Superheterodyne Receiver

Suppose a superheterodyne receiver's intermediate frequency (IF) is tuned to a specific frequency, such as 455 kHz. In this case, the receiver acts as a mixer, generating two different frequencies: $F_{LO} + F_{RF}$ and $F_{LO} - F_{RF}$, where:

• $F_{LO}$ is the Local Oscillator frequency.

• $F_{RF}$ is the Radio Frequency signal that is being received.

The Local Oscillator frequency $F_{LO}$ is typically set such that $F_{LO} = F_{RF} + F_{IF}$, where $F_{IF}$ is the Intermediate Frequency (e.g., 455 kHz in this case).

The intermediate frequency signal carries the same modulation (audio or other data) as the original $F_{RF}$, but at 455 kHz. This is similar to Double Sideband Suppressed Carrier (DSB-SC) or Single Sideband Suppressed Carrier (SSB-SC) modulation. In SSB-SC, the signal’s upper sideband (from $f_c$ to $f_c + f_m$) or lower sideband (from $f_c - f_m$ to $f_c$) contains the modulated information.

Superheterodyne Receiver: Concept and Example

A superheterodyne receiver converts an incoming radio frequency (RF) signal to a fixed intermediate frequency (IF) so that filtering and demodulation become easier.

 

1. Incoming Signal (RF)

The example uses a 1 MHz AM radio station:

$f$_RF = 1,000 kHz

 

2. Local Oscillator Frequency (LO)

The IF is fixed at 455 kHz for AM radios. The local oscillator follows:

$f$_LO = f_RF + f_IF

Substitute values:

$f$_LO = 1000 + 455 = 1455 kHz

 

3. Mixing (Heterodyning)

The mixer produces sum and difference frequencies:

$f$_sum = f_LO + f_RF

$f$_diff = | f_LO − f_RF |

Compute values:

$f$_sum = 1455 + 1000 = 2455 kHz

$f$_IF = 1455 − 1000 = 455 kHz

Only the IF (455 kHz) is kept.

 

4. AM IF Signal Equation

The intermediate frequency signal carries the same modulation (audio) but at 455 kHz:

$v\left(t\right)=[1+m(t\left)\right]·\cos (2\pi ·455000t)$

 

5. Demodulation

A. Rectification (Envelope Detection)

A diode removes the negative part:

$v$_rect(t) = | [1+m(t)] · cos(2π·455k) |

B. Low-Pass Filtering

A capacitor & resistor remove the 455 kHz carrier, leaving only the audio envelope:

$v$_audio(t) ≈ m(t)

 

6. Final Output

The recovered audio (e.g., a 1 kHz tone) is amplified and sent to the speaker.

 

Further Reading

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