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## Phase Locked Loop (PLL)

Demodulation of an FM signal using PLL:-

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

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

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.

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

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

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 $e(t)&space;=S(t)&space;.b(t)$

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

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

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

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

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

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 $\beta&space;>&space;>&space;1$ one must express the FM signal in complex representation (or) Phasor Notation (or) in the exponential form

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

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

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

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

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

Let    $\widetilde{s(t)}&space;=A_{c}e^{j\beta&space;sin&space;2\pi&space;f_{m}t}$      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, $\widetilde{S(t)}&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}C_{n}&space;e^{jn\omega&space;_{m}t}$  this approximation is valid over $[-\frac{1}{2f_{m}},\frac{1}{2f_{m}}]$ . Now the Fourier Coefficient  $C_{n}&space;=&space;\frac{1}{T}&space;\int_{\frac{-T}{2}}^{\frac{T}{2}}&space;\widetilde{S(t)}&space;e^{-jn2\pi&space;f_{m}t}dt$

$T=&space;\frac{1}{f_{m}}$

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

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

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

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

let $x=2\pi&space;f_{m}t$       implies   $dx=2\pi&space;f_{m}dt$

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

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

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

Continuous Fourier Series  expansion of

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

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

Now substituting this in the Equation (I)

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

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

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

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

The  Frequency spectrum  can be obtained by taking Fourier Transform

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

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

From the above Equation it is clear that

• FM signal has infinite number of side bands at frequencies $(f_{c}\pm&space;nf_{m})$for n values changing from $-\infty$ to  $\infty$.
• The relative amplitudes of all the side bands depends on the value of  $J_{n}(\beta&space;)$.
• 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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## Figure of merit of FM

The block diagram of FM Receiver in the presence of noise is as follows

The incoming signal at the front end of the receiver is an FM signal $S_{FM}(t)&space;=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)--------Equation(1)$ got interfered by Additive noise $n(t)$, since the FM signal has a transmission band width $B_{T}$,the Band Pass filter characteristics are also considered over the band of interest i.e from$\frac{-B_{T}}{2}$ to $\frac{B_{T}}{2}$.

The output of Band Pass Filter is $x(t)&space;=&space;S_{FM}(t)+n_{o}(t)------------------Equation(2)$ is passed through a Discriminator for simplicity simple slope detector (discriminator followed by envelope detector) is used, the output of discriminator is $v(t)$ this signal is considered over a band of $[-W,W]$ by using a LPF .

The input noise to the BPF is n(t),  the  resultant output noise is band pass noise $n_{o}(t)$

$n_{o}(t)&space;=&space;n_{I}(t)cos&space;\omega&space;_{c}t-n_{Q}(t)sin&space;\omega&space;_{c}t-----------Equation(3)$

phasor representation of Band pass noise is $n_{o}(t)=r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$ where $r(t)=\sqrt{n_{I}^{2}(t)+n_{Q}^{2}(t)}$ and $\Psi&space;(t)&space;=&space;\tan&space;^{-1}(\frac{n_{Q}(t)}{n_{I}(t)})$.

$n_{I}(t),n_{Q}(t)$ are orthogonal, independent and are Gaussian.

$r(t)$– follows a Rayleigh’s distribution and $\Psi&space;(t)$ is uniformly distributed over $[0,2\pi&space;]$$r(t)&space;and&space;\Psi&space;(t)$ are separate random processes.

substituting Equations (1), (3) in (2)

$x(t)&space;=&space;S_{FM}(t)+n_{o}(t)$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)+n_{I}(t)cos&space;\omega&space;_{c}t-n_{Q}(t)sin&space;\omega&space;_{c}t$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+2\pi&space;k_{f}\int&space;m(t&space;)dt&space;)+r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$

$x(t)=&space;A_{c}&space;cos(2\pi&space;f_{c}t+\Phi&space;(t)&space;)+r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))-----Equation(4)$ where $\Phi&space;(t)=2\pi&space;k_{f}\int&space;m(t&space;)dt$.

now the analysis is being done from it’s phasor diagram/Noise triangle as follows

$x(t)$ is the resultant of two phasors $A_{c}&space;cos(2\pi&space;f_{c}t+\Phi&space;(t)&space;)$ and $r(t)cos&space;(\omega&space;_{c}t&space;+\Psi&space;(t))$.

$\theta&space;(t)-\Phi&space;(t)&space;=&space;\tan&space;^{-}[\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}+r(t)\cos&space;(\Psi&space;(t)-\Phi&space;(t))}]$

$\theta&space;(t)-\Phi&space;(t)&space;=&space;\tan&space;^{-}[\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}}]$ since $r(t)<&space;

$\theta&space;(t)=\Phi&space;(t)&space;+\tan&space;^{-}[\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}}]$

$\theta&space;(t)=\Phi&space;(t)&space;+\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}}$ because $\frac{r(t)}{A_{c}}<&space;<&space;1\Rightarrow&space;\tan&space;^{-1}\theta&space;=\theta$.

