I_2 = 0.02 \cdot 10 + 0.15 \cdot 5 = 0.2 + 0.75 = 0.95 \, \text{A} I2=0.02⋅10+0.15⋅5=0.2+0.75=0.95 A I_2 = 0.02 \cdot 10 + 0.15 \cdot 5 = 0.2 + 0.75 = 0.95 \, \text{A}

Exercise 3: Conversion between Z-parameters and H-parameters

A quadrupole has the following Z-parameters:

  • Z_{11} = 4 \, \OmegaZ11=4 ΩZ_{11} = 4 \, \Omega
  • Z_{12} = 2 \, \OmegaZ12=2 ΩZ_{12} = 2 \, \Omega
  • Z_{21} = 1 \, \OmegaZ21=1 ΩZ_{21} = 1 \, \Omega
  • Z_{22} = 3 \, \OmegaZ22=3 ΩZ_{22} = 3 \, \Omega

Calculate the corresponding H parameters.

Solution:
The H parameters can be calculated from the Z parameters using the following relations:

H_{11} = Z_{11} \quad \text{e} \quad H_{12} = \frac{Z_{12}}{Z_{22}} H11=Z11eH12=Z12Z22 H_{11} = Z_{11} \quad \text{e} \quad H_{12} = \frac{Z_{12}}{Z_{22}}

H_{21} = \frac{Z_{21}}{Z_{22}} \quad \text{e} \quad H_{22} = \frac{1}{Z_{22}} H21=Z21Z22eH22=1Z22 H_{21} = \frac{Z_{21}}{Z_{22}} \quad \text{e} \quad H_{22} = \frac{1}{Z_{22}}

Let's calculate the H parameters:

  1. H_{11} = Z_{11} = 4 \, \OmegaH11=Z11=4 ΩH_{11} = Z_{11} = 4 \, \Omega
  2. H_{12} = \frac{Z_{12}}{Z_{22}} = \frac{2}{3}H12=Z12Z22=23H_{12} = \frac{Z_{12}}{Z_{22}} = \frac{2}{3}
  3. H_{21} = \frac{Z_{21}}{Z_{22}} = \frac{1}{3}H21=Z21Z22=13H_{21} = \frac{Z_{21}}{Z_{22}} = \frac{1}{3}
  4. H_{22} = \frac{1}{Z_{22}} = \frac{1}{3}H22=1Z22=13H_{22} = \frac{1}{Z_{22}} = \frac{1}{3}

So, the H parameters are:

  • H_{11} = 4H11=4H_{11} = 4
  • H_{12} = \frac{2}{3}H12=23H_{12} = \frac{2}{3}
  • H_{21} = \frac{1}{3}H21=13H_{21} = \frac{1}{3}
  • H_{22} = \frac{1}{3}H22=13H_{22} = \frac{1}{3}

Exercise 4: Calculating the response of a quadrupole

A quadrupole has the following T parameters:

  • T_{11} = 0.5T11=0.5T_{11} = 0.5
  • T_{12} = 0.2T12=0.2T_{12} = 0.2
  • T_{21} = 0.3T21=0.3T_{21} = 0.3
  • T_{22} = 0.4T22=0.4T_{22} = 0.4

If the voltage input V_2 = 12 \, \text{V}V2=12 VV_2 = 12 \, \text{V} and input current I_2 = 1 \, \text{A}I2=1 AI_2 = 1 \, \text{A}, calculate output voltage V_1V1V_1 and output current I_1I1I_1.

Solution:
We use the T-parameter equations:

V_1 = T_{11} V_2 + T_{12} I_2 V1=T11V2+T12I2 V_1 = T_{11} V_2 + T_{12} I_2

I_1 = T_{21} V_2 + T_{22} I_2 I1=T21V2+T22I2 I_1 = T_{21} V_2 + T_{22} I_2

Let's calculate V_1V1V_1:

V_1 = 0.5 \cdot 12 + 0.2 \cdot 1 = 6 + 0.2 = 6.2 \, \text{V} V1=0.5⋅12+0.2⋅1=6+0.2=6.2 V V_1 = 0.5 \cdot 12 + 0.2 \cdot 1 = 6 + 0.2 = 6.2 \, \text{V}

Let's calculate I_1I1I_1:

I_1 = 0.3 \cdot 12 + 0.4 \cdot 1 = 3.6 + 0.4 = 4.0 \, \text{A} I1=0.3⋅12+0.4⋅1=3.6+0.4=4.0 A I_1 = 0.3 \cdot 12 + 0.4 \cdot 1 = 3.6 + 0.4 = 4.0 \, \text{A}

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