Aksum University College of Engineering and Technology Department of electrical & computer engineering Introduction to communication Systems Chapter-2 Amplitude Modulation Lecture -1 By Yonas Desta (MSc. Candidate in Communication Engineering) Electronics & communication Engineering Stream Basic formulas(Trigonometry Identity) οΌ cos π + π =cos π cos π − sin π sin π οΌ sin π + π = sin π cos π + cos π sin π οΌ cos π cos π 1 1 = cos π − π + cos π + π 2 2 οΌ sin π sin π 1 = cos π − π 2 1 − cos π + π 2 οΌ sin π cos π 1 = sin π − π 2 1 + sin π + π 2 Cont… To find S π π’π π: οΆπ π‘ ↔π π οΆπ π2πππ π‘ ↔ πΏ π − ππ οΆπ −π2πππ π‘ ↔ πΏ π + ππ οΆexp(π2πππ π‘)π(π‘) ↔M(π − ππ ) οΆexp(−π2πππ π‘)π(π‘) ↔M(π + ππ ) Introduction ο Message signals /Baseband signals produced by various information sources are not always suitable for direct transmission over a channel. ο These signals are modified to facilitate transmission. ο This conversion process is modulation ο In amplitude modulation, the message signal m(t) is impressed on the amplitude of the carrier signal c(t). Amplitude Modulation οΌ A modulation process in which the amplitude of the carrier is varied in accordance with the instantaneous value of the modulating signal. οΌ Amplitude modulation is defined as the process in which the amplitude of the carrier signal is varied in accordance with the modulating signal or message signal. Illustration of amplitude modulation οΌ The amplitude of the carrier C(t) changes in accordance with the input modulating signal m(t). οΌ S(t) is the modulated waveform which is transmitted by the antenna Cont… Amplitude Modulation • AM is defined as a process in which the amplitude of the carrier wave c(t)is varied about a mean value, linearly with baseband signal m(t). baseband signal (message signal) =m(t) carrier wave/signal =c(t) = π΄π cos2πππ t Then the general expression for the AM signal is given by ππ΄π (t) = π΄π 1 + ππ π(π‘) cos2πππ t (1) Where, οΌ ππ =Amplitude sensitivity of AM modulator οΌ π΄π =Carrier amplitude Measured in volts οΌ π π‘ = πππ π πππ π πππππ ππ΄π (t) is the output AM-modulated carrier frequency Cont… Equation(1) can be written as ππ΄π (t) =π΄π cos2πππ t +π΄π ππ π(π‘) cos2πππ t Carrier signal (2) Modulated signal which does not have information contains the information, which has several spectral components, requiring further analysis to quantify them ο AM signal contains carrier in addition to actual modulated signal ο Because of this additional carrier ,transmission power will be wasted, but modulator become much simpler AM Spectrum and Bandwidth οΌ π(π‘) FT οΌ c(t) FT οΌ ππ΄π (t) FT Cont… • ππ΄π (t) BW =Highest +ve freq + low freq = ππ +π − ππ − π = ππ + W -ππ +W =2W Twice of message signal the spectral response of the AM-modulated signal. It has three spectral components: οΌ The carrier frequency : ππ οΌ Upper sideband: ππ +π οΌ Lower sideband: ππ − π NB: The actual message signal will be retained only by the Side bands. The bandwidth of each of the sideband is equal to the message BW. Single Tone AM οΆIn single tone AM, m(t) consists of one frequency component i.e. π(π‘) =π΄π cos2πππ t (3) Substitute equ (3) to Substitute equ (1) ππ΄π (t) = π΄π 1 + ππ π΄π cos2πππ t cos2πππ t (4) π Where , π =ππ π΄π is modulation factor /modulation index/modulation depth ππ΄π (t) = π΄π 1 + πcos2πππ t cos2πππ t (5) Cont… οΌ We can be represent equation (5) 1 1 By using πππ π΄πππ π΅ = cos(A-B) + cos(A+B) 2 2 ππ΄π (t) = π΄π 1 + πcos2πππ t cos2πππ t =π΄π cos2πππ t +ππ΄π cos2πππ tcos2πππ t 1 ππ΄π (t) =π΄π cos2πππ t + ππ΄π cos 2π ππ + ππ π‘ 2 Carrier 1 2 Upper sideband + ππ΄π cos 2π ππ − ππ π‘ (6) Lower sideband οΌ Equation(6) is the standard time domain equation for singletone AM signal Cont… οΌ We find that the Fourier transform of the AM wave ππ΄π (t) is given by: οΌ Using the following relation ship οΆ π π2πππ π‘ ↔ πΏ π − ππ 1 cos2πππ t ↔ 2 1 πΏ π − ππ + πΏ π + ππ 2 οΆ π −π2ππππ‘ ↔ πΏ π + ππ π(π‘) =π΄π cos2πππ t π΄π π(π) = 2 πΏ π − ππ + πΏ π + ππ c(π‘) =π΄π cos2πππ t and π΄π c(π) = 2 πΏ π − ππ + πΏ π + ππ and Cont… Equation (6) becomes π΄π ππ΄π (f) = πΏ π − ππ + πΏ π + ππ 2 ππ΄π + πΏ π −ππ −ππ + πΏ π +ππ +ππ 4 ππ΄π + 4 πΏ π −ππ +ππ + πΏ π +ππ −ππ Then the spectrum of ππ΄π (t) ππ΄π (f) ↔ (7) Cont… • The spectrum of AM becomes Cont… οΆAM spectrum consists of two impulse frequencies located π΄π at +ππ and weighted 2 οΆTwo USB’s at ππ + ππ and οΆTwo LSB’s at ππ − ππ and ππ΄π −ππ − ππ weighted 4 ππ΄π −ππ + ππ weighted 4 Note : οΌ For positive frequencies: The spectrum of an AM wave above fc is referred to as the upper sideband, below fc is referred to as the lower sideband. οΌ For negative frequencies: The upper sideband is below – fc and the lower sideband is above – fc. The Cont… οΆ AMπ΅ππ = ππ + ππ - ππ − ππ =2ππ Note: ο For positive frequencies, the highest frequency