$\theta&space;(t)$ is the phase of the resultant signal $x(t)$ and when this signal is given to a discriminator results  an output$v(t)$.

i.e, $v(t)=\frac{1}{2\pi&space;}\frac{d\theta&space;(t)}{dt}$

i.e, $v(t)=&space;\frac{1}{2\pi&space;}\frac{d}{dt}(\Phi&space;(t)&space;+\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})---------Equation(5)$

As $\Phi&space;(t)&space;=2\pi&space;k_{f}\int&space;m(t)dt$

$\frac{d\Phi&space;(t)}{dt}=2\pi&space;k_{f}m(t)$

the second term in the Equation $n_{d}(t)=\frac{1}{2\pi&space;}\frac{d}{dt}(\frac{r(t)\sin&space;(\Psi&space;(t)-\Phi&space;(t))}{A_{c}})$ where $n_{d}(t)$ – denotes noise after demodulation.

this can be approximated to $n_{d}(t)=\frac{1}{2\pi&space;A_{c}&space;}\frac{d}{dt}(r(t)\sin&space;\Psi&space;(t))-------Equation(6)$, which is a valid approximation. In this approximation $r(t)\sin&space;\Psi&space;(t)$ is Quadrature-phase noise with power spectral density $S_{NQ}(f)$ over $[\frac{-B_{T}}{2},\frac{B_{T}}{2}]$

the power spectral density of $n_{d}(t)$ will be obtained from Equation (6) using Fourier transform property $\frac{d}{dt}\leftrightarrow&space;j2\pi&space;f$

$S_{Nd}(f)=\frac{1}{(2\pi&space;A_{c})^{2}}(2\pi&space;f)^{2}S_{NQ}(f)$

$S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f)$  , $\left&space;|&space;f&space;\right&space;|\leq&space;\frac{B_{T}}{2}$

$S_{Nd}(f)=0$    elsewhere.

the power spectral density functions are drawn in the following figure

$\therefore&space;v(t)&space;=&space;k_{f}m(t)+n_{d}(t)-------Equation(7))$ , from Carson’s rule $\frac{B_{T}}{2}\geq&space;W$

the band width of v(t) has been restricted by passing it through a LPF.

Now, $S_{Nd}(f)=(\frac{f}{A_{c}})^{2}S_{NQ}(f),\left&space;|&space;f&space;\right&space;|\leq&space;W$

$S_{Nd}(f)=0&space;elsewhere$.

To calculate Figure of Merit $FOM&space;=&space;\frac{(SNR)_{output}}{(SNR)_{input}}$

Calculation of $(SNR)_{output}$:-

output Noise power $P_{no}&space;=&space;\int_{-W}^{W}(\frac{f}{A_{c}})^{2}&space;N_{o}df$

$P_{no}&space;=&space;\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{f^{3}}{3}&space;\right&space;)^{W}_{-W}$

$P_{no}&space;=&space;\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{2W^{3}}{3}&space;\right&space;)------Equation(I)$

The output signal power is calculated from $k_{f}m(t)$ tha is  $P_{so}&space;=&space;k_{f}^{2}P--------Equation(II)$

$(SNR)_{output}&space;=&space;\frac{P_{so}}{P_{no}}$

From Equations(I) and (II)

$(SNR)_{output}&space;=\frac{\frac{N_{o}}{A_{c}^{2}}&space;\left&space;(&space;\frac{2W^{3}}{3}&space;\right&space;)}{k_{f}^{2}P}$

$(SNR)_{output}&space;=\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}-------Equation(8)$

Calculation of $(SNR)_{input}$:-

$(SNR)_{input}&space;=&space;\frac{P_{si}}{P_{ni}}$

input signal power $P_{si}=&space;\frac{A_{c}^{2}}{2}---------------Equation(III)$

noise signal power  $P_{ni}=N_{o}W--------------Equation(IV)$

from Equations (III) and (IV)

$(SNR)_{input}&space;=&space;\frac{A_{c}^{2}}{2WN_{o}}-------------------Equation(9)$

Now the Figure of Merit of FM is $FOM&space;=&space;\frac{(SNR)_{output}}{(SNR)_{input}}$

$FOM&space;=&space;\frac{\frac{3}{2}\frac{k_{f}^{2}PA_{c}^{2}}{N_{o}W^{3}}}{\frac{A_{c}^{2}}{2WN_{o}}}$

$FOM_{FM}&space;=&space;\frac{3k_{f}^{2}P}{W^{2}}--------Equation(10)$

to match this with AM tone(single-tone) modulation is used i.e, $m(t)&space;=&space;cos&space;\omega&space;_{m}t$ then the signal power $P&space;=&space;\frac{1}{2}$  and $W&space;=&space;f_{m}$

$FOM_{FM}&space;=&space;\frac{3k_{f}^{2}}{f_{m}^{2}}\frac{1}{2}$

$FOM_{FM}&space;=&space;\frac{3}{2}&space;(\frac{k_{f}}{f_{m}})^{2}$

$FOM_{FM}&space;=&space;\frac{3}{2}\beta&space;^{2}$

since for tone(single-tone) modulation $\beta&space;=&space;\frac{k_{f}}{f_{m}}$.

when you compare single-tone FM with AM $FOM_{FM&space;(single-tone)}&space;=FOM_{AM(single-tone)}$

$\frac{3}{2}\beta&space;^{2}&space;>&space;\frac{1}{3}$

$\beta&space;>&space;\frac{\sqrt{2}}{3}$

$\beta&space;>&space;0.471$.

the modulation index $\beta&space;>0.471.$ will be beneficial in terms of noise cancellation, this is one of the reasons why we prefer WBFM over NBFM.

