Solution of ECE 342 Test 2 S12
1.
All questions regarding superheterodyne receivers refer to this diagram.
xc (t )
Antenna
BT < BRF < 2 f IF
fc
Other Signals
x RFi ( t )
RF
BIF = BT
fIF
Mixer
x RFo ( t )
x IFi ( t )
IF
Baseband
x IFo ( t )
Demodulator
x (t )
Local Oscillator
f LO = fc ± fIF
(a)
A superheterodyne receiver is designed to receive signals with carrier frequencies between 4 and 6 MHz
with transmitted bandwidths of 100 kHz each. Its IF frequency is 850 kHz. What range of local oscillator
frequencies is required using high-side injection ( fLO > fc ) ?
fLO = fc + fIF ⇒ 4.85 MHz ≤ fLO ≤ 6.85 MHz
(b)
The RF amplifier/filter is an RLC parallel tuned circuit with a -3 dB bandwidth of 200 kHz. It is tuned to
5 MHz and the IF frequency is 850 kHz with high-side injection. What is the numerical Q of the filter?
What is the numerical image frequency fc′ ? What is the image rejection in dB?
Q=
fo
5 × 10 6
=
= 25
BW 200 × 10 3
The image frequency is fc′ = fc + 2 fIF = 6.7 MHz .
H( f ) =
1
1
=
⎛ f
⎛
f
5 × 10 6 ⎞
f ⎞
−
1 + jQ ⎜ − 0 ⎟ 1 + j50 ⎜
6
f ⎟⎠
⎝ 5 × 10
f⎠
⎝ f0
H( f ) =
1
= 0.0672e− j1.5035
⎛ 6.7 × 10 6
5 × 10 6 ⎞
1 + j25 ⎜
−
6.7 × 10 6 ⎟⎠
⎝ 5 × 10 6
H ( f ) dB = 20 log10 ( 0.0672 ) = −23.45 dB
So the image rejection is 23.45 dB.
2.
All questions regarding phase-locked loops refer to these diagrams, the mixer is an analog multiplier, x c ( t ) is
sinusoidal and v ( t ) is sinusoidal when the loop is locked.
LPF
xc (t )
v(t )
(a)
Ka
y (t )
Φ(s)
VCO
Kv
E (s)
+
Φv ( s )
H (s)
Ka
Y( s )
−
2π K v / s
A phase-locked loop is locked with an incoming signal frequency of 1 MHz and a free-running VCO
frequency of 1 MHz and a VCO gain of 10 kHz/volt and an amplifier gain of 5. What is the numerical
phase shift between x c ( t ) and v ( t ) and which one is leading? If the input signal suddenly increases in
amplitude and the loop is still stable what effect does that have on the phase shift when the loop again
returns to a locked condition?
v ( t ) leads x c ( t ) by 90°
There is no effect on the phase shift. It is still 90°.
(b)
A phase-locked loop is locked with an incoming signal frequency of 100 kHz and a free-running VCO
frequency of 98 kHz and a VCO gain of 10 kHz/volt and an amplifier gain of 5. Assuming the LPF is
ideal and completely blocks the higher frequency from the mixer, what is the numerical output voltage
y ( t ) in volts? Is the phase shift between x c ( t ) and v ( t ) greater than or less than 90°? If the input signal
suddenly increases in amplitude and the loop is still stable what effect does that have on the phase shift
when the loop again returns to a locked condition?
y ( t ) = 0.2 V
The phase shift is less than 90°.
Effect on phase shift is to move it closer to 90°.
(c)
Y( s )
sK a H ( s )
where
=
Φ ( s ) s + 2π K a K v H ( s )
H ( s ) is the transfer function of the lowpass filter. Let K a = 10 and K v = 10 kHz/volt and let
The closed-loop transfer function of a linearized phase-locked loop is
H (s) =
1000
. Find the numerical values of the poles for the overall system. Is the system stable?
s + 1000
1000
10s
Y( s )
sK a H ( s )
10000s
s
+ 1000
=
=
= 2
Φ ( s ) s + 2π K a K v H ( s ) s + 2π × 10 5 1000
s + 1000s + 2π × 10 8
s + 1000
Poles are at s = −500 ± j25061 . The system is stable.
3.
What is the numerical frequency of the output signal from this frequency synthesizer (assuming the highpass filter
is ideal and all signals are sinusoids) ?
HPF
230 MHz
÷200
PD
VCO
Corner Frequency = 230 MHz
÷7
230 MHz
× 7 = 8.05 MHz . The two mixer output frequencies are therefore
200
238.05 MHz and 221.95 MHz. Only the higher one gets through the highpass filter. So the output frequency is
238.05 MHz.
The last mixer mixes 230 MHz with
4.
Three signals are to be frequency-division multiplexed. All three have baseband bandwidths of 7 kHz. Two
subcarrier frequencies are used. One signal DSB modulates the lower-frequency subcarrier, one of them LSSB
modulates the higher-frequency subcarrier and one is unmodulated (before the main modulation which is AM). A guard
band of 2 kHz is required to avoid crosstalk. Draw a sketch of the spectrum of the signals after they are added but before
the main modulation. Indicate on the sketch the numerical frequencies at the edges of the bands. The main carrier
frequency is 1.2 MHz. The final AM modulated signal lies between two frequency limits flow and fhigh in positive
frequency space. What are the numerical values of flow and fhigh ?
Baseband from -7 kHz to 7 kHz. Guard band from 7 kHz to 9 kHz. DSB modulated signal from 9 kHz to 23 kHz.
Guard band from 23 kHz to 25 kHz. LSSB modulated signal from 25 kHz to 32 kHz. So the double-sided spectrum
stretches from -32 kHz to 32 kHz. When this AM modulates a 1.2 MHz carrier, the bandpass signal that results lies
between 1.168 MHz and 1.232 MHz.