Asst. Prof. Dr. Ashwaq Q. Hameed Al Faisal UOT- Electrical … · 2018-01-19 · Communication...

Communication Systems I

Asst. Prof. Dr. Ashwaq Q. Hameed Al Faisal

UOT- Electrical Engineering Dept. Electronic Engineering branch

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Items :

1-Introduction to Communication Systems

2-Amplitude Modulation

3- Angle Modulation

4-Detectors and Receivers.

5- Pulse Modulation.

6-Noise in Communication Systems

7- Transmission Line Theory.

Reference :

1-B.P. Lathi,” Modern Digital and Communication Systems “ , Fourth Edition , Oxfodr university press, 2009.

2-Rodger E. Ziemer and William H. Tranter,” Principles of Communications systems , modulation and noise “, sixth edition , john Wiley and Sons, Inc., 2009.

3-Simon Haykin, “ Communication Systems “, fourth Edition , John Wiley and Sons, Inc., 2001.

4- Eerrel G. Steremler," Introduction to Communication Systems", second edition , part one, part two and part three,1982

- سيد احمد مرعي ," اساسيا ت االتصاالت ", مديرية دار الكتب للطباعة والنشر , جامعة الموصل , 5 .1989كانون الثاني -




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Item 1 : Introduction to Communication Systems

Transmitter (Tx). 1-The transmitter puts the information from the source (meant for the receiver) on to the channel. The information can be represented using the signals . A signal is formally defined as a function of one or more variables that conveys information on the nature of a physical phenomenon. The signals can be classified in various ways such as: a)Power or Energy. b) Deterministic or Random. c) Real or Complex. d) Periodic or a periodic . f) continues –time signal and discrete – time signal. h) Analog signal and digital signal.

Channel Encoder Modulation

Channel h(t)

Demodulation Channel Decoder

Noise and Distortion

Transmitted Message x(t)

Received Messages y(t)=x(t)×h(t)

Transmitter Tx

Receiver Rx

Communications system block diagram

Tx signal

Rx signal

Information Image video audio speech

message thought – put into words ( verbal symbols or any the symbolic form of

expression……)




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2-converts electrical signal into a form suitable for transmission through the channel 3- conversion is made through modulation : amplitude (AM), frequency (FM)and phase (PM). Example : AM, FM radio broadcast. 4-other function : filtering, amplification, radiation. Modulation:

Modulation is defined as the process by which some characteristics (i.e. amplitude, frequency, and phase) of a carrier are varied in accordance with a modulating wave.

Channel.

The channel is the medium connecting the transmitter and the receiver and the transmitted information travels on this channel until it reaches the destination. Channels can be of two types:

1) wired channels : telephone line • Twisted pair telephone channels • Coaxial cables • Fiber optic cable

2) wireless channels: radio • earth’s atmosphere (enabling the propagation of ground wave and sky wave • Satellite channel • Laser beam. • Radio wave. • Sea water etc.

For efficient radiation, the size of the antenna should be λ/ 10 or more (preferably around λ/ 4 ), where λ is the wavelength of the signal to be radiated. Take the case of audio, which has spectral components almost from DC up to 20 kHz. Assume that we are designing the antenna for the mid frequency; that is,10 kHz. Then the length of the antenna that is required, even for the λ/ 10 situation is:

c/ f×10=3 ×108 /10×104=3 ×103 meters, c being the velocity of light.




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Even an antenna of the size of 3 km, will not be able to take care of the entire spectrum of the signal because for the frequency components around 1 kHz, the length of the antenna would be λ/ 100. Challenges . a) the transmitted signals may have to travel long distances (there by undergoing severe attenuation) before they can reach the receiver. b) of imperfections of the channel over which the signals have to travel. c) of interference due to other signals sharing the same channel and d) of noise at the receiver input1.

The term noise is present unwanted signals that tend to disturb the transmission and processing of signals in communication systems and over which we have incomplete control. In practice, we find that there are many potential sources of noise in a communication system. The sources of noise :

1- external to the system ( atmospheric noise, galactic noise, man-made noise), 2- internal to the system.( white noise, thermal noise,…..) The second category includes an important type of noise that arises from spontaneous fluctuations of current or voltage in electrical circuits.

Receiver (Rx).

1- main function : to recover the message from the received signal. 2- Demodulation : inverse of the modulation . 3- Operates in the presence of noise and interference. 4- Filtering , suppression of noise and interference.

Demodulation is the reverse process of modulation, which is used to get back the original message signal. Modulation is performed at the transmitting end whereas demodulation is performed at the receiving end. In analog modulation sinusoidal signal is used as carrier where as in digital modulation pulse train is used as carrier.




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Type of Communication Systems. 1- Types of communication systems : wire line and wireless, RF and optical ,

digital and analog , point-to-point and broadcasting , low frequency / high frequency

2- Example : telephone , cell phone , TV, internet, hard disk in a PC.

Aim of communications Engineer.

To design transmitter and receivers that are :

1- Cost efficient 2- Bandwidth efficient. 3- Maximize information transfer (message at sink is a faithful representation

of the source message). 4- Power efficient (uses little power necessary)

Many of the above goals are contrary to one another : 1- For example , one way to improve message fidelity at the receiver is to

increase transmit power. 2- Therefore tradeoffs are required.




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3-modulation is a process in which the characteristics like frequency , time , amplitude and phase of a carrier signal is changed according to message signal.

Item 2 : Amplitude Modulation

Causes of modulation in communication systems :

1) To send a signal over long distance it requires more energy.

E = hv;

E = Energy of the signal,h = plancks constant,v = frequency of the signal

Modulation is the process of increasing the frequency content of a signal ( indirectly increasing the energy of the signal to enable it to travel long distance). Demodulation is the exactly opposite process to it (Decreasing the frequency content of the signal) 2) to decrease antenna height. For transmitting a signal of wavelength λ the antenna height must be λ/4. So if we want to send a 1 Hz (λ=3*10^8 m) signal ( very low frequency) using an antenna , its height must be 75,000 Km ( impossible to build such a huge antenna ). If the same signal is modulated to some high frequency say 88 MHZ ( λ = 3.4 m ) , antenna height needed is 0.8522 m (88 MHZ is the starting range of Frequency modulation which exists up to 108 MHZ).




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Modification of Amplitude Modulation (AM).

Amplitude modulation results when a DC bias A is added to m(t) prior to the modulation process. The result of the DC bias is that a carrier component is present in the transmitted signal. For AM, the transmitted signal is typically defined as

y(t) =x(t) cos(ωc t)

in which Ac is the amplitude of the un modulated carrier A cos (ωc t) is the normalized message signal to be discussed in the following paragraph, and the parameter a ≤ 1 is known as the modulation index.1 We shall assume that m(t) has zero DC value so that the carrier component in the transmitted signal arises entirely from the bias.

Multiplying a signal by a sinusoidal carrier signal is called amplitude modulation. The signal “modulates” the amplitude of the carrier.

cos(ωc t)

y(t) =x(t) cos(ωc t) Carrier signal

carrier frequency

x(t)

𝟐𝟐𝟐𝟐𝒇𝒇𝒄𝒄




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There are three types of linear modulation involving a single message signal: 1. Double sideband-suppressed carrier (DSB-SC) modulation, where only the upper and lower sidebands are transmitted. 2. Single sideband (SSB) modulation, where only one sideband (the lower sideband or the upper sideband) is transmitted. 3. Vestigial sideband (VSB) modulation, where only a vestige of one of the sidebands and a correspondingly modified version of the other sideband are transmitted.

AMPLITUDE MODULATION DOUBLE-SIDEBAND-SUPPRESSED CARRIER.(AM-DSB/SC)

This form of linear modulation is generated by using a product modulator that simply multiplies the message signal m(t) by the carrier wave Ac cos(2πfct).

(a) Block diagram of product modulator. (b) Baseband signal. (c) DSB-SC modulated wave




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As DSB-SC modulation involves just the multiplication of the message signal and the carrier, this scheme is also known as product modulation

s(t) = Acm(t) cos(2πfct) For double-sideband (DSB) modulation , A = 0 and

Sc(t) = m(t) cos 2πfct The Fourier transform of s(t) is obtained as

𝑠𝑠(𝑓𝑓) =12𝐴𝐴𝑐𝑐 [𝑀𝑀(𝑓𝑓 − 𝑓𝑓𝑐𝑐) + 𝑀𝑀(𝑓𝑓 + 𝑓𝑓𝑐𝑐)]

Where M(f): Fourier transform

Power calculation of DSB-SC . Total power PT=PLSB+ PUSB+Pc

Total power PT=Ac2 Am

2/8+ Ac2 Am

2/8+Pc

(a) Spectrum of baseband signal. (b) Spectrum of DSB-SC

× m(t) s(t)

Ac cos ωct

Lower Side Band

Lower Side Band

Upper Side Band

Lower Side Band




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Total power PSB=Ac2 Am

2/4

Exercise An antenna has an impedance of An un modulated AM signal produces a current of 4.8 A. The modulation is 90 percent. Calculate (a) the carrier power, (b) the total power, and (c) the sideband power.

Example1:

Draw the modulated output signal for a< 1 , of message signal as show in fig.(a) and the modulator show in fig.(c).

mn(t) amn(t) 1+amn(t) Xc(t)

a 1 Ac cos(2πfct)

(c)

× × +




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Example 2

Let m(t) be real signal with the spectrum M(f),fc =100KHz .Ac/2=1. Draw the shifted spectrum.

Example 3:

Let m(t) be a complex signal with M(f) as shown . draw the shifted spectrum at fc=100KHz.

Baseband spectrum (real signal)

Shifted spectrum

Baseband spectrum (complex)

Shifted spectrum

USB USB LSB LSB




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Any one of these two sidebands has the complete information about the message signal. As we shall see later, SSB modulation conserves the bandwidth by transmitting only one sideband and recovering the m(t ) with appropriate demodulation.

consider the scheme shown in Fig. The ideal HPF has the cutoff frequency at 10 kHz. Given that f1 = 10 kHz and f2 = 15 kHz, let us sketch Y (f ) for the X (f ) given at (b).

Example 4:

We have V (f ) = X (f − f1) + X (f + f1) , which is as shown in (a). The HPF eliminates the spectral components for | f | ≤ 10 kHz. Hence W (f ) is shown in (b)

Y (f ) = W (f − f2 ) + W (f + f2 ) . This is shown in (c).




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AMPLITUDE MODULATION SINGLE -SIDEBAND-SUPPRESSED CARRIER.(AM-SSB/SC)

the USB and LSB have even amplitude and odd phase symmetry about the carrier frequency Thus transmission of both sidebands is not necessary because the transmitted information in LSB same with transmitted information USB . when one of the sideband has been separated and the same information can be transmitted in ½ BW.

AM-SS/Sc Signal a)signal spectrum b) modulated signal spectrum




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VESTIGIAL –SIDE BAND MODULATION. In vestigial sideband (VSB) modulation, one of the sidebands is partially up pressed and a vestige of the other sideband is transmitted to compensate for that suppression. frequency discrimination method can be used to generating a VSB-modulated wave is to use the. We generate a DSB-SC modulated wave and then pass it through a band pass filter, as shown in Figure , it is the special design of the band-pass filter that distinguishes VSB modulation from SSB modulation. Assuming that a vestige of the lower sideband is transmitted. The frequency response H(f) of the band-pass filter takes the form shown in Figure .at the carrier frequency fc , the |H(fc) | = 1/2. The cutoff portion of the frequency response around the carrier frequency fc exhibits odd symmetry. That is, inside the transition interval fc - fv ≤ | f | ≤ fc + fv: 1. The sum of the values of the magnitude response | H(f) | at any two frequencies equally displaced above and below fc is unity. 2. The phase response |H(f) | is linear:

H(f – fc) +H ( f + fc)=1 for W ≤ f ≤ W




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The transmitted band width of VSB modulation is

BT=W +fv

W: message band width./

fv: the width of the vestigial sideband.