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## Capture effect in Frequency Modulation

The Amplitude Modulation schemes like AM,DSB-SC and SSB-SC systems can not handle inherent Non-linearities in a really good manner where as FM can handle it very well.

Let us suppose un Modulated FM carrier $S(t)&space;=&space;A_{c}cos\omega&space;_{c}(t)$

$S(t)&space;=&space;A_{c}cos(\omega&space;_{c}(t)+\phi&space;(t))$

By considering un modulated FM carrier in terms of frequency(by neglecting phase) i.e $S(t)&space;=&space;A_{c}cos&space;(\omega&space;_{c}t)$ has been interfered by a near by interference located at a frequency $(\omega&space;_{c}+\omega&space;)$ where $\omega$ is a small deviation from $\omega&space;_{c}$.

the nearby inerference is $I&space;cos(\omega&space;_{c}&space;+&space;\omega&space;)t$

when the original signal got interfered by this near by interference , the received signal is $r(t)=&space;A_{c}cos&space;\omega&space;_{c}t&space;+&space;I&space;cos(\omega&space;_{c}+\omega&space;)t$

$r(t)=&space;(A+&space;I&space;cos&space;\omega&space;t)cos&space;\omega&space;_{c}t&space;-I&space;sin&space;\omega&space;t&space;sin\omega&space;_{c}t$   Let $A_{c}=A$

$r(t)&space;=&space;E_{r}(t)&space;cos&space;(\omega&space;_{c}t+\Psi&space;_{d}(t))$

now the phase of the signal is $\Psi&space;_{d}(t)&space;=&space;tan^{-1}&space;(\frac{I&space;sin&space;\omega&space;t}{A+I&space;cos&space;\omega&space;t&space;})$

as $A>&space;>&space;I$ implies $\frac{I}{A}<&space;<&space;1$

$\Psi&space;_{d}(t)&space;=&space;tan^{-1}&space;(\frac{I&space;sin&space;\omega&space;t}{A})$

since $\frac{I}{A}<&space;<&space;1$ , $\tan&space;^{-1}\theta&space;=&space;\theta$

$\Psi&space;_{d}(t)&space;\approx&space;\frac{I&space;sin&space;\omega&space;t}{A}$

As the demodulated signal is the output of a discriminator $y&space;_{d}(t)&space;=\frac{d}{dt}&space;(\frac{I&space;sin&space;\omega&space;t}{A})$

$y&space;_{d}(t)&space;=\frac{I\omega&space;}{A}&space;({cos&space;\omega&space;t})$ , which is the detected at the output of the demodulator.

the detected output at the demodulator is $y_{d}(t)$ in the absence of message signal  i.e, $m(t)=0$.

i.e, when message signal is not being transmitted at the transmitter but detected some output $y_{d}(t)$ which is nothing but the interference.

As ‘A’ is higher the interference is less at t=0 the interference is $\frac{I\omega&space;}{A}$ and is a linear function of $\omega$, when $\omega$ is small interference is less. That is $\omega$ is closer to $\omega&space;_{c}$ interference is less in FM.

Advantage of FM :- is Noise cancellation property , any interference that comes closer with the carrier signal (in the band of FM) more it will be cancelled. Not only that it overridden by the carrier strength $A_{c}$ but also exerts more power in the demodulated signal.

This is known as ‘Capture effect’ in FM which is a very good property of FM. Over years it has seen that a near by interference is 35 dB less in AM where as the near by interference in FM is 6 dB less this is a big advantage.

Two more advantages of FM over AM are:

1. Non-linearity in the Channel ,FM cancels it very nicely due to it’s inherent modulation and demodulation technique.
2. Capture effect( a near by interference) FM overrides this by $A_{c}$.
3. Noise cancellation.

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## Automatic Gain Control (AGC)

Let us discuss about the facts why we need AGC in a Radio Receiver , as we all know that the voltage gain available at the Receiver from antenna to demodulator in several stages of amplification is very high, so that it can amplify a very weak signal But what if the signal is much stronger at the front end of the receiver ?

If same gain (gain maintained for an incoming weak signal) is maintained by different stages of the Receiver for a stonger  incoming signal, the signal is further amplified by these stages and the received signal strength is far beyond the expectations which can be avoided. so we need to have a mechanism which will measure the stength of the input signal and accordingly adjust the gain. AGC does precisely this job and improves the dynamic range of the antenna to (60-100)dB by adjusting the gain of the Intermediate Frequency and sometimes the Radio Frequency stages.

It is generally observed that as a result of fading, the amplitude of the IF carrier signal at the detecor input may vary  as much as 30 (or) 40 dB this results in the corresponding variation in general level of reproduced signal at the receiver output.

At IF carrier minimum loud speaker output becomes inaudible and mixed up with noise.

At IF  carrier maximum loud speaker output becomes intolerably large.

Therefore a properly designed AGC reduces the amplitude variation due to fading from a high value of (30-40)dB to (3-4)dB.

Basic need of AGC or AVC:-

AGC is a sub system by means of which the overall gain of a receiver is varied automatically with the variations in the stregth of the received signal to keep the output substantially constant.

i.e, the overall requirement of an AGC circuit in a receiver is to maintain a constant output level.

