Phase Locked Loop (PLL)

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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

]]> 1652 https://atomic-temporary-93025954.wpcomstaging.com/frequency-domain-representation-of-a-wide-band-fm/ Thu, 21 Feb 2019 13:48:15 +0000 https://atomic-temporary-93025954.wpcomstaging.com/?p=1639 Continue reading “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

$S_{WBFM}(f) = \frac{A_{c}}{2}\sum_{n=-\infty }^{\infty }J_{n}(\beta )$

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

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.