The VSB modulated wave in time domain :

𝑠𝑠(𝑡𝑡) =12𝐴𝐴𝑐𝑐𝑚𝑚

(𝑡𝑡) cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡) ∓12𝐴𝐴𝑐𝑐�̀�𝑚 (𝑡𝑡) sin(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡)

Spectra of AM signals.

Taking the Fourier transform:

[𝑠𝑠(𝑓𝑓)]𝐴𝐴𝑀𝑀 =𝐴𝐴𝑐𝑐2

[𝛿𝛿(𝑓𝑓 − 𝑓𝑓𝑐𝑐) + 𝛿𝛿(𝑓𝑓 + 𝑓𝑓𝑐𝑐)] +𝐴𝐴𝑐𝑐2 [𝑀𝑀(𝑓𝑓 − 𝑓𝑓𝑐𝑐) + 𝑀𝑀(𝑓𝑓 + 𝑓𝑓𝑐𝑐)]

Spectrum of the AM signal

Baseband message spectrum M(f )

Lower sideband

Lower sideband

Upper sideband

Upper sideband




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1-The spectrum has two sidebands, the USB [between fc to fc + W , and (− fc −W) to − fc] and the LSB[ ( fc − W to fc and − fc to (− fc + W) ]. 2-If the baseband signal has bandwidth W , then the AM signal has bandwidth 2W . That is, the transmission bandwidth BT , required for the AM signal is 2W . 3-Spectrum has discrete components at f = ± fc , indicated by impulses of area Ac/2.

4-In order to avoid the overlap between the positive part and the negative part of S(f ), fc > W (In practice, fc >> W , so that s (t ) is a narrowband signal) The discrete components at f = ± fc , do not carry any information and as such AM does not make efficient use of the transmitted power.

Example 5 :

For AM with tone modulation, let us find 𝜂𝜂 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠 𝑝𝑝𝑡𝑡𝑝𝑝𝑠𝑠𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑝𝑝𝑡𝑡𝑝𝑝𝑠𝑠𝑝𝑝

as a function

of modulation index µ.

for tone modulation

𝑠𝑠(𝑡𝑡) = 𝐴𝐴𝑐𝑐[1 + 𝜇𝜇 cos(𝜔𝜔𝑚𝑚𝑡𝑡)] cos(𝜔𝜔𝑚𝑚𝑡𝑡)

Carrier term =𝐴𝐴𝑐𝑐cos(𝜔𝜔𝑚𝑚𝑡𝑡)

Carrier power= 𝐴𝐴𝑐𝑐2

2

USB term= 𝐴𝐴𝑐𝑐𝜇𝜇2

cos(𝜔𝜔𝑐𝑐 +𝜔𝜔𝑚𝑚)𝑡𝑡

Power in USB=�𝐴𝐴𝑐𝑐𝜇𝜇2 �

2

2= 𝐴𝐴𝑐𝑐

2𝜇𝜇2

8

Power in LSB=power in USB

Total sideband power = 2 × 𝐴𝐴𝑐𝑐2 𝜇𝜇2

8= 𝐴𝐴𝑐𝑐2𝜇𝜇2

4

Total power = 𝐴𝐴𝑐𝑐2

2+ 𝐴𝐴𝑐𝑐2𝜇𝜇2

4= 𝐴𝐴𝑐𝑐2

2�1 + 𝜇𝜇2

2� = 𝐴𝐴𝑐𝑐2(2+𝜇𝜇2)

4




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𝜂𝜂 =𝐴𝐴𝑐𝑐2𝜇𝜇2

4𝐴𝐴𝑐𝑐2(2 + 𝜇𝜇2)

4

=𝜇𝜇2

2 + 𝜇𝜇2

μ

η

0.25 0.03 0.50 0.11 0.75 1.0

𝑠𝑠(𝑡𝑡) = 𝐴𝐴𝑐𝑐[1 + 𝜇𝜇 cos(𝜔𝜔𝑚𝑚𝑡𝑡)]𝑐𝑐𝑡𝑡𝑠𝑠(𝜔𝜔𝑚𝑚𝑡𝑡)

𝑅𝑅𝑠𝑠{𝐴𝐴𝑐𝑐𝑠𝑠𝑗𝑗𝜔𝜔 𝑐𝑐𝑡𝑡 +𝐴𝐴𝑐𝑐𝜇𝜇

2 �𝑠𝑠𝑗𝑗 (𝜔𝜔𝑐𝑐+𝜔𝜔𝑚𝑚 }𝑡𝑡 + 𝑠𝑠𝑗𝑗 (𝜔𝜔𝑐𝑐−𝜔𝜔𝑚𝑚 }𝑡𝑡 �}

[𝑠𝑠(𝑡𝑡)]𝑐𝑐𝑠𝑠 = 𝐴𝐴𝑐𝑐 +𝐴𝐴𝑐𝑐𝜇𝜇

2 𝑠𝑠𝑗𝑗𝜔𝜔𝑚𝑚 𝑡𝑡 +𝐴𝐴𝑐𝑐𝜇𝜇

2 𝑠𝑠−𝑗𝑗𝜔𝜔𝑚𝑚 𝑡𝑡

Pharos diagrams such as the one shown in Fig. are helpful in the study of unequal attenuation of the sideband components.

Pharos diagram for AM with tone modulation




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Example 6 :

Let Ac = 1, μ = 1 /2 and let the upper sideband be attenuated by a factor of 2. Let us find the expression for the resulting envelope, A(t ) .

the sidebands is no longer collinear with the carrier

[𝑠𝑠(𝑡𝑡)]𝑐𝑐𝑠𝑠 = 1 +18 �cos(𝜔𝜔𝑚𝑚𝑡𝑡) + 𝑗𝑗𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝑚𝑚𝑡𝑡)� +

14 �cos(𝜔𝜔𝑚𝑚𝑡𝑡) − 𝑗𝑗𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝑚𝑚𝑡𝑡)�

= 1 +38�cos(𝜔𝜔𝑚𝑚𝑡𝑡) − 𝑗𝑗

18 𝑠𝑠𝑠𝑠𝑠𝑠

(𝜔𝜔𝑚𝑚𝑡𝑡)�

𝐴𝐴(𝑡𝑡) = [(1 +38 cos(𝜔𝜔𝑚𝑚𝑡𝑡))2 + �

18 𝑗𝑗𝑠𝑠𝑠𝑠𝑠𝑠

(𝜔𝜔𝑚𝑚𝑡𝑡)�2

]12

The AM signal is really a composite of several signal voltages, namely, the carrier and the two sidebands, and each of these signals produces power in the antenna. The total transmitted power PT is simply the sum of the carrier power Pc and the power in the two sidebands PUSB and PLSB:

PT=Pc+ PUSB+ PLSB

𝑣𝑣𝐴𝐴𝑀𝑀 = 𝑉𝑉𝑐𝑐 sin 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 +𝑉𝑉𝑚𝑚2 cos 2𝜋𝜋𝑡𝑡( 𝑓𝑓𝑐𝑐 − 𝑓𝑓𝑚𝑚 ) −

𝑉𝑉𝑚𝑚2 cos 2𝜋𝜋𝑡𝑡( 𝑓𝑓𝑐𝑐 + 𝑓𝑓𝑚𝑚)

Phasor diagram for an AM signal with unequal sidebands

Carrier lower sideband upper sideband




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𝑉𝑉𝑐𝑐 and 𝑉𝑉𝑚𝑚 are peak values of the carrier and modulating sine waves. The rms carrier and sideband voltages :

𝑣𝑣𝐴𝐴𝑀𝑀 =𝑉𝑉𝑐𝑐

√2sin 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 +

𝑉𝑉𝑚𝑚2√2

cos 2𝜋𝜋𝑡𝑡( 𝑓𝑓𝑐𝑐 − 𝑓𝑓𝑚𝑚) −𝑉𝑉𝑚𝑚

2√2cos 2𝜋𝜋𝑡𝑡( 𝑓𝑓𝑐𝑐 + 𝑓𝑓𝑚𝑚)

The power in the carrier and sidebands can be calculated by using the power formula P=V2/R where P is the output power, V is the rms output voltage, and R is the resistive part of the load impedance, which is usually an antenna.

𝑃𝑃𝑇𝑇 =(𝑉𝑉𝑐𝑐

√2)2

𝑅𝑅 +( 𝑉𝑉𝑚𝑚 2√2

)2

𝑅𝑅 +( 𝑉𝑉𝑚𝑚 2√2

)2

𝑅𝑅 =𝑉𝑉𝑐𝑐

2

2𝑅𝑅 +𝑉𝑉𝑚𝑚

2

8𝑅𝑅 +𝑉𝑉𝑚𝑚

2

8𝑅𝑅

β=𝑉𝑉𝑚𝑚 𝑉𝑉𝑐𝑐

𝑉𝑉𝑚𝑚 = 𝑚𝑚𝑉𝑉𝑐𝑐

𝑃𝑃𝑇𝑇 =(𝑉𝑉𝑐𝑐 )2

𝑅𝑅 +(𝑚𝑚𝑉𝑉𝑐𝑐 )2

8𝑅𝑅 +(𝑚𝑚𝑉𝑉𝑐𝑐 )2

8𝑅𝑅 =𝑉𝑉𝑐𝑐

2

2𝑅𝑅 +𝑚𝑚2𝑉𝑉𝑐𝑐

2

8𝑅𝑅 +𝑚𝑚2𝑉𝑉𝑐𝑐

2

8𝑅𝑅 2

𝑉𝑉𝑐𝑐 2

2𝑅𝑅 = 𝑝𝑝𝑚𝑚𝑠𝑠 𝑐𝑐𝑡𝑡𝑝𝑝𝑝𝑝𝑠𝑠𝑠𝑠𝑝𝑝 𝑝𝑝𝑡𝑡𝑝𝑝𝑠𝑠𝑝𝑝 𝑃𝑃𝑐𝑐

𝑃𝑃𝑇𝑇 =(𝑉𝑉𝑐𝑐 )2

𝑅𝑅 (1 +(𝑚𝑚)2

4 +(𝑚𝑚)2

4

the total power in an AM signal when the carrier power and the percentage of modulation are known:

𝑃𝑃𝑇𝑇 = 𝑃𝑃𝑐𝑐(1 +(𝑚𝑚)2

2 )

PT=IT2R

𝐼𝐼𝑇𝑇 = 𝐼𝐼𝑐𝑐�(1 + 𝑚𝑚2/2)

𝑚𝑚 = �2 ��𝐼𝐼𝑇𝑇𝐼𝐼𝑐𝑐�

2

− 1�




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Generation of AM :

We shall discuss two methods of generating AM signals, one using a nonlinear element and the other using an element with time-varying characteristic. 1-Square law product

consider the scheme shown below :

For small variations of v around a suitable operating point, v2 (t )

𝑣𝑣𝑡𝑡(𝑡𝑡) = 𝛼𝛼1𝑣𝑣𝑠𝑠(𝑡𝑡) + 𝛼𝛼2𝑣𝑣𝑠𝑠2(𝑡𝑡)

𝛼𝛼1𝑡𝑡𝑠𝑠𝑠𝑠 𝛼𝛼2: constant

𝑣𝑣𝑠𝑠(𝑡𝑡) = 𝐴𝐴𝑐𝑐 cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) + 𝑚𝑚(𝑡𝑡)

A circuit with a nonlinear element

v-j characteristic of the diode

V1(t) V2(t)




21

𝑣𝑣𝑡𝑡(𝑡𝑡) = 𝛼𝛼1𝐴𝐴𝑐𝑐 �1 +2𝛼𝛼2

𝛼𝛼1𝑚𝑚(𝑡𝑡)� cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) + 𝛼𝛼1𝑚𝑚(𝑡𝑡) + 𝛼𝛼2𝑚𝑚2(𝑡𝑡)

+ 𝛼𝛼2𝐴𝐴2𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠2(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 )

component Spectrum 𝐴𝐴𝑐𝑐 cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) A

2𝛼𝛼2𝐴𝐴𝑐𝑐𝑚𝑚(𝑡𝑡)𝑐𝑐𝑡𝑡𝑠𝑠2(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) B 𝛼𝛼1𝑚𝑚(𝑡𝑡) C 𝛼𝛼2𝑚𝑚2(𝑡𝑡) D

𝛼𝛼2𝐴𝐴2𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠2(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) E

then the required AM signal would be available at the output of the filter. This is possible by placing a BPF with centre at fc and bandwidth 2W provided ( fc − W )> 2W or fc > 3W .

characteristics.

