Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

Let the input to PLL is an FM signal S(t) = A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt)

let  \Phi _{1} (t) = 2\pi k_{f} \int_{0}^{t}m(t)dt ------Equation (I)

 Now the signal at the output of VCO is FM signal (another FM signal, which is different from input FM signal) Since Voltage Controlled Oscillator is an FM generator.

\therefore b(t) = A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

the corresponding phase    \Phi _{2} (t) = 2\pi k_{v} \int_{0}^{t}v(t)dt ------Equation (II)

It is observed that S(t) and b(t) are out of phase by 90^{o}. Now these signals are applied to a phase detector , which is basically a multiplier

\therefore the error signal e(t) =S(t) .b(t)

e(t) =A_{c} \sin (2 \pi f_{c}t+2\pi k_{f} \int_{0}^{t}m(t)dt). A_{v} \cos (2 \pi f_{c}t+2\pi k_{v} \int_{0}^{t}v(t)dt)

e(t) =A_{c}A_{v} \sin (2 \pi f_{c}t+\phi _{1}(t)). \cos (2 \pi f_{c}t+\phi _{2}(t))

on further simplification , the product yields a higher frequency term (Sum) and a lower frequency term (difference)

e(t) =A_{c}A_{v}k_{m} \sin (4 \pi f_{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

e(t) =A_{c}A_{v}k_{m} \sin (2 \omega _{c}t+\phi _{1}(t)+\phi _{2}(t))- A_{c}A_{v}k_{m}\sin (\phi _{1}(t)-\phi _{2}(t))

This product e(t) is given to a loop filter , Since the loop filter is a LPF it allows the difference and term and rejects the higher frequency term.

the over all output of a loop filter is 

 

Frequency domain representation of a Wide Band FM

To obtain the frequency-domain representation of Wide Band FM signal for the condition \beta > > 1 one must express the FM signal in complex representation (or) Phasor Notation (or) in the exponential form

i.e, Single-tone FM signal is S_{FM}(t)=A_{c}cos(2\pi f_{c}t+\beta sin 2\pi f_{m}t).

Now by expressing the above signal in terms of  Phasor notation (\because \beta > > 1 , None of the terms can be neglected)

S_{FM}(t) \simeq Re(A_{c}e^{j(2\pi f_{c}t+\beta sin 2\pi f_{m}t)})

S_{FM}(t) \simeq Re(A_{c}e^{j2\pi f_{c}t}e^{j\beta sin 2\pi f_{m}t})

S_{FM}(t) \simeq Re(e^{j2\pi f_{c}t} A_{c}e^{j\beta sin 2\pi f_{m}t})-------Equation(I)

Let    \widetilde{s(t)} =A_{c}e^{j\beta sin 2\pi f_{m}t}      is the complex envelope of FM signal.

\widetilde{s(t)} is a periodic function with period \frac{1}{f_{m}} . This \widetilde{s(t)} can be expressed in it’s Complex Fourier Series expansion.

i.e, \widetilde{S(t)} = \sum_{n=-\infty }^{\infty }C_{n} e^{jn\omega _{m}t}  this approximation is valid over [-\frac{1}{2f_{m}},\frac{1}{2f_{m}}] . Now the Fourier Coefficient  C_{n} = \frac{1}{T} \int_{\frac{-T}{2}}^{\frac{T}{2}} \widetilde{S(t)} e^{-jn2\pi f_{m}t}dt

T= \frac{1}{f_{m}}

C_{n} = \frac{1}{\frac{1}{f_{m}}} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} \widetilde{S(t)} e^{-jn2\pi f_{m}t}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{j\beta sin 2\pi f_{m}t} e^{-jn2\pi f_{m}t}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{{j\beta sin 2\pi f_{m}t-jn2\pi f_{m}t}}dt

C_{n} = f_{m} \int_{\frac{-1}{2f_{m}}}^{\frac{1}{2f_{m}}} A_{c}e^{j({\beta sin 2\pi f_{m}t-n2\pi f_{m}t})}dt

let x=2\pi f_{m}t       implies   dx=2\pi f_{m}dt

as x\rightarrow \frac{-1}{2f_{m}} \Rightarrow t\rightarrow -\pi     and    x\rightarrow \frac{1}{2f_{m}} \Rightarrow t\rightarrow \pi

C_{n} = \frac{A_{c}}{2\pi } \int_{-\pi }^{\pi } e^{j({\beta sin x-nx})}dx

let J_{n}(\beta ) = \frac{1}{2\pi } \int_{-\pi }^{\pi } e^{j({\beta sin x-nx})}dx   as    n^{th}  order Bessel Function of first kind then   C_{n} = A_{c} J_{n}(\beta ).

