Homework #6 Solution Set

(32 points - 1 per problem)

1. List the five steps in the process of receiving a radio signal.

a) Acquisition of the radio signal

b) Selection of the appropriate carrier
c) Amplification of the selected RF signal
d) Recovery of the information (Detection)
e) Processing the recovered information signal

2. In the crystal radio receiver of figure 5-2, which of the five steps are being
performed?
D1
Long Wire Antenna

IN60 - Germanium

L1 C1
Tuning Coil - 250 uH 30 - 365 pF - Tuning
High Impedance
Headphones

Earth Ground

[Figure 5-2]

The following steps are being performed: Acquisition (Long-wire antenna),

Selection (Tuned circuit with L1 and C1), Detection (D1), and Processing of the
recovered information signal (Headphones).

3. Why is RF amplification needed in a radio receiver?

RF amplification is needed because the input signals at the antenna are of

extremely low voltage and power; much too low to be directly processed without
amplification.

4. How is a variable capacitor used in the tuning mechanism of a receiver?

A variable capacitor is used with an inductor to form a variable-frequency

bandpass filter. This filter is tuned to the frequency of the desired carrier signal.

5. A certain TRF radio receiver uses a variable capacitor with a range of 40 pF to 300
pF in combination with a fixed inductor of 200 µH. What is the tuning range of this
receiver?

To solve this problem, the resonance formula is used twice:

1 1
f min = = = 649.75 KHz
2π LC max 2π (200 µH )(300 pF )
 1 1
f max = = = 1779.41KHz
2π LC min 2π (200µH )(40 pF )

6. A receiver's bandpass filter can be tuned by changing the inductor instead of the
capacitor. Such an arrangement is called permeability tuning. If the selector circuit
of this receiver uses an inductor with a range of 20 µH to 200 µH in combination
with a fixed capacitance of 270 pF, what will the tuning range be?

This technique was once very popular in high-end communications receivers.

Again, the resonance formula is applied to find the operating frequencies:

1 1
f min = = = 684.89 KHz
2π Lmax C 2π (200 µH )(270 pF )

1 1
f max = = = 2165.82 KHz
2π Lmin C 2π (20 µH )(270 pF )

7. Compare and contrast an AM diode detector and a half-wave DC power supply.

The primary similarity between these stages is the use of a diode as a half-wave
rectifier. The primary difference between them is the length of the RC time
constant in the circuit. A DC power supply has a long RC time constant so that
relatively smooth (ripple-free) DC is produced. An AM detector has a short RC time
constant so that the information can pass (the information appears as "ripple" on
the resulting DC voltage).

8. What are the three frequency components present in the output of a diode detector
circuit?

The three frequency components are a DC level, the RF carrier frequency, and the
AF information signal.

9. Why is a low-pass filter circuit used in a diode AM detector circuit?

A low-pass filter allows the AF information signal to pass, but blocks the RF carrier
frequency component, which is no longer needed after detection.

10. What causes diagonal clipping distortion in an AM detector? What operating

conditions are most likely to produce this form of distortion?

Diagonal clipping distortion is caused by having a falling envelope slope that is

larger (faster) than the RC time constant used in the detector. The condition is
mostly likely to show up under high percentages of modulation.
11. What is the function of inductor L1 in figure 5-7?
R1

100K

+ C2 C3 R4 L1
2.2uF 0.1 uF 330K 500 uH
LOOPSTICK

U1
ZN414

2 3

1
C1
150 pF C9
TUNING 0.1uF

[Partial of Figure 5-7]

L1 performs two functions in this circuit. It acts as the receiving antenna as well as
part of the selector band-pass filter circuit.

12. Explain the operation of a loopstick antenna. What portion of the radio wave is such
an antenna sensitive to?

A loopstick antenna is a coil of fine wire wound on a rod of ferrite, which is a

magnetic material. The antenna essentially acts like a transformer secondary, as it
couples to the magnetic portion of the radio wave field.

13. Explain the difference between the sensitivity and selectivity of a radio receiver.
What circuit characteristic controls selectivity?

Sensitivity is the ability to process weak signals, while selectivity is the ability to
separate (select) signals that are closely-spaced in frequency. One parameter
describes an amplitude characteristic, while the other describes a frequency
characteristic. The bandwidth of the receiver controls its selectivity.

14. What are the two main problems with TRF receivers?

The two main problems with TRF receivers are (1) Bandwidth that varies with
operating frequency, and (2) Unsuitability for operation above approximately 2
MHz.

15. A certain tuned circuit has a resonant frequency of 710 KHz and a Q of 100. What is
its bandwidth?

f 710 KHz
BW = = = 7.1KHz
Q 100
16. What is the Q of an inductor having an inductive reactance XL of 10 Ohms an
equivalent parallel resistance RP of 100 Ohms?