The disadvantages. 1) The required square-law nonlinearity of a given device would be available only over a small part of the (v − i ) characteristic. Hence, it is possible to generate only low levels of the desired output. 2) If fc is of the order of 3W, then we require a BPF with very sharp cut off

Spectra of the components of v2 (t )

Spectrum of message signal Spectrum of AM signal




22

2- Switching modulator

In this method, diode will be used as a switching element. As such, it acts as a device with time-varying characteristic

𝑣𝑣1(𝑡𝑡) = 𝑐𝑐(𝑡𝑡) + 𝑚𝑚(𝑡𝑡)

= 𝐴𝐴𝑐𝑐 cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 ) + 𝑚𝑚(𝑡𝑡)

If we assume that | m(t) | max<< Ac

𝑣𝑣2(𝑡𝑡) = �𝑣𝑣1(𝑡𝑡) 𝑐𝑐(𝑡𝑡) > 00 𝑐𝑐(𝑡𝑡) ≤ 0

�

𝑣𝑣2(𝑡𝑡) = 𝑣𝑣1(𝑡𝑡)𝑥𝑥𝑝𝑝(𝑡𝑡)

Where 𝑥𝑥𝑝𝑝(𝑡𝑡) is the periodic rectangular pulse train

𝑥𝑥𝑝𝑝(𝑡𝑡) = �1 𝑠𝑠𝑓𝑓 𝑐𝑐(𝑡𝑡) > 0 (𝑝𝑝𝑡𝑡𝑠𝑠𝑠𝑠𝑡𝑡𝑠𝑠𝑣𝑣𝑠𝑠 ℎ𝑡𝑡𝑡𝑡𝑓𝑓 𝑐𝑐𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠 𝑡𝑡𝑓𝑓 𝑐𝑐(𝑡𝑡))0 𝑠𝑠𝑓𝑓 𝑐𝑐(𝑡𝑡) < 0( 𝑠𝑠𝑠𝑠𝑛𝑛𝑡𝑡𝑡𝑡𝑠𝑠𝑣𝑣𝑠𝑠 ℎ𝑡𝑡𝑡𝑡𝑓𝑓 𝑐𝑐𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠 𝑡𝑡𝑓𝑓 𝑐𝑐(𝑡𝑡))

�

When f0=fc

𝑥𝑥𝑝𝑝(𝑡𝑡) = � 𝑥𝑥𝑠𝑠𝑠𝑠𝑗𝑗2𝜋𝜋𝑠𝑠𝑓𝑓𝑐𝑐𝑡𝑡∞

𝑠𝑠=−∞

Where 𝑥𝑥𝑠𝑠 = 12

sin �𝑠𝑠2� 𝑥𝑥𝑠𝑠 = 0 𝑓𝑓𝑡𝑡𝑝𝑝 𝑠𝑠 = ±2, ±4 … ….

𝑥𝑥𝑝𝑝(𝑡𝑡) =12 +

2𝜋𝜋 cos( 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡) + �

(−1)𝑠𝑠−1

2𝑠𝑠 − 1

∞

𝑠𝑠=2

cos[2𝜋𝜋(2𝑠𝑠 − 1)𝑓𝑓𝑐𝑐𝑡𝑡

(a) Switching modulator

(b) Switching characteristic of the diode-load combination.




23

𝑣𝑣𝑡𝑡(𝑡𝑡) = 𝐴𝐴𝑐𝑐/2 �1 +4𝜋𝜋𝐴𝐴𝑐𝑐

𝑚𝑚(𝑡𝑡)� cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡) +𝑚𝑚(𝑡𝑡)

2 + 2𝐴𝐴𝑐𝑐/𝜋𝜋𝑐𝑐𝑡𝑡𝑠𝑠2(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡).

switching modulator advantages: a) Generated AM signals can have larger power levels. b) Filtering requirements are less stringent because we can separate the desired AM signal if fc > 2W .





24

balanced modulator

1-A balanced modulator is a circuit that generates a DSB signal, suppressing the carrier and leaving only the sum and difference frequencies at the output.

2-The output of a balanced modulator can be further processed by filters or phase-shifting circuitry to eliminate one of the sidebands, resulting in a SSB signal.

𝑆𝑆1(𝑡𝑡) = 𝐴𝐴𝑐𝑐[1 + 𝑘𝑘𝑡𝑡𝑚𝑚(𝑡𝑡)]𝑐𝑐𝑡𝑡𝑠𝑠 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡

𝑆𝑆2(𝑡𝑡) = 𝐴𝐴𝑐𝑐[1 − 𝑘𝑘𝑡𝑡𝑚𝑚(𝑡𝑡)]𝑐𝑐𝑡𝑡𝑠𝑠 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡

𝑆𝑆(𝑡𝑡) = 𝑆𝑆1(𝑡𝑡) − 𝑆𝑆2(𝑡𝑡) = 2𝐴𝐴𝑐𝑐𝑘𝑘𝑡𝑡𝑚𝑚(𝑡𝑡)𝑐𝑐𝑡𝑡𝑠𝑠 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡

𝑆𝑆(𝑓𝑓) = 𝑘𝑘𝑡𝑡𝐴𝐴𝑐𝑐[𝑀𝑀(𝑓𝑓 − 𝑓𝑓𝑐𝑐) + 𝑀𝑀(𝑓𝑓 + 𝑓𝑓𝑐𝑐)]





25

3- Transistor linear modulation

Low-Level AM: Transistor Modulator:

1-Transistor modulation consists of a resistive mixing network, a transistor, and an LC tuned circuit.

2-The emitter-base junction of the transistor serves as a diode and nonlinear device.

3-Modulation and amplification occur as base current controls a larger collector current.

4-The LC tuned circuit oscillates (rings) to generate the missing half cycle.

Simple transistor modulator

Carrier

Modulation signal

AM

+vcc




26

AM Transmitter Principles

An AM transmitter can be divided into two major sections according to the frequencies at which they operate, radio-frequency and audio-frequency units. 1-The radio-frequency unit is the section of the transmitter used to generate the radio-frequency carrier wave. oscillator stage where it is generated as a constant-amplitude, constant-frequency sine wave. The carrier is not of sufficient amplitude and must be amplified in one or more stages before it attains the high power required by the antenna. With the exception of the last stage, the amplifiers between the oscillator and the antenna are called INTERMEDIATE POWER AMPLIFIERS (IPA) The final stage, which connects to the antenna, is called the FINAL POWER AMPLIFIER (FPA).

2-The second section of the transmitter contains the audio circuitry. This section of the transmitter takes the small signal from the microphone and increases its amplitude to the amount necessary to fully modulate the carrier. The last audio stage is the MODULATOR. It applies its signal to the carrier in the final power amplifier. .

Block diagram of an AM Modulation




27

Advantages of Amplitude modulation:- • Generation and detection of AM signals are very easy • It is very cheap to build, due to this reason it I most commonly used in AM radio broad casting Disadvantages of Amplitude of modulation:- • Amplitude modulation is wasteful of power • Amplitude modulation is wasteful of band width Application of Amplitude modulation: - AM Radio Broadcasting

160m 20w AM Transmitter

Single Sideband Transmitter




28

UItem 3 : Angle Modulation

Introduction .

It is another method of modulating a sinusoidal carrier wave, namely, angle Modulation in which either the phase or frequency of the carrier wave is varied according to the message signal. there are two types of Angle modulation techniques namely: 1. Phase modulation 2. Frequency modulation

e(t) = A cos (ωt + φ)

Angle modulation :

phase- frequency modulation




29

Phase Modulation PM

𝜃𝜃𝑠𝑠(𝑡𝑡) = 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 𝑘𝑘𝑝𝑝𝑚𝑚(𝑡𝑡)

2𝜋𝜋𝑓𝑓𝑐𝑐 :phase of un modulated carrier signal

𝑘𝑘𝑝𝑝 : modulator constant rad/volt

𝑚𝑚(𝑡𝑡):modulation signal volt

𝑠𝑠(𝑡𝑡)𝑃𝑃𝑀𝑀 = 𝐴𝐴𝑐𝑐cos[2𝜋𝜋𝑓𝑓𝑐𝑐 + 𝑘𝑘𝑝𝑝𝑚𝑚(𝑡𝑡)]

Frequency Modulation FM

𝑓𝑓𝑠𝑠(𝑡𝑡) = 𝑓𝑓𝑐𝑐 + 𝑘𝑘𝑓𝑓𝑚𝑚(𝑡𝑡)]

𝑓𝑓𝑐𝑐 :frequency of un modulated carrier signal

𝑘𝑘𝑓𝑓 : modulator constant Hz/volt

𝜃𝜃𝑠𝑠(𝑡𝑡) = 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 2𝜋𝜋� 𝑘𝑘𝑓𝑓𝑚𝑚(𝑡𝑡)𝑡𝑡

0

𝑠𝑠(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐cos[2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 2𝜋𝜋𝑘𝑘𝑓𝑓 � 𝑚𝑚(𝑡𝑡)𝑠𝑠𝑡𝑡𝑡𝑡

0]

Generation of FM signal using phase modulator

PM signal

FM signal

Differentiator Frequency modulator

Integration Phase modulator

Modulating wave

Modulating wave

𝐴𝐴𝑐𝑐cos(2𝜋𝜋𝑓𝑓𝑐𝑐)

𝐴𝐴𝑐𝑐cos(2𝜋𝜋𝑓𝑓𝑐𝑐)

Generation of PM signal using frequency modulator




30

Single Tone Frequency Modulation

m(t) = Am cos (2πfmt)

𝑓𝑓𝑠𝑠(𝑡𝑡) = 𝑓𝑓𝑐𝑐 + ∆𝑓𝑓𝑐𝑐𝑡𝑡𝑠𝑠 (2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡)

∆𝑓𝑓:frequency deviation

∆𝑓𝑓 = 𝑘𝑘𝑓𝑓𝐴𝐴𝑚𝑚

𝜃𝜃𝑠𝑠(𝑡𝑡) = 2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 𝛽𝛽sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡) 𝛽𝛽 = ∆𝑓𝑓𝑓𝑓𝑚𝑚

𝑠𝑠(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐cos[𝜔𝜔𝑐𝑐𝑡𝑡 + 𝛽𝛽 sin(𝜔𝜔𝑚𝑚𝑡𝑡)]

Where β= ∆f/fm= modulation index of the FM wave 1- When β<<1 radian then it is called as narrowband FM consisting essentially of a carrier, an upper side-frequency component, and a lower side-frequency component. 2- When β>>1 radian then it is called as wideband FM which contains a carrier and an infinite number of side-frequency components located symmetrically around the carrier.

Multi Tone Frequency Modulation

Let 𝑚𝑚(𝑡𝑡) = 𝐴𝐴1 cos(𝜔𝜔1𝑡𝑡) + 𝐴𝐴2 cos(𝜔𝜔2𝑡𝑡)

Where 𝑓𝑓1 and 𝑓𝑓2 are arbitrary.