Continuous Fourier Series  expansion of  

\widetilde{S(t)} = \sum_{n=-\infty }^{\infty }C_{n} e^{jn\omega _{m}t}

\widetilde{S(t)} = \sum_{n=-\infty }^{\infty }A_{c} J_{n} (\beta )e^{jn\omega _{m}t} 

Now substituting this in the Equation (I)

S_{WBFM}(t) \simeq Re(e^{j2\pi f_{c}t} \sum_{n=-\infty }^{\infty }A_{c} J_{n} (\beta )e^{jn\omega _{m}t})

S_{WBFM}(t) \simeq A_{c} Re( \sum_{n=-\infty }^{\infty }J_{n} (\beta ) e^{j2\pi f_{c}t} e^{jn\omega _{m}t})

S_{WBFM}(t) \simeq A_{c} Re( \sum_{n=-\infty }^{\infty }J_{n} (\beta ) e^{j2\pi (f_{c}+nf _{m}t)})

\therefore S_{WBFM}(t) \simeq A_{c} \sum_{n=-\infty }^{\infty }J_{n} (\beta ) cos 2\pi (f_{c}+nf _{m}t)

The  Frequency spectrum  can be obtained by taking Fourier Transform 

S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{n}(\beta ) [\delta (f-(f_{c}+nf_{m}))+\delta (f+(f_{c}-nf_{m}))]

n value wide Band FM signal
0 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{0}(\beta ) [\delta (f-f_{c})+\delta (f+f_{c})]
1 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{1}(\beta ) [\delta (f-(f_{c}+f_{m}))+\delta (f+(f_{c}+f_{m}))]
-1 S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{-1}(\beta ) [\delta (f-(f_{c}-f_{m}))+\delta (f+(f_{c}-f_{m}))]
….

From the above Equation it is clear that 

  • FM signal has infinite number of side bands at frequencies (f_{c}\pm nf_{m})for n values changing from -\infty to  \infty.
  • The relative amplitudes of all the side bands depends on the value of  J_{n}(\beta ).
  • The number of significant side bands depends on the modulation index \beta.
  • The average power of FM wave is P=\frac{A_{c}^{2}}{2} Watts.

 

 

 

 

 

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Matched Filter, impulse response h(t)

Matched Filter can be considered as a special case of Optimum Filter. Optimum Filter can be treated as Matched Filter when the noise at the input of the receiver is White Gaussian Noise.

Transfer Function of Matched Filter:-

Transfer Function of Optimum filter is H(f)=\frac{k X^{*}(f)e^{-j2\pi fT}}{S_{ni}(f)}

if input noise is white noise , its Power spectral density (Psd) is S_{ni}(f)=\frac{N_{o}}{2}.

then H(f) becomes H(f)=\frac{k X^{*}(f)e^{-j2\pi fT}}{\frac{N_{o}}{2}}

H(f)=\frac{2k}{N_{o}}X^{*}(f)e^{-j2\pi fT}-----Equation(I)

From the properties of Fourier Transforms , by Conjugate Symmetry property  X^{*}(f) = X(-f)

Equation (I) becomes

H(f)=\frac{2k}{N_{o}}X(-f)e^{-j2\pi fT}------Equation(II)

From Time-shifting property of Fourier Transforms

x(t)\leftrightarrow X(f)

From Time-Reversal Property  x(-t)\leftrightarrow X(-f)

By Shifting the signal x(-t) by T Seconds in positive direction(time) ,the Fourier Transform is given by  x(T-t)\leftrightarrow X(-f)e^{-j2\pi ft}

Now the inverse Fourier Transform of the signal from the Equation(II) is

F^{-1}[H(f)]=F^{-1}[\frac{2k}{N_{o}}X(-f)e^{-j2\pi fT}]

h(t)=\frac{2k}{N_{o}}x(T-t)

Let the constant \frac{2k}{N_{o}} is set to 1, then the impulse response of Matched Filter will become h(t) = x(T-t).