component of the AM wave equals fc +W, and the lowest frequency component equals fc – W. ο The difference between these two frequencies defines the transmission bandwidth(BT) for an AM wave. Example: Given: Input modulating frequency ππ = 10 kHz Carrier frequency ππ = 400 kHz Spectral components ππ = 400kHz USB =ππ + ππ = 410 kHz, LSB =ππ − ππ =390kHz BW = 2ππ =20kHz Modulation index ο Modulation index (π =ππ π΄π ) specifies to what extent the carrier is modulated by the π(π‘). Resulting AM signal ο The envelope of s(t) has essentially the same shape as the baseband signal m(t). ο Modulation index defined as the ratio of amplitude of message signal to the amplitude of carrier signal. i.e. π΄π π= π΄π Cont… From the above figure the modulation index/depth is: π΄πππ₯ −π΄πππ π= π΄πππ₯ +π΄πππ π΄πππ₯ = π΄π 1 + π π΄πππ = π΄π 1 − π π΄π = π΄πππ₯ +π΄πππ 2 Types of modulation index (1) Under modulation (π < 1) π΄ π < π΄π (1) Critical/ideal modulation (π =1) π΄π = π΄π (1) Over modulation (π > 1) π΄π > π΄ π ο π΄π & π΄π denote he maximum and minimum values of the envelope of the modulated wave. Percentage modulation ο The maximum value of π multiplied by 100 is referred to as the percentage modulation π΄πππ₯ −π΄πππ π= x100% π΄πππ₯ +π΄πππ Total power in AM Analyzing AM spectrum we see that from 1 ππ΄π (t) =π΄π cos2πππ t + ππ΄π cos 2π ππ + ππ π‘ 2 1 + ππ΄π cos 2π ππ − ππ π‘ 2 ππ‘ = ππ + ππΏππ΅ + ππππ΅ οΌ The power in the carrier and sidebands can be calculated by π£2 using the power formula P= where P is the output power, V is π the rms output voltage, and R is the resistive part of the load impedance, which is usually an antenna. or οΌ Power of any signal is equal to the mean square value of the signal We know that: 2 π£πππ π= π π£ and π£πππ = π΄π or where π£π΄π is the peak value 2 Note: For power calculations, rms values must be used for the voltages. We can convert from peak to rms by dividing the peak value by 2 2 π£π there fore ππ = 2π π΄2πΆ = 2π ο Assume that the AM signal is dissipated in antenna load of ‘R’ Ω i. e . normalized power which is average power normalized across 1Ω resistor π΄2πΆ ππ = 2 Cont… 2 π£πππ But, π ππΏππ΅ =ππππ΅ π£π = π£πππ = 2 Then ππΏππ΅ =ππππ΅ = ππ΄π 2 2 2π π2 π΄2πΆ π2 π΄2πΆ = = 8π 8 So, ππ‘ = ππ + ππΏππ΅ +ππππ΅ π΄2πΆ π2 π΄2πΆ π2 π΄2πΆ = + + 2π 8π 8π π΄2πΆ π2 π΄2πΆ π2 π΄2πΆ = + + due to 2 8 8 Cont… π΄2πΆ ππ‘ = 2π 2π2 π΄2πΆ π΄2πΆ + = 8π 2π π2 π΄2πΆ π΄2πΆ + = 4π 2π π2 1+ 2 wats But due to normalized power R=1 πΊ π΄2πΆ ππ‘ = 2 π2 1+ 2 wats π2 ππ‘ =ππ 1 + 2 π2 =ππ +ππ 2 π2 π΄2πΆ π2 = ππΏππ΅ = ππππ΅ = = 8 4 So, π2 π2 π2 πππ΅ = ππΏππ΅ +ππππ΅ = ππ + ππ = ππ 4 4 2 π΄2πΆ 2 π2 =ππ 4 Cont… οΌ ππ is independent of modulation index (π) οΌ As π , πππ΅ , hence ππ‘ Case(i) If π =0 No modulation π2 ππ‘ = ππ +ππ = ππ and ππ‘ 2 Case(ii) If π =100% i.e. π =1 π2 ππ‘ = ππ +ππ =ππ 2 π2 1+ 2 12 =ππ 1 + 2 =1.5ππ Note: if π from 0 to 1 the AM power increase by 50% Cont… For π =1 ππ‘ = 1.5ππ ππ‘ ππ = =0.667ππ‘ and 1.5 π2 From ππ‘ = ππ +ππ = ππ + πππ΅ 2 πππ΅ = ππ‘ - ππ =ππ‘ - 0.666ππ‘ =0.333ππ‘ π2 ππ‘ 1 ππΏππ΅ = ππππ΅ =ππ =( )( ) = 0.666ππ‘ 4 1.5 4 0.25 = 0.166ππ‘ Note: for 100% modulation ππ =66.66% of ππ‘ & πππ΅ =33.33% of ππ‘ Cont… ο which simply means that the share of power(p) by the carrier gets reduced from 100% to 66.66% this is the biggest drawback of AM . Modulation efficiency π οΌ It specifies the share of the sideband(SB) power in the total power οΌ For efficient power distribution, π should be high πππ΅ π= = ππ‘ π2 ππ 2 π2 ππ 1+ 2 = π2 2 π2 1+ 2 If π =0,then π=0, πππ΅ =0 π2 π2 = = 2 π2 2+π 2 1+ 2 ππ‘ = ππ π. π. ππ 100% ππ‘ Cont… 0.25 οΌ If π =0.5,then π= = 0.11 & πππ΅ =11% ππ ππ‘ and 2.25 ππ = 89%of ππ‘ 1 οΌ If π =100%,then π = =0.33 & πππ΅ =33.33% ππ ππ‘ and 3 ππ = 66.66% of ππ‘ Note:ππ is independent of π and share of ππ in ππ‘ as π Current & Voltage relations in AM Current relations in AM We know that π2 ππ‘ = ππ 1 + 2 πΌπ‘2 R = πΌπ2 π πΌπ‘ =πΌπ ο πΌπ is independent of π π2 1+ 2 π2 1+ 2 Cont… Voltage relations in AM We know that π2 ππ‘ = ππ 1 + 2 ππ‘2 ππ2 = π π π2 1+ 2 ππ‘ =ππ π2 1+ 2 ο ππ is independent of π Cont… Multitone modulation ο± If the message signal is having multiple frequencies , then the modulation is corresponds to multiple modulation Lets assume π(π‘) =π΄π1 cos2πππ1 t+π΄π2 cos2πππ2 t π΄π1 > π΄π2 & ππ2 > ππ1 The general expression for AM is ππ΄π (t) = π΄π 1 + ππ π(π‘) cos2πππ t ππ΄π (t) = π΄π 1 + ππ π΄π1 πππ 2πππ1 π‘ + ππ π΄π2 πππ 2πππ2 π‘ cos2πππ t Let ππ π΄π1 =π1 ππ π΄π2 =π2 Cont… ππ΄π (t) = π΄π 1 + π1 πππ 2πππ1 π‘ + π2 πππ 2πππ2 π‘ cos2πππ t = [π΄π cos2πππ t + π΄π π1 πππ 2πππ1 π‘cos2πππ t + π΄π π2 πππ 2πππ2 π‘cos2πππ t] 1 1 By using πππ π΄πππ π΅ = cos(A-B) + cos(A+B) the above 2 2 equation becomes πππ΅1 π΄π π1 ππ΄π (t) = π΄π cos2πππ t + cos2π ππ + ππ1 t π΄π π1 π΄π2π2 + cos2π ππ − ππ1 t+ cos2π ππ + ππ2 t 2 2 πΏππ΅1 π΄π π2 + cos2π ππ − ππ2 t 2 πΏππ΅2 πππ΅2 Cont… Then the spectrum of ππ΄π (t) becomes π΄π ππ΄π (f) = πΏ π − ππ + πΏ π + ππ 2 π1 π΄π + πΏ π −ππ −ππ1 + πΏ π +ππ +ππ1 4 π1 π΄π + πΏ π −ππ +ππ1 + πΏ π +ππ −ππ1 4 π2 π΄π + πΏ π −ππ −ππ2 + πΏ π +ππ +ππ2 4 π2 π΄π + πΏ π −ππ +ππ2 + πΏ π +ππ −ππ2 4 Cont… BW = 2max freq of m(t) if we assume π1 >π2 & ππ2 >ππ1 BW = 2ππ2 The AM spectrum becomes Cont… Total power of AM for Multitone Modulation (for two message signals) ππ‘ = ππ + ππππ΅ + ππΏππ΅ ππ‘ = ππ + ππππ΅1 + ππππ΅2 + ππΏππ΅1 +ππΏππ΅2 π΄2πΆ π12 π΄2πΆ π22 π΄2πΆ π12 π΄2πΆ π22 π΄2πΆ = + + + + 2π 8π 8π 8π 8π π΄2πΆ π12 π22 π12 π22 π΄2πΆ π12 π22 = 1+ + + + = 1+ + 2π 4 4 4 4 2π 2 2 π΄2πΆ π12 +π22 = 1+ , 2π 2 but ππ‘2 = π12 + π22 π‘ππ‘ππ ππππ’πππ‘πππ πππππ₯=ππ‘ = π12 + π22 ππ‘2 ππ‘ =ππ 1 + 2 Cont… οΆSimilarly: ππ‘2 π= 2 + ππ‘2 πΌπ‘ =πΌπ ππ‘2 1+ 2 ππ‘ =ππ ππ‘2 1+ 2 Generation of AM There are two methods of generation of AM (1) Square Law Modulator (2) Switching Modulator (1) Square Law Modulator οΌ Consists of three stages Adder, Non linearity(NL) devices and BPF for extracting the desired message signal. Cont… οΆA non linear device (diode) is suitably biased and operated in a restricted portion of its characteristic curve Transfer characteristics' of diode Cont… ο When a nonlinear element such as a diode is suitably biased and operated in a restricted portion of its characteristic curve, that is ,the signal applied to the diode is relatively weak, we find that transfer characteristic of diode-load resistor combination can be represented closely by a square law : π£0 (t)=π1 π£π (t)+π2 π£ 2π (t)+….(1) Where π1 , π2 , π3 are constants & π£0 (t) is out put voltage Now, the input voltage Vi (t) is the sum of both carrier and message signals i.e., π£π (t) =π£1 (t) = π π‘ + π π‘ π£π (t) = π π‘ +Ac cos2 πππ t (2) ο The π π‘ sholud be such that the peak voltage of π£1 must be less than the cut in voltage( π£πΉ ) of the diode so that the diode exhibits Square law characteristic Cont… • Substitute (2) in (1) π£0 (t)=π1 π π‘ + π΄π cos2πππ t + π2 π π‘ + π΄π cos2πππ t π π‘ + π΄π cos2πππ t =π1 π π‘ +π1 π΄π cos2πππ t +π2 π2 π‘ +2π2 π π‘ π΄π cos2πππ t +π2 π΄2π πππ 2 2πππ t (3) By collecting the same terms we obtain AM wave 2π2 π1 π΄π 1 + π π‘ cos2πππ t +π1 π π‘ +π2 π2 π‘ π1 +π2 π΄2π πππ 2 2πππ t (4) Cont… οΆThe first bracketed term is desired AM wave at carrier ππ with 2π2 ππ = π1 Then the bracketed term in (4) becomes ππ΄π (t)= π1 π΄π 1 + ππ π π‘ cos2πππ t οΆRemaining terms are at base band or at 2ππ and are filtered out by BPF οΆBPF has center frequency ππ band width of 2w οΆThe πππ 2 term has components at baseband and at ππ becouse 1 2 οΆπππ π₯ = 2 1 + πππ 2π₯ The Fourier transform of ππ΄π (t) or the spectrum of ππ΄π (t) becomes ππ΄π (t)= π1 π΄π 1 + ππ π π‘ cos2πππ t ππ΄π (t)= π1 π΄π cos2πππ t + π1 π΄π ππ π π‘ cos2πππ t π1 π΄π ππ΄π (f) = 2 πΏ π − ππ + πΏ π + ππ π π − ππ + π π + ππ π1 ππ π΄π + 2 Cont… Spectrum of Message signal π(π‘) ↔ Spectrum of AM signal ππ΄π (f) ↔ Cont… BPF: ο The pass band of the BPF is such that it has to allow only the frequency component of AM The out put of the diode π£2 (t)=π1 π£1 (t)+π2 π£ 21 (t) +π3 π£ 23 (t)+… We can’t consider the higher order terms b/c they are not allowed by the BPF. Cont… So, π£2 (t)=π1 π π‘ + π΄π cos2πππ t 2 1 +π2 π2 π‘ +π΄2π πππ 2 2πππ t+ 2π π‘ . π΄π cos2πππ t 3 4 5 Note: οΌ π π‘ , π2 π‘ & π΄2π πππ 2 2πππ t not allowed by BPF οΌ π΄π cos2πππ t , & 2π π‘ π΄π cos2πππ allowed by BPF Cont… Spectrum of π£2 (t) π£2 (f) ↔ Cont… To avoid un desired frequency components at the O/p of BPF ππ β« 3π€ Thus, the O/p of BPF ππ΄π (t) = π1 π΄π cos2πππ t +2π2 π π‘ .π΄π os2πππ t 2π2 = π1 π΄π 1 + π π‘ cos2πππ t π1 but, ππ = 2π2 π1 ,then ππ΄π (t) =π1 π΄π 1 + ππ π π‘ cos2πππ t (2) Switching Modulator We use diode as switching device I/p of diode = π£1 (t) =π π‘ + π π‘ = π π‘ + π΄π cos2πππ t οΌ The message signal is practically of low strength οΌ The carrier signal is generated with high frequency οΌ So the operation of diode is mainly controlled by carrier π£1 (t), π π‘ > 0, πππππ ππ ππ , π‘βππ π£2 (t) = π£1 (t) π£2 (t) ≈ 0, π π‘ < 0, πππππ ππ πππ , π‘βππ π£2 (t) = 0 οΌ Thus the O/p of diode is switch b/n ′π£1 ’ and 1 ‘0’periodically at time interval of or ππ οΌ load voltage π2 (t) varies periodically between the values π1 (t) and zeros at a rate equal to the carrier frequency fc. Then π£2 (t) ≈ π£1 (t) .ππ0 π‘ = π π‘ + π΄π cos2πππ t ππ0 π‘ Where ππ0 π‘ 1 is pulse train with π0 = ππ From trigonometric Fourier series, The pulse train is (1) Substitute (2) in (1) we obtain 1 2 = + 2 π π−1 −1 ∞ π=1 2π−1 ππ0 π‘ πππ 2πππ t 2π − 1 (3) Then π£2 (t) ≈ π£1 (t) .