𝑠𝑠𝑝𝑝𝑠𝑠 (𝑡𝑡) = 𝐴𝐴𝑐𝑐�𝑠𝑠𝑗𝑗𝛽𝛽1sin(𝜔𝜔1𝑡𝑡)]��𝑠𝑠𝑗𝑗𝛽𝛽1 sin(𝜔𝜔2𝑡𝑡)]�𝑠𝑠𝑗𝑗𝜔𝜔𝑐𝑐𝑡𝑡

𝛽𝛽1 = 𝐴𝐴1𝑘𝑘𝑓𝑓𝑓𝑓1

and 𝛽𝛽2 = 𝐴𝐴2𝑘𝑘𝑓𝑓𝑓𝑓2

𝑠𝑠𝑝𝑝𝑠𝑠 (𝑡𝑡) = 𝐴𝐴𝑐𝑐 ��𝐽𝐽𝑚𝑚(𝛽𝛽1)𝑚𝑚

𝑠𝑠𝑗𝑗𝑚𝑚𝜔𝜔 1𝑡𝑡][ �� 𝐽𝐽𝑠𝑠(𝛽𝛽2)𝑠𝑠

𝑠𝑠𝑗𝑗𝑠𝑠𝜔𝜔 2𝑡𝑡]}𝑠𝑠𝑗𝑗𝜔𝜔𝑐𝑐𝑡𝑡

𝑠𝑠𝑝𝑝𝑠𝑠 (𝑡𝑡) = 𝐴𝐴𝑐𝑐 ��𝐽𝐽𝑚𝑚𝑠𝑠

(𝛽𝛽1)𝐽𝐽𝑠𝑠(𝛽𝛽2)𝑐𝑐𝑡𝑡𝑠𝑠𝑚𝑚

[(𝜔𝜔𝑐𝑐 + 𝑚𝑚𝜔𝜔1 + 𝑠𝑠𝜔𝜔2)𝑡𝑡]��




31

1- Narrow Band Frequency Modulation (NBFM).

𝑥𝑥(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐 cos(𝜔𝜔𝑡𝑡) cos[𝛽𝛽 sin(𝜔𝜔𝑚𝑚𝑡𝑡)] − 𝐴𝐴𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝑐𝑐𝑡𝑡)sin[𝛽𝛽sin( 𝜔𝜔𝑚𝑚𝑡𝑡)]

For study case , assume :

cos[𝛽𝛽 sin(𝜔𝜔𝑚𝑚𝑡𝑡)] ≅ 1

sin[𝛽𝛽 sin(𝜔𝜔𝑚𝑚𝑡𝑡)] ≅ 𝛽𝛽 sin(𝜔𝜔𝑚𝑚𝑡𝑡)

𝑥𝑥(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐 cos(𝜔𝜔𝑐𝑐𝑡𝑡)−𝛽𝛽𝐴𝐴𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝑐𝑐𝑡𝑡)sin( 𝜔𝜔𝑚𝑚𝑡𝑡)

𝑥𝑥(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐 cos(𝜔𝜔𝑐𝑐𝑡𝑡) +12𝛽𝛽𝐴𝐴𝑐𝑐[cos(𝜔𝜔𝑐𝑐 + 𝜔𝜔𝑚𝑚) 𝑡𝑡 − 𝑐𝑐𝑡𝑡𝑠𝑠( 𝜔𝜔𝑐𝑐 − 𝜔𝜔𝑚𝑚)𝑡𝑡]

𝑥𝑥(𝑡𝑡)𝐴𝐴𝑀𝑀 = 𝐴𝐴𝑐𝑐 cos(𝜔𝜔𝑐𝑐𝑡𝑡) +12𝛽𝛽𝐴𝐴𝑐𝑐[cos(𝜔𝜔𝑐𝑐 + 𝜔𝜔𝑚𝑚) 𝑡𝑡 + 𝑐𝑐𝑡𝑡𝑠𝑠( 𝜔𝜔𝑐𝑐 − 𝜔𝜔𝑚𝑚)𝑡𝑡]

ʃdt Parallel modulator

Phase shift 90o

Oscillator

+ NBFM

Generation of NBFM signal

m(t) FM

PM

cos(𝜔𝜔𝑐𝑐𝑡𝑡)




32

A pharos comparison of narrowband FM and AM waves for sinusoidal modulation. (a) Narrowband FM wave. (b) AM wave.




33

2-Wide Band Frequency Modulation (WBFM).

The spectrum of single-tone FM signal , the modulation index is β.

𝑠𝑠(𝑡𝑡) = 𝐴𝐴𝑐𝑐cos[2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡)]

𝛽𝛽 =∆𝑓𝑓𝑓𝑓𝑚𝑚

We assume that the carrier frequency fc is large enough compared to the bandwidth of FM signal.

s(t)FM = Re[Ac exp (j2πfct + jβ sin(2πfmt )] = Re[s(t) exp(j2πfct)] where s(t) is the complex envelope of the FM signal s(t)FM

𝑠𝑠(𝑡𝑡) = �𝑐𝑐𝑠𝑠 𝑠𝑠𝑥𝑥𝑝𝑝(𝑗𝑗2𝜋𝜋𝑓𝑓𝑚𝑚

∞

−∞

𝑡𝑡)

where the complex Fourier coefficient cn is defined by

𝑐𝑐𝑠𝑠 = 𝑓𝑓𝑚𝑚𝐴𝐴𝑐𝑐 � exp[𝑗𝑗𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡) − 𝑗𝑗2𝜋𝜋𝑠𝑠𝑓𝑓𝑚𝑚𝑡𝑡]𝑠𝑠𝑡𝑡 12𝑓𝑓𝑚𝑚

− 12𝑓𝑓𝑚𝑚




34

Assume 𝑥𝑥 = 2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡 𝑇𝑇 = 2𝜋𝜋𝜔𝜔𝑚𝑚

𝑐𝑐𝑠𝑠 =𝐴𝐴𝑐𝑐2𝜋𝜋� exp[𝑗𝑗𝛽𝛽 sin 𝑥𝑥 − 𝑠𝑠𝑥𝑥]𝑠𝑠𝑥𝑥

12𝑓𝑓𝑚𝑚

− 12𝑓𝑓𝑚𝑚

𝐽𝐽𝑠𝑠(𝛽𝛽) =1

2𝜋𝜋� exp[𝑗𝑗𝛽𝛽 sin 𝑥𝑥 − 𝑠𝑠𝑥𝑥]𝑠𝑠𝑥𝑥𝜋𝜋

− 𝜋𝜋

cn=AcJn(β)

𝑠𝑠(𝑡𝑡)𝐹𝐹𝑀𝑀 = 𝐴𝐴𝑐𝑐 .𝑅𝑅𝑠𝑠[ � 𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑠𝑠𝑥𝑥𝑝𝑝(𝑗𝑗2𝜋𝜋(𝑓𝑓𝑐𝑐 + 𝑠𝑠𝑓𝑓𝑚𝑚 )∞

𝑠𝑠=−∞

𝑡𝑡)

This is the desired form for the Fourier series representation of the single-tone FM signal s(t) for an arbitrary value of β. The discrete spectrum of s(t) is obtained by taking the Fourier transforms of both sides .

Plots of Bessel functions 𝑱𝑱𝒏𝒏(𝜷𝜷 )of the first kind for varying d




35

𝑠𝑠(𝑓𝑓)𝐹𝐹𝑀𝑀 =𝐴𝐴𝑐𝑐2 � 𝐽𝐽𝑠𝑠(𝛽𝛽)[𝛿𝛿(𝑓𝑓 − 𝑓𝑓𝑐𝑐 − 𝑠𝑠𝑓𝑓𝑚𝑚 ) +

∞

𝑠𝑠=−∞

𝛿𝛿((𝑓𝑓 + 𝑓𝑓𝑐𝑐 + 𝑠𝑠𝑓𝑓𝑚𝑚 )]

𝑱𝑱𝒏𝒏(𝜷𝜷 ) properties : 1-𝐽𝐽𝑠𝑠(𝛽𝛽) = (−1)𝑠𝑠𝐽𝐽−𝑠𝑠(𝛽𝛽) for all n, both positive and negative 2-for small values of the modulation index 𝛽𝛽 :

�𝐽𝐽0(𝛽𝛽) ≈ 1

𝐽𝐽1(𝛽𝛽) ≈𝛽𝛽2

𝐽𝐽𝑠𝑠(𝛽𝛽) ≈ 0, 𝑠𝑠 > 2

�

∑ 𝐽𝐽𝑠𝑠2(𝛽𝛽)∞𝑠𝑠=−∞ = 1

Example 7:

The investigate variations in the amplitude and frequency of a sinusoidal modulating signal affect the spectrum of the FM signal. Discrete amplitude spectra of an PM signal, normalized with respect to the carrier amplitude, for the case of sinusoidal modulation of fixed frequency and varying amplitude. Only the spectra for positive frequencies are shown.




36

Discrete amplitude spectra of an FM signal, normalized with respect to the carrier amplitude, for the case of sinusoidal modulation of varying frequency and fixed amplitude. Only the spectra for positive frequencies are shown.




37

Example 8:

Compare AM to FM for x(t)= cos (ωmt).

The advantages of FM:

1-constant power

2-no need to transmit carrier ( unless DC important)

3-bandwidth




38

Generation of FM signals There are two distinct methods of generating WBFM signals: a) Direct FM b) Indirect FM. Details on their generation are as follows. a) Indirect FM (Armstrong’s method). In this method - attributed to Armstrong - first a narrowband FM signal is generated. This is then converted to WBFM by using frequency multiplication. This is shown schematically in Fig. 5.14. The generation of NBFM has already been described. A frequency multiplier is a nonlinear device followed by a BPF. A nonlinearity of order n can give rise to frequency multiplication by a factor of n. consider a square law device with output y (t ) = x2 (t ) where x (t ) is the input. Let x (t ) be the FM signal.

𝑥𝑥(𝑡𝑡) = cos[𝜃𝜃(𝑡𝑡)] ,𝑝𝑝ℎ𝑠𝑠𝑝𝑝𝑠𝑠 𝜃𝜃(𝑡𝑡) = 𝜔𝜔𝑐𝑐𝑡𝑡 + 2𝜋𝜋𝑘𝑘𝑡𝑡 � 𝑚𝑚(𝛼𝛼)𝑠𝑠𝛼𝛼𝑡𝑡

−∞

𝑐𝑐(𝑡𝑡) = cos2[𝜃𝜃(𝑡𝑡)]

= 12

[1 + cos[2𝜃𝜃(𝑡𝑡)]

𝑐𝑐(𝑡𝑡) =12 +

12 cos[2𝜔𝜔𝑐𝑐𝑡𝑡] + 4𝜋𝜋𝑘𝑘𝑡𝑡 � 𝑚𝑚(𝛼𝛼)𝑠𝑠𝛼𝛼

𝑡𝑡

−∞

The DC term can be filtered out to give an FM output .

Carrier frequency:2fc

Frequency deviation of NBFM:∆f

Generation of WBFM (Armstrong method)




39

Frequency deviations: ∆f,2:∆f,…………n:∆f

The multiplication scheme used in FM transmitter

The carrier frequency of the NBFM signal fc1, is 200 kHz with the corresponding Δf1 = 25 Hz. Desired FM output is to have the frequency deviation Δf4 = 75 kHz and a carrier (fc 4 ) of 91.2 MHz. To obtain Δf4 = 75 kHz starting from Δf1 = 25 Hz, we require a total frequency multiplication of (75×103)/25=3000 fc1=200kHz total multiplication factor =64×48=3072 carrier frequency (fc4) =200×3072=614.4 MHz the final required carrier frequency is 91.2 MHz Δf2 = Δf3 = 1.6 kHz

Frequency convertor f× 64 f× 48

fc1=200 kHz

∆f1=25 Hz

fc2=12.8 MHz

∆f2=1.6 Hz

fc3=200 kHz

∆f3=25 Hz fc1=200 kHz

∆f1=25 Hz

fc4=200 kHz

∆f4=25 Hz

Crystal oscillator 10.9 MHz




40

these six cannot be used as a single cascade because, that would result in a carrier frequency equal to 20 × 103 × 36 = 14.58MHz.

