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Communication system Communication is the process of exchanging information between two points
Elements of communication system:
source
Transduce
r
Modulator
&
Transmitter
Receiver
&
Demodulator
Transducer Destinatio
n
channel
Audio electrical signal electrical signal audio
Fig: communication system
Source: The information which has to be transmitted is generated by source ex: audio video text
etc.
Transducer: Transducer is a device which converts one form of energy into another form of
energy here transducer convert audio signal to electrical signal and vice versa.
Modulator and Transmitter: Here signal is modulated and transmitted over long distance.
Channel: The channel is a medium through which signals are transmitted to receiver channel
may be wired or wireless.
Receiver and demodulator: Here signals are received and information signal is detected by
demodulator
Destination: signal is received in original form i.e. (Audio).
Frequency ranges with application:
Frequency range Application
Super high frequencies 3GHⱬ-30GHⱬ Radar.
Ultra high frequncies300MHⱬ-3GHⱬ Communication satellites cellular phones
personal communication system.
Very high frequncies30MHⱬ-300MHⱬ TV and FM broadcast.
High frequncies3MHⱬ-30MHⱬ Short-wave broadcast commercial.
Medium frequencies300KHⱬ-3MHⱬ AM broadcast.
Low frequencies30KHⱬ-300KHⱬ Navigation, submarine communication.
Very low frequncies3KHⱬ-30KHⱬ Sub marine communication
Voice frequencies300Hⱬ-3KHⱬ Audio, submarine communication
Extremely low frequencies30Hⱬ-300Hⱬ Power Transmission
Modulation: Modulation is the process of varying carrier in accordance with instantaneous value
of information signal or modulating signal.
The carrier signal, c(t)=VcSinWct
𝑉𝑐=peak voltage 𝑊𝑐=2πfc 𝑓𝑐=carrier frequency
Need of modulation :
1. Modulation process helps to transmit the signal to longer distance.
2. Reduce the height of antenna.
Height of antenna , ℓ = 𝜆 4⁄ =
𝐶
4𝑓 𝜆 =𝐶
𝑓⁄
Ex: 1 f = 1 KHz
ℓ = 𝜆 4⁄ =
𝐶
4𝑓 =
3×108
4×1𝑘 = 0.75× 105=75000m
Ex: 2 f=1MHz
ℓ = 𝜆 4⁄ =
𝐶
4𝑓 =
3×108
4×1×106 = 75m
By above example it is clear that transmitting frequency is increased height of antenna is
decreased.
3. We can send multiple numbers of signals through signal communication channel having
wider bandwidth
4. The designing and processing of signal becomes easier for transmitter and receiver
5. Modulation Process reduces the effect of noise which is added in communication channel
Three types of modulations:
1. Amplitude modulation (AM)
2. Frequency modulation(FM)
3. Phase modulation(PM)
Amplitude modulation: Amplitude modulation is a process of varying amplitude of carrier
signal in accordance with instantaneous value of message signal keeping frequency and phase
constant.
Let message signal m (t) is
m (t)=𝑉𝑚 Sin𝑊𝑚 t----------(1)
𝑉𝑚=peak voltage of message signal
𝑊𝑚=2πfm
𝑓𝑚=modulating frequency
Lt carrier signal C(t) is
C(t)=𝑉𝑐 sinWct-------------(2)
Vc=peak voltage of carrier signal
Wc=2πfc
𝑓𝑐=carrier frequency
\From eqn(1) & (2) amplitude modulated signal is given by,
V(t)=(Vc+Vm Sin Wmt)sinWct
V(t)=𝑉𝑐sinWct +Vm sinWct Sin Wmt
[using Sin A. SinB=1
2[Cos(A-B)-Cos(A+B)]
V(t)=𝑉𝑐sinWct+ Vm1
2[Cos(Wct-Wmt) - Cos(Wct +Wmt)]