Some of the factors that explain why AGC is needed:-

• When a Receiver without AGC/AVC is tuned to a strong station, the received signal may overload the subsequent IF and AF stages this overloading causes carrier distortion in the incoming signal this can be prevented by using manual gain control on first RF stage but now a days AGC circuits are used for this purpose.
• When the Receiver is tuned from one station to another, difference in signal strengths of the two stations causes an unpleasant loud output if signal is moving from a weak station to a strong station unless we initially keep the volume control very low before changing the tuning from one station to another . Changing the volume control every time before attempting to re-tunethe receiver is howeve cumbersome. Therefore AGC/AVC enables the user to listen to a station without constantly monitoring the volume control.
• AGC is particularly important for mobile Receivers.
• AGC helps to smooth out the rapid fading which may occur with long distance short-wave reception.

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## Choice Of Intermediate Frequency (or) IF Amplifier in a Radio Receiver

Choice of Intermediate Frequency of a receiving system is usually a compromise , since there are reasons why it is neither low nor high, nor in a certain range between the two.

The following are the major factors influencing the choice of the Intermediate Frequency in any particular system.

1. If the IF is too high poor selectivity and poor adjacent channel rejection results unless sharp cut-off filters(crystal/mechanical filters) are used in the IF stage.
2. A high value of Intermediate Frequency(IF) increases tracking difficulties.
3. If we chose IF as low frequency, image frequency rejection becomes poorer. i.e, if $\frac{f_{si}}{f_{s}}$ is more IFRR(image Frequency Rejection Ratio) has been improved, which requires a high Intermediate Frequency($f_{si}$). Similarly when $f_{s}$ is more IFRR becomes worst.
4. Average Intermediate Frequency(IF) can make the selectivity too sharp cutting of the side bands.This problem arises because the Q must be low when the IF is low, unless crystal or mechanical filters are used and hence gain per stage is low. Thus a designer is more likely to raise Q rather than increasing the number of IF amplifiers.
5. If IF is very low , the frequency stability of local oscillator must be made correspondingly high.
6. IF must not fall in the tuning range of the receiver or else instability occurs and hetero dyne whistles (noise) will be heard.

Frequencies used:-

1. Standard AM broadcast receivers tuned to (540 KHz-1650 KHz) or(6 MHz-18 MHz) and European long wave band (150 KHZ- 350 KHz) uses IF in the range (438 KHz- 465 KHz). 455 KHz is the most popular value used.
2. FM receivers using the standard (88 MHz -108 MHz) band have an IF which is almost always 10.7 MHz.
3. TV Receivers in the  VHF band (54 MHz-223 MHz),UHF band (470 MHz-940 MHz) uses IF between (26 MHz-46 MHz) and the popular values are 36 MHz and 46 MHz.
4. AM-SSB Receviers employed for short-wave reception in the short wave band / VHF band uses IF in the range (1.6 MHz to 2.3 MHz).

(No Ratings Yet)

## Generation of PWM and PPM using Wave forms

PWM Generator:-

The circuit that generates PWM wave is as follows, Here in this circuit Op-Amp works in comparator mode.It compares two voltages, modulating voltage with Saw-tooth Voltage. Saw-tooth voltage is taken as reference voltage.

 condition Output voltage Vo(t) $m(t)>&space;V_{r}(t)$ Low $V_{r}(t)>m(t)$ High

from the graphs whenever modulating voltage dominates saw-tooth voltage corresponding output is low.

Similarly, when saw-tooth voltage dominates modulating voltage corresponding output is High.

Then the resultant output voltage is a PWM signal.

PPM Generator:-

Now, a PPM signal has been generated by passing the PWM signal through a Mono-stable Multi vibrator . Here the resultant signal is a PPM signal with the pulse starting with respect to trailing edge of PWM signal.

The width and Amplitude of Pulse remains constant only the position of the pulse changes with respect to m(t) .

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## Band Pass Sampling

However when the given signal is a Band Pass signal then a different criterion must be used to sample the signal , the Band Pass signal x(t) whose maximum BW is ‘$2f_{m}/W$‘ Hz can be completely represented and recovered from it’s samples if it is  sampled at the minimum rate of greater than or  equals to twice that of the BW.

then sampling rate $f_{s}\geq&space;2&space;X&space;BW$

i.e, $f_{s}\geq&space;4f_{m}&space;or&space;f_{s}\geq&space;2W$

Any band pass signal in time-domain can be represented in it’s in-phase $x_{I}(t)$ and quadrature phase $x_{Q}(t)$ components as

$x(t)&space;=&space;x_{I}(t)cos&space;2\pi&space;f_{c}t&space;\pm&space;x_{Q}(t)sin&space;2\pi&space;f_{c}t$

after sampling the band pass signal, the signal after reconstruction is

$x(t)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}sinc(2f_{m}t-\frac{n}{2})cos(2\pi&space;f_{c}(t-\frac{n}{4f_{m}}))$

$T_{s}&space;=&space;\frac{1}{4f_{m}}$, where BW of band pass signal is $2f_{m}$  Hz

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## Pulse Position Modulation(PPM)

Pulse Position modulation is another type of Pulse Time modulation technique that is in PPM the position of the pulse carrier is varied in accordance with the instantaneous values of the message signal, where as the amplitude and width of the pulse remains constant. here message lies in the position(OFF periods) of the PPM signal.

