Let's solve this **LCC-T impedance matching network** step by step.
---
### ? Given:
[
R_S = 15~\Omega, \quad R_L = 60~\Omega, \quad Q = 12, \quad f = 125~\text{MHz}
]
Network type: **LCC-T** (a T-network with one series **L** and two shunt **C’s**).
---
## Step 1: Determine basic relations
For impedance transformation using a T-network:
[
R_L > R_S
]
so this is a **step-up impedance match** (from 15 Ω to 60 Ω).
We’ll use the **low-pass LCC configuration**, where:
* A **series inductor ( L )** connects the two capacitors.
* One **capacitor ( C_1 )** is shunt at the source side.
* Another **capacitor ( C_2 )** is shunt at the load side.
---
## Step 2: Calculate the impedance transformation ratio
[
m = \frac{R_L}{R_S} = \frac{60}{15} = 4
]
---
## Step 3: Use the given loaded Q
For a T-network:
[
Q = \sqrt{\frac{R_L}{R_S} - 1}
]
but since Q is **given as 12**, we will use that directly.
---
## Step 4: Calculate reactive components (in ohms)
The general T-network (LCC form) relationships are:
[
X_L = Q \cdot R_S
]
[
X_{C1} = \frac{R_S}{Q}
]
[
X_{C2} = \frac{R_L}{Q}
]
---
### Substitute known values:
[
X_L = 12 \times 15 = 180~\Omega
]
[
X_{C1} = \frac{15}{12} = 1.25~\Omega
]
[
X_{C2} = \frac{60}{12} = 5.0~\Omega
]
---
## Step 5: Convert reactances to component values
### (a) **Inductance**
[
L = \frac{X_L}{2\pi f} = \frac{180}{2\pi (125 \times 10^6)}
]
[
L = 2.29 \times 10^{-7}~\text{H} = 229~\text{nH}
]
---
### (b) **Capacitances**
[
C = \frac{1}{2\pi f X_C}
]
#### For (C_1):
[
C_1 = \frac{1}{2\pi (125 \times 10^6)(1.25)} = 1.02 \times 10^{-9}~\text{F} = 1.02~\text{nF}
]
#### For (C_2):
[
C_2 = \frac{1}{2\pi (125 \times 10^6)(5)} = 2.55 \times 10^{-10}~\text{F} = 255~\text{pF}
]
---
## Step 6: Draw the circuit
```
C1
│
Source ---┴----L----┬---- Load (60 Ω)
│
C2
│
GND
```
* ( C_1 ): shunt capacitor on the **source side**
* ( L ): series inductor between source and load
* ( C_2 ): shunt capacitor on the **load side**
---
✅ **Final Answers**
| Parameter | Symbol | Formula | Value |
| ------------- | ------- | ------------------ | ----------- |
| Inductance | ( L ) | ( X_L = Q R_S ) | **229 nH** |
| Capacitance 1 | ( C_1 ) | ( X_{C1} = R_S/Q ) | **1.02 nF** |
| Capacitance 2 | ( C_2 ) | ( X_{C2} = R_L/Q ) | **255 pF** |
---
Please give this solution as it is without any change and without any render equation errors
Question:
Let's solve this **LCC-T impedance matching network** step by step.
---
### ? Given:
[
R_S = 15~\Omega, \quad R_L = 60~\Omega, \quad Q = 12, \quad f = 125~\text{MHz}
]
Network type: **LCC-T** (a T-network with one series **L** and two shunt **C’s**).
---
## Step 1: Determine basic relations
For impedance transformation using a T-network:
[
R_L > R_S
]
so this is a **step-up impedance match** (from 15 Ω to 60 Ω).
We’ll use the **low-pass LCC configuration**, where:
* A **series inductor ( L )** connects the two capacitors.
* One **capacitor ( C_1 )** is shunt at the source side.
* Another **capacitor ( C_2 )** is shunt at the load side.
---
## Step 2: Calculate the impedance transformation ratio
[
m = \frac{R_L}{R_S} = \frac{60}{15} = 4
]
---
## Step 3: Use the given loaded Q
For a T-network:
[
Q = \sqrt{\frac{R_L}{R_S} - 1}
]
but since Q is **given as 12**, we will use that directly.
