Question

The bandlimited signals m(t) and m, (t) of bandwidths 1 kHz and 2kHz are transmitted over
channel using Time division multiplexing, the minimum sampling rate required is,

a.6khz
b.3khz
c.4khz
d.5khz​

Answer 1

Twelve voice signals, each band-limited to 3 kHz, are frequency -multiplexed using 1 kHz guard bands between channels and between the main carrier and the first channel. The modulation of the main carrier is AM. Calculate the bandwidth of the composite channel
if the subcarrier modulation is (a) DSB, (b) LSSB.
(a) With DSB each 3 kHz channel becomes a 6 kHz bandwidth when modulating a subcarrier. So each channel occupies 7 kHz for a total bandwidth of 84 kHz before modulating the main carrier. The main carrier is AM so the bandwidth requirement for that signal is double
84 kHz or 168 kHz.
(b) With LSSB each 3 kHz channel becomes a 3 kHz bandwidth when modulating a subcarrier. So each channel occupies 4 kHz for a total bandwidth of 48 kHz before modulating the main carrier. The main carrier is AM so the bandwidth requirement for that signal is double
48 kHz or 96 kHz.

The signal x(t)= cos(2000πt) is used to modulate a 5 kHz carrier. Sketch the time waveforms and line spectra if the modulation used is (a) DSB, (b) AM with μ = 0.5, (c) USSB, (d) LSSB.
(a) xc (t)= Ac cos(2000πt)cos(10000πt)
1 0.5
-0.5 -1
0.5
1 1.5
|Xc( f )|
0.25
0.2
0.15
0.1
0.05 5
2 t(ms)
=(Ac /2)⎡⎣cos(8000πt)+cos(12000πt)⎤⎦ xc(t)
-10 -5
10f (kHz)

(b) x (t)= A ⎡1+ 0.5cos(2000πt)⎤cos(10000πt) cc⎣⎦
=Ac cos(10000πt)+(Ac /4)⎡⎣cos(8000πt)+cos(12000πt)⎤⎦ xc(t)
21
-1 -2
0.5 1 1.5
|Xc( f )|
0.5 0.4 0.3 0.2 0.1
2 t(ms)
-10-5 5
10 f (kHz)

(c) xc(t)=(Ac /2)cos(12000πt) xc(t)
0.6
0.2 t (ms) -0.2 0.5 1 1.5 2
-0.6
-10 -5
|Xc( f )|
0.25
0.2
0.15
0.1
0.05 5
10f (kHz)

(d) xc (t)=(Ac /2)cos(8000πt)

Show that the system below acts as an envelope detector for a bandpass signal. Verify that the system can indeed demodulate an AM wave. (Hint: Consider a general bandpass signal
xc (t)cos(ωct)+ xs (t)sin(ωct).
Show that the output is the envelope (A / 2) x2 (t)+ x2 (t). Assume a cs
narrowband signal.)

x(t)= xc (t)cos(ωct)+ xs (t)sin(ωct) On the top path:
x (t)cos(ω t)+x (t)sin(ω t)⎯⎯⎯→A⎡xc(t)cos(ωct) ⎤cos(ω t+θ) ccsc⎢⎥c
A⎡⎣xc (t)cos(ωct)cos(ωct +θ)+ xs (t)sin(ωct)cos(ωct +θ)⎤⎦ =
Mixer ⎢+x (t)sin(ω t)⎥ ⎣s c⎦
(A / 2) x (t)⎡cos(θ)+ cos(2ω t +θ)⎤ + x (t)⎡sin(−θ)+ sin(2ω t +θ)⎤ ⎯⎯→ {}
c⎣c⎦s⎣c⎦ ⎧x (t)cos(θ) ⎫ ⎧x2 (t)cos2 (θ)+ x2 (t)sin2 (θ)⎫
LPF
Squarer 2
(A/2)⎨ c ⎬⎯⎯⎯→(A /4)⎨ c s ⎬
⎩−xs (t)sin(θ)⎭ ⎪⎩−2xc (t)xs (t)cos(θ)sin(θ) ⎪⎭