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Mutual Information I(X ; Y) Properties

Property 1:- Mutual Information is Non-Negative

Mutual Information is given by equation I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2} \frac{P(\frac{x_{i}}{y_{j}})}{P(x_{i})}---------Equation(I)

we know that P(\frac{x_{i}}{y_{j}})=\frac{P(x_{i}, y_{j})}{P(y_{j})}-------Equation(II)

Substitute Equation (II) in Equation (I)

I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i}, y_{j})}{P(x_{i})P(y_{j})}

The above Equation can be written as 

I(X ; Y) =-\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i})P(y_{j})}{P(x_{i}, y_{j})}

-I(X ; Y) =\sum_{i=1}^{m}\sum_{j=1}^{n}P(x_{i}, y_{j})\log _{2}\frac{P(x_{i})P(y_{j})}{P(x_{i}, y_{j})}------Equation(III)

we knew that \sum_{k=1}^{m} p_{k}\log _{2}(\frac{q_{k}}{p_{k}})\leq 0---Equation(IV)

This result can be applied to Mutual Information I(X ; Y) , If p_{k} = P(x_{i}, y_{j}) and q_{k} be P(x_{i}) P( y_{j}), Both p_{k} and q_{k} are two probability distributions on same alphabet , then Equation (III) becomes 

-I(X ; Y) \leq 0

i.e, I(X ; Y) \geq 0  , Which implies that Mutual Information is always Non-negative (Positive).

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Example Problems in Electro Magnetic Theory Wave propagation

  1. A medium like Copper conductor which is characterized by the parameters \bg_black \sigma = 5.8 X 10^{7} Mho's/meter and \epsilon _{r}=1,\mu _{r}=1 uniform plane wave of frequency 50 Hz. Find \alpha ,\beta ,v,\eta  and \lambda.

Ans.  Given \bg_black \bg_black \sigma = 5.8 X 10^{7} Mho's/meter  ,     \bg_black \epsilon _{r}=1,\mu _{r}=1    and \bg_white f= 50 Hz

\bg_white \alpha =? ,\beta =? ,v = ?,\eta =? and \bg_white \lambda =?

Find the Loss tangent \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{2 \pi X50X\epsilon _{o}\epsilon _{r}}

                                              \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = \frac{5.8X 10^{7}}{100\pi X\epsilon _{o}}

                                            \bg_white \bg_white \frac{\sigma }{\omega \epsilon } = 2.08 X 10 ^{16}> > 1

So given medium is a Conductor (Copper)

then \bg_white \alpha (or) \beta =\sqrt{\frac{\omega \mu \sigma }{2}}

                         \bg_white =\sqrt{\frac{5.8X10^{7}X2\pi X 50X\mu _{o}}{2}}

                      \bg_white \alpha = 106.99  , \bg_white \beta =106.99.

\bg_white v_{p}=\frac{\omega }{\beta }  \bg_white =\frac{2\pi X50}{106.99}\bg_white =2.936 meters/Sec.

\bg_white \lambda =\frac{2\pi }{\beta }=\frac{2\pi }{106.99}=0.0587 meters.

\bg_white \eta =\sqrt{\frac{j\omega \mu }{(\sigma +j\omega \epsilon )}}

    \bg_white =\sqrt{\frac{jX2\pi X50X\mu _{o}}{(5.8X10^{7}+j2\pi X50X\epsilon _{o})}}

    \bg_white = \sqrt{\frac{j 3.947 X10^{-4}}{(5.8X10^{7}+j 2.78 X10^{-9})}}

    \bg_white = \sqrt{\frac{ 3.947 X10^{-4}\angle 90^{o}}{(5.8X10^{7}\angle -2.74 X 10^{-15})}}

    \bg_white = \sqrt{0.68 X10 ^{-11}}\angle \frac{90-(2.74 X 10^{-5})}{2}

\eta = 2.6 X 10^{-6}\angle 45^{o}.

2. If \bg_white \epsilon _{r}=9,\mu =\mu _{o} for a medium in which a wave with a frequency of \bg_white f= 0.3 GHz is propagating . Determine the propagation constant and intrinsic impedance of the medium when \bg_white \sigma =0.

Ans: Given \bg_white \epsilon _{r}=9,  \bg_white \mu =\mu _{o} , \bg_white f=0.3GHz and \bg_white \sigma =0.

\bg_white \gamma =?,\eta =?

Since \bg_white \sigma =0, the given medium is a lossless Di-electric.

which implies \bg_white \alpha = \frac{\sigma }{2}\sqrt{\frac{\mu }{\epsilon }} =0.

\bg_white \beta = \omega \sqrt{\mu \epsilon }

    \bg_white =2\pi X o.3X10^{9}\sqrt{\mu _{o}X9\epsilon _{o}}

    \bg_white = 18.86.

\bg_white \eta = \sqrt{\frac{\mu }{\epsilon }}

\bg_white \eta = \sqrt{\frac{\mu_{o}\mu _{r} }{\epsilon_{o}\epsilon _{r} }}

\bg_white \eta = \sqrt{\frac{\mu_{o} }{9\epsilon_{o} }}

\bg_white \eta = \frac{120\pi }{3}

\bg_white \eta = 40 Ω.

 

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