AC lab material EC05032Notes-35.pdf

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ANALOG COMMUNICATIONS UNIT-IV LECTURE NOTES-35 D.TIRUMALA RAO ECE GMRIT Page 1 of 4 Ref: Analog & Digital Communications Simon Haykins UNIT-IV ANGLE MODULATION Notes-35 Phase Locked Loop (PLL):      The Phase locked loop (PLL) is a negative feedback system that consists of three major components : a multiplier, a loop filter and a voltage controlled oscillator (VCO) as shown in fig. below.    The voltage controlled oscillator (VCO) is a sinewave generator and its frequency is determined by a voltage applied to it.    When control voltage is zero initially, assume VCO satisfies two following conditions. i. the frequency of the VCO is precisely set at the unmodulated carrier frequency fc and ii. the VCO output has a 900 phase shift wrt the unmodulated carrier wave.    Consider the fm input signal applied to the PLL is ( ) [ ] )1(2sin)( 1  −+=  t t  f  At S  cC   φ π   where AC →  Carrier amplitude.    With modulating wave m(t), we have )2()(2)( 01  −= ∫ t  f   dt t mK t   π φ   where Kf is the frequency sensitivity of the frequency modulator.    Now VCO output is given by ( ) [ ] )3(2cos)( 2  −+=  t t  f  At r  cV   φ π   where AV →  is the amplitude    With a control voltage V(t) applied to the VCO, we have )4()(2)( 02  −= ∫ t V   dt t vK t   π φ   Where K V  is the frequency sensitivity of the VCO,    Now on applying an fm wave S(t) and VCO output r(t) to the multiplier, two components are produced.    That is by [ ] [ ] )(2cos)(2sin)()( 21  t t  f  At t  f  AK t r t S  cV cC m  φ π φ π   +⋅+=⋅   [ ] [ ] { } )()(sin)()(4sin 2121  t t t t t  f  A AK  cV C m  φ φ φ φ π   −+++=      In the above expression, the first term is a high frequency component and it is eliminated by passing it through lowpass filter.    Therefore the input to the loop filter is given by www.jntuworld.com www.jntuworld.com www.jwjobs.net   ANALOG COMMUNICATIONS UNIT-IV LECTURE NOTES-35 D.TIRUMALA RAO ECE GMRIT Page 2 of 4 Ref: Analog & Digital Communications Simon Haykins )5()(sin)(  −=  t e A AK t e V C m  φ   Where φ e (t) →  Phase error given by ∫  −−=−= t V e  dt t V K t t t t  0121 )6()(2)()()()(  π φ φ φ φ       The loop filter operates on its input e(t) to produce An output ∫ ∞∞− =  et V  )( Where h(t) is the impulse response of the filter.    Now from (5), (6) and (7), we have, with differentiation, [ ] [ ] [ ] )()()( 21  t dt d t dt d t dt d  e  φ φ φ   −=   [ ][ ] [ ][ ] ∫∫∫ ∞∞− −=−=−=  −= (2)( )(2)()(2)( )(2)( 110101 eK t dt d t V K t  dt d dt t V K t  dt d dt t V K  dt d t dt d  V V t V t V  π φ π φ π φ  π φ    [ ] [ ]  ( ) ∫ ∞∞− −=  deht  AKlAK t  dt d  eV C v )(sin2)( 1  φ π φ    [ ] [ ]  ( ) [ ]  ( ) ∫ ∞∞− −== )8(sin2)()( 01  deht eK t  dt d t dt d  e  φ π φ φ   Where K0 →  loop parameter )9( 0  −= V C V m  A AK K K   As per the equation (8), we can draw the model where multiplier and VCO are replaced by subtractor with non linearity and integrator respectively.    When the phase error φ e (t) is zero, the PLL is said to be in phase lock.    Consider all the times, φ e (t) is small compared to one radian, so that [ ] )10()()(  −≅  t t Sin ee  φ φ       Therefore with above approximation, it can be drawn to linearized model, as shown below www.jntuworld.com www.jntuworld.com www.jwjobs.net   ANALOG COMMUNICATIONS UNIT-IV LECTURE NOTES-35 D.TIRUMALA RAO ECE GMRIT Page 3 of 4 Ref: Analog & Digital Communications Simon Haykins    According this linearized model, the equation (8) is given by [ ]  ( ) ( )  [ ] ∫ ∞∞− −=−+ )11()(2)( 10  t dt d det ht eK t  dt d  e  φ φ π φ       Applying Fourier Transform, we have )12()( )(21 1)( 1  −+=  t  f  f  e  φ φ   )13()()( 0  −=  jf  f  H K  f  Lwhere  Where H(f) is the transfer function of the loop filter. And L(f) is called open-loop transfer function of the PLL.    Consider for all values of f, L(f) is very large compared to unity.    Then φ e(f) becomes zero    That means the phase of VCO asymptotically equal to the phase of the incoming wave and the same is said to phase locked.    From fig(c) )()()( 0 t t hK K t V  eV  φ  =      Now Fourier transform of this, is )14()()()( 0 −=  f  f  H K K  f V  eV  φ   From equation (13) Or )15()()()(  −=  f  f  LK  jf  f V  eV  φ       Now from equation (11), we have ( ) ( ) ( )( ) )16(1 / )( 1 −+=  f  Lt  f  LK  jf  f V  r  φ       Now by assumption L(f) >> 1, we have )17()()( 1  −=  t K  jf  f V  V  φ       Now time-domain relation is obtained by taking inverse Fourier transform, www.jntuworld.com www.jntuworld.com www.jwjobs.net   ANALOG COMMUNICATIONS UNIT-IV LECTURE NOTES-35 D.TIRUMALA RAO ECE GMRIT Page 4 of 4 Ref: Analog & Digital Communications Simon Haykins i.e., ( ) [ ] )18(21)( 1  −≅  t dt d K t V  v φ π       Thus if L(f) is very large for all frequencies, PLL may eb modeled as a differentiator with output scaling factor V  K  π  21as shown in the following diagram.    So, when the input signal is an FM wave, the phase related to the modulating wave m(t), from equation (2), we have )()( )(221)( 0 t mK K t V t d t mK  dt d K t V  r  f t  f v =  ≅ ∫ π π       Thus the output of the phase locked loop is approximately same, except the scaling factor Kf/Kv, as the srcinal message signal m(t) and the frequency demodulation is accomplished. www.jntuworld.com www.jntuworld.com www.jwjobs.net

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