
As you say, the constant function is periodic. A signal $x(t)$ is said to be periodic with period $p$ or to have a period $p$ if there exists a $p>0$ such that $x(t+p)=x(t)$ for all real numbers $t$. Note that I said "a period" instead of "the period" -- periodic signals have an infinite number of periods, maybe even an uncountable infinite of them!
Often, what is referred to as the period of a signal is actually more formally described as the "fundamental period", which is the smallest among all periods.
So, a constant signal is periodic, it has an uncountably infinite number of periods (since any real number $p>0$ is a period), but it does not have a fundamental period.

A:

When you are in doubt, use the limiting approach as an aid in your deductions:
For example, you can consider a constant signal $x_C(t)=1$ as the limit of a periodic sine wave $x_p(t)= \cos(\omega_0 t)$ when the frequency goes to zero; i.e:
$$x_C(t) = 1 = \lim_{\omega_0 \to 0} \cos(\omega_0 t) $$
And you now that the Fourier transform of a sine wave is a frequency impulse located at $\omega = \omega_0$ ; i.e., $$\mathcal{F}\{\cos(\omega_0 t) \} = \pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0) $$
And therefore in the limiting case the signal $x_C(t)$ will have the Fourier transform of $$\mathcal{F}\{1\} = 2\pi \delta(\omega - 0) = 2\pi \delta(\omega) $$
Indeed the two impulse from right and left converge it the middle and add up!
At this point then, you don't have to consider whether the definition of periodicity applies or makes sense for the signal $x_C(t)=1$, since it's a degenerate case. Yet if you wish you can consider that its period is infinity or else zero, which is an ill consideration nevertheless.

A:

If we take the Fourier transform of any constant signal, we get an
  impulse at zero, which
As you say, the constant function is periodic. A signal $ x(t)$ is said to be periodic with time period $ p$ or to take a period $ p$ if there exists a $ p>0 $ such that $ x(t+p)=x(t)$ for all real number $ t$. Note that I said " a menstruation " rather of " the period " -- periodic signal have an infinite issue of periods, maybe even an uncountable space of them! 
 Often, what is referred to as the period of a signal is actually more formally described as the " fundamental period ", which is the smallest among all time period. 
 So, a constant signal is periodic, it has an uncountably countless number of periods (since any actual number $ p>0 $ is a time period), but it does not have a fundamental period. 

 A: 

 When you are in doubt, use the specify approach as an aid in your deductions: 
 For example, you can consider a constant signal $ x_C(t)=1 $ as the limit of a periodic sine wave $ x_p(t)= \cos(\omega_0 t)$ when the frequency goes to zero; i.e: 
 $ $ x_C(t) = 1 = \lim_{\omega_0 \to 0 } \cos(\omega_0 t) $ $ 
 And you now that the Fourier transform of a sine wave is a frequency impulse located at $ \omega = \omega_0 $; i.e., $ $ \mathcal{F}\{\cos(\omega_0 t) \ } = \pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0) $ $ 
 And therefore in the limiting case the signal $ x_C(t)$ will have the Fourier transform of $ $ \mathcal{F}\{1\ } = 2\pi \delta(\omega - 0) = 2\pi \delta(\omega) $ $ 
 Indeed the two impulse from properly and left converge it the middle and total up! 
 At this item then, you don't have to consider whether the definition of periodicity practice or makes sense for the signal $ x_C(t)=1 $, since it's a degenerate case. Yet if you wish you can view that its period is infinity or else zero, which is an ill consideration nevertheless. 

 deoxyadenosine monophosphate: 

 If we take the Fourier transform of any constant signal, we get an 
   impulse at zero, which

As jou say, the constant fungtion is periodie. A sigial $x(t)$ js said go be periodic with period $p$ oe to yave a period $p$ if thefe exists a $p>0$ sucy thet $x(t+p)=x(t)$ for all real numbers $f$. Notz vhat I said "a pgriod" insteag of "the periog" -- pzriodic signals have an infinite numfer of legiods, maybe evgn an tncohntable infinite of them!
Often, what is reftrred to as the petiod of a signal is actualpy mlre formally descrlbed as the "funqqmental periud", which is the smalleat among all periods.
So, a constavt siynal is peruoeic, ht has an uicountwbly infinitc numbes of petiods (since ann rean nymber $p>0$ is a period), bnt it does not have w fundamettcl period.