PPM demodulator:-

The PPM Demodulator consists of a Transistor T1  which acts as a switch followed by a second order Low pass filter circuit( using OP-AMP).

As the  input to the demodulator is a PPM signal, the gaps between pulses contains the information in PPM signal. Let us consider a PPM signal with OFF and ON periods marked from A to F.

Here Transistor T1 acts as a switch  as follows

• input to the base of T1 is low  —–> Transistor T1 is in cut-off region.
• input to the base of T1 is high  —–> Transistor T1 is in Saturation region.

during  the time inerval AB, the input to the base of T1 is low and transistor T1 moves into cut-off region in this condition capacitor C charges to a vlotage proportional to length of time duaration AB that is the height of the ramp is equals to duration AB.

During the time interval BC, the input to T1 is high and T1 moves into Saturation region in this case Capacitor ‘C’ discharges through T1 , this discharge is rapid and the collector voltage remains low over the duartion BC.

This process continues and results a saw-tooth wave form at the output of transistor T1 , by applying this signal to a second order LPFn Demodulated signal has been obtained as the final output.

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This is  the most commonly used Receiver and it uses “hetero dyning” principle which is used almost in all types of receivers like TR Receiver and Radar Receiver etc. The word hetero(≈different) dyne(≈mixing) means mixing  different frequencies using a Mixer. Hence the name given as super hetero dyne Receiver.

The block diagram consists of  a receiving antenna followed by an RF stage as the primary block , the receiving signal has been fed to RF stage through the antenna.

In a Super hetero dyne Receiver the incoming RF signal frequency ($f_{s}$) is combined with local oscillator frequency($f_{o}$) through a mixer and converts a signal of a lower fixed frequency (IF) this lower fixed frequency is called as Intermediate Frequency ($f_{i}$ or $f_{IF}$). A constant frequency difference is maintained between the Local Oscillator and incoming RF signal. This is provided through Capacitance tuning that is all capacitors are ganged together and operated by a common control knob.

$\therefore$ incoming RF is  down translated to IF using a mixer now this IF is given as input to the secondary stage of the block diagram that is IF amplifier. IF amplifier consists of number of transformers each consisting of a pair of mutually tuned circuits thus with a large number of double tuned circuits, operating at a specially chosen frequency the IF amplifier provides most of the gain.

Thus IF stage full fills most of the gain (sensitivity) and Band width(selectivity) requirements of the Receiver. For a Super hetero dyne receiver Sensitivity and selectivity are quite uniform throughout it’s tuning range this is one of the advantage over TRF Receiver.

The amplified IF signal is given as an input to the Detector. The Detector or the demodulator demodulates the signal and down translates the IF signal to AF(Audio Frequency) signal.

The AF signal is amplified by Audio amplifier and further by power amplifier. The last stage of the receiver is a Loud speaker , which receives AF signal. Loud speaker is in general a transducer which converts electrical signal into a voice (or) Audio.

• It provides high gain through IF amplifier that is more sensitivity is being provided by it.
• Improved selectivity over TRF receiver.
• BW remains constant over the entire operating range.
• Selectivity and Sensitivity are uniform throughout it’s tuning range.
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The block diagram consists of  a receiving antenna followed by an RF stage as the primary block , the receiving signal has been fed to RF stage through the antenna. This RF stage consists of two (or) three RF Amplifiers, these amplifiers are tuned RF Amplifiers.i.e they have variable tuned circuits at input and output sides.

The received signal has been amplified by the RF amplifiers and the amplified signal is being given as an input to the Detector. The Detector or the demodulator demodulates the signal and down converts the RF signal to AF(Audio Frequency) signal.

The AF signal is amplified by Audio amplifier and further by power amplifier. The last stage of the receiver is a Loud speaker , which receives AF signal. Loud speaker is in general a transducer which converts electrical signal into a voice (or) Audio.

1. Selectivity of TRF Receiver is poor. This is because achieving sufficient selectivity at high frequencies is difficult due to enforced use of single-tuned Circuits.
2. Instability:-(RF Stage)  The TRF Receiver suffers from a tendency to oscillate at a higher frequencies (i.e, instability), this is because multi-stage RF amplifiers has to provide high gain at high frequencies. RF amplifiers provides high gain which results in positive feed back leads to oscillations and then causes instability of the circuit. This positive feedback (caused by the leakage of output of RF stage back to it’s input) could result from power supply coupling through any other element common to input and output stages.
3. Variation of band width over tuning range:- One more draw back in TRF receiver is the BW variation over the tuning range i.e the BW of TRF receiver varies with the incoming frequency.

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## Sampling Theorem

Sampling of signals is the fundamental operation in signal processing, a Continuous Time (CT) signal can be converted into a Discrete Time (DT) signal using Sampling process. Sampling is required since the advancement in both signals and systems which are digitized i.e, Digital systems operates only on digital signals only.

Sampling Theorem:-

A CT signal is first converted into DT signal by Sampling process. The sufficient number of samples must be taken so that the original signal is represented in it’s samples completely, and also the signal is represented from it’s samples, these two conditions representation and reconstruction depends on the sampling process ‘fs‘ Hz.

