Chapter 5

Tutorial 5

Question 1

A superhetrodyne receiver is tuned to a frequency of 5 MHz when the local oscillator frequency is 6.65 MHz. What is the IF? Which type of injection is in use?

Solution

Solution

a. fIF = 6.6 MHz – 5 MHz = 1.6 MHz.

b. High injection, because the local oscillator frequency is greater than the signal frequency.

Question 2

A superhetrodyne receiver has an IF of 9 MHz and tunes the frequency range from 50 to 60 MHz. The mixer uses low-side injection of the local oscillator signal. Calculate the range of local oscillator frequencies.

Solution

Solution

For a signal at 50 Mhz, fLO = 50 MHz – 9 MHz = 41 MHz.

For a signal at 60 Mhz, fLO = 60 MHz – 9 MHz = 51 MHz.

Question 3

A receiver uses low-side injection for the local oscillator with an IF of 1750 kHz. The local oscillator is operating at 15.750 MHz. To what frequency is the receiver tuned? What is the image frequency?

Solution

ImageLORFIFFrequency

fIFfIF

2fIF

ImageLORFIFFrequency

fIFfIF

2fIF

ImageLORFIFFrequency

fIFfIF

2fIF

RF LO ImageIF Frequency

fIFfIF

2fIF

High-side injection

Low-side injection

Solution

a. fsig = fLO + fIF = 15.7 MHz + 1.75 MHz = 17.5 MHz.

b. fim = fsig - 2fIF = 17.5 MHz - 2×1.75 MHz = 14 MHz.

Question 4 An AM broadcast receiver with high-side injection and

an IF of 455 kHz is tuned to a station at 910 kHz. What is the local oscillator frequency? What is the image frequency?

Solution

455 Lo RF IF

Rf

f f F

f kHz

535 1605 kHz to kHz

995 2055 kHz to kHz

450 455 460

Solution

a. fLo = fsig + fIF = 910 KHz + 455 KHz = 1.365 MHz.

b. fimage = fsig + 2fIF = 910 KHz - 2×455 KHz = 1.82 MHz.

Noncoherent SSB-BFO ReceiversQuestion 5

An SSBSC signal has a suppressed carrier frequency in the IF of a receiver, of 2 MHz. It is modulated with two audio tones having frequencies of 1500 Hz and 2200 Hz. The receiver is mistuned so that the BFO is at 1.9995 MHz. What will be the output frequencies of the demodulator if the signal is LSB?

1.9995 MHz

1.9995 MHz

1.9978MHz 1.9985 MHz 2MHz

1000 Hz 1700 Hzf

1000 Hz 1700 Hzf

Solution

The incoming signal has components at:

2MHz – 1500 Hz = 1.9985 MHz, and 2MHz – 2200 Hz = 1.9978 MHz when these components mix with BFO frequency of 1.9995 MHz, the output frequencies are: 1.9995 MHz – 1.9985 MHz = 1000 Hz, and 1.9995 MHz – 1.9978 MHz = 1700 Hz

Question 6 A double-conversion receiver covers the frequency range

from 100 to 200 MHz. The first IF is 30 MHz, and the first mixer uses high-side injection of the local-oscillator signal. The second IF is 1 MHz, and the second mixer uses low-side injection. Calculate the operating frequency or frequency range for

both local oscillators. Draw a block diagram for this receiver.

Solutiona. 1st L. O. ranges from 100 MHz + 30 MHz = 130 MHz to

200 MHz + 30 MHz = 230 MHz. 2nd L. O. is fixed at 30 MHz – 1 MHz = 29 MHz

b.

Ganged tuning

Noncoherent SSB-BFO ReceiversQuestion 7

For the BFO receiver a received RF frequency band of 30 MHz to 30.005 MHz, RF local oscillator frequency of 20 MHz, and a BFO frequency of 10 MHz, determine

Demodulated information frequency band. Demodulated information frequency band if the RF local oscillator

frequency drifts down 0.001%

Ganged tuning

30 MHz 30.005 MHz

Ganged tuning

20 MHz 10 MHz

0 5000 Hz10 MHz 10.005 MHz30 MHz 30.005 MHz 0 5000 Hz

Solution

30 MHz 30.005 MHz

Ganged tuning

fLO= 20 MHz – 200 Hz

= 19.9998 MHz

10 MHz

200 Hz 5200 Hz10.0002 MHz 10.0052 MHz

Solution