Example 10 ;

Armstrong’s method is to be used to generate a WBFM signal. The NBFM signal has the carrier frequency fc1 = 20 kHz. The WBFM signal that is required must have the parameters fc = 6 MHz and Δf = 10 kHz . Only frequency triples are available. Draw the schematic block diagram of this example. Total frequency multiplication required =6×106/20×103=300 Frequency triples=35=243 36=729

Generation of WBFM from NBFM of example the final frequency deviation required is 10 kHz

NBFM ∆f1=10×103/729=13.71Hz After the frequency conversion stage, we have one more stage multiplication by 3. the carrier frequency at the mixer output fc3

fc3×3=6MHz fc3=2 MHz

fLO =6,86 MHz




41

Exercise In the indirect FM scheme shown in Fig. 5.17, find the values of fc,i and Δfi for i = 1, 2 and 3 . What should be the centre frequency, f0 , of the BPF. Assume that fLO > fc,2 . Direct FM Method (1) A common method used for generating FM directly is to vary the inductance or capacitance of a tuned electronic oscillator . If L and C are the inductance and capacitance , of a simple tuned circuit . The oscillation frequency f0 of a parallel tuned circuit with inductance L and capacitance C is given by:

𝑓𝑓𝑡𝑡 =1

2𝜋𝜋√𝐿𝐿𝐿𝐿

𝜔𝜔𝑡𝑡 =1

√𝐿𝐿𝐿𝐿

Let C be varied by the modulating signal m(t ) , as given by

𝑐𝑐(𝑡𝑡) = 𝑐𝑐𝑡𝑡 − 𝑘𝑘𝑚𝑚(𝑡𝑡)




42

k:constant 𝜔𝜔𝑠𝑠(𝑡𝑡) = 1

�𝐿𝐿𝐿𝐿𝑡𝑡[1 + 𝑘𝑘𝑚𝑚 (𝑡𝑡)

2𝑐𝑐𝑡𝑡] 𝑘𝑘𝑚𝑚 (𝑡𝑡)

2𝑐𝑐𝑡𝑡≪ 1

𝜔𝜔𝑐𝑐 =1

�𝐿𝐿𝑐𝑐𝑡𝑡

𝜔𝜔𝑠𝑠(𝑡𝑡) = 𝜔𝜔𝑐𝑐 + 𝑐𝑐𝑓𝑓𝑚𝑚(𝑡𝑡)

𝑐𝑐𝑓𝑓 =𝑘𝑘𝜔𝜔𝑐𝑐2𝑐𝑐𝑡𝑡

𝑐𝑐𝑠𝑠 = 𝑐𝑐𝑡𝑡 − 𝑘𝑘𝑚𝑚(𝑡𝑡) c(t ) of the oscillator circuit is:

𝑐𝑐(𝑡𝑡) = (𝑐𝑐1 + 𝑐𝑐𝑡𝑡) − 𝑘𝑘𝑚𝑚(𝑡𝑡) = 𝑐𝑐𝑡𝑡 − 𝑘𝑘𝑚𝑚(𝑡𝑡) Where 𝑐𝑐𝑡𝑡 = 𝑐𝑐1 + 𝑐𝑐𝑡𝑡

𝑐𝑐𝑠𝑠 =100

�1 + 2𝑣𝑣𝑠𝑠𝑝𝑝𝑓𝑓

Example 11:

Consider the circuit of Fig. 5.20 for the direct generation of FM. The diode a pacitance Cd , is related to the reverse bias as,

vd: the voltage across the varactor vB=4v m(t)=0.054 sin [( 10 π×103)t] c1=250 pF the circuit resonates at 2 MHz when m(t)=0

𝑐𝑐𝑠𝑠 =10−10

3 − 0.2 × 10−12sin[(10𝜋𝜋 × 103)𝑡𝑡] 𝑓𝑓𝑠𝑠(𝑡𝑡) = 2 × 106 + 705.86 sin[(10𝜋𝜋 × 103)𝑡𝑡]




43

Method ( 2)

The electronic switch is designed such that it is in position 1 when y (t ) = V0 and goes to position 2 when y ( t) =- V0 For 0 < t ≤ t1

𝑥𝑥(𝑡𝑡) = −𝐸𝐸0 +1𝑅𝑅𝐿𝐿� 𝑣𝑣0𝑠𝑠𝑑𝑑

1

0

When t=t1,let x(t) become E0 .then y(t) = -𝑣𝑣0 and the electronic switch assume positio2.

𝐸𝐸0 = −𝐸𝐸0 +1𝑅𝑅𝐿𝐿� 𝑣𝑣0𝑠𝑠𝑑𝑑

𝑡𝑡1

0

𝑡𝑡1 =2𝑅𝑅𝐿𝐿𝐸𝐸0

𝑣𝑣0

The output x (t ) decreasing to t=t2 when x(t)=- 𝐸𝐸0




44

𝑡𝑡2 − 𝑡𝑡1 =2𝑅𝑅𝐿𝐿𝐸𝐸0

𝑣𝑣0

The x(t) and y(t) are periodic with period 4𝑅𝑅𝐿𝐿𝐸𝐸0

𝑣𝑣0

The fundamental frequency of these waveforms

𝑓𝑓0 = �2𝑅𝑅𝐿𝐿𝐸𝐸0

𝑣𝑣0�−1

When v0=E0

𝑓𝑓0 =1

4𝑅𝑅𝐿𝐿 Direct and Indirect FM Transmitter. How to transmit a signal with frequency ranging in (-5KHz,5KHz) using a channel operating in (100KHz,110KHz)? What should be the carrier frequency ? Draw the block diagrams for the modulator and demodulator, and sketch the spectrum of the modulation and demodulation




45




46

FDM Receiver

Block diagram of FDM system.




47

Application of modulation and FDM 1-AM Radio (535KHz--1715KHz): – Each radio station is assigned 10 KHz, to transmit a mono-channel audio (band limited to 5KHz) – Using Amplitude modulation to shift the baseband signal 2- FM Radio (88MHz--108 MHz): – Each radio station is assigned 200 KHz, to transmit a stereo audio. – The left and right channels (each limited to 15KHz) are multiplexed into a single baseband signal using amplitude modulation – Using frequency modulation to shift the baseband signals 3- TV broadcast (VHF: 54-88,174-216MHz, UHF:470-890MHz) – Each station is assigned 6 MHz – The three color components and the audio signal are multiplexed into a single baseband signal – Using vestigial sideband AM to shift the baseband signals.

Modulation steps in an FDM system




48

Angle modulation spectrum assume that 𝜃𝜃(𝑡𝑡) = 𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡) 𝛽𝛽: modulation index and is the maximum value of phase deviation for both FM and PM. The signal: 𝑠𝑠(𝑡𝑡) = 𝐴𝐴𝑐𝑐cos[2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡)]

𝛽𝛽 =∆𝑓𝑓𝑓𝑓𝑚𝑚

We assume that the carrier frequency fc is large enough compared to the bandwidth of FM signal.

xc(t) = Ac cos[( 2πfct + jβ sin(2πfmt )] = Re[Ac exp (jβsin(2πfmt ))exp(j2πfct)] The signal can be expressed as: 𝑥𝑥𝑐𝑐(𝑡𝑡) = 𝑅𝑅𝑠𝑠[𝑥𝑥�𝑐𝑐(𝑡𝑡)𝑠𝑠𝑗𝑗2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡] the complex envelope of the modulated carrier signal 𝑥𝑥�𝑐𝑐(𝑡𝑡) = 𝐴𝐴𝑐𝑐𝑠𝑠𝑗𝑗𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚 𝑡𝑡) The complex envelope is periodic with frequency fm and can therefore be expanded in a Fourier series. The Fourier coefficients are given by

𝑓𝑓𝑚𝑚� 𝑠𝑠𝑗𝑗𝛽𝛽 sin(2𝜋𝜋𝑓𝑓𝑚𝑚 𝑡𝑡)1/2𝑓𝑓𝑚𝑚

−1/2𝑓𝑓𝑚𝑚𝑠𝑠−𝑗𝑗2𝜋𝜋𝑠𝑠𝑓𝑓𝑚𝑚 𝑡𝑡𝑠𝑠𝑡𝑡 =

12𝜋𝜋� 𝑠𝑠−[𝑗𝑗𝑠𝑠𝑥𝑥 −𝛽𝛽sin(𝑥𝑥)

𝜋𝜋

−𝜋𝜋𝑠𝑠𝑥𝑥

The integral is a function of n and b and is known as the Bessel Function of the first kind of order n and argument 𝛽𝛽 the Fourier series for the complex envelope can be written as:




49

𝑠𝑠𝑥𝑥𝑝𝑝𝑗𝑗𝛽𝛽sin(2𝜋𝜋𝑓𝑓𝑚𝑚𝑡𝑡) = � 𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑠𝑠𝑥𝑥𝑝𝑝(𝑗𝑗2𝜋𝜋𝑓𝑓𝑚𝑚

∞

𝑠𝑠=−∞

𝑡𝑡)

𝑥𝑥𝑐𝑐(𝑡𝑡) = 𝑅𝑅𝑠𝑠[(𝐴𝐴𝑐𝑐 � 𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑠𝑠𝑥𝑥𝑝𝑝(𝑗𝑗2𝜋𝜋𝑠𝑠𝑓𝑓𝑚𝑚𝑡𝑡)∞

𝑠𝑠=−∞

𝑠𝑠𝑥𝑥𝑝𝑝(𝑗𝑗2𝜋𝜋𝑠𝑠𝑓𝑓𝑐𝑐𝑡𝑡))

𝑥𝑥𝑐𝑐(𝑡𝑡) = 𝐴𝐴𝑐𝑐 � 𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑐𝑐𝑡𝑡𝑠𝑠(2𝜋𝜋(𝑓𝑓𝑐𝑐 + 𝑠𝑠𝑓𝑓𝑚𝑚)∞

𝑠𝑠=−∞

𝑡𝑡

𝐽𝐽−𝑠𝑠(𝛽𝛽) = 𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑠𝑠 𝑠𝑠𝑣𝑣𝑠𝑠𝑠𝑠

𝐽𝐽−𝑠𝑠(𝛽𝛽) = −𝐽𝐽𝑠𝑠(𝛽𝛽) 𝑠𝑠 𝑡𝑡𝑠𝑠𝑠𝑠 relationship between values of Jn(𝛽𝛽 )for various values of n is the recursion formula

𝐽𝐽𝑠𝑠+1(𝛽𝛽) =2𝑠𝑠𝛽𝛽 𝐽𝐽𝑠𝑠(𝛽𝛽) + 𝐽𝐽𝑠𝑠−1(𝛽𝛽)

Spectra of an angle-modulated signal. (a) Single-sided amplitude spectrum. (b) Single-sided phase




50

Amplitude spectrum of an FM complex envelope signal for increasing b and decreasing




51

Item 4 : detectors and Receivers

Principle of AM Detector .

Envelope Detectors.

assume the circuit as shown :

the diode D to be ideal. When it is forward biased, it acts as a short circuit and thereby, making the capacitor C charge through the source resistance Rs . When D is reverse biased, it acts as an open circuit and C discharges through the load resistance RL . As the operation of the detector circuit depends on the charge and discharge of the capacitor C

The envelope detector circuit

Envelope detector waveforms

Vout (t) after DC block

V1(t) before DC block




52

𝑅𝑅𝑠𝑠𝐿𝐿 ≪ 1𝑓𝑓𝑐𝑐

1𝑓𝑓𝑐𝑐≪ 𝑅𝑅𝐿𝐿𝐿𝐿 ≪

1𝑊𝑊

Synchronous Detectors.