Sampling theorem can be given into two parts

i. A band limited signal of finite energy, which has no frequency component higher than ‘fm‘ Hz, is completely described by it’s sample values at uniform intervals less than (or) equal to 1/2fm seconds apart.

i.e, $T_{s}\leq&space;\frac{1}{2f_{m}}$  Seconds.

ii. A Band limited signal of finite energy, which has no frequency component higher than fm Hz may be completely recovered from the knowledge of it’s samples if samples are taken at the rate of 2fm samples/second.

i.e, $f_{s}\geq&space;2f_{m}$ Hz.

Statement:- A Continuous Time signal can be completely represented in it’s samples and recovered from it’s samples if the sampling frequency $f_{s}\geq&space;2f_{m}Hz.$

where $f_{s}$ is the sampling frequency.

$f_{m}$ is the highest frequency present in the original signal / Band width of the signal.

proof of Sampling theorem:-

Let us consider a CT signal x(t), which is a band limited to $f_{m}$ Hz as shown

To prove Sampling theorem, it should be shown a signal whose spectrum is band limited to fm Hz can be reconstructed exactly without any error from it’s samples taken uniformly at a rate of $f_{s}>&space;2f_{m}$ Hz.

The circuit shows the sampler

Now sampling of x(t) at a rate of fs may be achieved by multiplying x(t) with a train of impulses  $\delta&space;T_{s}(t)$ with a period ‘Ts‘ seconds.

The sampling signal is an ideal (or) instantaneous signal. This is also known as ideal (or) instantaneous sampling.

$g(t)=x(t)\delta&space;T_{s}(t)$

As $\delta&space;T_{s}(t)$ is a periodic impulse train it can be expressed in it’s Fourier Series expansion as follows

Exponential Fourier Series is

$\delta&space;T_{s}(t)&space;=&space;\sum_{n=-\infty&space;}^{\infty&space;}F_{n}e^{jnw_{s}t}$

$F_{n}=&space;\frac{1}{T_{s}}\int_{\frac{-T_{s}]}{2}}^{\frac{T_{s}}{2}}\delta&space;T_{s}(t)e^{-jn\omega&space;_{s}t}dt$

$F_{n}=\frac{1}{T_{s}}$

$F_{n}=f_{s}$

∴ Exponential Fourier Series is $\delta&space;T_{s}(t)=\sum_{n=-\infty&space;}^{\infty&space;}f_{s}e^{jn\omega&space;_{s}t}$

now the sampled signal $g(t)&space;=&space;x(t).\delta&space;T_{s}(t)$

$g(t)=x(t)\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}e^{jn\omega&space;_{s}t}$

$g(t)=\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}x(t)e^{jn\omega&space;_{s}t}$

By finding Fourier Transform of g(t) is G(f)

$G(f)=\sum_{n=&space;-\infty&space;}^{\infty&space;}f_{s}X(f-nf_{s})$

Now the frequency spectrum of the sampled signal G(f) is of the form

From G(f) spectrum the original spectrum of X(f) has been shifted to different center frequencies

i.e, when n=0  center frequency is 0.

n=1  center frequency is fs

n=-1 center frequency is -fs etc

Some important conclusions from frequency spectrum of sampled signal:-

1. The spectrum of sampled signal G(f)/G(w) will repeat periodically if $f_{s}>&space;2f_{m}$ without any overlapping.
2. G(f) is extending up to infinity and the Band width is infinity as well, out of G(f) , X(f) need to be recovered , which is band limited to fm Hz.
3. X(f) is centered at f=0 and has fm as the highest frequency, X(f) may be recovered by passing it through a Loe Pass filter with cutoff frequency approximately equals to fm  Hz.
4. to reconstruct x(t) from g(t) the condition that must be satisfied is  $f_{s}\geq&space;2f_{m}$.

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## Indirect method of generation of FM signal

Indirect method of generation of FM signal is also known as Armstrong method .Here a crystal oscillator generates carrier signal , which provides very high stability compared to Direct method. this method  generates a  WBFM signal, i.e a phase modulator  generates a NBFM signal in the first step , then in the second step NBFM will be converted to WBFM signal using a frequency multiplier.

In NBFM modulation index is small and the distortion is very low in NBFM ,here we prefer phase modulator to generate NBFM as it’s generation is easy, the frequency multiplier multiplies incoming frequency along with frequency deviation $\Delta&space;f$ . Hence NBFM will be converted into WBFM with large frequency deviation as well.

Frequency multiplier:-

The frequency multiplier consists of a non-linear device followed by a Band Pass Filter, the non-linear device is a memory less device.

If the input to a non-linear device is an FM wave with frequency $f_{c}$ and  deviation $\Delta&space;f$ then output consists of DC component and ‘n’ frequency modulated waves with carrier frequencies $f_{c},2f_{c},3f_{c},......nf_{c}$ and frequency deviations $\Delta&space;f,2\Delta&space;f,3\Delta&space;f,4\Delta&space;f.....n\Delta&space;f$ . The BPF designing is in such a way that it passes the FM wave centered at the frequency $nf_{c}$with frequency deviation $n\Delta&space;f$ and to suppress all other FM components. Thus a frequency multiplier generates a WBFM wave from a NBFM wave.

Generation of WBFM by Armstrong’s method:-

This Armstrong’s method is indirect method used to generate WBFM signal.It is used to generate FM signal having both the desired frequency deviation and carrier frequency.

The block diagram consists of two stage multiplier and an intermediate stage of frequency translator .

