Question

3. A satellite communication receiver is designed to receive a transmission at 136.5 MHz. The receiver uses two heterodyne operations (dual-conversion receiver) with intermediate frequencies of 25 MHz Texas A&M University-Kingsville, Electrical Engineering and Computer Science Department & 12 MHz, as shown below. The first local oscillator is designed to operate above the incoming carrier frequency, and the second local oscillator is to operate below the first intermediate frequency Mixer # 1 Mixer #2 Output RF X Filter & IF, Amplifier X Filter & IF, Amplifier (12 MHz) Detector and Demodulator Amplifier (25 MHZ) (12 MHZ) Demoddiator First Local Oscillator Second Local Oscillator a) The first local oscillator is operating at MHz b) The second local oscillator is operating at MHz. c) If the filters in RF amplifier, IF, and IF2 are all non-ideal filters, determine all possible image frequencies. d) Describe how you can eliminate the image frequencies if possible.

Answer 1

At the hetrodyne receiver , it has two design choices for local oscillator,

1)Flo = Frf + Fif (Called High side injecion, where Flo>Frf)

2)Flo = Frf - Fif (Called low side injection,where Flo<Frf)

Usually for AM and FM waves we choose that oscillator which can provide higher frequency than desired RF frequency that is ,

Flo = Frf + Fif

We will use above formula

a)

So for Fif1 = 25 MHz and Frf = 136.5 MHz

Flo = 25 + 136.5 = 161.5 MHZ

So Frequency of First Local osclllator = 161.5 MHz

So this frequency is greater than the incoming frequency

b)

So as the input to the second oscillator is the Intermediate Frequency 1 so fromula becoms

So for Fif2 = 12 MHz and Fif1 = 25 MHz

Flo = Fif1 - Fif2

Flo = 25 - 12 = 13MHz

So Frequency of Second Local osclllator = 13 MHz

So this frequency is smaller than incoming Intermediate Frequency 2

c)

To find image frequencies they are given by fromula

fimage = Frf + 2*Fif (IF Flo > Frf)

fimage = Frf - 2*Fif (IF Flo < Frf)

Fimage1 due to IF1

fimage1 = 136.5 + 50 = 186.5 MHz (+ sign is used because here flo>frf)

Fimage1 due to IF2

fimage1 = 136.5 - 24 = 112.5 MHz (- sign because Here also flo < Fif1)

d)

These image frequencies defines above are the extra harmonics or we can say high frequency components which get introduce in the signal and tend to degrade the signal if not filtered out.

One way is there that we use a Tunable RF Bandpass filter after Intermediate frequency, So as to filter out these frequencies at their defined frequency level and pass the others carrying original information.

This filter shoulf filter out the frequencies at these ( (Frf + 2*Fif ) and (Frf - 2*Fif )) levels and pass the other ones .

This way we can filter out these unwanted frequency components.