𝑣𝑣(𝑡𝑡) =12𝐴𝐴𝑐𝑐�̅�𝐴𝑐𝑐 cos(4𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡 + 𝜑𝜑)𝑚𝑚(𝑡𝑡) +

12𝐴𝐴𝑐𝑐�̅�𝐴𝑐𝑐 cos(𝜑𝜑)𝑚𝑚(𝑡𝑡)

𝑣𝑣0(𝑡𝑡) =12𝐴𝐴𝑐𝑐�̅�𝐴𝑐𝑐 cos(𝜑𝜑)𝑚𝑚(𝑡𝑡)

Coherent detector for demodulating DSB-SC modulated

Illustrating the spectrum of a product modulator output with a




53

Costas PLLs

systems utilizing feedback can be used to demodulate angle-modulated carriers. A feedback system also can be used to generate the coherent demodulation carrier necessary for the demodulation of DSB signals. One system that accomplishes this is the Costas PLL illustrated in Figure 3.53. The input to the loop is the assumed DSB signal

x(t)=m(t)cos(2πfct )

The lowpass filter preceding the VCO is assumed sufficiently narrow so that the output is K sinð2uÞ, essentially the DC value of the input. This signal drives the VCO such that u is reduced. For sufficiently small u, the output of the top lowpass filter is the demodulated output.




54

Low pass filter

×

×

× Low pass filter

VCO

Low pass filter

Demodulated output

90o phase shift

12𝑚𝑚

2(𝑡𝑡)sin(2𝜃𝜃) 𝐾𝐾sin(2𝜃𝜃)

m(t)cos(θ)

m(t)sin(𝜃𝜃)

𝑥𝑥(𝑡𝑡) = 𝑚𝑚(𝑡𝑡) cos𝜔𝜔𝑐𝑐𝑡𝑡

2 sin(𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃)

2 cos(𝜔𝜔𝑐𝑐𝑡𝑡 + 𝜃𝜃)




55

Demodulation of FM Signals.

Balanced Slope Detector.

There are three tuned circuits: two on the secondary side of the input transformer and one on the primary. The resonant circuit on the primary is tuned to fc whereas the two resonant circuits on the secondary side are tuned to two different frequencies, one above fc and the other, below fc . The outputs of the tuned circuits on the secondary are envelope detected separately; the difference of the two envelope detected outputs would be proportional to m(t ) .




56

𝑉𝑉𝑃𝑃𝑅𝑅𝐼𝐼 = 𝐼𝐼𝑃𝑃𝑅𝑅𝐼𝐼𝑋𝑋𝑃𝑃𝑅𝑅𝐼𝐼 = 𝐼𝐼𝑃𝑃𝑅𝑅𝐼𝐼𝑗𝑗𝜔𝜔𝐿𝐿𝑃𝑃𝑅𝑅𝐼𝐼

1 − � 𝜔𝜔𝜔𝜔0�

2

the width of linear frequency response is about 3B , where 2B is the width of the 3-dB bandwidth of the individual tuned circuits) and does not require any DC bock (The two resonant frequencies of the secondary are appropriately selected so that output of the discriminator is zero for f = fc ), it suffers from the disadvantage that the three tuned circuits are to be maintained at three different frequencies

Foster Seely Discriminator

𝒗𝒗𝟐𝟐𝟐𝟐 = 𝒊𝒊𝒔𝒔(−𝒋𝒋𝒙𝒙𝒄𝒄𝟐𝟐)

=𝑗𝑗𝑀𝑀𝐿𝐿1

𝑣𝑣𝑠𝑠𝑠𝑠𝑥𝑥𝑐𝑐2

𝑅𝑅2 + 𝑗𝑗𝑥𝑥2

The voltage applied to D1:

𝒗𝒗𝟔𝟔𝟐𝟐 = 𝑣𝑣𝑠𝑠𝑠𝑠 +12𝒗𝒗𝟐𝟐𝟐𝟐

The voltage applied to D2:




57

𝒗𝒗𝟔𝟔𝟐𝟐 = 𝑣𝑣𝑠𝑠𝑠𝑠 −12𝒗𝒗𝟐𝟐𝟐𝟐

Reponses curve of the Foster- Seely discriminator




58

Ratio Detector.

Comparing the ratio detector circuit with that of the Foster-Seely discriminator, we find the following differences: direction of D2 is reversed, a parallel RC combination consisting of (R5 + R6 ) and C5 has been added and the output Vout is taken across a different pair of points. We shall now briefly explain the operation of the circuit.

𝑣𝑣𝑡𝑡𝑜𝑜𝑡𝑡 = 𝑣𝑣64 + 𝑣𝑣47 = 𝑣𝑣64 − 𝑣𝑣74

𝑣𝑣𝑡𝑡𝑜𝑜𝑡𝑡 = 𝑣𝑣64 −12 𝑣𝑣54

= 𝑘𝑘[|𝑣𝑣62| − |𝑣𝑣63|]

the Foster-Seely discriminator and the ratio detector have been the work horses of the FM industry. Companies such as Motorola have built high quality FM receivers using the Foster-Seely discriminator and the ratio detector

FM Pre emphasis and De emphasis concept:




59

Ultimately recovery of m(t) from an FM signal involves differentiation always worries signals engineers because it is a high frequency boost out to all frequencies

TROOBLE : that boosts the noise from the channel in the receiver, but only just restore the signal . The receiver FM signal:

𝜑𝜑𝐹𝐹𝑀𝑀 (𝑡𝑡) = 𝐴𝐴𝑐𝑐𝑡𝑡𝑠𝑠 [𝜔𝜔𝑐𝑐(𝑡𝑡) + 𝑘𝑘𝐹𝐹 ∫ 𝑚𝑚(𝑑𝑑)𝑠𝑠𝑑𝑑 +𝑡𝑡0 𝜑𝜑𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡)]

The derivate of the phase of 𝜑𝜑𝐹𝐹𝑀𝑀 (𝑡𝑡):

= 𝑘𝑘𝑓𝑓𝑚𝑚(𝑡𝑡) +𝑠𝑠𝑠𝑠𝑡𝑡 𝜑𝜑𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡)

𝑠𝑠𝑠𝑠 𝑓𝑓𝑝𝑝𝑠𝑠𝑓𝑓𝑜𝑜𝑠𝑠𝑠𝑠𝑐𝑐𝑐𝑐 𝑠𝑠𝑡𝑡𝑚𝑚𝑡𝑡𝑠𝑠𝑠𝑠 𝑘𝑘𝑓𝑓𝑀𝑀(𝜔𝜔) + 𝑗𝑗𝜔𝜔. 𝑁𝑁(𝜔𝜔) → 𝒏𝒏𝒏𝒏𝒊𝒊𝒔𝒔𝒏𝒏 − 𝒖𝒖𝒏𝒏𝒖𝒖𝒏𝒏𝒔𝒔𝒊𝒊𝒖𝒖𝒏𝒏𝒖𝒖

Desired boost to the noise

Preemphasis and Deemphasis filters.

this system has been used in commercial broadcasting as shown in figure below :

the pre emphasis ( before modulation ) and de emphasis ( after modulation ) filter Hp(ω) and Hd (ω) the frequency f1 is 2.1 kHz and f2 is typically 30 kHz or more .

these filter can be realized by simple RC circuits . the pre emphasis transfer function is :

𝐻𝐻𝑝𝑝(𝜔𝜔) = 𝐾𝐾𝑗𝑗𝜔𝜔 + 𝜔𝜔1𝑗𝑗𝜔𝜔 + 𝜔𝜔2

𝐻𝐻𝑝𝑝(𝜔𝜔) =𝜔𝜔2𝜔𝜔1

𝑗𝑗𝜔𝜔 + 𝜔𝜔1𝑗𝑗𝜔𝜔 + 𝜔𝜔2

K: the gain =𝜔𝜔2𝜔𝜔1




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For ω « ω1,

𝐻𝐻𝑝𝑝(𝜔𝜔) ≈ 1

For ω1 « ω « ω2

𝐻𝐻𝑝𝑝(𝜔𝜔) ≈𝑗𝑗𝜔𝜔𝜔𝜔1

The preemphasizer acts as a differentiator at intermediate frequency 2.1-15kHz , which effectively makes the scheme PM over these frequencies . this mean that FM with PDE is FM over the modulating signal frequency range of 0 – 2.1 kHZ and is nearly PM over the range of 2.1- 15 kHz as desired . The deemphasis filter Hd (ω) is :

𝐻𝐻𝑠𝑠(𝜔𝜔) =𝜔𝜔1

𝑗𝑗𝜔𝜔 + 𝜔𝜔1

For ω « ω2,

𝐻𝐻𝑝𝑝(𝜔𝜔) ≈𝑗𝑗𝜔𝜔 + 𝜔𝜔1𝜔𝜔1

𝐻𝐻𝑝𝑝(𝜔𝜔)𝐻𝐻𝑠𝑠(𝜔𝜔) ≈ 1 𝑡𝑡𝑣𝑣𝑠𝑠𝑝𝑝 𝑡𝑡ℎ𝑠𝑠 𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠 0 − 15𝑘𝑘𝐻𝐻𝑘𝑘

Preemphasis filter

deemphasis filter




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a) FM stereo transmitter ,b) spectrum of a baseband signal , c) FM stereo receiver




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Receivers.

Radio receiver is an electronic equipment which pick ups the desired signal, reject the unwanted signal and demodulate the carrier signal to get back the original modulating signal.

Function of Radio Receivers.

1-Intercept the incoming modulated signal

2-Select desired signal and reject unwanted signals

3-Amplify selected R.F signal

4-Detect modulated signal to get back original modulating signal

5-Amplify modulating frequency signal

Design of Receiver: Requirements:

Has to work according to application as for AM or FM signals

Tune to and amplify desired radio station

Filter out all other stations

Demodulator has to work with all radio stations regardless of carrier frequency

Classification of Radio Receivers.

1-Depending upon application

a) AM Receivers - receive broadcast of speech or music from AM transmitters which operate on long wave, medium wave or short wave bands.

b) FM Receivers – receive broadcast programs from FM transmitters which operate in VHF or UHF bands.

c) Communication Receivers - used for reception of telegraph and short wave telephone signals.




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d) Television Receivers - used to receive television broadcast in VHF or UHF bands.

e) Radar Receivers – used to receive radio detection and ranging signals.

2-Depending upon fundamental aspects

a) Tuned Radio Frequency (TRF)Receivers b) Super-heterodyne Receivers

Typical receiver circuits include: RF amplifiers, IF amplifiers, AGC,AFC and Special circuits

TRF (Tuned Radio frequency) RECEIVER.

Advantages of TRF:

TRF receivers are simple to design and allow the broadcast frequency 535 KHz to 1640 KHz.

High sensitivity.




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Disadvantages of TRF:

1-At the higher frequency, it produces difficulty in design.

2-It has poor audio quality.

3-Instability:1- reactance of stray capacitances decreases at higher frequencies resulting in increased feedback.

2-Due to high frequency, multi stage amplifiers are susceptible to breaking into oscillation.

3-gain of RF amplifier is very high ,a small feedback from output to input with correct phase can lead to oscillations.

4-Variation in BW: 1-The bandwidth is inconsistent and varies with the center frequency when tuned over a wide range of input frequencies.

2-As frequency increases, the bandwidth ( f/Q) increases. Thus, the selectivity of the input filter changes over any appreciable range of input frequencies.

5-Poor Selectivity: 1-The gains are not uniform over a very wide frequency range.

2-Due to higher frequencies ability to select desired signal is

affected.

Superheterodyne Receiver

The shortcomings of the TRF receiver are overcome by the super heterodyne or superhet receiver. Basically, the receiver consists of a radio-frequency (RF) section, a mixer and local oscillator, an intermediate-frequency (IF) section, demodulator, and power amplifier. Typical frequency parameters of commercial AM and FM radio receivers are listed in Table

Low side fLO=fRF –fIF

High side= fLO=fRF +fIF




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AM Radio FM Radio RF carrier range 0.535-1.605 MHz 88-108 MHz Midband frequency of IF section 0.455 MHz 10.7 MHz IF bandwidth 10 kHz 200 kHz The Figure shows the block diagram of a superheterodyne receiver for amplitude modulation using an envelope detector for demodulation.