ππ0 π‘ = π π‘ + π΄π cos2πππ t ππ0 π‘ (4) substitute (3) to(4) we obtain π£2 (t) ≈ π π‘ + π΄π cos2πππ t 1 2 + 2 π π−1 −1 ∞ π=1 2π−1 πππ 2πππ t 2π − 1 Taking π =1,2,3…. we obtain 1 2 2 π£2 (t) ≈ π π‘ + π΄π cos2πππ t + cos2πππ t − cos2π(3ππ )t + β― 2 π 3π only odd harmonics are present 1 2 π£2 (t) ≈ π π‘ + 1 - π΄π 2 2π΄ cos2πππ t + π π‘ cos2πππ t + π πππ 2 2πππ t 2 π π 3 2 2 2π΄ π π‘ cos2π(3ππ )t − π π π‘ cos2πππ t. cos2π(3ππ )t 3π 3π 5 6 4 (5) οΌ Now by designing the BPF with center frequency ππ & pass band of 2w, the result would be the required AM signal i.e. BPF allow 2 & 3 π΄π 2 οΌ ππ΄π (t) = cos2πππ t + π π‘ cos2πππ t 2 π π΄π 4 = 1+ π π‘ cos2πππ t 2 ππ΄π 4 but, ππ = ,then ππ΄π π΄π ππ΄π (t) = 2 1 + ππ π π‘ cos2πππ t Unwanted component, the spectrum of which contains ο Delta function at 0, ±2fc, ±4fc and so on. ο Occupy frequency intervals of width 2W centered at 0, ±3fc, ±5fc and soon, where W is the message bandwidth. ο Be removed by a band filter with mid frequency f and using band-pass mid-band fc bandwidth 2W, provide that fc >2W. The Fourier transform of ππ΄π (t) or the spectrum of ππ΄π (t) becomes π΄π ππ΄π (t) = 1 + ππ π π‘ cos2πππ t 2 π΄π π΄π ππ΄π (t) = cos2πππ t + ππ π π‘ cos2πππ t 2 2 π΄π ππ΄π (f) = πΏ π − ππ + πΏ π + ππ 4 π΄ π ππ + π π − ππ + π π + ππ 4 Spectrum of Message signal π(π‘) ↔ Spectrum of AM signal ππ΄π (t) ↔ Demodulation of AM οΌ The process of demodulation is used to recover the original modulating wave from the incoming modulated wave. οΌ There are three methods of demodulation of AM signal/wave (1) Square law Demodulator (2) Envelope detector (3) Synchronous detector (1) Square law Demodulator As π decrease it has certain drawbacks ππ΄π (t) =π΄π 1 + ππ π π‘ cos2πππ t ππ΄π (t) =π΄π cos2πππ t +ππ . π π‘ π΄π cos2πππ t (1) But we know that π£2 (t) =π1 π£1 (t)+π2 π£ 21 (t) so, π£2 (t) = π1 . ππ΄π (t) + π2 ππ΄π 2 (t) (2) Then substitute (2) in (1) π£2 (t)=π1 π΄π cos2πππ t +ππ . π π‘ π΄π cos2πππ t + π2 [ π΄π cos2πππ t +ππ . π π‘ π΄π cos2πππ t (π΄π cos2πππ t 1 2 We know that: πππ π₯ = 2 1 + πππ 2π₯ π£2 (t) =π1 π΄π cos2πππ t +π1 ππ . π π‘ π2 π΄2π (1 + cos2πππ t + 2 π π To reconstruct message β« 1 and if <1 then message π π can’t be reconstructed π π2 π΄2π ππ π π‘ 2 = 2 2 2 = π π2 π΄π ππ π (t) ππ π π‘ (1) 2 Assuming π π‘ =π΄π cos2πππ t (2) Substitute (2) in (1) ,we obtain π 2 = π ππ .π΄π cos2πππ t (3) But ππ . π΄π =π (4) Substitute (4) in (3) we obtain π 2 = π πcos2πππ t Assuming cos2πππ t =1, then π 2 = π π π 2 For β« 1, β« 1 or π βͺ 2 , π π π βͺ 2 π π Note :to reconstruct message , π should be small, which means π is below 1. hence this method is not practically suitable (2) Envelope Detector ο Envelope detector is simple and cheap ο Envelope detector is used to detect high level modulated levels, where as square-law detector is used to detect low level modulated signals (i.e., below 1v) ο It is also based on the switching action or switching characteristics of a diode. It consists of a diode and a resistor-capacitor(RC) filter. We know diode is forward biased diode reverse biased diode Closed switch Open switch Operation of envelope detector οΆ On a positive half cycle of the input signal, the diode is forward biased and the capacitor C charges up rapidly to the peak value of the input signal. When the input signal falls below this value, the diode becomes reverse biased and the capacitor C discharges slowly through the load resistor RL . οΆ The discharging process continues until the next positive half cycle. When the input signal becomes greater than the voltage across the capacitor, the diode conducts again and the process is repeated. The charging time constant π π C is very small when compared to the carrier period 1/ ππ i.e., π π C << 1/ ππ Where π π = internal resistance of the voltage source. • C = capacitor • fc = carrier frequency i.e., the capacitor C charges rapidly to the peak value of the signal. οΌ The discharging time constant RLC is very large when compared to the charging time constant i.e., 1/ ππ << RLC << 1/W Where RL= load resistance value W = message signal bandwidth i.e., the capacitor discharges slowly through the load resistor. Advantages: οΌ It is very simple to design οΌ It is inexpensive οΌ Efficiency is very high when compared to Square Law detector Disadvantage: ο Due to large time constant, some distortion occurs which is known as diagonal clipping i.e., selection of time constant is some what difficult Application: οΆ It is most commonly used in almost all commercial AM Radio receivers. (3) Synchronous detector ο§ For perfect reconstruction of message ,LO O/p should be perfectly synchronized both in frequency and phase transmitted with respect to carrier ο§ Frequency synchronization can be easily achieved but achieving phase synchronization is difficult and hence needs additional complex circuitry, and detection becomes complex/expensive ππ΄π (t) =π΄π 1 + ππ π π‘ cos2πππ t =π΄π cos2πππ t +π΄π ππ . π π‘ cos2πππ t (1) Case (i) assume Lo O/p having perfect synchronization i.e. Lo =π΄π cos2πππ t There fore o/p of multiplier= ππ΄π (t). Lo = π΄π cos2πππ t +π΄π ππ . π π‘ cos2πππ t π΄π cos2πππ t = π΄2π πππ 2 2πππ t + π΄2π ππ . π π‘ πππ 2 2πππ t 1 2 By using πππ π₯ = 1 + πππ 2π₯ 2 π΄2π π΄2π ππ .