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Sketching Amplitude Spectrum?

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fouadalnoor

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Hello guys,

Can anyone please explain to me how to sketch the amplitude spectrum of a message signal as shown below:

Vm(t) = 3cos(20Πt+Π/4)+2sin(60Πt)-cos(100Πt)

I know how to do it if we only have one trig function such as:

Vm(t) = Vmcos(2Πfmt)

But I just can't think of a way to draw the one above. Do I have to superimpose all the trigonometric functions into one function and THEN do it?

An example is attached.

Hope you can help.

Fouad.
 

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Last edited:
Like this?

You can do it by inspection. Note that voltage vs time is just the instantaneous sum of three sinusoids, at 10Hz, 30Hz and 50Hz, so the spectrum contains only those three peaks. The relative amplitude of the peaks is proportional to the coefficients of the three sinusoids.
 

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Like this?

Em, that's probably how you do it using software. If you look at the attached picture you can see how they have done it. In Figure 3.9 you can see they have a line representing what frequency and amplitude the message signal (Vm) has. Since the message signal has a maximum frequency fm at an amplitude of Vm then they drew one line corresponding to that, while since it has a minimum frequency at f1 and an amplitude of V1 at that frequency they drew a line showing that.

Can you see it?

I am basically trying to do the same thing with my message signal, only this time it has 3 components...
 
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Hello again,

You insert the signal into the formula:

Vam=(Vc+Kam*Vm(t))*cos(2*pi*fc*t)
and then reduce to trig terms that sum. You end up with sum and differences terms of the carrier freq and the Vm frequency (or frequencies). You then draw a single line at each of those sum and difference frequencies who's amplitude equals the peak multiplier of that term, divided by 2 because that's how the trig functions reduce.
For the example they gave, you can see this in detail as they solved it completely. For a single cos signal you would only have one sum and difference to do, plus the carrier as they did in the example.
Here's an example with three cosine sources of different frequencies:
Vc*cos(2*fc*pi*t)
+(V3*kam*cos(2*fc*pi*t+2*f3*pi*t))/2+(V3*kam*cos(2*fc*pi*t-2*f3*pi*t))/2
+(V2*kam*cos(2*fc*pi*t+2*f2*pi*t))/2+(V2*kam*cos(2*fc*pi*t-2*f2*pi*t))/2
+(V1*kam*cos(2*fc*pi*t+2*f1*pi*t))/2+(V1*kam*cos(2*fc*pi*t-2*f1*pi*t))/2

Note that for the third one here, we would plot a line at the sum and difference frequencies with a height of kam*V3/2. If you examine the above you can see the pattern.
 
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Sorry, I didn't get that you were modulating a carrier, just "explain to me how to sketch the amplitude spectrum of a message signal as shown below:

Vm(t) = 3cos(20Πt+Π/4)+2sin(60Πt)-cos(100Πt)
"
 
Sorry, I didn't get that you were modulating a carrier, just "explain to me how to sketch the amplitude spectrum of a message signal as shown below:

Vm(t) = 3cos(20Πt+Π/4)+2sin(60Πt)-cos(100Πt)
"

The second part is modulating the carrier, but the first part is simply drawing a Amplitude Spectrum of the message signal itself.

So they want us to draw two Amplitude Spectrum's. One for the message signal Vm(t) and another one for the Amplitude Modulated signal (AM). But its the first part I dont get. How do I draw a Amplitude Spectrum of the message signal when I have 3 trig terms instead of 2 as they have in their example?

Do I just plot the amplitude against the frequency for each term, so I will have 3 lines?
 
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Oh, yeah that's when I want to modulate the signal. I want to plot a Amplitude Spectrum for the message signal itself, just as they have done. And then AM modulate it and plot a Amplitude Spectrum of the AM signal as well.

I think I know how to do it, just not sure. I will do it now and post it up so you can take a look.

Thanks.
 
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Oh, yeah that's when I want to modulate the signal. I want to plot a Amplitude Spectrum for the message signal itself, just as they have done. And then AM modulate it and plot a Amplitude Spectrum of the AM signal as well.

I think I know how to do it, just not sure. I will do it now and post it up so you can take a look.

Thanks.


Hello again,

You insert the signal into the formula:

Vam=(Vc+Kam*Vm(t))*cos(2*pi*fc*f)
and then reduce to trig terms that sum. You end up with sum and differences terms of the carrier freq and the Vm frequency (or frequencies). You then draw a single line at each of those sum and difference frequencies who's amplitude equals the peak multiplier of that term, divided by 2 because that's how the trig functions reduce.
For the example they gave, you can see this in detail as they solved it completely. For a single cos signal you would only have one sum and difference to do, plus the carrier as they did in the example.
Here's an example with three cosine sources of different frequencies:
Vc*cos(2*fc*pi*t)
+(V3*kam*cos(2*fc*pi*t+2*f3*pi*t))/2+(V3*kam*cos(2*fc*pi*t-2*f3*pi*t))/2
+(V2*kam*cos(2*fc*pi*t+2*f2*pi*t))/2+(V2*kam*cos(2*fc*pi*t-2*f2*pi*t))/2
+(V1*kam*cos(2*fc*pi*t+2*f1*pi*t))/2+(V1*kam*cos(2*fc*pi*t-2*f1*pi*t))/2

Note that for the third one here, we would plot a line at the sum and difference frequencies with a height of kam*V3/2. If you examine the above you can see the pattern.


It shouldnt be hard to figure out using the above patterns. You'd just have one such set, along with the carrier.
 
Ok, I have now done it I think. The only thing I'm not entirely sure about is the -cos(100*pi*t) term. I am guessing the amplitude is just 1 since I am doing |Vm(t)|.

Also, I am pretty sure I got the mathematics for the second part wrong somewhere. I know that the identity for multiplying two sinusoids is:

cos(A)*cos(B) = 1/2cos(A-B) + 1/2cos(A+B)

But does this mean that the order of multiplication matters? also, what happens to the phase.. umm.. donno just not sure. On Wolfram Alpha I get the term:

0.5cos(2000*pi*t)*3cos(20*pi*t+pi/4) = 0.75 cos(pi/4-1980 pi t)+0.75 cos(2020 pi t+pi/4)

But why is it "0.75 cos(pi/4-1980 pi t)"?

I got:

0.75 cos(1980*pi*t-pi/4)"?

and the term

0.5cos(2000*pi*t)*2cos(60*pi*t-pi/2)

I got totally wrong...
 
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Ok, I have now figured it out I think, please take a look at the last picture 2 pictures uploaded.
 

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Ok, I have now attached the questions and solutions to Lectures 3 and 4. I have only reached up to Q3 on L4.

Thanks for all your help.

UPDATE:

I have now attempted Q2-Q6...

Hope you can help.

Thanks.
 

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Ok, I have now figured it out I think, please take a look at the last picture 2 pictures uploaded.

Hi,


That spectrum looks good to me, I got the same result.
 
Hi again,

Yeah i figured that ha ha.
Your L3 looks good too. I guess we are on to L4 next then.
 
Hi again,

Ok it's cleared now. It fills up too fast i think, maybe they should allow more messages :)
I tried to save to a new folder too but couldnt find out how to do that.
 
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