x(t)= xc (t)cos(ωct)+ xs (t)sin(ωct) On the bottom path:
x (t)cos(ω t)+x (t)sin(ω t)⎯⎯⎯→A⎡xc(t)cos(ωct) ⎤sin(ω t+θ) ccsc⎢⎥c
A⎡⎣xc (t)cos(ωct)sin(ωct +θ)+ xs (t)sin(ωct)sin(ωct +θ)⎤⎦ =
Mixer ⎢+x (t)sin(ω t)⎥ ⎣s c⎦
(A / 2) x (t)⎡sin(θ)+ sin(2ω t +θ)⎤ + x (t)⎡cos(θ)+ cos(2ω t +θ)⎤ ⎯⎯→ {}
c⎣c⎦s⎣c⎦ ⎧x (t)sin(θ) ⎫ ⎧x2 (t)sin2 (θ)+ x2 (t)cos2 (θ)⎫
LPF
Squarer 2
(A/2)⎨ c ⎬⎯⎯⎯→(A /4)⎨ c s ⎬
⎩+xs (t)cos(θ)⎭ ⎪⎩+2xc (t)xs (t)cos(θ)sin(θ) ⎪⎭

⎧x2 (t)cos2 (θ)+ x2 (t)sin2 (θ)⎫ On the top path: (A2 / 4)⎨ c s ⎬ ⎪⎩−2xc (t)xs (t)cos(θ)sin(θ) ⎪⎭
⎧x2 (t)sin2 (θ)+ x2 (t)cos2 (θ)⎫ On the bottom path: (A2 / 4)⎨ c s ⎬ ⎪⎩+2xc (t)xs (t)cos(θ)sin(θ) ⎪⎭
Adding the two signals we get
Square
⎧x2 (t)cos2 (θ)+ x2 (t)sin2 (θ)⎫
(A2 /4)⎪⎨ c s ⎪⎬=(A2 /4)⎡⎣x2 (t)+x2 (t)⎤⎦
⎪⎩x2c (t)sin2 (θ)+x2s (t)cos2 (θ)⎪⎭ c s
⎯⎯⎯→(A/2) x (t)+x (t) cs
Rooter 2 2

This is a circuit that performs the square-root function. It is taken from a National Semiconductor collection of op-amp circuits.

Show that a squaring circuit followed by a lowpass filter followed by a square rooter acts as an envelope detector for an AM wave. Show that
if a DSB signal x(t)cos(ωct) is demodulated by this scheme the output
willbe x(t)/ 2.
AM: x(t)=A⎡1+μx(t)⎤cos(ωt) cc⎣⎦c
x2 (t)= A2 ⎡1+μx(t)⎤2 cos2 (ω t)=(A2 /2)⎡1+μx(t)⎤2 ⎡1+cos(2ω t)⎤ cc⎣⎦cc⎣⎦⎣c⎦
2 LPF 2
(A /2)⎡1+μx(t)⎤2 ⎡1+cos(2ω t)⎤⎯⎯→(A /2)⎡1+μx(t)⎤2
c⎣⎦⎣c⎦c⎣⎦ ( ) Square ( )
2 Rooter
A /2 ⎡1+μx(t)⎤2 ⎯⎯⎯→ A / 2 ⎡1+μx(t)⎤
c⎣⎦c⎣⎦ DSB: xc (t)= x(t)cos(ωct)
x2 (t)= x2 (t)cos2 (ω t)= x2 (t)(1/ 2)⎡1+ cos(2ω t)⎤ cc⎣c⎦
2 ⎣c⎦LPF2
x (t)(1/2)⎡1+cos(2ω t)⎤⎯⎯→x (t)/2
Square
2 Rooter 2
x(t)/2⎯⎯⎯→ x(t)/2=x(t)/ 2

Twenty-five radio stations are broadcasting in the band between 3 MHz and 3.5 MHz. You wish to modify an AM broadcast receiver to receive the broadcasts. Each audio signal has a maximum frequency fm = 10 kHz. Describe in detail the changes you would have to make to the standard broadcast superheterodyne receiver in order to receive the broadcast.
For standard AM, each channel has a transmitted bandwith of BT = 10kHz. After demodulation that becomes a baseband bandwidth of W = 5kHz. We need here a bandwidth of 10kHz for the demodulated baseband signal. So we must increase the IF bandwidth by a factor of two. The RF range of
a standard AM receiver is 540 to 1700 kHz with an IF of 455khz. The
local oscillator range is then 995 to 2155 kHz. We need to change that passband at fIF = 3.3965MHz is
fLO − fc . Therefore, fLO = 7.225MHz + 3.3965MHz = 10.6215MHz. The image frequency is fL