V(t)=𝑉𝑐sinWct+ Vm
2[Cos(Wc-Wm)t-Cos(Wc+Wm)t]
V(t)=𝑉𝑐sinWct+ Vm
2Cos(Wc-Wm)t-
Vm
2cos(Wc+Wm)t-----------(A)
Modulation index m=Vm
𝑉𝑐 =
peak voltage of msg signal1
peak voltage of carrier
m𝑉𝑐=Vm
Eqn (A) = >
V(t)=𝑉𝑐sinWct+ 𝑚𝑉𝑐
2 Cos(Wc-Wm)t-
𝑚𝑉𝑐
2cos(Wc+m)t
Carrier lower sideband upper sideband
Bandwidth=Wc+Wm-(Wc-Wm)
= Wc+Wm-Wc+Wm
B.W =2Wm
Fig: AM spectrum
Modulation index in terms of 𝑽𝒎𝒂𝒙&𝑽𝒎𝒊𝒏:
Modulation index is defined as ratio of peak amplitude of message signal to peak amplitudes of
carrier signal
m𝑉𝑐 =Vm
Fig: Amplitude modulated signal
𝑉𝑚𝑎𝑥 =𝑉𝑐 + Vm
𝑉𝑚𝑎𝑥 = 𝑉𝑐 + mVc
𝑉𝑚𝑎𝑥 = 𝑉𝑐 [1+ m]-----------(1)
𝑉𝑚𝑖𝑛 = 𝑉𝑐 -Vm
𝑉𝑚𝑖𝑛 = 𝑉𝑐 -mVc
𝑉𝑚𝑖𝑛 = 𝑉𝑐[1-m]------------- (2)
(1)
(2) =
𝑉𝑚𝑎𝑥
𝑉𝑚𝑖𝑛=
𝑉𝑐[1+m]
𝑉𝑐[1−m]
𝑉𝑚𝑎𝑥 [1-m]= 𝑉𝑚𝑖𝑛 [1+m]
𝑉𝑚𝑎𝑥 - m𝑉𝑚𝑎𝑥 =𝑉𝑚𝑖𝑛+m 𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥 – 𝑉𝑚𝑖𝑛 = m𝑉𝑚𝑖𝑛+m 𝑉𝑚𝑎𝑥
𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛 = m[𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛] 𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛 = m
m = Vm
𝑉𝑐
𝑉𝑚𝑎𝑥
𝑉𝑚𝑖𝑛=
1+m
1−m
m = 𝑉𝑚𝑎𝑥– 𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+ 𝑉𝑚𝑖𝑛
Spectrums of amplitude modulated signals are:
Fig: AM modulated signal spectrum
Fig: DSB-SC spectrum Fig: SSB Spectrum
DSB-SC: Double side band –suppressed carrier here carrier signal is suppressed only side hands
are transmitted
SSB: Single side band here carrier and lower side band are not transmitted only upper side band
is transmitted
Efficiency (Ƞ) :( % of power)
Ƞ = 𝑃𝐿𝑆𝐵 +𝑃𝑈𝑆𝐵
𝑃𝑡 𝑃𝐿𝑆𝐵=𝑃𝑈𝑆𝐵 =
𝑚2𝑉𝑐2
8&𝑃𝑐=
𝑉𝑐2
2
Ƞ =
𝑚2𝑉𝑐2 +
8
𝑚2𝑉𝑐2
8
𝑃𝑐[1+𝑚2
2]
=
2𝑚2𝑉𝑐2
8
𝑃𝑐[1+𝑚2
2]
=𝑚2𝑉𝑐2
4
𝑃𝑐[1+𝑚2
2]
Ƞ =
𝑚2𝑉𝑐2
4
𝑉𝑐2
2[1+
𝑚2
2] =
𝑚2
2
[2+𝑚2
2] =
𝑚2
2+𝑚2
Ƞ = 𝑚2
𝑚2+2
Spectrum power :(Total transmitted in AM wave )
Power, p = 𝑉2
𝑟𝑚𝑠
𝑅 =
(𝑉
√2)
2
𝑅 =
𝑉2
2
Assume R =1Ω
The total power of amplitude modulated wave ,
𝑃𝑡= 𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵---------- (1)
= (𝑉𝑐
√2)
2+(
𝑚𝑉𝑐2
√2)
2
+(𝑚𝑉𝑐
2
√2)
2
𝑃𝑐 = (𝑉𝑐
√2)
2 =
𝑉𝑐2
2
= 𝑉𝑐2
2+
𝑚2𝑉𝑐2
4
2 +
𝑚2𝑉𝑐2
4
2 𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=(
𝑚𝑉𝑐2
√2)
2
=𝑚2𝑉𝑐
2
8
= 𝑉𝑐
2
2 +
𝑚2𝑉𝑐2
4×
1
2+
𝑚2𝑉𝑐2
4×
1
2 LSB=lower side band
USB=upper side band
= 𝑉𝑐
2
2 +
𝑚2𝑉𝑐2
8 +
𝑚2𝑉𝑐2
8
= 𝑉𝑐
2
2 +
2𝑚2𝑉𝑐2
8
= 𝑉𝑐
2
2 +
𝑚2𝑉𝑐2
4
𝑃𝑡= 𝑉𝑐
2
2[1+
𝑚2
2]
𝑃𝑡= 𝑃𝑐[1+𝑚2
2]
If modulation index m=1,
𝑃𝑡= 𝑃𝑐[1+12
2] = 𝑃𝑐[
1+1
2]
𝑃𝑡 = 𝑃𝑐[3
2]
or 50% more than 𝑃𝑐 𝑃𝑡=1.5 𝑃𝑐
AM Detection [ AM Demodulation]:
(a) Demodulation circuit
(b)demodulated waveform
The process of recovery of message signal from amplitude modulated signal is called am
demodulation or detection
Fig (a) shows demodulation circuit and fig(3) demodulated waveform
As v(t) rises to peak the diode conducts and capacitor charges through diode
When v(t) falls below peak the diode stops conducting and capacitor begins to discharge
through resistor R
This process repeat for all cycles
The time constant T = RC
The Conduction Tc<<Rc<<Tm
1fc <<Rc<<1 fm
Tc = time period of carrier
Fc = carrier frequency
Fm = modulating signal frequency
Tm = Time period of modulating signal
Frequency modulation :[FM]
The frequency modulation is the process of varying the frequency of carrier signal in
accordance with instantaneous value of message signal keeping amplitude and phase
constant.