π π‘ 1 + cos4πππ t + 1 + cos4πππ t 2 2 π΄2π π΄2π π΄2π ππ .π π‘ π΄2π ππ .π π‘ + cos4πππ t + + cos4πππ t 2 2 2 2 οΌ From the above we neglect the dc term frequency b/c LPF is desired for π π‘ π΅π€ οΌ So, the result is and carrier π΄2π ππ .π π‘ O/p = 2 Case(ii) Assume no phase synchronization of Lo with transmitted carrier i.e. Lo =π΄π cos(2πππ t +π) o/p of multiplier = ππ΄π (t). Lo = π΄π cos2πππ t +π΄π ππ . π π‘ cos2πππ t π΄π cos(2πππ t +π = π΄2π cos2πππ tcos(2πππ t +π)+ + π΄2π ππ . π π‘ cos(2πππ t +π) By using 1 1 = cos π − π + cos π + π 2 2 cos π cos π cos −π₯ =cos(π₯) o/p of multiplier = ππ΄π (t). Lo π΄2π cos 4πππ t + π 2 π΄2π ππ .π π‘ + cosπ 2 π΄2π π΄2π ππ .π π‘ + cosπ + 2 2 we obtain and cos 4πππ t + π οΌ From the above we neglect the dc term and carrier frequency b/c LPF is desired for π π‘ π΅π€ π΄2π ππ .π π‘ So, the result is O/p = 2 cosπ οΌ For reconstruction of message ‘π’ should be constant. οΌ To maintain this ,additional circuit is to be used hence, result is complex detector Advantages of AM ο Its system is relatively cheap to build: Use simple and low cost circuit; we don’t need any special equipment and complex circuits that are used in frequency modulation. The reason that AM radio broadcasting has been popular for so long. ο Zero crossing in Amplitude modulation is equidistant. ο Modulation & Demodulation is simple ο Used for long distance communication due to amplitude modulation wave length Disadvantage/Limitations of AM AM is wasteful of power: οΌ The carrier wave c(t) and baseband signal m(t) are independent. The carrier wave represents a waste of power, which means that in AM only a fraction of the total transmitted power is actually affect by m(t). AM is wasteful of bandwidth: οΌ The upper and lower sidebands of an AM wave are uniquely related to each other by virtue of their symmetry about the carrier frequency. οΌ AM is also susceptible to interference, since it affects the amplitude of the carrier. To overcome the limitations of AM, we trade off system complexity for improved utilization of communication resources. Three modified forms of amplitude modulation: These methods are (1) Double-sideband suppressed carrier (DSBSC)AM (2) Single sideband (SSB)AM and (3) Vestigial-sideband (VSB)AM. Double-sideband Suppressed Carrier (DSBSC)AM ο known as product modulator, is an AM signal that has a suppressed /removed /attenuated carrier. ο The result is less power is transmitted with perfect communication, thus the advantage of DSBSC over AM is transmitter power will be saved. ο Product of the message signal m(t) and the carrier wave c(t): We know that ππ΄π (t) = 1 + ππ π(π‘) π(π‘) assume ππ =1 = π(π‘) + π(π‘) π π‘ Hence DSBSC is suppress of carrier The general expression for DSBSC is ππ·ππ΅ππΆ (t) =π π‘ . π(π‘) (1) But, π(π‘) =π΄π cos2πππ t (2) Substitute (2) in (1) becomes ππ·ππ΅ππΆ (t) =π΄π cos2πππ t . π π‘ (3) Then The Fourier transform of ππ·ππ΅ππΆ (t) or the spectrum of ππ·ππ΅ππΆ (t) becomes οΆ By using (recalling shifting property) οΆ exp(π2πππ π‘)π(π‘) ↔M(π − ππ ) οΆ exp(−π2πππ π‘)π(π‘) ↔M(π + ππ ) ππ·ππ΅ππΆ (t)=π΄π cos2πππ t . π π‘ π΄π ππ₯π −j2πππ t π π‘ + ππ₯π j2πππ t π π‘ 2 π΄π We obtain ππ·ππ΅ππΆ (f) = π π − ππ + π π + ππ 2 As usual π π‘ ↔ ππ·ππ΅ππΆ (t) ↔ Single tone DSBSC Assume π π‘ =π΄π cos2πππ t (1) (single message frequency) Then, ππ·ππ΅ππΆ (t) =π΄π cos2πππ t . π π‘ (2) Substitute (2) in (1) the result becomes ππ·ππ΅ππΆ (t) =π΄π cos2πππ tπ΄π cos2πππ t (3) By Using cos π cos π 1 1 = cos π − π + cos π + π 2 2 Equ(3) becomes ππ·ππ΅ππΆ (t) π΄π π΄π = cos 2π ππ + ππ π‘ 2 π΄π π΄π + cos[2π(ππ − 2 Then The Fourier transform of π(t) & π(t) or the spectrum of m(t) or π(t) becomes π π‘ =π΄π cos2πππ t π΄π π΄π cos2πππ t = 2 π π2ππππ‘ + π −π2πππ π‘ By using π π2πππ π‘ ↔ πΏ π − ππ π −π2πππ π‘ ↔ πΏ π + ππ 1 cos2πππ t ↔ 2 1 πΏ π − ππ + πΏ π + ππ 2 π΄π M(f) = πΏ π − ππ + πΏ π + ππ similarly 2 π΄π πΆ(f) = πΏ π − ππ + πΏ π + ππ 2 π΄ π π΄π ππ·ππ΅ππΆ (f) = + πΏ π −ππ −ππ + πΏ π +ππ +ππ 4 π΄π π΄π + πΏ π −ππ +ππ + πΏ π +ππ −ππ 4 Spectrum of Message signal π π‘ ↔ Spectrum of Carrier signal c π‘ ↔ Spectrum of ππ·ππ΅ππΆ signal ππ·ππ΅ππΆ π‘ ↔ but , Then 2 π£πππ π ππΏππ΅ =ππππ΅ = From equ(1) Power of DSBSC signal/wave ππ‘ = πππ΅ ππ‘ = ππΏππ΅ + ππππ΅ (1) π£ = π£πππ = π΄π 2 π΄π π΄π 2 2 2π π΄2πΆ π΄2π π΄2π π΄2π = = 8π 8 due to normalized power R=1πΊ π΄2πΆ π΄2π π΄2πΆ π΄2π ππ‘ = ππΏππ΅ + ππππ΅ = + 8π 8π π΄2πΆ π΄2π = 4π Efficiency of DSBSC signal/wave πππ΅ π= ππ‘ = 2 π΄2 πΆ π΄π 4π 2 π΄2 πΆ π΄π 4π =1 =100% Multitone DSBSC/ modulation Assume π π‘ =π΄π1 cos2πππ1 t +π΄π2 cos2πππ2 t (1) c π‘ =π΄π cos2πππ t (2) Then ππ·ππ΅ππΆ π‘ = π π‘ . c π‘ = π΄π1 cos2πππ1 t +π΄π2 