Frequency on double sideband (Amplitude )modulation

[simbyak]

Hi guys,i have attached the question..im stuck at the point where we have 2 terms with cos added in the brackets,,and then multplied by the carrier signal..im more conversant with a signal in the form
Ac(1+m cos(ωt))cos ωt where this eqn represents a full AM signal..the ω in the bracket is the modulation freq and the ω outside the bracket is carrier frequency..was unable to put f and c as subsccripts on both ω thus explaining it here.thnx

 

[Electronworks]

Remembering back 20 years ago when I last did this, it looks like you have a 1500rad/s tone (of amplitude '2') added to a 3000rad/s tone (of amplitude '10') added to a dc offset (of amplitude '20') all modulated with an hf carrier of frequency 10^5 Hz.

This should represent itself as spectral spurs of 10^5 Hz, 1500rad/s and 3000rad/s...

Does this make any sense...?

 

[simbyak]

Remembering back 20 years ago when I last did this, it looks like you have a 1500rad/s tone (of amplitude '2') added to a 3000rad/s tone (of amplitude '10') added to a dc offset (of amplitude '20') all modulated with an hf carrier of frequency 10^5 Hz.

This should represent itself as spectral spurs of 10^5 Hz, 1500rad/s and 3000rad/s...

Does this make any sense...?

Well,it does make some sense especially the breakdown intepretation..
Your interpretation of the question is absolutlely spot on.i agree with the spectral spur at 10^5,however for the 1500rad/s and 3000rad/s... ..it confuses me because the spur of 10^5 Hz is the carrier and since its DSB signal.,,the carrier freq should be in between the 2 sideband freq..which doesnt apply here,,...also how would gt the modulatio effiiciency..thnx

 

[Hayato]

Simbyak.

Do not forget that you have to use the cosine laws over there.

You have the modulating wave ( cos(2*pi*10^5 Hz) ) multiplying a sum of 2 cosine waves over there one with 1500*PI rad/s and 3000*PI rad/s + the DC level.

The DC is gonna give you the carrier component at 10^5 Hz.
The 1500*PI rad/s cosine is going to give you 2 impulses, one at 10^5 + 750 Hz and another at 10^5 - 750 Hz.
The 3000*PI rad/s cosine is going to give you 2 impulses, one at 10^5 + 1500 Hz and another at 10^5 - 1500 Hz.

 

[simbyak]

Simbyak.

Do not forget that you have to use the cosine laws over there.

You have the modulating wave ( cos(2*pi*10^5 Hz) ) multiplying a sum of 2 cosine waves over there one with 1500*PI rad/s and 3000*PI rad/s + the DC level.

The DC is gonna give you the carrier component at 10^5 Hz.
The 1500*PI rad/s cosine is going to give you 2 impulses, one at 10^5 + 750 Hz and another at 10^5 - 750 Hz.
The 3000*PI rad/s cosine is going to give you 2 impulses, one at 10^5 + 1500 Hz and another at 10^5 - 1500 Hz.

Haayato,thanks for your contribution. but for me this is where thngs get kind of messy,..well the wave given is amplitude modulated..so to get the frequency spectra..wht i did is multiply each individual sine wave and the Dc level by cos(2*pi*10^5 Hz)..i.e we have 3 product terms now.
So i go on to use the property of modulation for fourier ttransform(attached below)..i am able to get the Dc as the carrier component at 10^5 Hz,however the transform for the 2 sine waves is where it gets me mixed up..
(and for the implses in your answer id expect the freq to be 1500 and 3000 respecctuvely,if u go by the properties attched)
thnx
P.S by saying i gottta use the cosine law..are u referring to this cos(A-B)=cos A cos B + sin A sin B?...hhmm if you do,doubt it will work coz looks like we gonna end up with a product term of cos which doesnt make lif any easier..
I have looked through my lecture notes and other soursces online and seems like..the spectra is
The DC is gonna give you the carrier component at 10^5 Hz.with a magnitude of 10.
The 1500*PI rad/s cosine is going to give you 2 impulses, with magnitude of 0.5 at 10^5 +1500 Hz and another at 10^5 -1500 Hz.
The 3000*PI rad/s cosine is going to give you 2 impulses .with magnitude of 1, one at 10^5 + 3000 Hz and another at 10^5 -3000 Hz.
Now im stuck at finding the modualtion efficiecny because the modulation is given by (amplitude of modulating wave/amplitude of carrier wave)....so what do we use as the amplitde of the modulating wave given tht our modulating is a sum of 2 waves..

 

[Hayato]

Santa Claus is here.

[latex]$\small x(t)=[20 + 2cos(3000\pi t) + 10cos(6000\pi t)].cos(2.10^{5}\pi t)\\
x(t)=20cos(2.10^{5}\pi t) + 2cos(3000\pi t)cos(2.10^{5}\pi t) + 10cos(6000\pi t)cos(2.10^{5}\pi t)\\
f(t).g(t)\overset{\mathcal{F}}{\rightarrow}F(f)\ast G(f)\\
f(t)+g(t)\overset{\mathcal{F}}{\rightarrow}F(f)+G(f)\\\\
X(f)= 20.\frac{1}{2}[\delta (f\pm 10^{5})] + [2\frac{1}{2}\delta (f\pm 1500)\ast\frac{1}{2}\delta (f\pm 10^{5})]+[10\frac{1}{2}\delta (f\pm 3000)\ast\frac{1}{2}\delta (f\pm 10^{5})]\\\\
X(f)= 10[\delta (f\pm 10^{5})] + \frac{1}{2}[\delta (f\pm 1500)\ast\delta (f\pm 10^{5})]+5\frac{1}{2}[\delta (f\pm 3000)\ast\delta (f\pm 10^{5})]\\\\
X(f)= 10[\delta (f\pm 10^{5})] + \frac{1}{2}[\delta (f\pm10^{5}\pm 1500)]+5\frac{1}{2}[\delta (f\pm10^{5}\pm 3000)]\\\\
Positive\;Spectrum:\\
X(f)= 10[\delta (f-10^{5})] + \frac{1}{2}[\delta (f-10^{5}\pm 1500)]+5\frac{1}{2}[\delta (f-10^{5}\pm 3000)]\\\\
Negative\;Spectrum:\\
X(f)= 10[\delta (f+10^{5})] + \frac{1}{2}[\delta (f+10^{5}\pm 1500)]+5\frac{1}{2}[\delta (f+10^{5}\pm 3000)]\\\\$ [/latex]

 

[simbyak]

Thanks Santa..xmas does come early sometimes...hehehehe.///
Any idea on part ii) Determinee the modulation efficienvcy of the signal?
really appreciate youe help..
i.e Your solution of the negative and positive spectrum cleared up some confusion on the delay of the pulse being on the positve spectrum..Thnx a bunch and would realy appreciate if you have any clue on how to go abt part ii)

 

[Hayato]

The power efficience of the modulation is the ammount of power used to carry information over the total modulated wave power.