Typical frequency parameter of AM and FM radio receivers

RF

section IF

section Mixer Envelop

detector

Antenna

Audio amplifier

Loud speaker

Local oscillator

Tuning

Superheterodyne Receiver




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Disadvantages :

1-Stability – as high frequency is down converted to IF the reactance of stray capacitances will not decrease as it was at higher frequencies resulting in increased feedback.

2- No variation in BW- as IF range is 438 to 465 KHz (in case of AM receivers) mostly 455KHz ,appropriate for Q limit (120).

3-Better selectivity- as no adjacent channels are picked due to variation in BW.

Typical spectrum at the input to the RF stage of a superhet

Spectrum at the input of the IF stage of a stage of a superhet




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Item 5 : Pulse Modulation




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Sampling Theorem

Let g(t) is analog signal

Let 𝒈𝒈𝜹𝜹(𝒕𝒕) denote the ideal sampled signal

𝑛𝑛𝛿𝛿(𝑡𝑡) = � 𝑛𝑛(𝑠𝑠𝑇𝑇𝑠𝑠)𝛿𝛿(𝑡𝑡 − 𝑠𝑠𝑇𝑇𝑠𝑠)∞

𝑠𝑠=−∞

Where 𝑇𝑇𝑠𝑠: sampling period

𝑓𝑓𝑠𝑠=1𝑇𝑇𝑠𝑠

sampling rate

𝐺𝐺𝑠𝑠(𝑓𝑓) = � 𝑛𝑛�𝑠𝑠

2𝑊𝑊� exp(−𝑗𝑗𝜋𝜋𝑠𝑠𝑓𝑓𝑊𝑊 )

∞

𝑠𝑠=−∞

The Fourier transform of 𝑛𝑛𝛿𝛿(𝑡𝑡):

𝐺𝐺𝑠𝑠(𝑓𝑓) = 𝑓𝑓𝑠𝑠𝐺𝐺(𝑓𝑓) � 𝐺𝐺(𝑓𝑓 −𝑚𝑚𝑓𝑓𝑠𝑠)∞

𝑚𝑚=−∞

1-𝐺𝐺(𝑓𝑓) = 0 𝑓𝑓𝑡𝑡𝑝𝑝 |𝑓𝑓| ≥ 𝑊𝑊




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2-𝑓𝑓𝑠𝑠 = 2𝑊𝑊 𝑁𝑁𝑐𝑐𝑓𝑓𝑜𝑜𝑠𝑠𝑠𝑠𝑡𝑡 𝑝𝑝𝑡𝑡𝑡𝑡𝑠𝑠 𝑇𝑇𝑠𝑠 = 1

2𝑊𝑊 𝑁𝑁𝑐𝑐𝑓𝑓𝑜𝑜𝑠𝑠𝑠𝑠𝑡𝑡 𝑠𝑠𝑠𝑠𝑡𝑡𝑠𝑠𝑝𝑝𝑣𝑣𝑡𝑡𝑡𝑡

𝐺𝐺(𝑓𝑓) =1

2𝑊𝑊𝐺𝐺𝑠𝑠(𝑓𝑓),−𝑊𝑊 < 𝑓𝑓 < 𝑊𝑊

𝐺𝐺(𝑓𝑓) =1

2𝑊𝑊 � 𝑛𝑛�𝑠𝑠

2𝑊𝑊� exp�−𝑗𝑗𝜋𝜋𝑠𝑠𝑓𝑓𝑊𝑊 � ,−𝑊𝑊 < 𝑓𝑓 < 𝑊𝑊

∞

𝑠𝑠=−∞

The inverse Fourier transform:

𝑛𝑛(𝑡𝑡) = ∫ 𝐺𝐺(𝑓𝑓) exp(𝑗𝑗2𝜋𝜋𝑓𝑓𝑡𝑡)𝑠𝑠𝑓𝑓∞−∞

𝑛𝑛(𝑡𝑡) = � 𝑛𝑛 �𝑠𝑠

2𝑊𝑊�1

2𝑊𝑊� exp �𝑗𝑗2𝜋𝜋𝑓𝑓 �𝑡𝑡 −𝑠𝑠

2𝑊𝑊��𝑠𝑠𝑓𝑓𝑊𝑊

−𝑊𝑊

∞

−∞

The integral term :

𝑛𝑛(𝑡𝑡) = � 𝑛𝑛�𝑠𝑠

2𝑊𝑊�sin(2πWt− nπ)

2πWt− nπ

∞

𝑠𝑠=−∞

= ∑ 𝑛𝑛 � 𝑠𝑠2𝑊𝑊� sinc(∞

𝑠𝑠=−∞ (2πWt− nπ) −∞ < 𝑡𝑡 < ∞




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To combat the effects of aliasing in practice, we may use two corrective measures:

1-Prior. to sampling, a low-pass anti-aliasing filter is used to attenuate those high frequency components of the signal that are not essential to the information being conveyed by the signal. 2-The filtered signal is sampled at a rate slightly higher than the Nyquist rate.

Spectrum of an under sampled version of the signal exhibiting the aliasing phenomenon

Spectrum of signal

Magnitude response of reconstruction filter

Anti- alias filtered spectrum of an information signal




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Example12 :

The time function f(t)= 5 cos 2π (1000)t cos 2π(300)t is sampled at 2100 cycle rate , somewhat higher than required minimum sampling rate for unique reconstruction . it samples for ∆t second . reconstruction takes place by passing the sample signal through a 1600Hz low pass filter having a unity gain . determine its sampled and filtered spectrum.

Solution:

𝑠𝑠𝑡𝑡𝑚𝑚𝑝𝑝𝑡𝑡𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑛𝑛𝑠𝑠𝑡𝑡𝑡𝑡 = 𝑠𝑠𝑠𝑠𝑠𝑠 𝜔𝜔0

∆𝑡𝑡2

𝜔𝜔0∆𝑡𝑡2

1-𝑓𝑓(𝑓𝑓) = 52

[𝑚𝑚 (1000 + 300)𝑡𝑡 + 𝑚𝑚(1000− 300)𝑡𝑡]

=52

[𝑚𝑚 (1300)𝑡𝑡 + 𝑚𝑚(700)𝑡𝑡]

2- 𝑓𝑓(𝑓𝑓) = 54

[𝑚𝑚 (1300 + 2100)𝑡𝑡 + 𝑚𝑚(1300− 2100)𝑡𝑡 + 𝑚𝑚(700 + 2100)𝑡𝑡 +𝑚𝑚700−2100𝑡𝑡

=54

[𝑚𝑚 (3400)𝑡𝑡 + 𝑚𝑚(800)𝑡𝑡 + 𝑚𝑚(2800)𝑡𝑡 + 𝑚𝑚(1400)𝑡𝑡]

-3400 -2800 -2100 -1400 -1300 -800 -700 0 700 800 1300 1400 2100 2800 3400 f

X(f) (T/∆t)

5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4

Sampled spectrum




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𝒚𝒚(𝒕𝒕) = 𝟓𝟓∆𝒕𝒕𝑻𝑻 𝒄𝒄𝒏𝒏𝒔𝒔 𝟐𝟐𝟐𝟐(𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏)𝒕𝒕 𝒄𝒄𝒏𝒏𝒔𝒔 𝟐𝟐𝟐𝟐(𝟐𝟐𝟏𝟏𝟏𝟏)𝒕𝒕

+ 𝟓𝟓∆𝒕𝒕𝑻𝑻 𝒄𝒄𝒏𝒏𝒔𝒔 𝟐𝟐𝟐𝟐(𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏)𝒕𝒕 𝒄𝒄𝒏𝒏𝒔𝒔 𝟐𝟐𝟐𝟐(𝟐𝟐𝟏𝟏𝟏𝟏)𝒕𝒕

Exercise

-1400 -1300 -800 -700 0 700 800 1300 1400 f

y(f) (T/∆t)

5/4 5/4 5/4 5/4 5/4 5/4 5/4 5/4

filter spectrum




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Type of Pulse Modulation.

Pulse Amplitude Modulation PAM.

A PAM waveform consists of a sequence of flat-topped pulses designating sample values. The amplitude of each pulse corresponds to the value of the message signal m(t) at the leading edge of the pulse. The essential difference between PAM and the sampling operation discussed in the previous chapter is that in PAM we allow the sampling pulse to have finite width. The finite width pulse can be generated from the impulse-train sampling function by passing the impulse train samples through a holding circuit. The impulse response of the ideal holding circuit is given by:

Holding network Impulse response of

holding network

Amplitude response of holding network

Phase response of holding network

Generation of PAM




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Op-amp 2 :is a high input-impedance voltage follower capable of driving low-impedance loads . The resistor R: is used to limit the output current of op-amp 1 when the FET is “on” and provides a voltage division with rd of the FET. (rd, the drain-to-source resistance, is low but not zero)

ℎ(𝑡𝑡) = ∏�𝑡𝑡 − 1

2 𝑑𝑑𝑑𝑑 �

The holding circuit transforms the impulse function samples:

𝑚𝑚𝛿𝛿(𝑡𝑡) = � 𝑚𝑚(𝑠𝑠𝑇𝑇𝑠𝑠)𝛿𝛿(𝑡𝑡 − 𝑠𝑠𝑇𝑇𝑠𝑠)∞

𝑠𝑠=−∞

𝑚𝑚𝑐𝑐(𝑡𝑡) = � 𝑚𝑚(𝑠𝑠𝑇𝑇𝑠𝑠)∏(𝑡𝑡 − (𝑠𝑠𝑇𝑇𝑠𝑠 + 1

2 𝑑𝑑𝑑𝑑 )

∞

𝑠𝑠=−∞

ℎ(𝑓𝑓) = 𝑑𝑑𝑠𝑠𝑠𝑠𝑠𝑠𝑐𝑐 (𝑓𝑓𝑑𝑑)𝑠𝑠−𝑗𝑗𝜋𝜋𝑓𝑓𝑑𝑑

the holding network does not have a constant amplitude response over the bandwidth of m(t), amplitude distortion results. This amplitude distortion, which can be significant unless the pulse width t is very small, can be removed by passing the samples, prior to reconstruction of m(t), through a filter having an amplitude response equal to 1/H(f) over the band width of m(t).

Pulse Width Modulation PWM

1-In pulse width modulation (PWM), the width of each pulse is made directly proportional to the amplitude of the information signal. 2-In pulse position modulation, constant-width pulses are used, and the position or time of occurrence of each pulse from some reference time is made directly proportional to the amplitude of the information signal. 3-PWM and PPM are compared and contrasted to PAM as shown in the Figure .




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a PWM modulator circuit is show below . This circuit is simply a high-gain comparator that is switched on and off by the saw tooth waveform derived from a very stable-frequency oscillator.




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1-Notice that the output will go to +Vcc the instant the analog signal exceeds the saw tooth voltage. 2-The output will go to -Vcc the instant the analog signal is less than the saw tooth voltage. With this circuit the average value of both inputs should be nearly the same. 3-This is easily achieved with equal value resistors to ground. Also the +V and –V values should not exceed Vcc.

Pulse Position Modulation PPM

APPM signal consists of a sequence of pulses in which the pulse displacement from a specified time reference is proportional to the sample values of the information-bearing signal.

𝒙𝒙(𝒕𝒕) = � 𝑛𝑛(𝑡𝑡 − 𝑡𝑡𝑠𝑠)∞

𝑠𝑠=−∞




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where g(t)represents the shape of the individual pulses, and the occurrence times𝑡𝑡𝑠𝑠 are related to the values of the message signal m (t) at the sampling instants nTs. The spectrum of a PPM signal is very similar to the spectrum of a PWM signal.

PPM demodulator.