cos2πππ2 t π΄π cos2πππ t =π΄π1 cos2πππ1 tπ΄π cos2πππ t +π΄π2 cos2πππ2 tπ΄π cos2πππ t 1 1 By Using cos π cos π = cos π − π + cos π + π 2 2 π΄π π΄π1 π΄π π΄π1 ππ·ππ΅ππΆ π‘ = cos 2π ππ + ππ1 π‘ + cos 2π ππ − ππ1 2 2 + USB1 π΄π π΄π2 cos 2π ππ + ππ2 2 USB2 + π΄π π΄π2 cos 2π ππ − ππ2 2 LSB2 LSB1 Then The Fourier transform of ππ·ππ΅ππΆ π‘ or the spectrum of ππ·ππ΅ππΆ π‘ becomes π΄π π΄π1 ππ·ππ΅ππΆ π =+ πΏ π −ππ −ππ1 + πΏ π +ππ +ππ1 4 π΄π π΄π1 + πΏ π −ππ +ππ1 + πΏ π +ππ −ππ1 4 π΄π π΄π2 + πΏ π −ππ −ππ2 + πΏ π +ππ +ππ2 4 π΄π π΄π2 + πΏ π −ππ +ππ2 + πΏ π +ππ −ππ2 4 Assume ππ2 > ππ1 and π΄π1 > π΄π2 Spectrum of Message signal ππ·ππ΅ππΆ π‘ ↔ BW= 2π₯βππβππ π‘ πππππ’ππππ¦ = 2π₯ππ2 Power of DSBSC signal/wave for multitone modulation ππ‘ = πππ΅1 + πππ΅2 ππ‘ = ππΏππ΅1 + ππΏππ΅2 + ππππ΅1 +ππππ΅2 but , 2 π£πππ π (1) π£ = π£πππ = π2 Then ππΏππ΅1 =ππππ΅1 = ππΏππ΅2 =ππππ΅2 = π΄π π΄π2 2 2 2π π΄π π΄π1 2 2 2π π΄2πΆ π΄2π1 π΄2π π΄2π1 = = 8π 8 π΄2πΆ π΄2π2 π΄2π π΄2π22 = 8π = 8 due to normalized power R=1πΊ From equ(1) ππ‘ = ππΏππ΅1 + ππΏππ΅2 + ππππ΅1 +ππππ΅2 = π΄2πΆ π΄2π1 π΄2πΆ π΄2π1 π΄2πΆ π΄2π2 π΄2πΆ π΄2π2 π΄2πΆ π΄2π1 π΄2πΆ π΄2π2 + 8π + 8π + 8π = 4π + 4π = 8π π΄2πΆ ππ‘ = 4π π΄2π1 + π΄2π2 Efficiency of DSBSC signal/wave π= π΄2 πΆ πππ΅ 4π ππ‘ π΄2πΆ 4π π΄2π1 +π΄2π2 π΄2π1 +π΄2π2 =1 =100% Generation of DSBSC There are two methods of generation of DSBSC (1) Balanced Modulator (2) Ring Modulator Balanced Modulator ο±Two AM modulator are arranged in a balanced configuration to suppress the carrier ο±Assume two modulators are identical except for sign reversal of π π‘ From the figure ππ΄π1 (π‘) = π΄π 1 + ππ π(π‘) cos2πππ t ππ΄π2 (π‘) = π΄π 1 − ππ π(π‘) cos2πππ t ππ·ππ΅ππΆ π‘ = ππ΄π1 (π‘) - ππ΄π2 (π‘) = π΄π cos2πππ t +ππ π(π‘) π΄π cos2πππ - π΄π cos2πππ t − ππ π(π‘) π΄π cos2πππ =2ππ π΄π π(π‘) cos2πππ , let 2π΄π ππ = π΄π ′ then 2ππ π΄π π(π‘) cos2πππ = π΄π ′ π(π‘) cos2πππ which is standard DSBSC signal 2ππ π΄π cos2πππ π(π‘) =2ππ c(π‘) π(π‘) Ring Modulator (Mixer) οΌ In this 4 diodes are conducted in the form of a ring to generate DSBSC signal. Assume diodes are ideal and transformers are center tapped. οΌ The diodes are controlled by a square-wave carrier c(t) of frequency ππ , which is applied longitudinally by means of two center-tapped transformers. The operation of the circuit. οΆAssuming that the diodes have a constant forward resistance ππ when switched on and a constant backward resistance ππ when switched off. And they switch as the carrier wave c(t) goes through zero. οΆOn one half-cycle of the carrier wave, the outer diodes are switched to their forward resistance ππ and the inner diodes are switched to their backward resistance ππ . On other half-cycle of the carrier wave, the diodes operate the half in the opposite condition. Case(i) If c(π‘) is +ve i.e. c(π‘) >0 οΌ Top(D1) & Bottom(D2) diodes conduct i.e. • D1 &D2 are switch case • D3 &D4 are open case • With c(π‘) > 0, π0 (π‘) = π(π‘) Case(ii) If c(π‘) is −ve i.e. c(π‘) < 0 οΌ D3 & Bottom D4 diodes conduct i.e. • D1 &D2 are open case • D3 &D4 are switch case • With c(π‘) <0, π0 (π‘) = −π(π‘) The output voltage has the same magnitude as the output voltage, but they have opposite polarity. Square-wave carrier c(t) can be represented by a Fourier series: The ring modulator output is therefore It is sometimes referred to as a double-balanced modulator, because it is balanced with respect to both the baseband signal and the square-wave carrier. Detection of DSBSC Coherent /Synchronous Detection οΌ It is assumed that the local oscillator signal is exactly coherent or synchronized, in both frequency and phase, with carrier wave c(t) used in the product modulator to generate s(t). This method of demodulation is known as coherent detection or synchronous demodulation. case(i) Assume Lo O/p π΄π cos2πππ t is in perfect synchronization with transmitted carriers O/p Multiplier = ππ·ππ΅ππΆ π‘ . πΏπ = π΄π π(π‘)cos2πππ π‘.π΄π cos2πππ t = π΄2π π(π‘)πππ 2 2πππ π‘ (1) 1 1 + πππ 2π₯ equ (1) becomes 2 π΄2π π(π‘) ππ·ππ΅ππΆ π‘ . πΏπ = 1 + cos4πππ π‘ 2 π΄2π π(π‘) π΄2π π(π‘) = + cos4πππ π‘ 2 2 By using πππ 2 π₯ = Hence LPF allows only message BW& reject others π΄2π π(π‘) The O/p of ππ·ππ΅ππΆ π‘ . πΏπ = 2 case(ii) Assuming Lo o/p having only frequency synchronization but not phase synchronization i.e. Lo O/p =π΄π cos(2πππ t + π) O/p Multiplier = ππ·ππ΅ππΆ π‘ . πΏπ = π΄π π(π‘)cos2πππ π‘.π΄π cos(2πππ t + π) = π΄2π π(π‘) cos2πππ π‘.cos(2πππ t + π) (1) By using cos π cos π 1 1 = cos π − π + cos π + π 2 2 equ (1) becomes π΄2π π(π‘) πππ 4πππ t + π 2 but cos(−π) = πππ π π΄2π π(π‘) − πππ π 2 (2) The o/p of equ(2) becomes π΄2π π(π‘) πππ 4πππ t + π 2 π΄2π π(π‘) + πππ π 2 due to LPF π΄2π π(π‘) The O/p of ππ·ππ΅ππΆ π‘ . πΏπ = πππ π 2 Advantage of DSBSC οΌ Transmitter power will be saved i.e.