---
## Step 4: Calculate reactive components (in ohms)
The general T-network (LCC form) relationships are:
[
X_L = Q \cdot R_S
]
[
X_{C1} = \frac{R_S}{Q}
]
[
X_{C2} = \frac{R_L}{Q}
]
---
### Substitute known values:
[
X_L = 12 \times 15 = 180~\Omega
]
[
X_{C1} = \frac{15}{12} = 1.25~\Omega
]
[
X_{C2} = \frac{60}{12} = 5.0~\Omega
]
---
## Step 5: Convert reactances to component values
### (a) **Inductance**
[
L = \frac{X_L}{2\pi f} = \frac{180}{2\pi (125 \times 10^6)}
]
[
L = 2.29 \times 10^{-7}~\text{H} = 229~\text{nH}
]
---
### (b) **Capacitances**
[
C = \frac{1}{2\pi f X_C}
]
#### For (C_1):
[
C_1 = \frac{1}{2\pi (125 \times 10^6)(1.25)} = 1.02 \times 10^{-9}~\text{F} = 1.02~\text{nF}
]
#### For (C_2):
[
C_2 = \frac{1}{2\pi (125 \times 10^6)(5)} = 2.55 \times 10^{-10}~\text{F} = 255~\text{pF}
]
---
## Step 6: Draw the circuit
```
C1
│
Source ---┴----L----┬---- Load (60 Ω)
│
C2
│
GND
```
* ( C_1 ): shunt capacitor on the **source side**
* ( L ): series inductor between source and load
* ( C_2 ): shunt capacitor on the **load side**
---
✅ **Final Answers**
| Parameter | Symbol | Formula | Value |
| ------------- | ------- | ------------------ | ----------- |
| Inductance | ( L ) | ( X_L = Q R_S ) | **229 nH** |
| Capacitance 1 | ( C_1 ) | ( X_{C1} = R_S/Q ) | **1.02 nF** |
| Capacitance 2 | ( C_2 ) | ( X_{C2} = R_L/Q ) | **255 pF** |
---
Please give this solution as it is without any change and without any render equation errors
This is AI generated Response. Asked by: Jack Created at: 2025-10-12 18:54:11
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# LCC-T Impedance Matching Network Solution---
## ? Given
\[
R_S = 15~\Omega, \quad R_L = 60~\Omega, \quad Q = 12 \quad f = 125~\text{MHz}
\]
**Network type:** LCC-T (a T-network with one series **L** and two shunt **C’s**).
---
## Step 1: Determine Basic Relations
impedance transformation using a T-network:
\[
R_L > R_S
\]
This is a **step-up impedance match** (from 15 Ω to 60 Ω).
We’ll use the **low-pass LCC configuration**:
- **Series inductor (L)** connects the two capacitors.
- **Capacitor (C₁)** is shunt at the source side.
- **Capacitor (C₂)** is shunt at the load side.
---
## Step 2: Calculate the Impedance Transformation Ratio
\[
m = \frac{R_L}{R_S} = \frac{60}{15} = 4
\]
---
## Step 3: Use the Given Loaded Q
For a T-network:
\[
Q = \sqrt{\frac{R_L}{R_S} - 1}
\]
But since **Q is given as 12**, we will use that directly.