Example 13:Draw the PAM , PWM and PPM of the analog signal ( sin wave ) .




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Exercise Draw the PMA pulse triangle , PWM and PPM that generated from this circuit .




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Exercise Draw the PMA pulse triangle , PWM and PPM that generated from this circuit .




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Multiplexing.

Time- Division Multiplexing

Multiplexing is a modulation method which improves channel bandwidth utilisation.

TDM is another form of multiplexing based on sampling which is a modulation

technique. In TDM, samples of several analogue message symbols, each one

sampled in turn, are transmitted in a sequence (time slots).




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the total number of baseband samples in a T-s interval is

𝑠𝑠𝑠𝑠 = ∑ 2𝑊𝑊𝑠𝑠𝑇𝑇𝑁𝑁𝑠𝑠=1

Assuming that the baseband is a low pass signal of bandwidth B, the required sampling rate is 2B. In a T-s interval, we then have 2BT total samples.

𝑠𝑠𝑠𝑠 = 2𝐵𝐵𝑇𝑇 = �2𝑊𝑊𝑠𝑠𝑇𝑇𝑁𝑁

𝑠𝑠=1

𝐵𝐵 = �𝑊𝑊𝑠𝑠

𝑁𝑁

𝑠𝑠=1

Time –division multiplexing of two signal




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Item 6 : Noise in Communication Systems

Electrical noise may be said to be the introduction of any unwanted energy, which tend to interfere with the proper reception and reproduction of transmitted signals.

1- Unwanted Signals that tend to disturb the Transmission and Processing of Signals in Communication System and over which we have incomplete control.

2-Noise is a general term which is used to describe an unwanted signal which affects a wanted signal.

3-These unwanted signals arise from a variety of sources.

The Noise Parameters:

1-Signal to noise ratio

2-Noise factor

3-Noise equivalent band width

4-Effective noise temperature

Sources of noise :External and Internal

External :

1-Atmosphere disturbance (e.g. electric storms, lighting, ionospheric effect etc), so called ‘Sky Noise’

2-Cosmic noise which includes noise from galaxy, solar noise

3-‘Hot spot’ due to oxygen and water vapour resonance in the earth’s atmosphere.

Internal :

1-Electronic communication systems are made up of circuit elements such




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as R , L and C , and devices like diodes, transistors, etc. All these components give rise to what is known as internal circuit noise. It is this noise that sets the fundamental limits on communication of acceptable quality. 2-This type of noise is the basic limiting factor of employing more complex Electrical Circuits in Communication System.

3-Most Common Internal Noises are: Shot Noise and Thermal Noise .

Effective Noise Temperature

Assume

v(t): voltage at the open ends of conductor

R: value of resistance

T : temperature of R (Kelvin) K=C +273

N : noise power (watts)

k: Boltzmann’s constant (1.38 10 -23 J/K)

B : total noise factor (hertz).

Sv(f) :spectral density of v(t) (volt)2/ Hz

Sv(f)=2RkT

Resistor as noise source The noise source driving a load




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𝑇𝑇 =𝑁𝑁𝑘𝑘𝐵𝐵

𝑇𝑇𝑠𝑠 = 𝑇𝑇(𝐹𝐹 − 1)

𝐹𝐹 = 1 +𝑇𝑇𝑠𝑠𝑇𝑇

Te = equivalent noise temperature

F = noise factor (unit less)

If this open circuit voltage is measured with the help of a true RMS voltmeter of bandwidth B (frequency range: − B to B), then the reading on the instrument would be√2𝑅𝑅𝑘𝑘𝑇𝑇. 2𝐵𝐵 v. Thermal noise sources are also characterized in terms of available noise PSD.

𝑃𝑃𝑆𝑆𝑃𝑃 =𝑆𝑆𝑣𝑣(𝑓𝑓)𝑅𝑅

|𝐻𝐻(𝑓𝑓)|2

=2𝑅𝑅𝑘𝑘𝑇𝑇

4𝑅𝑅 =𝑘𝑘𝑇𝑇2 𝑝𝑝𝑡𝑡𝑡𝑡𝑡𝑡𝑠𝑠/𝐻𝐻𝑘𝑘

the available power in a bandwidth of B Hz is

𝑝𝑝𝑠𝑠 =𝑘𝑘𝑇𝑇2 . 2𝐵𝐵 = 𝑘𝑘𝑇𝑇𝐵𝐵

Noise Figure .

The noise performance of a receiver is described by a figure of merit called the noise figure (NF).

𝑁𝑁𝐹𝐹 =𝑆𝑆𝑁𝑁𝑅𝑅𝑠𝑠𝑠𝑠 𝑆𝑆𝑁𝑁𝑅𝑅𝑡𝑡𝑜𝑜𝑡𝑡 =

𝑆𝑆𝑠𝑠𝑠𝑠𝑁𝑁𝑠𝑠𝑠𝑠�

𝑆𝑆𝑡𝑡𝑜𝑜𝑡𝑡𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡�

=𝑅𝑅𝑝𝑝 + 𝑅𝑅𝑠𝑠𝑅𝑅𝑝𝑝

𝑆𝑆𝑠𝑠𝑠𝑠 𝑁𝑁𝑠𝑠𝑠𝑠� = 𝑉𝑉𝑠𝑠2

4𝑅𝑅𝑝𝑝𝐾𝐾𝑇𝑇0𝐵𝐵 𝑆𝑆𝑡𝑡𝑜𝑜𝑡𝑡 𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡� = 𝑉𝑉𝑠𝑠2

4(𝑅𝑅𝑝𝑝+𝑅𝑅𝑠𝑠 )𝐾𝐾𝑇𝑇𝐵𝐵




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𝑁𝑁𝐹𝐹 =𝑁𝑁𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑜𝑜𝑡𝑡𝑝𝑝𝑜𝑜𝑡𝑡 𝑡𝑡𝑓𝑓 𝑡𝑡𝑐𝑐𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡 𝑝𝑝𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠𝑣𝑣𝑠𝑠𝑝𝑝 𝑁𝑁𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑜𝑜𝑡𝑡𝑝𝑝𝑜𝑜𝑡𝑡 𝑡𝑡𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑡𝑡 𝑝𝑝𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠𝑣𝑣𝑠𝑠𝑝𝑝 =

𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡𝐺𝐺𝑁𝑁𝑠𝑠𝑠𝑠

Where G is antenna gain.

𝑡𝑡𝑝𝑝 𝑠𝑠𝑠𝑠 𝑠𝑠𝐵𝐵: 10 log �𝑁𝑁𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑜𝑜𝑡𝑡𝑝𝑝𝑜𝑜𝑡𝑡 𝑡𝑡𝑓𝑓 𝑡𝑡𝑐𝑐𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡 𝑝𝑝𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠𝑣𝑣𝑠𝑠𝑝𝑝 𝑁𝑁𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡𝑜𝑜𝑡𝑡𝑝𝑝𝑜𝑜𝑡𝑡 𝑡𝑡𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑡𝑡𝑡𝑡 𝑝𝑝𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠𝑣𝑣𝑠𝑠𝑝𝑝 � = 10 𝑡𝑡𝑡𝑡𝑛𝑛

G:power gain of the device

𝐺𝐺 =the signal power at the outputthe signal power at the input

=𝑃𝑃𝑡𝑡𝑜𝑜𝑡𝑡𝑃𝑃𝑠𝑠𝑠𝑠

𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡 = 𝑁𝑁𝐹𝐹 𝐺𝐺𝑁𝑁𝑠𝑠𝑠𝑠

𝑁𝑁𝑠𝑠𝑠𝑠 = 𝐾𝐾𝑇𝑇𝐵𝐵

Assume T=T0

𝑁𝑁𝐹𝐹 =𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡𝐺𝐺𝐾𝐾𝑇𝑇0𝐵𝐵

𝑁𝑁𝑡𝑡𝑜𝑜𝑡𝑡 = 𝑁𝑁𝐹𝐹𝐺𝐺𝐾𝐾𝑇𝑇0𝐵𝐵

The noise in amplifier :

𝑁𝑁𝑡𝑡𝑚𝑚𝑝𝑝𝑡𝑡𝑠𝑠𝑓𝑓𝑠𝑠𝑝𝑝𝑠𝑠 = 𝑁𝑁𝐹𝐹𝐾𝐾𝑇𝑇0𝐵𝐵 − 𝐾𝐾𝑇𝑇0𝐵𝐵

= (𝑁𝑁𝐹𝐹 − 1)𝐾𝐾𝑇𝑇0𝐵𝐵




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Noise in Cascaded Systems

The noise figure :

𝑁𝑁𝐹𝐹 = 1 + (𝑁𝑁𝐹𝐹1 − 1) +(𝑁𝑁𝐹𝐹2 − 1)

𝐴𝐴𝑝𝑝1+ ⋯+

(𝑁𝑁𝐹𝐹𝑘𝑘 − 1)𝐴𝐴𝑝𝑝1 …𝐴𝐴𝑝𝑝𝑘𝑘

When all input and output impedances equal Rs:

𝑁𝑁𝐹𝐹 = 1 + (𝑁𝑁𝐹𝐹1 − 1) +(𝑁𝑁𝐹𝐹2 − 1)𝐴𝐴𝑣𝑣1−𝑡𝑡

2 + ⋯+(𝑁𝑁𝐹𝐹𝑘𝑘 − 1)

𝐴𝐴𝑣𝑣1−𝑡𝑡2 …𝐴𝐴𝑣𝑣𝑘𝑘−𝑡𝑡2




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Narrow Band Noise.

1- The narrowband noise is defined in terms of a pair of components called the in-phase and quadrature components .

2- The narrowband noise is defined in terms of two other components called the envelope and phase.

Let n(t) is the Narrowband Noise with Bandwidth 2B centered at fc.

𝑠𝑠(𝑡𝑡) = 𝑠𝑠𝑠𝑠(𝑡𝑡) cos(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡) − 𝑠𝑠𝑓𝑓(𝑡𝑡) sin(2𝜋𝜋𝑓𝑓𝑐𝑐𝑡𝑡)

ni(t) : Phase Component) = ∑ �2η0∆fi cos(2𝜋𝜋(𝑓𝑓𝑠𝑠 − 𝑓𝑓𝑐𝑐)𝑡𝑡 + 𝜃𝜃𝑠𝑠(𝑡𝑡))

nq(t) :Quadrature Component) from n(t). = ∑ �2η0∆fi sin(2𝜋𝜋(𝑓𝑓𝑠𝑠 − 𝑓𝑓𝑐𝑐)𝑡𝑡+𝜃𝜃𝑠𝑠(𝑡𝑡))

Fig (a) spectral components of NB Noise concentrates about +fc

Fig (b) sample function n(t) of such process appears somewhat similar to a sinusoidal wave of frequency fc




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NARROWBAND NOISE ANALYSER

NARROWBAND NOISE Construction




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Exercise :

A transistor amplifier has measured S/R of 10 at its input and 5 at its output.

A) Calculate the NR

B) Calculate the NF

Example 15:

A mixer stage has a noise figure of 20 dB and this is preceded by an amplifier that has a noise figure of 9 dB and an available power gain of 15 dB . calculate the overall noise figure refer red to the input . F2 = 20 dB = 100 / 1 F1 = 9dB = 7.94 / 1 G1 = 15 dB = 31.62 / 1

𝑁𝑁𝐹𝐹 = 𝑁𝑁𝐹𝐹1 +𝑁𝑁𝐹𝐹2 − 1𝐺𝐺1

= 7.94 +100 − 1

31.62 = 11.07

𝑁𝑁𝐹𝐹𝑠𝑠𝐵𝐵 = 10 log10 11.07

=10.44dB

Asst. Prof. Dr. Ashwaq Q. Hameed Al Faisal UOT- Electrical … · 2018-01-19 · Communication...

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Transcript of Asst. Prof. Dr. Ashwaq Q. Hameed Al Faisal UOT- Electrical … · 2018-01-19 · Communication...