π =100% οΌ Used for long distance communication Disadvantage of DSBSC οΆ Demodulation is complex οΆ Needs high transmission bandwidth Application οΆ Uses in Quadrature carrier multiplexing and optical fiber systems Single –sideband Suppressed Carrier(SSBSC) οΌ The advantage of SSBSC over AM & DSBSC is that both transmitter power(power efficient) & transmission bandwidth(bandwidth conservation) οΌ In DSBSC, the two sidebands carry redundant information thus we can eliminate one side-band to get SSB-SC usually referred to as SSB. οΌ SSB with carrier is called SSB-AM οΌ ππππ΅ππΆ π‘ ↔ or οΌ BW of SSBSC =W=BW of message signal οΌ In SSB ,compared to DSBSC , 50% of BW & 50% of transmission power will be saved Single tone SSBSC Modulation Assuming π(π‘) = π΄π cos2πππ t (1) ππ΄π1 (π‘) = π΄π 1 + ππ π΄π cos2πππ t cos2πππ t (2) but ππ π΄π =π , π‘βππ πππ’ 2 πππππππ ππ΄π1 (π‘) = π΄π 1 + πcos2πππ t cos2πππ t But we know that , ππ·ππ΅ππΆ π‘ π΄π π΄π = cos 2π ππ + ππ π‘ 2 π΄π π΄π + cos[2π(ππ − 2 So. ππππ΅ππΆ π‘ ππππ΅ππΆ π‘ π΄π π΄π = cos 2π ππ + ππ π‘ 2 π΄π π΄π = cos 2π ππ − ππ π‘ 2 for USB or for LSB Then The Fourier transform of ππππ΅ππΆ π‘ or the spectrum of ππππ΅ππΆ π‘ becomes ππππ΅ππΆ π π΄π π΄π = 4 πΏ π −ππ −ππ + πΏ π +ππ +ππ for USB or ππππ΅ππΆ π π΄ π π΄π =+ 4 πΏ π −ππ +ππ + πΏ π +ππ −ππ for LSB Spectrum of ππππ΅ππΆ signal ππππ΅ππΆ π‘ ↔ For USB or Spectrum of ππππ΅ππΆ signal ππππ΅ππΆ π‘ ↔ For LSB Power of SSBSC signal/wave ππ‘ = ππΏππ΅ =ππππ΅ but , 2 π£πππ π π£π΄π = π£πππ = 2 π΄π π΄π 2 2 π΄2πΆ π΄2π π΄2π π΄2π = = 8π 8 Then ππΏππ΅ =ππππ΅ = due to 2π normalized power R=1πΊ Efficiency of DSBSC signal/wave πππ΅ π= ππ‘ = 2 π΄2 πΆ π΄π 8π 2 π΄2 πΆ π΄π 8π =1 =100% General expression for SSBSC ππππ΅ππΆ π‘ π΄π π΄π = cos 2π ππ ± ππ π‘ 2 (1) i.e. +for USB or – for LSB By using cos π ± π =cos π cos π β sin π sin π equ(1) becomes π΄π π΄π π΄π π΄π ππππ΅ππΆ π‘ = cos 2πππ tcos 2πππ tβ sin2πππ tsin 2 2 2πππ t i.e. here - for USB & +for LSB We know π(π‘) = π΄π cos2πππ t , 90° πβππ π π βπππ‘ ππ π(π‘)=π(t)= π΄π sin2πππ t Where π(t) is Hilbert transform of π(π‘) there fore ππππ΅ππΆ π‘ π΄π π΄π = π(π‘)cos 2πππ tβ π(t) sin2πππ t 2 2 Percentage of power saved in DSBSC &SSBSC as compared to Am (i) %Power saved in DSBSC as compared to AM π‘ππ‘ππ πππ€ππ π ππ£ππ ππ π·ππ΅ππΆ ππ π ππ = = π= π‘ππ‘ππ πππ€π ππ π΄π ππ‘ ππ‘ π 1+π2 π = 2 1 2 = π2 = 2 2+π 1+ 2 %Power saved in SSBSC as compared to AM π2 ππ ππ 4 π2 ππ 1+ 2 π‘ππ‘ππ πππ€ππ π ππ£ππ ππ πππ΅ππΆ ππ +ππΏππ΅ ππππππ΅ = = π‘ππ‘ππ πππ€π ππ π΄π ππ‘ % saved power= π2 1+ 4 π2 1+ 2 4+π2 = 2 2+π2 + π2 ππ 1+ 4 π2 ππ 1+ 2 = = Generation of SSBSC There are two methods of generation of SSBSC (1) Frequency(Filter) discrimination method (2) Phase discrimination method Frequency(Filter) discrimination method οΆIn this method ,DSBSC signal is passed through BPF to generate SSBSC We know that ππ·ππ΅ππΆ π‘ = π΄π π(π‘) cos2πππ t οΆO/p of BPF = ππππ΅ππΆ π‘ O/p of BPF for USB & ππππ΅ππΆ π‘ ↔ O/p of BPF for LSB & ππππ΅ππΆ π‘ ↔ Phase discrimination method Phase discrimination method οΌ Hilbert transform: is nothing more than a filter that shifts the π of −π all the frequency components by 2 −π οΌ i.e. Hilbert transform of x(t) is a 2 phase shifted οΌ Therefore We know π(π‘) = π΄π cos2πππ t , 90° πβππ π π βπππ‘ ππ π(π‘)=π(t)= π΄π sin2πππ t Where π(t) is Hilbert transform of π(π‘) there fore ππππ΅ππΆ π‘ π΄π π΄π = π(π‘)cos 2πππ tβ π(t) sin2πππ t 2 2 Drawbacks of Frequency/Filter discriminator ο BPF is not ideal practically , the resulting SSB will contain undesired frequencies & so the message signal cannot be reconstructed perfectly. ο Because of this SSB is limited only to voice transmission and fails for audio/radio signal(Band-gap) Drawbacks of phase discriminator ο This method is available/suitable only for generation of single tone SSBSC ο For multi-tone SSBSC , frequency discriminator is available Demodulation of SSBSC Coherent /Synchronous Detection We know ππππ΅ππΆ π‘ π΄π π΄π = π(π‘)cos 2πππ tβ π(t) sin2πππ t 2 2 Case(i) ο Considering Lo o/p in perfect synchronization with the transmitter carrier ο O/p Multiplier = ππππ΅ππΆ π‘ π₯ π΄π cos2πππ t π΄π π΄π = π(π‘)cos 2πππ t β π(t) sin2πππ t 2 2 π΄π cos2πππ t π΄2π π΄2π 2 = π(π‘)πππ 2πππ t β π(t)sin2πππ tcos2πππ t (1) 2 2 1 π ππ2π₯ 2 By using πππ π₯ = 1 + πππ 2π₯ and π πππ₯πππ π₯ = 2 2 equ (1)becomes π΄2π π(π‘) 1 + cos 4πππ t 4 π΄2π β π(t)sin4πππ t 4 Due to LPF we reject carrier frequency component π΄2π O/p LPF = π(π‘) 4 Case(ii) Considering Lo o/p in having no phase synchronization with the transmitter carrier O/p Multiplier = ππππ΅ππΆ π‘ π₯ π΄π cos(2πππ t+π) = π΄π π΄π π(π‘)cos 2πππ t β π(t) sin2πππ t π΄π cos(2πππ t+π) 2 2 π΄2π = π(π‘) cos 2πππ tcos(2πππ t+π) + 2 2 π΄π π(t) sin2πππ tcos(2πππ t+π) (1) 2 By using cos π cos π 1 1 = cos π − π + cos π + π 2 2 sin π cos π 1 = sin π − π 2 1 + sin π + π 2 Then equ(1) becomes π΄2π = π(π‘) 4 cosπ+cos(4πππ t+π sinπ+sin(4πππ t+π π΄2π β π(π‘) 4 