Message signal m(t)
m(t)=VmSinWmt
The frequency modulated signal
f(t) = A Sin [Wc +Kf m(t)]t-----------(1)
modulation index : from eqn(1) f = fc + Kf m(t)
f = fc + KfVmsinWmt
Δf = kfvm =peak frequency deviation
Modulation index = peak frequncy deviation
modulating frequency
m = Δf
fm
Band width: ( BW)fm = 2 [Δf+fm]
Δf=frequency deviation.
fm=frequency of modulating signal.
Phase modulation[PM]: Phase modulation is process of varying phase of carrier signal in
accordance with instantaneous value of message signal keeping amplitude &phase constant
The phase modulated signal is given by
P(t) = A Sin[Wct + Kpm(t)]
Where m(t) = Vm Sin Wmt
NOTE :FM& PM are angle modulation
Information signal message signal modulating signal all are same
Comparison of amplitude and frequency modulation:
characteristics AM FM
1 Bandwidth B W =2 fm
Lesser bandwidth
B W =2 [Δf + fm]
larger bandwidth
2 Operating carrier frequency
Am uses lower carrier
frequency
Fm uses higher carrier
frequency
3 Transmission efficiency Less efficient More efficient
4 Area of reception Am covers more distance Fm covers limited distance
5 Noise performance poor better
6 channel Small channel is sufficient Wider channel is required
7 Common channel interference Distortion occurs Less Distortion
8 Tuning Tuning is not
required
Tuning is required
Problems :
1. Determine the power content of the carrier and each of sideband for an AM signal having a
percent modulation of 80% and totl power of 2500w
m= 80% m=80\100 =8\10=0.8 𝑃𝑡 =2500
𝑃𝑡= 𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵----------(1)
𝑃𝑡= 𝑃𝑐[1+𝑚2
2]
2500=𝑃𝑐[1+(0.8)2
2]
2500=𝑃𝑐[1.32]
𝑃𝑐=2500
1.32 =1893.9W
𝑃𝑡=𝑃𝑐+𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵
𝑃𝑡-𝑃𝑐 = 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵
𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 =2500-1893.9
𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 =606.1W
𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=606.1
2 W
𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=303.50W
2.The total power control of an AM Signal is 1000W determine the power being transmitted at
carrier frequency and at each of sidebands when percent modulation is 100%
𝑃𝑐=666.67W 𝑃𝑈𝑆𝐵 =𝑃𝐿𝑆𝐵 =166.66W
3.An amplitude modulated wave has a power content of 800W at its carrier frequency.
Determine the power content of each sidebands for a 90% modulation
Given :𝑃𝑐= 800W
m=90% =90\100 =0.9
𝑃𝑡= 𝑃𝑐[1+𝑚2
2]
𝑃𝑡=800[1+(0.9)2\2] = 1124W
𝑃𝑡=𝑃𝑐 + 𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵
𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 = 𝑃𝑡 –𝑃𝑐 = 1124-800
𝑃𝑈𝑆𝐵+𝑃𝐿𝑆𝐵 = 324W
𝑃𝑈𝑆𝐵 = 𝑃𝐿𝑆𝐵= 324
2 =162W
4 .The total power content of an AM wave is 600W. Determine the percent modulation of the
signal if each of sidebands contains 75W
Given𝑃𝑡=600W
𝑃𝑈𝑆𝐵=𝑃𝐿𝑆𝐵=75W
m = ?
𝑃𝑡 = 𝑃𝑐 + 𝑃𝑈𝑆𝐵 + 𝑃𝐿𝑆𝐵
𝑃𝑐 = 𝑃𝑡 − 𝑃𝑈𝑆𝐵 − 𝑃𝐿𝑆𝐵
𝑃𝑐 = 600-75-75
𝑃𝑐 =450W
𝑃𝑡= 𝑃𝑐[1+𝑚2
2]
600 =450[1+𝑚2
2]
600
450= 1 +
𝑚2
2
1.33-1= 𝑚2
2
0.33×2 =𝑚2
0.66 =𝑚2
m=√0.66 =0.816
m=0.816 % m = 0.816×100 =81.6%
Question paper questions :
1. Write the block diagram of communication system explain each element
2. What are commonly used frequency range in communication system? Mention application
of each range
3. What is modulation ?explain need of modulation
4. Define amplitude modulation?
Show that𝑉𝑚𝑎𝑥
𝑉𝑚𝑖𝑛=
1+𝑚
1−𝑚
5. Explain amplitude modulation with expression
6. Derive pt= 𝑝𝑐[1 +−2
𝑚2]
7. Derive total power transmitted in AM
8. Write short note on (a )frequency modulation
(b) phase modulation
9. Explain AM detection circuit
10. List the difference between AM & FM
11. what is modulation index for AM ?
Show that m =𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛
𝑉𝑚𝑎𝑥+𝑉𝑚𝑖𝑛