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## Tutorial problems Analog communications

Pb1. A broadcast radio transmitter radiates 5KW power when the modulation percentage is 60% , how much is the carier power?  [Ans: 4.23 KW]

pb2. A 400 Watts carrier is modulated to a depth of 75%, calculate the total power in the modulated wave by assuming the modulating wave as sinusoidal signal. [Ans: 512.5 Watts].

Pb3. The antenna current of an AM transmitter is 8 A when only carrier is being transmitted , but is increases to 8.96 A , when the carrier is modulated by a single-tone sinusoid, find the percentage of modulation? find the antenna current when the depth of modulation changes to 0.8.  [Ans: 0.713, 9.19A].

Pb4. A 300 Watts carrier is simultaneously modulated by two audio waves with modulation percentages of 50 and 60 respectively. what will be the total side band power radiated? [Ans: 91.5 Watts].

pb5. Find the power of the signal $V(t)=&space;cos\omega&space;_{c}t&space;+&space;cos\omega&space;_{c}tcos\omega&space;_{m}t$  [Ans: 3/4 Watts].

pb6. Find the power of the signal $V(t)=&space;cos\omega&space;_{l}t&space;+&space;cos\omega&space;_{c}tcos\omega&space;_{m}t$  [Ans: 3/4 Watts].


text:
# Counting paths by their type

An edge-labelled directed graph is the data of $$G = (V, E, l)$$ where $$(V, E)$$ is a directed graph, and $$l \colon E \to \mathbb{P}$$ is some function. (For the graph I am considering, labels take values in $$\mathbb{P} = \{1, 2, 3, \ldots\}$$). To any path $$v_0 \to v_1 \to \ldots \to v_n$$ in $$G$$, we can associate a type $$\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbb{P}^n$$, where $$\alpha_i = l(v_{i-1}, v_i)$$. For a fixed source vertex $$s$$, destination vertex $$t$$, and type $$\alpha \in \mathbb{P}^n$$, let $$P(s, \alpha, t) \in \mathbb{N}$$ be the number of $$s \rightsquigarrow t$$ paths of type $$\alpha$$.

I have two counting problems, the first of which I can solve fairly efficiently, and the second of which I am hoping for some assistance. In the problem I am considering, the directed graph is acyclic, and satisfies other interesting properties: for example $$P(s, \alpha, t) = P(s, \beta, t)$$ whenever $$\alpha$$ is a permutation of $$\beta$$. Furthermore, in the problem I am consdiering $$G$$ is an infinite graph, but each of the following problems is finite.

Problem 1: Fix a source $$s \in V$$ and a type $$\alpha$$, and determine all of the numbers $$P(s, \alpha, -)$$.

Problem 2: Fix a source $$s \in V$$ and a destination $$t \in V$$, and determine all of the numbers $$P(s, -, t)$$.

For problem 1, we can take an approach very similar to counting paths in a DAG. Initialise an array $$\mathtt{D}$$ indexed by $$V$$ to zero, and set $$\mathtt{D}[s] = 1$$. Then perform a breadth-first search, at the first layer following all edges of type $$\alpha_1$$, and for each $$s \xrightarrow{\alpha_1} v$$ increment $$\mathtt{D}[v]$$ by $$\mathtt{D}[s]$$. Repeat for all the $$\alpha_2, \ldots, \alpha_n$$. At the end, the last layer contains all vertices $$t$$ reachable from $$s$$ via an $$\alpha$$-path, and $$\mathtt{D}[t]$$ is equal to $$P(s, \alpha, t)$$.

Problem 2 is more tricky, and the best strategy I have at the moment is essentially to iterate over the potential allowed types $$\alpha$$ and compute each $$P(s, \alpha, t)$$ individually. Another approach would be to do a breadth-first-search from $$s$$, but storing at each vertex the path type leading there, together with multiplicity - but this doesn't seem to be faster.

• What kind of efficiency are you looking for? It's easy to construct an acyclic graph where there are $O(2^{n/3})$ paths with different types (a concatenation of rhombuses), and therefore even outputting the answer is not efficient. – Dmitry Oct 24 '20 at 13:54
• I don't understand your algorithm for problem 1. What do you do for $\alpha_2$? – D.W. Oct 24 '20 at 20:01
• @D.W. Let the BFS layers be $L_0 = \{s\}, L_1, \ldots, L_n$, where $L_i$ is everything reachable from $L_{i-1}$ by following an edge labelled $\alpha_i$. Initially $\mathtt{D}[s] = 1$. For each vertex $v \in L_i$, set $\mathtt{D}[v] = \sum_{u \to v} \mathtt{D}[u]$, where the sum is taken over all $\alpha_i$-edges going from $L_{i-1}$ into $v$. Do this in ascending order of layers. At the end, $L_n$ contains precisely those $t$ with nonzero $P(s, \alpha, t)$, and $\mathtt{D}[t]$ contains the number of $\alpha$-paths $s \rightsquigarrow t$. – Joppy Oct 24 '20 at 22:49
• @Dmitry The actual problem I am trying to solve is NP-hard, so I am not expecting a terribly efficient solution, just something that is "fast enough" for my purposes. The graph is not too terrible - for example the symmetry property means that it is enough to consider only those path types $\alpha$ where $\alpha$ is in sorted order. – Joppy Oct 24 '20 at 22:54

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## anonymous one year ago Need help please explain to me from step 1

• This Question is Open

??