π΄2π π΄2π = π(π‘)cosπ+ π(π‘)cos(4πππ t+π 4 4 π΄2π π΄2π β π(π‘)sinπ + π(π‘) sin(4πππ t+π 4 4 Due to LPF we reject carrier frequency component π΄2π π΄2π O/p LPF = π(π‘)cosπ β π(π‘)sinπ 4 4 π΄2π If π =0, O/p LPF = π(π‘) 4 π΄2π If π =90°, O/p LPF = β π(π‘) 4 Advantages of SSBSC ο§ Transmitter power is saved(π=100%) ο§ Transmitter BW is saved Drawbacks of SSBSC ο± Demodulation is complex ο± Limited only to voice signal transmission Application of SSBSC οΌ Preferred for voice signal οΌ Two way radio communications οΌ Frequency division multiplexing οΌ Up conversion in numerous telecommunication systems Vestigial sideband (VSB) ο VSB is similar to SSB but it retains a small portion (a vestige) of the unneeded sideband. This reduces DC distortion. ο VSB signals are generated using standard AM or DSB-SC modulation, then passing modulated signal through a sideband shaping filter. ο VSB modulation with envelope detection are used to modulate image in analog TV signals. (The audio signal is modulated using FM.) ο VSB is derived by filtering DSBSC such that one SB is passed completely say(USB) while only a vestige remains of the other SB (LSB) this small portion is called Vestige & known as VSB. VSB Spectrum οΌ In vestigial sideband, part of the lower sideband is retained. οΌ A non ideal band pass filter is used to cut off the lower sideband gradually. οΌ The special design of BPF (sideband shaping filter) distinguishes VSB from SSBSC USB We know ππππ΅ππΆ π‘ π΄π π΄π = π(π‘)cos 2πππ tβ π(t) sin2πππ t 2 2 LSB ο To generate a VSB AM signal we begin by generating a DSB-SC AM signal and passing it through a sideband filter with frequency response H( f ) as shown in ο In the time domain the VSB signal may be expressed as ππππ΅ π‘ = π΄π π(π‘) cos2πππ t β π‘ where h(t) is the impulse response of the VSB filter. In the frequency domain, the corresponding expression is π΄π ππππ΅ π = 2 π π − ππ + π π + ππ H(f) ο To determine the frequency-response characteristics of the filter, let us consider the demodulation of the VSB signal ππππ΅ (t). We multiply ππππ΅ (t) by the carrier component π΄π cos2πππ t and pass the result through an ideal low pass filter as shown in fig below and the product signal is O/p LPF =V(t) = ππππ΅ π‘ π΄π cos2πππ t = π΄π π(π‘) cos2πππ t π΄π cos2πππ t β π‘ π΄π V(f) = 2 ππππ΅ (π) π − ππ + ππππ΅ π π + ππ then we obtain H(f) π΄2π V(f) = 4 π΄2π H(π − ππ ) + 4 π π − 2ππ + π π π π + π π + 2ππ H(π + ππ ) The low pass filter rejects the double-frequency terms and passes only the components in the frequency range | f | ≤ W. Hence, the signal spectrum at the output of the ideal is π΄2π V(f) = = π π 4 H(π − ππ ) + H(π + ππ ) οΌ We require that the message signal at the output of the low pass filter be undistorted. Hence, the VSB filter characteristic must satisfy the condition. H(π − ππ ) + H(π + ππ ) =constant, | f | ≤ W οΌ This condition is satisfied by a filter that has the frequency-response characteristic as shown below. We note that H( f ) selects the upper sideband and a vestige of the lower sideband. οΌ It has odd symmetry about the carrier frequency fc, in the frequency range ππ − ππ < f < ππ + ππ , where fr is a conveniently selected frequency (vestige frequency.) that is some small fraction of W, i.e., ππ βͺW. οΌ The response of the filter should be symmetric about ππ οΌ BW =DSBSC>VSB >SSBSC οΌ Power=AM>DSBSC>VSB &SSBSC οΌ The practical difficulty is in building a filter ππ to isolate the SB completely οΌ VSB is a compromise b/n DSBSC & SSBSC οΌ In VSB, instead of rejecting one sideband completely as in SSB, a gradual cutoff of one sideband is accepted. οΌ All of the one sideband is transmitted and a small amount (vestige) of the other sideband is transmitted as well. ππππ΅ π‘ ↔ ππππ΅ π‘ ↔ BW =w+ ππ Where ,w is message bandwidth & ππ is vestige frequency Application of VSB ο Frequency spectrum of VSB modulated signal for practical Tv transmitter Frequency Conversion • The basic operation involved in single-sideband modulation is in fact a form of frequency/conversion translation. • SSB modulation is sometimes referred to as frequency changing, mixing, or heterodyning. • The mixer consists a product modulator followed by a band-pass filter. • Band-pass filter bandwidth: equal to that of the modulated signal π1 (t) used as input • Due to frequency translation performed by the mixer : We may set π2= π1 + ππ ππ= π2 − π1 assume π2 >π1 Translated upward π2= π1 − ππ ππ= π1 − π2 assume π1 >π2 Translated downward Or π1 (t)π₯ π΄π πππ 2πππ π‘ (1) but π1 (t)=π(π‘) πππ 2ππ1 π‘ Then equ(1) becomes π(π‘) πππ 2ππ1 π‘ π₯ π΄π πππ 2πππ π‘ (2) by using cos π cos π 1 1 = cos π − π + cos π + π 2 2 equ(2) becomes π΄ 2 = π π(π‘) πππ 2π π1 + ππ π‘ + πππ 2π π1 − ππ π‘ οΌ The band-pass filter rejects the unwanted frequency and keeps the desired οΌ Mixing is a linear operation
