# Calculate: AM, SSB Reception and Modulation

1. You are asked to build an AM receiver with a 15 micro H inductor which must tune from 500 to 880 kHz. Find the required variable capacitance and the bandwidth at 770 kHz, if the ideal bandwidth at 380 kHz is 10 kHz.

2.A receiver tunes from 20 to 30 MHz using a 10.7 MHz IF. What is the range of the oscillator and image frequencies?

3. A super-heterodyne receiver tunes the band of frequencies from 4-10 MHz. The IF is 1.8 MHz. The double-ganged capacitor used has a 325 pF maximum capacitance per section. The tuning capacitors are at the maximum value (325 pF) when the RF frequency is 4 MHz. Calculate the required RF and local oscillator coil inductance and the required tuning capacitor values when the receiver is tuned to receive 4 MHz.

4. A single side band transmission consists of a 770 Hz sine wave and its second and fourth harmonics. Suppose that the receiver's de-modulator has drifted 10 Hz. Find the resulting speaker output frequencies.

5. A 1000 W single side band signal is fully modulated. What is the power transmitted?

6. A SSB transmitter uses a 9 MHz crystal RF frequency oscillator and a 4 MHz medium frequency oscillator. What are the output frequencies of the medium frequency balanced modulator? What filter Q is required in the linear power amplifier? The intelligence signal is a 4 kHz tone.

7. An SSB signal is generated around a 200-kHz carrier. Before filtering, the upper and lower side bands are separated by 200 Hz. Calculate the filter Q required to obtain 40-dB suppression.

© BrainMass Inc. brainmass.com August 16, 2018, 1:23 pm ad1c9bdddf#### Solution Preview

1. Use the expression for tuned frequency fc in terms of tuning inductance (L) & tuning capacitance (C).

fc = 1/{2pi*Sqrt(LC)} (1)

Re-arrange (1) in terms of the tuning capacitance required.

C = 1/(2pi*fc)^2*L (2)

Now put in values for lower tuned frequency fc = 500 kHz and solve for the tuning capacitance required to derive this frequency using (2).

C = 1/{4pi^2 x (5 x 10^5)^2 x 15 x 10^-6}

C= 6.75 nF

Similarly we find the tuning capacitance to derive a frequency of fc = 880 kHz.

C = 1/{4pi^2 x (8.8 x 10^5)^2 x 15 x 10^-6}

C = 2.18 nF

Thus the tuner should have a variable capacitance that ranges between 2.18 and 6.75 nF so as to be able to implement the tuning range 500 to 880 kHz.

Also we are asked to derive the tuning capacitance to derive a frequency of 770 kHz. Clearly we know that this should lay in the variable capacitance range 2.18 to 6.75 nF. We calculate this as before using (2).

C = 1/{4pi^2 x (7.7 x 10^5)^2 x 15 x 10^-6}

C = 2.85 nF

Ideal Bandwidth(BW) at 380 kHz is 10 kHz. The Q of the tuner is given by (3).

Q = fc /BW (3)

Q of tuner in this example is therefore,

Q = 380 x 10^3/10^4

Q = 38

Now we assume that the Q is constant for the tuning range so at a frequency of 770 kHz we can say that (3) applies, so the BW at 770 kHz is given as BW770 as below.

38 = 770 x 10^3/BW770

Re-arranging

BW770 = 770 x 10^3/38 = 20,263.16 Hz

BW770 ~ 20.263 kHz

2. We use the identity for the resultant IF frequency {f(IF)} in terms of the local oscillator frequency {f(LO)} and tuned frequency {f(c)} given by ...

#### Solution Summary

This solution provides an organized, step by step solution illustrating how to solve a number of questions regarding AM, SSB reception and modulation calculations. All equations and variables required are included.