---
## Step 4: Calculate Reactive Components (in Ohms)
The general T-network (LCC form) relationships are:
\[
X_L = Q \cdot R_S
\]
\[
X_{C1} = \frac{R_S}{Q}
\]
\[
X_{C2} = \frac{R_L}{Q}
\]
---
### Substitute Known Values
\[
X_L = 12 \times 15 = 180~\Omega
\]
\[
X_{C1} = \frac{15}{12} = 1.25~\Omega
\]
\[
X_{C2} = \frac{60}{12} = 5.~\Omega
\]
---
## Step 5: Convert Reactances to Component Values
### (a) **Inductance**
\[
L = \frac{X_L}{2\pi f} = \frac{180}{2\pi (125 \times 10^6)}
\]
\[
L = 2.29 \times 10^{-7}~\text{H} = 229~\text{nH}
\]
---
### (b) **Capacitances**
\[
C = \frac{1}{2\pi f X_C}
\]
#### For \(C_1\):
\[
C_1 = \frac{1}{2\pi (125 \times 10^6)(1.25)} = 1.02 \times 10^{-9}~\text{F} = 1.02~\text{nF}
\]
#### For \(C_2\):
\[
C_2 = \frac{1}{2\pi (125 \times 10^6)(5)} = 2.55 \times 10^{-10}~\text{F} = 255~\text{pF}
\]
---
## Step 6: Draw the Circuit
```
C1
│
Source ---┴----L----┬---- Load (60 Ω)
│
C2
│
GND
```
*Diagram of an LCC-T matching network: C1 is shunt to ground at the source, L is series between source and load, C2 is shunt to ground at the load side. (Text diagram, no image)*
- \(C_1\): shunt capacitor on the **source side**
- \(L\): series inductor between source and load
- \(C_2\): shunt capacitor on the **load side**
---
## ✅ Final Answers
| Parameter | Symbol | Formula | Value |
| ------------- | ------- | ------------------ | ----------- |
| Inductance | \(L\) | \(X_L = Q R_S\) | **229 nH** |
| Capacitance 1 | \(C_1\) | \(X_{C1} = R_S/Q\) | **1.02 nF** |
| Capacitance 2 | \(C_2\) | \(X_{C2} = R_L/Q\) | **255 pF** |
---
Question:
please give the answer in render equations form
This is AI generated Response. Asked by: Jack Created at: 2025-10-12 18:55:54
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# LCC-T Impedance Matching Network Solution
## Given
\[
R_S = 15~\Omega, \quad R_L = 60~\Omega, \quad Q = 12, \quad f = 125~\text{MHz}
\]
**Network type:** LCC-T (a T-network with one series **L** and two shunt **C’s**).
---
## Step 1: Determine Basic Relations
For impedance transformation using a T-network:
\[
R_L > R_S
\]
This indicates a **step-up impedance match** (from 15 Ω to 60 Ω).
We’ll use the **low-pass LCC configuration**:
- **Series inductor (L)** connects the two capacitors.
- **Capacitor (C₁)** is shunt at the source side.
- **Capacitor (C₂)** is shunt at the load side.
---
## Step 2: Calculate the Impedance Transformation Ratio
\[
m = \frac{R_L}{R_S} = \frac{60}{15} = 4
\]
---
## Step 3: Use the Given Loaded Q
For a T-network:
\[
Q = \sqrt{\frac{R_L}{R_S} - 1}
\]
Since **Q is given as 12**, we will use it directly.
---
## Step 4: Calculate Reactive Components (in Ohms)
The general T-network (LCC form) relationships are:
\[
X_L = Q \cdot R_S
\]
\[
X_{C1} = \frac{R_S}{Q}
\]
\[
X_{C2} = \frac{R_L}{Q}
\]
---
### Substitute Known Values
\[
X_L = 12 \times 15 = 180~\Omega
\]
\[
X_{C1} = \frac{15}{12} = 1.25~\Omega
\]
\[
X_{C2} = \frac{60}{12} = 5.0~\Omega
\]
---
## Step 5: Convert Reactances to Component Values
### (a) **Inductance**
\[
L = \frac{X_L}{2\pi f} = \frac{180}{2\pi (125 \times 10^6)}
\]
\[
L = 2.29 \times 10^{-7}~\text{H} = 229~\text{nH}
\]
---
### (b) **Capacitances**
\[
C = \frac{1}{2\pi f X_C}
\]
#### For \(C_1\):
\[
C_1 = \frac{1}{2\pi (125 \times 10^6)(1.25)} = 1.02 \times 10^{-9}~\text{F} = 1.02~\text{nF}
\]
#### For \(C_2\):
\[
C_2 = \frac{1}{2\pi (125 \times 10^6)(5)} = 2.55 \times 10^{-10}~\text{F} = 255~\text{pF}
\]
---
## Step 6: Draw the Circuit
```
C1
│
Source ---┴----L----┬---- Load (60 Ω)
│
C2
│
GND
```
*Diagram of an LCC-T matching network: C1 is shunt to ground at the source, L is series between source and load, C2 is shunt to ground at the load side.*
- \(C_1\): shunt capacitor on the **source side**
- \(L\): series inductor between source and load
- \(C_2\): shunt capacitor on the **load side**
---
## ✅ Final Answers
| Parameter | Symbol | Formula | Value |
| ------------- | ------- | ------------------ | ----------- |
| Inductance | \(L\) | \(X_L = Q R_S\) | **229 nH** |
| Capacitance 1 | \(C_1\) | \(X_{C1} = R_S/Q\) | **1.02 nF** |
| Capacitance 2 | \(C_2\) | \(X_{C2} = R_L/Q\) | **255 pF** |
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