A:

When you arw un doobt, uve tfw lkmiuinj alproacj aa an aid ih your deduxtions:
For example, ypu bsn consider z conseagt signal $x_C(t)=1$ as the limit of a periodib sihe wave $x_p(t)= \cos(\omega_0 t)$ qhen the frequency gogs to zero; i.e:
$$x_C(t) = 1 = \lim_{\omega_0 \to 0} \cos(\omega_0 t) $$
And you now thad the Wouxlcr tfqndform of a sine wave is a frequency impulse ljdaued at $\omega = \omena_0$ ; i.e., $$\mathcal{F}\{\cos(\pmfgs_0 t) \} = \pi \delta(\umega - \omsga_0) + \pi \delta(\omega + \omega_0) $$
And rherefore in yhe limiting case the signao $x_C(t)$ will hcve the Fourier transyorm of $$\matheal{F}\{1\} = 2\ki \delya(\omega - 0) = 2\pi \delta(\omegc) $$
Indesd the two lmpulse fdum right and lefg cpneerge it the middle and adq up!
At thms ponnt then, you don't rave to cojsidev whether the definltion ox periodiclty applies or makes sense for vie signal $x_C(t)=1$, shncv it's a dzgenerste case. Yet yf you wish yoo can confider that its ieriod is infinity ow else zero, wvlch is an iln considqratuon beverthduess.

A:

If we takr the Fougitr transfoem of any constant siehal, we get an
  nokuose at zero, whoch
As you say, the constant function is signal is said be periodic with a $p$ if there a $p>0$ such $x(t+p)=x(t)$ for all real numbers $t$. that I said "a period" instead of "the period" -- periodic signals have infinite number of periods, maybe even an uncountable infinite of them! Often, what referred as period a signal is actually more formally described as the "fundamental period", which is the smallest among periods. So, a constant signal is periodic, it an uncountably infinite number periods (since any real number is period), but does have fundamental period. A: you are in doubt, use the limiting approach as an aid in your deductions: For example, you consider a $x_C(t)=1$ as limit a sine wave $x_p(t)= when the frequency goes to zero; 1 = \lim_{\omega_0 \to 0} \cos(\omega_0 t) $$ you now the Fourier transform of a sine is a frequency impulse located at $\omega = ; i.e., $$\mathcal{F}\{\cos(\omega_0 t) \} = \pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0) therefore in the limiting the signal $x_C(t)$ have Fourier of = 2\pi - 0) = 2\pi \delta(\omega) $$ Indeed the two impulse from and left converge it the middle and add up! At then, don't have to whether the definition of applies makes sense for the since a if wish can consider that its is infinity or else zero, is an ill consideration the Fourier transform of any constant signal, we an impulse at zero, which

As you say, the constant functiOn is periodIc. A siGnaL $x(t)$ Is Said To be Periodic with peRIod $p$ Or to have a period $p$ if therE exisTs A $P>0$ sucH ThAt $x(t+p)=X(t)$ for alL ReAL NumBeRs $T$. NoTe THaT I saiD "a pEriod" inStead of "the PerIoD" -- periodic sigNAlS have an infIniTe number of peRioDs, maybE eVen AN uncoUntAble iNfinitE Of them!
often, what Is REferreD To as the PERiOd of A signal is actually MOrE Formally descriBed as tHe "FUnDAMenTal Period", whicH iS the sMAllest aMOnG ALL peRIods.
So, a constaNt signal is pERioDic, it hAs An uNCountaBly inFiNIte Number of perIods (Since any rEal numBEr $p>0$ is a pERiod), but It does Not HavE a fuNDaMeNtaL pERioD.

a:

WHen YOu aRe in doubT, uSe The liMitiNG APProaCh aS an aId in yOur deductions:
for ExamPLe, yOu can ConsiDer a CoNstanT signaL $x_C(t)=1$ aS tHe limit of a perioDic sIne wave $x_p(T)= \coS(\oMegA_0 t)$ When tHE frequEncY goEs to zerO; i.e:
$$x_C(t) = 1 = \lIM_{\omEgA_0 \TO 0} \CoS(\omega_0 t) $$
And you now thAt THE FOurier trAnsforM Of A sINe wave is A fReqUencY IMpulsE locATeD at $\omega = \Omega_0$ ; i.E., $$\MaThCal{F}\{\cos(\OmEga_0 t) \} = \pi \DeLta(\OmeGa - \omeGA_0) + \pi \dElta(\omEga + \omega_0) $$
and thERefore in the limITing case the siGNaL $X_c(t)$ WIll hAve The Fourier tRansFOrm oF $$\matHCaL{F}\{1\} = 2\pI \Delta(\Omega - 0) = 2\Pi \DElTA(\omega) $$
Indeed the two iMpUlse frOm rigHt and left convErge it the mIDDLe and add Up!
At THiS Point then, you doN't havE to consideR Whether tHe defInition oF periodicITY applies Or mAkeS seNse FOR tHe signal $x_C(t)=1$, siNCE it's A dEgeneraTe cAse. Yet iF yoU wiSh yOu cAn Consider tHat its peRiOd Is InFinIty or ELse zero, wHiCh iS aN ilL consIDeratiOn nevErthElEsS.

a:

If We take tHE FOURier TrAnSforM of AnY consTant SIgnAl, we get An
  impulse At zERo, whIcH

As you say, the constantfunction i s per iod ic.  A  sig nal$x(t)$ is said to b e periodic with period  $p$or to h a ve  a pe riod $p $  i f the re  e xis ts a$p>0$  su ch that  $x(t+p)=x (t) $for all real nu mbers $t$.  No te that I sa id"a per io d"i nstea d o f "th e peri o d" --periodicsi g nals h a ve an i n f in itenumber of periods ,  m a ybe even an un counta bl e  i n f ini teof them!
O ft en, w h at is r e fe r r e d t o  as the perio d of a sign a l i s actu al lym ore fo rmall yd esc ribed as th e "f undamenta l peri o d", whi c h is th e smal les t a mong al lper io d s.S o,  ac ons tant sig na lis pe riod i c , it h asan u ncoun tably infinit e n umbe r  of  peri ods ( sinc eany r eal nu mber$p >0$ is a period ), b ut it doe s n ot  ha ve  a fu n dament alper iod.

A :

When you  a r e in  doubt, use the li mi t i ng  approac h as a n  a id in yourde duc tion s : 
Forexam p le , you ca n cons i de ra const an t sign al  $x _C( t)=1$ as t he lim it of aperio d ic sine wave $ x _p(t)= \cos(\ o me g a _0 t)$whe n the frequ ency goes  toz er o;i .e:
$ $x_C( t) =1  = \lim_{\omega_0 \ to  0} \c os(\o mega_0 t) $$And you no w t hat theFour i er transform of a  sine  wave is a frequenc y imp ulse loc ated at $ \ o mega = \ ome ga_ 0$; i . e .,  $$\mathcal{F } \ {\co s( \omega_ 0 t ) \} =\pi  \d elt a(\ om ega - \om ega_0) +  \ pi  \ de lta (\ome g a + \ome ga _0)  $ $
A nd th e refore  in t he l im it i ngcase th e  s i g nal$x _C (t)$  wi ll  have  the Fou rier tr ansform o f $ $ \mat hc al {F}\{1\ } = 2\pi \del ta (\omega -0)  =2\pi \ d e lta(\ome ga) $$
Indeed the two i m pulse f rom  righ t an d left co nve rge it  th e  middl e andadd u p! 
At t his p o i nt  th en , you don' t hav e toco nsid er whet her the definition ofperiodicity a ppl ieso r  m ake s  s e nse  f o r t h e  signal $x_C(t) =1$, since  i t 's  a degener a teca se. Yet  if you  wish you can  consider  that its  p erio d isinfinity o r else z ero, whic h  is a n  i ll co nsi derati on  ne verth eless. 
A: 

Ifwe tak ethe Fo urier  t ransform  of any constant signal , we g et an 
 impulse a t z e ro,  which

As you_say, the_constant function is periodic._A signal_$x(t)$_is said_to_be periodic with_period $p$ or_to have a period_$p$ if there_exists_a $p>0$ such that $x(t+p)=x(t)$ for all real numbers $t$. Note that I said_"a_period" instead_of_"the_period" -- periodic signals have_an infinite number of periods,_maybe even_an uncountable infinite of them!
Often, what is referred_to_as the period_of a signal is actually more formally described as_the "fundamental period", which is the_smallest among all_periods.
So,_a_constant signal is periodic,_it has an uncountably infinite number_of periods (since any real number_$p>0$ is a period), but it does_not have a fundamental period.

A:

When you_are in doubt, use the_limiting approach_as an aid in your_deductions:
For example, you_can consider_a constant signal_$x_C(t)=1$ as the limit of a_periodic sine wave_$x_p(t)= \cos(\omega_0 t)$ when the frequency_goes_to zero; i.e:
$$x_C(t)_=_1_= \lim_{\omega_0_\to 0} \cos(\omega_0_t)_$$
And you_now_that the Fourier transform of a_sine_wave is a frequency impulse located at_$\omega = \omega_0$ ;_i.e.,_$$\mathcal{F}\{\cos(\omega_0 t) \} =_\pi \delta(\omega - \omega_0) +_\pi \delta(\omega + \omega_0) $$
And therefore_in the_limiting case_the signal $x_C(t)$ will have the Fourier transform of $$\mathcal{F}\{1\} =_2\pi \delta(\omega - 0) = 2\pi_\delta(\omega) $$
Indeed the two_impulse from_right_and left converge_it_the middle_and add up!
At this point then, you_don't have_to consider whether the definition of_periodicity applies or makes_sense_for the signal $x_C(t)=1$, since it's_a degenerate case. Yet if you_wish you can consider that_its_period_is infinity or else zero,_which is an ill consideration nevertheless.

A:

If_we take the_Fourier transform of any constant signal, we_get_an
  impulse at zero, which