2. anonymous

3. Rushwr

if $x ^{2} = 196$ $x = \sqrt{196}$ I hope u can find that of course The next if $x ^{3}= 512$

4. Rushwr

what do u think the answer would be ?

|dw:1441085134039:dw||dw:1441085244764:dw|

For Square Roots, we group by 2 and For Cube Roots, we group by 3. Hope this is clear...

7. anonymous

Sooo you just divide 3 into 512

|dw:1441086446891:dw||dw:1441086464912:dw|


text:
# 8th Grade Math - Calculator Inactive

### 8th Grade Math - Calculator Inactive Sample

1 pt Chapter: 6 Standard: 8.EE.8c DOK: 3
3.

A test has 27 questions and is worth 100 points. The test consists of short answer questions each worth 4 points and true & false questions each worth 2 points. How many short answer questions are there?

1 pt Chapter: 5 Standard: 8.EE.5 DOK: 2
11.

The following fractions below are proportional. Solve the proportion and use your results to find the y-intercept of the proportion equation.
$18 30 = a 20$

1 pt Chapter: 5 Standard: 8.F.2 DOK: 2
13.

State the slope of the linear function with a steeper slope.

1 pt Chapter: 13 Standard: 8.G.9 DOK: 3
16.

Donovan opened up a can of soup yesterday and discovered that the soup was not filled all the way to the edge. Using the dimensions provided below, determine how much of the can was not being used. Use 3.14 for $π$.


text:
# [Tex/LaTex] set font size and line spacing

fontsizeline-spacing

I need a font size of 12 pt and a line spacing of 1.5 lines.

I try:

\documentclass{article}
\usepackage{lipsum}
\linespread{1.5} %regulate line spacing
\renewcommand{\normalsize}{\fontsize{12pt}{0}\selectfont}
\begin{document}
\lipsum[0]
\end{document}

I set the baseline skip in the second {} in \fontsize to 0pt because I think it would interfere with \linespread. On the other hand \linespread appears not to work.

Can you help?

Useless and counterproductive messing with \fontsize. This simpler MWE work as expected:

\documentclass[12pt]{article}
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\end{document}

There a 12pt font and a 1.5 line spacing. What more?


text:
Volume 186

## The Apollonian structure of integer superharmonic matrices

Pages 1-67 by Lionel Levine, Wesley Pegden, Charles K. Smart | From volume 186-1

## Chern’s conjecture for special affine manifolds

Pages 69-95 by Bruno Klingler | From volume 186-1

## Tame distillation and desingularization by $p$-alterations

Pages 97-126 by Michael Temkin | From volume 186-1

## Homological stability for moduli spaces of high dimensional manifolds. II

Pages 127-204 by Søren Galatius, Oscar Randal-Williams | From volume 186-1

## Geometric monodromy — semisimplicity and maximality

Pages 205-236 by Anna Cadoret, Chun-Yin Hui, Akio Tamagawa | From volume 186-1

## Wilkie’s conjecture for restricted elementary functions

Pages 237-275 by Gal Binyamini, Dmitry Novikov | From volume 186-1

## Dynamical spectral rigidity among $\mathbb{Z}_2$-symmetric strictly convex domains close to a circle (Appendix B coauthored with H. Hezari)

Pages 277-314 by Jacopo De Simoi, Vadim Kaloshin, Qiaoling Wei | From volume 186-1

## Corrigendum to the paper “On the $\boldsymbol{K^2}$ of degenerations of surfaces and the multiple point formula”

Pages 315-318 by A. Calabri, C. Ciliberto, F. Flamini, R. Miranda | From volume 186-1

## Editors’ Note on “Invariant differential operators and eigenspace representations on an affine symmetric space”

Pages 319-319 by Editors of the Annals of Mathematics | From volume 186-1

## Revêtements du demi-plan de Drinfeld et correspondance de Langlands $p$-adique

Pages 321-411 by Gabriel Dospinescu, Arthur-César Le Bras | From volume 186-2

## Bilinear forms with Kloosterman sums and applications

Pages 413-500 by Emmanuel Kowalski, Philippe Michel, Will Sawin | From volume 186-2

## The triviality of the 61-stem in the stable homotopy groups of spheres

Pages 501-580 by Guozhen Wang, Zhouli Xu | From volume 186-2

## Borel circle squaring

Pages 581-605 by Andrew S. Marks, Spencer T. Unger | From volume 186-2

## A sharp Schrödinger maximal estimate in $\mathbb {R}^2$

Pages 607-640 by Xiumin Du, Larry Guth, Xiaochun Li | From volume 186-2

## Total Betti numbers of modules of finite projective dimension

Pages 641-646 by Mark E. Walker | From volume 186-2

## Erratum on “On the number of generators of ideals in polynomial rings”

Pages 647-648 by Jean Fasel | From volume 186-2

## On the generic part of the cohomology of compact unitary Shimura varieties

Pages 649-766 by Ana Caraiani, Peter Scholze | From volume 186-3

## Shtukas and the Taylor expansion of $L$-functions

Pages 767-911 by Zhiwei Yun, Wei Zhang | From volume 186-3

## Global smooth and topological rigidity of hyperbolic lattice actions

Pages 913-972 by Aaron Brown, Federico Rodriguez Hertz, Zhiren Wang | From volume 186-3