For a parallel reactive circuit, Q is found by:

Rp 100Ω
Qp = = = 10
Xp 10Ω

17. A certain TRF tuning circuit consists of a 300 µH loopstick, in combination with a
30 to 300 pF variable capacitor for tuning. The equivalent tank parallel resistance is
100K Ohms. (See figure 5-9)
a) What is the tuning range of the receiver?
b) What is the circuit bandwidth when tuned to 540 KHz?
c) What is the bandwidth at 1500 KHz?

a) To determine the tuning range, the resonant frequency formula is used:

1 1
f min = = = 530.52 KHz
2π LC max 2π (300µH )(300 pF )

1 1
f max = = = 1677.64 KHz
2π LC min 2π (300 µH )(30 pF )

b) The bandwidth at 540 KHz is determined as follows:

X p = X L = 2πfL = 2π (540 KHz )(300 µH ) = 1018Ω

Rp 100 KΩ
Qp = = = 98.24
Xp 1018Ω
f 540 KHz
BW = = = 5.5KHz
Q 98.24

c) At 1500 KHz, the bandwidth becomes:

X p = X L = 2πfL = 2π (1500 KHz )(300 µH ) = 2827Ω

Rp 100 KΩ
Qp = = = 35.37
Xp 2827Ω
f 1500 KHz
BW = = = 42.4 KHz
Q 35.37
18. How does a superheterodyne receiver achieve a constant bandwidth?

A superhet receiver has a constant bandwidth because the primary bandpass

filtering occurs in the fixed frequency tuned circuits of the IF amplifier.

19. What is the function of the converter section in a superhet receiver?

The function of the converter is to convert the incoming RF carrier frequency up or

down to the IF frequency.

20. An ideal mixer is being supplied with the frequencies 600 KHz and 400 KHz. What
output frequencies will be produced?

The following frequencies will be produced: (a) Originals, 600 KHz and 400 KHz;
(b) Sum, 1000 KHz ; (c) Difference, 200 KHz.

NB: A balanced mixer does not allow the original frequencies to pass, ideally.
Balanced mixers are used in high-grade communication transmitters and receivers.

21. A certain mixer is producing the following output frequencies: 760 KHz, 1760 KHz,
1000 KHz, and 240 KHz. What are the two most likely input frequencies?

The two most likely input frequencies are 760 KHz and 1000 KHz. These two
frequencies appear at the output and also could be responsible for producing the
other two frequencies by sum (1760 KHz) and difference (240 KHz).

22. What is the standard IF frequency for AM broadcast receivers?

The standard IF frequency for AM broadcast receivers is 455 KHz.

NB: Other IF frequencies have been used; for example, 262.5 KHz was once used in
many early broadcast receivers. Modern DSP-based receivers may use a final IF
frequency of only 10 KHz to 20 KHz.

23. In order to convert a 870 KHz carrier down to a 455 KHz IF frequency using high-
side injection, what local oscillator frequency is required?

Because high side-injection is being used, equation (5-1) is used:

f lo = f c + f if = 870 KHz + 455KHz = 1325 KHz

24. What local oscillator frequency would be needed in question 23 if low-side injection
was used?

For low-side injection, the local oscillator runs below the carrier frequency, and we
get:

f lo = f c − f if = 870 KHz − 455 KHz = 415 KHz

25. The superhet receiver of figure 5-15 is tuned to a carrier frequency of 1080 KHz.
a) What frequency is the preselector tuned to?
b) What frequency is the local oscillator tuned to?
c) List all the frequencies produced by the mixer.
d) Which of the mixer output frequencies is amplified by the IF amplifier?

a) The preselector is tuned to the carrier frequency fc, or 1080 KHz.

b) f lo = f c + f if = 1080 KHz + 455KHz = 1535 KHz
c) The mixer will output the following frequencies: 1080 KHz, 1535 KHz, 455 KHz,
and 2615 KHz.

26. The superhet receiver of figure 5-15 is tuned to an unknown carrier frequency. The
preselector tank consists of a 300 µH inductor and a 30 to 300 pF variable capacitor,
which is set for a capacitance of 88 pF.
a) What frequency is the receiver tuned to?
b) What frequency should the local oscillator be producing?
c) If the local oscillator has a 250 µH inductor, what capacitance will set the local
oscillator to the correct frequency?

1 1
a) f pre = = = 979.53KHz
2π LC max 2π (300 µH )(88 pF )
b) f lo = f c + f if = 979.53KHz + 455KHz = 1434.53KHz
c) The resonant frequency formula must be solved for C to determine this answer:

1 1
C= = = 49.24 pF
(2π ) 2 2
f L (2π ) (1434.53KHz ) 2 (250 µH )
2

27. An AM superhet receiver is tuned to 1270 KHz and uses a 455 KHz IF. If high-side
injection is being used, what is the image frequency?

f image = f c + 2 f if = 1270 KHz + 2(455KHz ) = 2180 KHz

28. What section of a superhet receiver is responsible for eliminating the image
frequency?

The preselector band-pass filter is responsible for eliminating (actually,

attenuating) the image frequency.

29. Why is AGC needed in AM receivers?

AGC is needed to compensate for varying signal strength because AM receivers are
greatly affected by signal strength.
30. What stage of an AM receiver develops the AGC signal?

The detector develops the AGC signal. The DC component of the detector's output
is used to develop the AGC control voltage.

31. What stages of an AM receiver are controlled by the AGC signal?

The RF and IF stages are controlled by the AGC signal.

32. Why are C1A and C1B mechanically ganged together in the receiver of figure 5-16?

These two capacitors are ganged to provide tracking between the preselector and
local oscillator circuits, so that a constant difference of fif (455 KHz) is maintained
as the receiver is tuned.